DP (Grade 11 & 12) · Maths AI HL
Number & Algebra
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Number Skills 50 questions
QUESTION 1
4 marks
Easy
A processor performs $4.5 \times 10^{9}$ operations per second. It runs continuously, at this constant rate, for $2.4 \times 10^{3}$ seconds.
Find the total number of operations performed, giving your answer in the form $a \times 10^{k}$ where $1 \le a < 10$ and $k \in \mathbb{Z}$.
Show complete worked solution
Using the formula for total operations = rate $\times$ time:
$$\text{Total operations} = (4.5 \times 10^{9}) \times (2.4 \times 10^{3})$$
Multiplying the coefficients and adding the powers of 10:
$$= (4.5 \times 2.4) \times 10^{9+3} = 10.8 \times 10^{12}$$
Since $10.8$ is not in the range $1 \le a < 10$, rewrite in standard form:
$$10.8 \times 10^{12} = 1.08 \times 10^{13}$$
$$\boxed{1.08 \times 10^{13} \text{ operations}}$$
QUESTION 2
5 marks
Easy
The mass of a single virus particle is $9.6 \times 10^{-18}$ kg. A laboratory sample contains $3.2 \times 10^{6}$ virus particles.
(a) Find the total mass of the particles in the sample, giving your answer in the form $a \times 10^{k}$ kg, correct to 3 significant figures.
(b) Given that $1$ nanogram $= 1 \times 10^{-12}$ kg, express this total mass in nanograms, correct to 3 significant figures.
Show complete worked solution
(a) Using mass = (mass per particle) $\times$ (number of particles):
$$\text{Total mass} = (9.6 \times 10^{-18}) \times (3.2 \times 10^{6})$$
$$= (9.6 \times 3.2) \times 10^{-18+6} = 30.72 \times 10^{-12}$$
Converting to standard form ($1 \le a < 10$):
$$30.72 \times 10^{-12} = 3.072 \times 10^{-11}$$
$$\boxed{3.07 \times 10^{-11} \text{ kg (3 s.f.)}}$$
(b) Using the conversion $1$ ng $= 1 \times 10^{-12}$ kg, and the unrounded value from (a):
$$\text{mass in ng} = \frac{3.072 \times 10^{-11}}{1 \times 10^{-12}}$$
$$= 3.072 \times 10^{1} = 30.72$$
$$\boxed{30.7 \text{ ng (3 s.f.)}}$$
QUESTION 3
5 marks
Medium
A two-stage rocket carries fuel for its first stage of mass $2.35 \times 10^{5}$ kg and fuel for its second stage of mass $8.60 \times 10^{3}$ kg.
(a) Find the total mass of fuel carried by the rocket, giving your answer in the form $a \times 10^{k}$ kg, correct to 3 significant figures.
(b) Find the percentage of the total fuel mass that is carried in the second stage, correct to 3 significant figures.
Show complete worked solution
(a) To add two numbers in scientific notation, first write them with the same power of 10.
$$8.60 \times 10^{3} = 0.0860 \times 10^{5}$$
Substituting into the sum:
$$\text{Total fuel} = 2.35 \times 10^{5} + 0.0860 \times 10^{5} = (2.35+0.0860)\times10^5$$
$$= 2.436 \times 10^{5}$$
$$\boxed{2.44 \times 10^{5} \text{ kg (3 s.f.)}}$$
(b) Using percentage $= \dfrac{\text{part}}{\text{whole}} \times 100$, with the unrounded total from (a):
$$\text{Percentage} = \frac{8.60 \times 10^{3}}{2.436 \times 10^{5}} \times 100$$
$$= \frac{8600}{243600} \times 100$$
$$= 3.5304\ldots\%$$
$$\boxed{3.53\% \text{ (3 s.f.)}}$$
QUESTION 4
5 marks
Medium
A data centre transmits a total of $7.2 \times 10^{15}$ bytes of data over a period of $3.6 \times 10^{4}$ seconds.
(a) Find the average rate of data transmission in bytes per second, giving your answer in the form $a \times 10^{k}$, correct to 3 significant figures.
(b) Given that $1$ gigabyte (GB) $= 1 \times 10^{9}$ bytes, express this rate in gigabytes per second.
Show complete worked solution
(a) Using rate = total bytes / total time:
$$\text{Rate} = \frac{7.2 \times 10^{15}}{3.6 \times 10^{4}}$$
Dividing the coefficients and subtracting the powers of 10:
$$= \frac{7.2}{3.6} \times 10^{15-4} = 2 \times 10^{11}$$
$$\boxed{2.00 \times 10^{11} \text{ bytes/s}}$$
(b) Using the conversion $1$ GB $= 1 \times 10^{9}$ bytes:
$$\text{Rate in GB/s} = \frac{2 \times 10^{11}}{1 \times 10^{9}} = 2 \times 10^{2}$$
$$\boxed{200 \text{ GB/s}}$$
QUESTION 5
13 marks
Hard
This question investigates distances, times and masses on an astronomical scale, all of which are conveniently expressed in scientific notation.
A particular spiral galaxy is $2.537 \times 10^{19}$ km from Earth. Light travels at $3.00 \times 10^{5}$ km/s.
(a) Find the time, in seconds, for light to travel from the galaxy to Earth. Give your answer in the form $a \times 10^{k}$, correct to 3 significant figures. [3]
(b) Given that $1$ year $\approx 3.156 \times 10^{7}$ seconds, convert your answer to part (a) into years. Give your answer in the form $a \times 10^{k}$, correct to 3 significant figures (this is the distance to the galaxy expressed in light-years). [4]
(c) The mass of this galaxy is estimated to be $1.5 \times 10^{12}$ times the mass of the Sun. Given that the mass of the Sun is $1.989 \times 10^{30}$ kg, find the mass of the galaxy in kg, giving your answer in the form $a \times 10^{k}$, correct to 3 significant figures. [3]
(d) A telescope can detect a light source only if its received energy flux is at least $4.0 \times 10^{-15}$ W/m$^2$. The energy flux received on Earth from this galaxy is $2.6 \times 10^{-13}$ W/m$^2$. Determine how many times greater this flux is than the minimum detectable flux, and hence state, with a reason, whether the galaxy is detectable by this telescope. [3]
Show complete worked solution
(a) Using time = distance / speed:
$$t = \frac{2.537 \times 10^{19}}{3.00 \times 10^{5}}$$
Dividing the coefficients and subtracting the powers of 10:
$$t = \frac{2.537}{3.00} \times 10^{19-5} = 0.845\overline{6} \times 10^{14}$$
Converting to standard form:
$$t = 8.4567\ldots \times 10^{13}$$
$$\boxed{t \approx 8.46 \times 10^{13} \text{ s (3 s.f.)}}$$
(b) Using the unrounded value from (a), $t = 8.45667 \times 10^{13}$ s, and the conversion $1$ year $\approx 3.156\times10^{7}$ s:
$$\text{time in years} = \frac{8.45667 \times 10^{13}}{3.156 \times 10^{7}}$$
$$= \frac{8.45667}{3.156}\times10^{13-7} = 2.6798\ldots \times 10^{6}$$
$$\boxed{2.68 \times 10^{6} \text{ years (3 s.f.)}}$$
i.e. the galaxy is about $2.68$ million light-years from Earth.
(c) Using mass of galaxy = (number of solar masses) $\times$ (mass of Sun):
$$m = (1.5 \times 10^{12}) \times (1.989 \times 10^{30})$$
$$= (1.5 \times 1.989) \times 10^{12+30} = 2.9835 \times 10^{42}$$
$$\boxed{m \approx 2.98 \times 10^{42} \text{ kg (3 s.f.)}}$$
(d) Using ratio = flux received / minimum detectable flux:
$$\text{ratio} = \frac{2.6 \times 10^{-13}}{4.0 \times 10^{-15}}$$
$$= \frac{2.6}{4.0}\times10^{-13-(-15)} = 0.65 \times 10^{2} = 65$$
$$\boxed{\text{The received flux is 65 times the minimum detectable flux}}$$
Since $65 > 1$, the received flux exceeds the telescope's minimum threshold, so the galaxy is
$$\boxed{\text{detectable by this telescope}}$$
QUESTION 6
4 marks
Easy
Solve the equation $3^{x+1} = 20$ for $x$, giving your answer correct to 3 significant figures.
Show complete worked solution
Take $\log_{10}$ of both sides:
$$\log(3^{x+1}) = \log(20)$$
Using the power law of logarithms:
$$(x+1)\log 3 = \log 20$$
Solving for $x+1$:
$$x+1 = \frac{\log 20}{\log 3} = \frac{1.30103\ldots}{0.47712\ldots} = 2.72683\ldots$$
Solving for $x$:
$$x = 2.72683\ldots-1 = 1.72683\ldots$$
$$\boxed{x \approx 1.73 \text{ (3 s.f.)}}$$
QUESTION 7
4 marks
Easy
Given that $\log_2 5 = a$ and $\log_2 3 = b$, express $\log_2 45$ in terms of $a$ and $b$.
Show complete worked solution
Write $45$ in terms of factors of $3$ and $5$:
$$45 = 9\times5 = 3^2\times5$$
Using the product law $\log(xy)=\log x+\log y$:
$$\log_2 45 = \log_2(3^2\times5) = \log_2(3^2)+\log_2 5$$
Using the power law $\log(x^n)=n\log x$:
$$= 2\log_2 3+\log_2 5$$
Substituting $\log_2 3=b$ and $\log_2 5=a$:
$$= 2b+a$$
$$\boxed{\log_2 45 = a+2b}$$
QUESTION 8
6 marks
Medium
The concentration of a drug in a patient's bloodstream, in mg/L, $t$ hours after an injection, is modelled by
$$C(t) = 80e^{-0.15t}$$
(a) Find the concentration of the drug in the bloodstream 4 hours after the injection, correct to 3 significant figures.
(b) Find the time taken for the concentration to fall to 10 mg/L, correct to 3 significant figures.
Show complete worked solution
(a) Substituting $t=4$ into $C(t)=80e^{-0.15t}$:
$$C(4) = 80e^{-0.15(4)}$$
$$= 80e^{-0.6}$$
$$= 80\times0.548812\ldots$$
$$= 43.9049\ldots$$
$$\boxed{C(4) \approx 43.9 \text{ mg/L (3 s.f.)}}$$
(b) Setting $C(t)=10$:
$$80e^{-0.15t} = 10$$
Dividing both sides by 80:
$$e^{-0.15t} = \frac{10}{80} = 0.125$$
Taking the natural logarithm of both sides:
$$-0.15t = \ln(0.125) = -2.079441\ldots$$
Solving for $t$:
$$t = \frac{-2.079441\ldots}{-0.15} = 13.8629\ldots$$
$$\boxed{t \approx 13.9 \text{ hours (3 s.f.)}}$$
QUESTION 9
6 marks
Medium
A colony of bacteria in a laboratory culture is modelled by
$$N(t) = 200 \times k^t$$
where $N(t)$ is the number of bacteria present $t$ hours after the start of the experiment, and $k$ is a constant.
After 5 hours, the population is 5400.
(a) Find the value of $k$, correct to 3 significant figures.
(b) Using the unrounded value of $k$, find the population predicted by the model after 12 hours, correct to the nearest whole number of bacteria.
Show complete worked solution
(a) Substituting $t=5$, $N=5400$ into $N(t)=200\times k^t$:
$$200k^5 = 5400$$
Dividing both sides by 200:
$$k^5 = \frac{5400}{200} = 27$$
Taking the fifth root of both sides (the positive real root, since $k$ must be positive for a growth model):
$$k = 27^{1/5} = 1.93318\ldots$$
$$\boxed{k \approx 1.93 \text{ (3 s.f.)}}$$
(b) Substituting the unrounded value of $k$ and $t=12$ into $N(t)=200k^t$:
$$N(12) = 200(1.93318\ldots)^{12}$$
$$= 200 \times 2724.41\ldots$$
$$= 544882.7\ldots$$
$$\boxed{N(12) \approx 544{,}883 \text{ bacteria}}$$
QUESTION 10
14 marks
Hard
The loudness of a sound, measured in decibels (dB), is defined by
$$L = 10\log_{10}\left(\dfrac{I}{I_0}\right)$$
where $I$ is the sound intensity in W/m$^2$ and $I_0 = 1.0 \times 10^{-12}$ W/m$^2$ is a reference intensity (approximately the quietest sound a human can hear).
(a) A jet engine at close range produces a sound intensity of $I = 1.0 \times 10^{1}$ W/m$^2$. Find the corresponding decibel level. [3]
(b) A quiet library has a decibel level of 35 dB. Find the sound intensity $I$ in the library, giving your answer in the form $a \times 10^{k}$ W/m$^2$, correct to 3 significant figures. [3]
(c) Show, using the laws of logarithms, that multiplying the sound intensity $I$ by a factor of 10 always increases the decibel level $L$ by exactly 10 dB, regardless of the starting intensity. [4]
(d) Two sounds are produced simultaneously in the same location: sound A has a decibel level of 70 dB and sound B has a decibel level of 65 dB. When two sounds combine, the resulting total intensity is the sum of their individual intensities. Find the combined decibel level of both sounds together, correct to 1 decimal place. [4]
Show complete worked solution
(a) Using $L=10\log_{10}\left(\dfrac{I}{I_0}\right)$ with $I=1.0\times10^{1}$, $I_0=1.0\times10^{-12}$:
$$L = 10\log_{10}\left(\frac{1.0\times10^{1}}{1.0\times10^{-12}}\right)$$
$$= 10\log_{10}(1.0\times10^{13})$$
$$= 10\times13 = 130$$
$$\boxed{L = 130 \text{ dB}}$$
(b) Substituting $L=35$ into $L=10\log_{10}\left(\dfrac{I}{I_0}\right)$:
$$35 = 10\log_{10}\left(\frac{I}{1.0\times10^{-12}}\right)$$
Dividing both sides by 10:
$$\log_{10}\left(\frac{I}{1.0\times10^{-12}}\right) = 3.5$$
Rewriting in exponential form:
$$\frac{I}{1.0\times10^{-12}} = 10^{3.5}$$
Solving for $I$:
$$I = 10^{3.5}\times1.0\times10^{-12} = 10^{-8.5} = 3.16228\ldots\times10^{-9}$$
$$\boxed{I \approx 3.16\times10^{-9} \text{ W/m}^2 \text{ (3 s.f.)}}$$
(c) Let the original intensity be $I$, giving level $L=10\log_{10}\left(\dfrac{I}{I_0}\right)$. Multiplying the intensity by 10 gives new intensity $10I$, and new level:
$$L' = 10\log_{10}\left(\frac{10I}{I_0}\right) = 10\log_{10}\left(10\times\frac{I}{I_0}\right)$$
Using the product law $\log(xy)=\log x+\log y$:
$$= 10\left[\log_{10}10+\log_{10}\left(\frac{I}{I_0}\right)\right]$$
Since $\log_{10}10=1$:
$$= 10\left[1+\log_{10}\left(\frac{I}{I_0}\right)\right] = 10+10\log_{10}\left(\frac{I}{I_0}\right) = 10+L$$
So $L'=L+10$ for any starting value of $I$, i.e. multiplying $I$ by 10 always adds exactly 10 dB to the level.
$$\boxed{L' = L+10} \quad \blacksquare$$
(d) Find each intensity from its decibel level using $I=I_0\times10^{L/10}$:
$$I_A = 1.0\times10^{-12}\times10^{70/10} = 1.0\times10^{-12}\times10^{7} = 1.0\times10^{-5} \text{ W/m}^2$$
$$I_B = 1.0\times10^{-12}\times10^{65/10} = 1.0\times10^{-12}\times10^{6.5} = 3.16228\ldots\times10^{-6} \text{ W/m}^2$$
Since the total intensity is the sum of the individual intensities:
$$I_{tot} = I_A+I_B = 1.0\times10^{-5}+3.16228\times10^{-6} = 1.31623\times10^{-5} \text{ W/m}^2$$
Substituting into the decibel formula:
$$L_{tot} = 10\log_{10}\left(\frac{1.31623\times10^{-5}}{1.0\times10^{-12}}\right)$$
$$= 10\log_{10}(1.31623\times10^{7})$$
$$= 10(7.11933\ldots) = 71.1933\ldots$$
$$\boxed{L_{tot} \approx 71.2 \text{ dB (1 d.p.)}}$$
QUESTION 11
4 marks
Easy
The radius of a spherical ball bearing is measured, to the nearest centimetre, as 8 cm, and its volume is estimated using this rounded value. The true radius of the ball bearing is actually 8.3 cm.
Using the formula $V = \dfrac{4}{3}\pi r^3$, find the percentage error in the estimated volume (based on $r=8$) compared to the actual volume (based on $r=8.3$), correct to 3 significant figures.
Show complete worked solution
Using $V=\dfrac{4}{3}\pi r^3$:
Estimated volume ($r=8$):
$$V_{est} = \frac{4}{3}\pi(8)^3 = \frac{4}{3}\pi(512) = 2144.66\ldots \text{ cm}^3$$
Actual volume ($r=8.3$):
$$V_{act} = \frac{4}{3}\pi(8.3)^3 = \frac{4}{3}\pi(571.787) = 2395.10\ldots \text{ cm}^3$$
Using percentage error $=\left|\dfrac{V_{est}-V_{act}}{V_{act}}\right|\times100$:
$$\text{Percentage error} = \frac{|2144.66-2395.10|}{2395.10}\times100$$
$$= \frac{250.44}{2395.10}\times100 = 10.4562\ldots$$
$$\boxed{\text{Percentage error} \approx 10.5\% \text{ (3 s.f.)}}$$
QUESTION 12
4 marks
Easy
A rectangular field has a length of 45 m and a width of 28 m, each measured correct to the nearest metre.
Find the lower bound and the upper bound for the perimeter of the field.
Show complete worked solution
Since each measurement is correct to the nearest metre, the true value lies within half a unit of the rounded value:
$$44.5 \le \text{length} < 45.5$$
$$27.5 \le \text{width} < 28.5$$
Using $P=2(\text{length}+\text{width})$:
The perimeter is smallest when both length and width take their lower bounds:
$$P_{min} = 2(44.5+27.5) = 2(72) = 144$$
$$\boxed{P_{min} = 144 \text{ m}}$$
The perimeter is largest when both length and width take their upper bounds:
$$P_{max} = 2(45.5+28.5) = 2(74) = 148$$
$$\boxed{P_{max} = 148 \text{ m}}$$
QUESTION 13
6 marks
Medium
A solid metal cube has a measured mass of 850 g, correct to the nearest 10 g, and a measured side length of 5.0 cm, correct to the nearest 0.1 cm.
(a) Write down the lower and upper bounds for the mass and for the side length.
(b) Hence find the lower and upper bounds for the density of the metal, in g/cm$^3$, correct to 3 significant figures. (Density $= \dfrac{\text{mass}}{\text{volume}}$, and the cube's volume is $(\text{side length})^3$.)
Show complete worked solution
(a) Since the mass is given to the nearest 10 g:
$$845 \le m < 855$$
Since the side length is given to the nearest 0.1 cm:
$$4.95 \le s < 5.05$$
$$\boxed{845 \le m < 855 \text{ g}, \quad 4.95 \le s < 5.05 \text{ cm}}$$
(b) Using $V=s^3$, cubing the side-length bounds (volume increases with side length):
$$V_{min} = (4.95)^3 = 121.287\ldots \text{ cm}^3$$
$$V_{max} = (5.05)^3 = 128.788\ldots \text{ cm}^3$$
Using density $=\dfrac{\text{mass}}{\text{volume}}$: density is smallest when mass is smallest and volume is largest, and largest when mass is largest and volume is smallest.
$$\rho_{min} = \frac{m_{min}}{V_{max}} = \frac{845}{128.788\ldots} = 6.5612\ldots$$
$$\rho_{max} = \frac{m_{max}}{V_{min}} = \frac{855}{121.287\ldots} = 7.0494\ldots$$
$$\boxed{6.56 \text{ g/cm}^3 \le \rho \le 7.05 \text{ g/cm}^3 \text{ (3 s.f.)}}$$
QUESTION 14
6 marks
Medium
A surveyor wants to estimate the height, $h$, of a tower using the formula $h = d\tan\theta$, where $d = 124$ m is the measured horizontal distance from the surveyor to the base of the tower, and $\theta = 38°$ is the measured angle of elevation to the top of the tower. To simplify the calculation by hand, the surveyor first rounds $\tan 38°$ to 2 decimal places (giving 0.78) before multiplying by 124.
(a) Calculate the surveyor's estimated height using the rounded value of $\tan 38°$.
(b) Calculate the accurate height using the unrounded value of $\tan 38°$ from your GDC, giving your answer correct to 3 significant figures.
(c) Find the percentage error in the surveyor's estimate from part (a), compared to the accurate value found in part (b), correct to 3 significant figures.
Show complete worked solution
(a) Using $h=d\tan\theta$ with the rounded value $\tan38^\circ \approx 0.78$:
$$h_{est} = 124\times0.78$$
$$= 96.72$$
$$\boxed{h_{est} = 96.72 \text{ m}}$$
(b) Using the unrounded GDC value $\tan38^\circ = 0.781286\ldots$:
$$h_{acc} = 124\times0.781286\ldots$$
$$= 96.8795\ldots$$
$$\boxed{h_{acc} \approx 96.9 \text{ m (3 s.f.)}}$$
(c) Using percentage error $=\dfrac{h_{est}-h_{acc}}{h_{acc}}\times100$:
$$\text{Percentage error} = \frac{96.72-96.8795\ldots}{96.8795\ldots}\times100$$
$$= -0.16455\ldots$$
$$\boxed{\text{Percentage error} \approx -0.165\% \text{ (3 s.f.)}}$$
(i.e. the rounded estimate underestimates the true height by about 0.165%)
QUESTION 15
13 marks
Hard
A factory manufactures cylindrical titanium rods. Each rod has a specified radius of 3.20 cm and a specified length of 45.0 cm, each correct to 3 significant figures. The density of titanium is 4.51 g/cm$^3$.
(a) Write down the lower and upper bounds for the radius and for the length of a rod. [2]
(b) Hence find the lower and upper bounds for the volume of a rod, using $V = \pi r^2 h$, correct to 4 significant figures. [4]
(c) Hence find the lower and upper bounds for the mass of a rod, correct to 4 significant figures. (Mass $=$ density $\times$ volume.) [3]
(d) The factory calculates the mass it will charge customers for using the specified (unrounded) values $r = 3.20$ cm and $h = 45.0$ cm directly, without considering bounds. Find this calculated mass, correct to 4 significant figures. Hence find the maximum possible percentage error between this calculated mass and the true mass of an individual rod, correct to 3 significant figures. [4]
Show complete worked solution
(a) Since $r=3.20$ cm is given to 3 s.f. (precision $\pm0.005$ cm):
$$3.195 \le r < 3.205$$
Since $h=45.0$ cm is given to 3 s.f. (precision $\pm0.05$ cm):
$$44.95 \le h < 45.05$$
$$\boxed{r: [3.195,3.205) \text{ cm}, \quad h: [44.95,45.05) \text{ cm}}$$
(b) Using $V=\pi r^2h$: since $V$ increases as both $r$ and $h$ increase, the minimum volume uses the lower bounds of both, and the maximum volume uses the upper bounds of both.
$$V_{min} = \pi(3.195)^2(44.95)$$
$$= \pi(10.2080\ldots)(44.95) = 1441.52\ldots \text{ cm}^3$$
$$V_{max} = \pi(3.205)^2(45.05)$$
$$= \pi(10.2720\ldots)(45.05) = 1453.79\ldots \text{ cm}^3$$
$$\boxed{V_{min} \approx 1442 \text{ cm}^3, \quad V_{max} \approx 1454 \text{ cm}^3 \text{ (4 s.f.)}}$$
(c) Using mass $=$ density $\times$ volume, treating the density (4.51 g/cm$^3$) as exact, so the mass bounds come directly from the volume bounds:
$$m_{min} = 4.51\times1441.52\ldots = 6501.26\ldots \text{ g}$$
$$m_{max} = 4.51\times1453.79\ldots = 6556.58\ldots \text{ g}$$
$$\boxed{m_{min} \approx 6501 \text{ g}, \quad m_{max} \approx 6557 \text{ g (4 s.f.)}}$$
(d) Using the specified values directly:
$$V_{calc} = \pi(3.20)^2(45.0) = \pi(10.24)(45.0) = 1447.65\ldots \text{ cm}^3$$
$$m_{calc} = 4.51\times1447.65\ldots = 6528.88\ldots \text{ g}$$
$$\boxed{m_{calc} \approx 6529 \text{ g (4 s.f.)}}$$
The true mass of an individual rod could lie anywhere between $m_{min}$ and $m_{max}$. Finding the percentage error between the calculated mass and each bound:
At the lower bound: $\dfrac{m_{min}-m_{calc}}{m_{calc}}\times100 = \dfrac{6501.26-6528.88}{6528.88}\times100 = -0.4230\ldots\%$
At the upper bound: $\dfrac{m_{max}-m_{calc}}{m_{calc}}\times100 = \dfrac{6556.58-6528.88}{6528.88}\times100 = 0.4242\ldots\%$
The larger magnitude occurs at the upper bound.
$$\boxed{\text{Maximum possible percentage error} \approx 0.424\% \text{ (3 s.f.)}}$$
QUESTION 16
4 marks
Easy
Express $2\log_5 3 + \log_5 4 - \log_5 6$ as a single logarithm $\log_5 k$, and hence find the value of k.
Show complete worked solution
Using the power law $n\log_a x = \log_a(x^n)$:
$$2\log_5 3 = \log_5(3^2) = \log_5 9$$
Using the product and quotient laws $\log_a x+\log_a y=\log_a(xy)$ and $\log_a x-\log_a y=\log_a\left(\dfrac{x}{y}\right)$:
$$\log_5 9+\log_5 4-\log_5 6 = \log_5\left(\frac{9\times4}{6}\right)$$
$$= \log_5\left(\frac{36}{6}\right) = \log_5 6$$
$$\boxed{2\log_5 3+\log_5 4-\log_5 6 = \log_5 6, \text{ so } k=6}$$
QUESTION 17
4 marks
Easy
Given that $\log_2 x = 3$ and $\log_2 y = 5$, find the value of $\log_2\left(\dfrac{x^2}{y}\right)$.
Show complete worked solution
Using the laws of logarithms, $\log_a\left(\dfrac{x^2}{y}\right) = \log_a(x^2)-\log_a y = 2\log_a x-\log_a y$:
$$\log_2\left(\frac{x^2}{y}\right) = 2\log_2 x-\log_2 y$$
Substituting $\log_2 x=3$ and $\log_2 y=5$:
$$= 2(3)-5$$
$$= 1$$
$$\boxed{\log_2\left(\frac{x^2}{y}\right) = 1}$$
QUESTION 18
6 marks
Medium
Solve the equation $\log_3(x+4) + \log_3(x-2) = 3$ for x, showing that you have found all solutions valid within the domain of the original equation.
Show complete worked solution
First state the domain: since the arguments of both logarithms must be positive, we require $x+4>0$ and $x-2>0$, i.e. $x>2$.
Using the product law $\log_a x+\log_a y=\log_a(xy)$:
$$\log_3(x+4)+\log_3(x-2) = \log_3[(x+4)(x-2)] = 3$$
Rewriting in exponential form:
$$(x+4)(x-2) = 3^3 = 27$$
Expanding the left-hand side:
$$x^2+2x-8 = 27$$
Rearranging into standard quadratic form:
$$x^2+2x-35 = 0$$
Factorising:
$$(x+7)(x-5) = 0$$
$$x=-7 \text{ or } x=5$$
Since the domain requires $x>2$, the solution $x=-7$ is rejected.
$$\boxed{x = 5}$$
QUESTION 19
6 marks
Medium
The Richter-type magnitude M of an earthquake relates to the amplitude A of the seismic waves it produces by
$M = \log_{10}\left(\dfrac{A}{A_0}\right)$
where $A_0$ is a fixed reference amplitude.
Two earthquakes have magnitudes $M_1 = 5.2$ and $M_2 = 6.8$, producing amplitudes $A_1$ and $A_2$ respectively.
Using the laws of logarithms, find the value of $\dfrac{A_2}{A_1}$, i.e. how many times greater the amplitude of the second earthquake is compared to the first, correct to 3 significant figures.
Show complete worked solution
Using the definition $M=\log_{10}\left(\dfrac{A}{A_0}\right)$ for each earthquake:
$$M_2-M_1 = \log_{10}\left(\frac{A_2}{A_0}\right)-\log_{10}\left(\frac{A_1}{A_0}\right)$$
Using the quotient law $\log_a x-\log_a y=\log_a\left(\dfrac{x}{y}\right)$:
$$M_2-M_1 = \log_{10}\left(\frac{A_2/A_0}{A_1/A_0}\right) = \log_{10}\left(\frac{A_2}{A_1}\right)$$
(the reference amplitude $A_0$ cancels)
Substituting $M_1=5.2$, $M_2=6.8$:
$$\log_{10}\left(\frac{A_2}{A_1}\right) = 6.8-5.2 = 1.6$$
Rewriting in exponential form:
$$\frac{A_2}{A_1} = 10^{1.6} = 39.8107\ldots$$
$$\boxed{\frac{A_2}{A_1} \approx 39.8 \text{ (3 s.f.)}}$$
QUESTION 20
15 marks
Hard
Sound intensity level, L, in decibels (dB), is defined in terms of sound intensity I (in W m$^{-2}$) by
$L = 10\log_{10}\left(\dfrac{I}{I_0}\right)$
where $I_0 = 1\times10^{-12}$ W m$^{-2}$ is the threshold of hearing.
(a) A vacuum cleaner produces a sound intensity of $I_1 = 3.2\times10^{-6}$ W m$^{-2}$. Find its sound intensity level $L_1$, correct to 3 significant figures. [4]
(b) A second machine, operating alone, produces a sound intensity level of $L_2 = 85$ dB. Find its sound intensity $I_2$ in W m$^{-2}$, giving your answer in the form $a\times10^{k}$, where $1 \le a < 10$. [5]
(c) When both machines operate together, the total sound intensity is $I_1 + I_2$ (sound intensities add, but decibel levels do not simply add). Use the laws of logarithms and your answers to (a) and (b) to find the combined sound intensity level $L_{total}$ when both machines operate at the same time, correct to 3 significant figures. Comment on the size of the increase from $L_2$ alone. [6]
Show complete worked solution
(a) Substituting $I_1=3.2\times10^{-6}$ into $L=10\log_{10}\left(\dfrac{I}{I_0}\right)$:
$$L_1 = 10\log_{10}\left(\frac{3.2\times10^{-6}}{1\times10^{-12}}\right)$$
$$= 10\log_{10}(3.2\times10^{6})$$
$$= 65.0515\ldots$$
$$\boxed{L_1 \approx 65.1 \text{ dB (3 s.f.)}}$$
(b) Substituting $L_2=85$:
$$85 = 10\log_{10}\left(\frac{I_2}{1\times10^{-12}}\right)$$
Dividing both sides by 10:
$$8.5 = \log_{10}\left(\frac{I_2}{1\times10^{-12}}\right)$$
Rewriting in exponential form:
$$\frac{I_2}{1\times10^{-12}} = 10^{8.5}$$
Solving for $I_2$:
$$I_2 = 10^{8.5}\times10^{-12} = 10^{-3.5} = 3.16227\ldots\times10^{-4}$$
$$\boxed{I_2 \approx 3.16\times10^{-4} \text{ W m}^{-2}}$$
(c) Since sound intensities add, the combined intensity is $I_1+I_2$:
$$I_{total} = I_1+I_2 = 3.2\times10^{-6}+3.16228\times10^{-4}$$
$$= 3.19428\times10^{-4} \text{ W m}^{-2}$$
Substituting into the decibel formula:
$$L_{total} = 10\log_{10}\left(\frac{3.19428\times10^{-4}}{1\times10^{-12}}\right)$$
$$= 85.0437\ldots$$
$$\boxed{L_{total} \approx 85.0 \text{ dB (3 s.f.)}}$$
Comment: this is only about $0.04$ dB higher than $L_2=85$ dB alone. Because $I_1$ is far smaller than $I_2$ (the vacuum cleaner is much quieter than the second machine), adding $I_1$ makes almost no difference to the total intensity; and because the decibel scale is logarithmic, this tiny change in intensity produces an even smaller, essentially imperceptible, change in the decibel level.
QUESTION 21
4 marks
Easy
Simplify fully, giving your answer in the form $a^{p/q}$:
$\dfrac{a^{5/6} \times a^{1/3}}{a^{1/4}}$, where $a > 0$.
Show complete worked solution
Using the laws of exponents - when multiplying powers with the same base the exponents add, and when dividing they subtract:
$$\text{exponent} = \frac{5}{6}+\frac{1}{3}-\frac{1}{4}$$
Using a common denominator of 12:
$$= \frac{10}{12}+\frac{4}{12}-\frac{3}{12}$$
$$= \frac{11}{12}$$
$$\boxed{\frac{a^{5/6}\times a^{1/3}}{a^{1/4}} = a^{11/12}}$$
QUESTION 22
4 marks
Easy
Evaluate $32^{3/5} \times 4^{-3/2}$, showing your method using the laws of exponents.
Show complete worked solution
Write each base as a power of 2, so the laws of exponents can be applied directly.
Since $32=2^5$:
$$32^{3/5} = (2^5)^{3/5} = 2^{5\times3/5} = 2^3 = 8$$
Since $4=2^2$:
$$4^{-3/2} = (2^2)^{-3/2} = 2^{2\times(-3/2)} = 2^{-3} = \frac{1}{8}$$
Multiplying the two results:
$$32^{3/5}\times4^{-3/2} = 8\times\frac{1}{8}$$
$$= 1$$
$$\boxed{32^{3/5}\times4^{-3/2} = 1}$$
QUESTION 23
6 marks
Medium
Biologists model the resting metabolic rate R (in watts) of a mammal in terms of its body mass m (in kg) using the allometric scaling law
$R = 3.5m^{3/4}$
A particular animal has a resting metabolic rate of 140 W.
(a) Rearrange the formula to express m in terms of R. [2]
(b) Hence find the body mass of this animal, correct to 3 significant figures. [4]
Show complete worked solution
(a) Starting from $R=3.5m^{3/4}$, isolate $m^{3/4}$:
$$m^{3/4} = \frac{R}{3.5}$$
Raising both sides to the power $\dfrac{4}{3}$ (the reciprocal of $\dfrac{3}{4}$):
$$\boxed{m = \left(\frac{R}{3.5}\right)^{4/3}}$$
(b) Substituting $R=140$:
$$m = \left(\frac{140}{3.5}\right)^{4/3}$$
$$= 40^{4/3}$$
Using a calculator:
$$m = 136.798\ldots$$
$$\boxed{m \approx 137 \text{ kg (3 s.f.)}}$$
QUESTION 24
6 marks
Medium
Simplify fully:
$\left(\dfrac{x^{2/3}y^{-1/2}}{x^{1/6}y^{3/2}}\right)^{2}$, where $x, y > 0$.
Give your answer in the form $x^p y^q$.
Show complete worked solution
First simplify inside the brackets, subtracting exponents of like bases.
For $x$: $\dfrac{2}{3}-\dfrac{1}{6}=\dfrac{4}{6}-\dfrac{1}{6}=\dfrac{1}{2}$
For $y$: $-\dfrac{1}{2}-\dfrac{3}{2}=-2$
So:
$$\frac{x^{2/3}y^{-1/2}}{x^{1/6}y^{3/2}} = x^{1/2}y^{-2}$$
Now raise this result to the power 2, multiplying each exponent by 2:
$$\left(x^{1/2}y^{-2}\right)^2 = x^{1}y^{-4}$$
$$\boxed{\left(\frac{x^{2/3}y^{-1/2}}{x^{1/6}y^{3/2}}\right)^{2} = xy^{-4}}$$
QUESTION 25
14 marks
Hard
The power output P (in watts) of a wind turbine depends on its blade radius r (in metres) and the wind speed v (in m s$^{-1}$) according to the model
$P = kr^2v^3$
where k is a positive constant depending on air density and turbine efficiency.
A turbine with blade radius 40 m, operating in a wind speed of 12 m s$^{-1}$, produces a power output of 1 200 000 W.
(a) Find the value of k, correct to 3 significant figures. [4]
(b) By rearranging the formula, express r in terms of P, v and k, writing your answer using a rational exponent. [4]
(c) A new turbine design is required to produce a power output of 2 000 000 W in the same wind speed of 12 m s$^{-1}$. Using an unrounded value of k from part (a) and your formula from part (b), find the required blade radius, correct to 3 significant figures. [6]
Show complete worked solution
(a) Rearranging $P=kr^2v^3$ for $k$:
$$k = \frac{P}{r^2v^3}$$
Substituting $P=1{,}200{,}000$, $r=40$, $v=12$:
$$k = \frac{1{,}200{,}000}{40^2\times12^3}$$
$$= \frac{1{,}200{,}000}{1600\times1728}$$
$$= \frac{1{,}200{,}000}{2{,}764{,}800}$$
$$= 0.434027\ldots$$
$$\boxed{k \approx 0.434 \text{ (3 s.f.)}}$$
(b) Starting from $P=kr^2v^3$, isolate $r^2$:
$$r^2 = \frac{P}{kv^3}$$
Raising both sides to the power $\dfrac{1}{2}$:
$$\boxed{r = \left(\frac{P}{kv^3}\right)^{1/2}}$$
(c) Substituting $P=2{,}000{,}000$, $v=12$, and the unrounded value $k=0.4340277\ldots$ from part (a):
$$r = \left(\frac{2{,}000{,}000}{0.4340277\ldots\times12^3}\right)^{1/2}$$
$$= \left(\frac{2{,}000{,}000}{750.0}\right)^{1/2}$$
$$= (2666.67\ldots)^{1/2}$$
$$= 51.6397\ldots$$
$$\boxed{r \approx 51.6 \text{ m (3 s.f.)}}$$
QUESTION 26
5 marks
Easy
The population of a particular country is recorded as $41\,382\,600$ people.
(a) Round this population to 3 significant figures. [2]
(b) Round the original population to the nearest million. [1]
(c) Write your answer to part (a) in the form $a \times 10^{k}$, where $1 \le a < 10$ and $k \in \mathbb{Z}$. [2]
Show complete worked solution
(a) To round $41\,382\,600$ to 3 significant figures, keep the first three significant digits ($4$, $1$, $3$) and look at the next digit ($8$) to decide whether to round up.
Since the next digit is $8 \ge 5$, round the third significant figure up from $3$ to $4$:
$$\boxed{41\,382\,600 \approx 41\,400\,000 \text{ (3 s.f.)}}$$
(b) To round to the nearest million, look at the digit in the hundred-thousands place. The population is $41\,382\,600$, and the digit after the millions place is $3$ (i.e. $382\,600 < 500\,000$), so round down:
$$\boxed{41\,382\,600 \approx 41\,000\,000 \text{ (nearest million)}}$$
(c) Writing $41\,400\,000$ in the form $a \times 10^{k}$ with $1 \le a < 10$, move the decimal point 7 places to the left:
$$41\,400\,000 = 4.14 \times 10^{7}$$
$$\boxed{4.14 \times 10^{7}}$$
QUESTION 27
5 marks
Easy
The diameter of a single red blood cell is approximately $0.0000075$ metres.
(a) Write this diameter in the form $a \times 10^{k}$ metres, where $1 \le a < 10$ and $k \in \mathbb{Z}$. [2]
(b) A scientist places $2.0 \times 10^{5}$ red blood cells in a single row, edge to edge, on a microscope slide. Find the total length occupied by the cells, in metres, giving your answer in the form $a \times 10^{k}$. [3]
Show complete worked solution
(a) Moving the decimal point 6 places to the right in $0.0000075$ gives a coefficient between 1 and 10:
$$0.0000075 = 7.5 \times 10^{-6}$$
$$\boxed{7.5 \times 10^{-6} \text{ m}}$$
(b) Using total length = (diameter of one cell) $\times$ (number of cells):
$$\text{Total length} = (7.5 \times 10^{-6}) \times (2.0 \times 10^{5})$$
Multiplying the coefficients and adding the powers of 10:
$$= (7.5 \times 2.0) \times 10^{-6+5} = 15 \times 10^{-1}$$
Since $15$ is not in the range $1 \le a < 10$, rewrite in standard form:
$$15 \times 10^{-1} = 1.5 \times 10^{0} = 1.5$$
$$\boxed{1.5 \times 10^{0} \text{ m (i.e. } 1.5 \text{ m)}}$$
QUESTION 28
5 marks
Easy
(a) By rounding each number in the following calculation to 1 significant figure, estimate the value of
$$\frac{587 \times 3.12}{0.213}$$
without using a calculator. Show the rounded values you use. [3]
(b) Use your GDC to find the accurate value of the calculation, correct to 3 significant figures. [2]
Show complete worked solution
(a) Rounding each number to 1 significant figure:
$$587 \approx 600, \qquad 3.12 \approx 3, \qquad 0.213 \approx 0.2$$
Substituting these rounded values into the calculation:
$$\text{Estimate} = \frac{600 \times 3}{0.2} = \frac{1800}{0.2}$$
$$= 9000$$
$$\boxed{\text{Estimate} \approx 9000}$$
(b) Using a GDC with the unrounded values:
$$\frac{587 \times 3.12}{0.213} = \frac{1831.44}{0.213} = 8598.30\ldots$$
$$\boxed{8600 \text{ (3 s.f.)}}$$
QUESTION 29
4 marks
Easy
The length of a garden path is measured and recorded as $12.4$ m, correct to the nearest $0.1$ m.
Write down the lower bound and the upper bound of the actual length, $L$, of the path, and express your answer as an error interval using inequality notation.
Show complete worked solution
Since the length is given correct to the nearest $0.1$ m, the true value lies within half of $0.1$ m (i.e. $0.05$ m) of the recorded value $12.4$ m.
Lower bound: $12.4 - 0.05 = 12.35$ m
Upper bound: $12.4 + 0.05 = 12.45$ m
The lower bound is attainable (a true length of exactly $12.35$ m would round to $12.4$ m), but the upper bound is not attainable (a true length of exactly $12.45$ m would round up to $12.5$ m), so the interval is closed at the lower end and open at the upper end:
$$\boxed{12.35 \le L < 12.45 \text{ (m)}}$$
QUESTION 30
5 marks
Easy
A student estimates the area of a square field by measuring its side length as $20$ m and calculating the area as $400$ m$^2$. A more precise survey shows the actual side length of the field is $20.6$ m.
Find the percentage error in the student's estimated area, compared to the actual area (based on the accurate side length of $20.6$ m), correct to 3 significant figures.
Show complete worked solution
Using $A = (\text{side length})^2$, the actual area (based on the accurate side length):
$$A_{act} = (20.6)^2 = 424.36 \text{ m}^2$$
The student's estimated area was $A_{est} = 400$ m$^2$.
Using percentage error $= \left|\dfrac{A_{est}-A_{act}}{A_{act}}\right|\times 100$:
$$\text{Percentage error} = \frac{|400-424.36|}{424.36}\times100$$
$$= \frac{24.36}{424.36}\times100 = 5.7404\ldots$$
$$\boxed{\text{Percentage error} \approx 5.74\% \text{ (3 s.f.)}}$$
QUESTION 31
4 marks
Easy
The width of a single transistor on a computer chip is $1.4 \times 10^{-8}$ m. A particular chip contains $6.0 \times 10^{9}$ of these transistors, arranged in a single row, edge to edge.
Find the total length occupied by the transistors, giving your answer in the form $a \times 10^{k}$ m, correct to 2 significant figures.
Show complete worked solution
Using total length = (width of one transistor) $\times$ (number of transistors):
$$\text{Total length} = (1.4 \times 10^{-8}) \times (6.0 \times 10^{9})$$
Multiplying the coefficients and adding the powers of 10:
$$= (1.4 \times 6.0) \times 10^{-8+9}$$
$$= 8.4 \times 10^{1}$$
This is already in the form $a \times 10^{k}$ with $1 \le a < 10$, and $8.4$ is already correct to 2 significant figures.
$$\boxed{8.4 \times 10^{1} \text{ m (i.e. } 84 \text{ m)}}$$
QUESTION 32
4 marks
Easy
A digital kitchen scale truncates every reading to the nearest whole gram below the true mass (it always rounds down, never up). A packet of rice placed on the scale displays a mass of $350$ g.
Write down the lower bound and the upper bound of the actual mass, $m$, of the packet, and express your answer as an error interval using inequality notation.
Show complete worked solution
Truncation is different from ordinary rounding: a truncated reading of $350$ g means the true mass, once its decimal part is cut off (discarded), gives exactly $350$.
This means the true mass could be exactly $350$ g, or any value up to (but not including) $351$ g, since any mass from $350.000\ldots$ up to $350.999\ldots$ g would truncate down to display as $350$ g.
$$\boxed{350 \le m < 351 \text{ (g)}}$$
QUESTION 33
4 marks
Easy
The table below shows four measurements recorded during a science experiment. Copy and complete the table by rounding each measurement to the stated number of significant figures (s.f.).
| Measurement | Round to | Rounded value |
|---|---|---|
| $0.048273$ | 2 s.f. | ? |
| $5286.5$ | 3 s.f. | ? |
| $0.0090351$ | 2 s.f. | ? |
| $719\,382$ | 4 s.f. | ? |
Show complete worked solution
In each case, identify the first significant figures required and use the next digit to decide whether to round up or down.
$0.048273$: the first 2 significant figures are $4$ and $8$ (leading zeros are not significant); the next digit is $2$, so round down:
$$0.048273 \approx 0.048 \text{ (2 s.f.)}$$
$5286.5$: the first 3 significant figures are $5$, $2$, $8$; the next digit is $6 \ge 5$, so round up:
$$5286.5 \approx 5290 \text{ (3 s.f.)}$$
$0.0090351$: the first 2 significant figures are $9$ and $0$; the next digit is $3$, so round down:
$$0.0090351 \approx 0.0090 \text{ (2 s.f.)}$$
$719\,382$: the first 4 significant figures are $7$, $1$, $9$, $3$; the next digit is $8 \ge 5$, so round up:
$$719\,382 \approx 719\,400 \text{ (4 s.f.)}$$
| Measurement | Round to | Rounded value |
|---|---|---|
| $0.048273$ | 2 s.f. | $\boxed{0.048}$ |
| $5286.5$ | 3 s.f. | $\boxed{5290}$ |
| $0.0090351$ | 2 s.f. | $\boxed{0.0090}$ |
| $719\,382$ | 4 s.f. | $\boxed{719\,400}$ |
QUESTION 34
4 marks
Easy
A country has a population of $6.72 \times 10^{7}$ people and a total land area of $2.4 \times 10^{5}$ km$^2$.
Find the population density of the country, in people per km$^2$, correct to 3 significant figures.
Show complete worked solution
Using density = population / area:
$$\text{Density} = \frac{6.72 \times 10^{7}}{2.4 \times 10^{5}}$$
Dividing the coefficients and subtracting the powers of 10:
$$= \frac{6.72}{2.4}\times10^{7-5} = 2.8\times10^{2}$$
$$= 280$$
$$\boxed{280 \text{ people per km}^2 \text{ (3 s.f.)}}$$
QUESTION 35
5 marks
Easy
A shopper buys 8 items, each priced at approximately \$4.95, and 3 items, each priced at approximately \$11.80.
(a) By rounding each price to the nearest whole dollar, estimate the total cost of the shopping. [3]
(b) State, with a reason, whether this estimate is an overestimate or an underestimate of the actual total cost. [2]
Show complete worked solution
(a) Rounding each price to the nearest whole dollar:
$$\$4.95 \approx \$5, \qquad \$11.80 \approx \$12$$
Using estimated total = (number of items) $\times$ (rounded price), summed over both types of item:
$$\text{Estimate} = 8\times\$5 + 3\times\$12$$
$$= \$40 + \$36$$
$$\boxed{\text{Estimate} = \$76}$$
(b) The actual total cost, using the unrounded prices, is:
$$\text{Actual} = 8\times\$4.95 + 3\times\$11.80 = \$39.60+\$35.40 = \$75.00$$
Since $\$76 > \$75.00$, the estimate is an overestimate. This makes sense because both individual prices were rounded up (\$4.95 rounded up to \$5, and \$11.80 rounded up to \$12), so the estimated total is larger than the true total.
$$\boxed{\text{The estimate is an overestimate, since both prices were rounded up}}$$
QUESTION 36
6 marks
Medium
Two wooden planks have lengths measured as $3.4$ m and $1.8$ m, each correct to the nearest $0.1$ m. The shorter plank is cut from the longer plank, leaving an offcut piece.
(a) Write down the lower and upper bounds for the length of each plank. [2]
(b) Find the lower bound and upper bound for the length of the offcut (the length remaining after the shorter plank's length is subtracted from the longer plank's length). [4]
Show complete worked solution
(a) Since each length is given correct to the nearest $0.1$ m, the true value lies within $0.05$ m of the recorded value.
For the longer plank, length $L$:
$$3.35 \le L < 3.45 \text{ (m)}$$
For the shorter plank, length $S$:
$$1.75 \le S < 1.85 \text{ (m)}$$
$$\boxed{3.35 \le L < 3.45, \quad 1.75 \le S < 1.85 \text{ (m)}}$$
(b) The offcut length is $O = L - S$. This is smallest when $L$ is as small as possible and $S$ is as large as possible (subtracting the most), and largest when $L$ is as large as possible and $S$ is as small as possible (subtracting the least):
$$O_{min} = L_{min} - S_{max} = 3.35 - 1.85 = 1.50$$
$$O_{max} = L_{max} - S_{min} = 3.45 - 1.75 = 1.70$$
$$\boxed{1.50 \text{ m} \le O \le 1.70 \text{ m}}$$
QUESTION 37
6 marks
Medium
A rectangular solar panel has a length of $165$ cm and a width of $99$ cm, each measured correct to the nearest cm.
(a) Write down the lower and upper bounds for the length and for the width of the panel. [2]
(b) Find the lower bound and upper bound for the area of the panel, in cm$^2$, correct to 4 significant figures. [4]
Show complete worked solution
(a) Since each measurement is correct to the nearest cm, the true value lies within $0.5$ cm of the recorded value.
For the length, $l$: $$164.5 \le l < 165.5 \text{ (cm)}$$
For the width, $w$: $$98.5 \le w < 99.5 \text{ (cm)}$$
$$\boxed{164.5 \le l < 165.5, \quad 98.5 \le w < 99.5 \text{ (cm)}}$$
(b) Using $A = l \times w$: since $A$ increases as both $l$ and $w$ increase, the minimum area uses the lower bounds of both, and the maximum area uses the upper bounds of both.
$$A_{min} = 164.5 \times 98.5 = 16203.25 \text{ cm}^2$$
$$A_{max} = 165.5 \times 99.5 = 16467.25 \text{ cm}^2$$
$$\boxed{A_{min} \approx 16200 \text{ cm}^2, \quad A_{max} \approx 16470 \text{ cm}^2 \text{ (4 s.f.)}}$$
QUESTION 38
6 marks
Medium
A high-resolution satellite image file has a size of $4.8 \times 10^{9}$ bytes. The satellite can transmit data to a ground station at a constant rate of $1.5 \times 10^{7}$ bytes per second.
(a) Find the time taken to transmit one image, in seconds, correct to 3 significant figures. [3]
(b) Given that there are $3.6 \times 10^{3}$ seconds in one hour, find the number of complete images that could be transmitted in one hour of continuous transmission. [3]
Show complete worked solution
(a) Using time = (file size) / (transmission rate):
$$t = \frac{4.8 \times 10^{9}}{1.5 \times 10^{7}}$$
Dividing the coefficients and subtracting the powers of 10:
$$= \frac{4.8}{1.5}\times10^{9-7} = 3.2\times10^{2}$$
$$\boxed{t = 320 \text{ s (3 s.f.)}}$$
(b) Using the unrounded values, the total number of bytes transmitted in one hour is:
$$\text{Total bytes} = (1.5\times10^{7})\times(3.6\times10^{3}) = 5.4\times10^{10} \text{ bytes}$$
Dividing by the size of one image:
$$\text{Number of images} = \frac{5.4\times10^{10}}{4.8\times10^{9}} = 11.25$$
Since only complete images can be counted, round down to the nearest whole number:
$$\boxed{11 \text{ complete images}}$$
QUESTION 39
7 marks
Medium
A cafe owner wants to quickly estimate her weekly revenue. In one week she sells approximately 218 cups of coffee at \$3.85 each, and 76 pastries at \$2.60 each.
(a) By rounding the number of cups to the nearest 200 and the price per cup to the nearest whole dollar, and by rounding the number of pastries to the nearest 100 and the price per pastry to the nearest whole dollar, estimate her total weekly revenue. [3]
(b) Calculate the actual total weekly revenue, using the unrounded figures, correct to 2 decimal places. [2]
(c) Find the percentage error in the estimate from part (a), compared to the actual revenue in part (b), correct to 3 significant figures. [2]
Show complete worked solution
(a) Rounding each quantity as instructed:
$$218 \approx 200 \text{ cups}, \quad \$3.85 \approx \$4, \quad 76 \approx 100 \text{ pastries}, \quad \$2.60 \approx \$3$$
Using estimated revenue = (rounded number) $\times$ (rounded price), summed over both products:
$$\text{Estimate} = 200\times\$4 + 100\times\$3$$
$$= \$800+\$300$$
$$\boxed{\text{Estimate} = \$1100}$$
(b) Using the unrounded figures:
$$\text{Actual} = 218\times\$3.85 + 76\times\$2.60$$
$$= \$839.30+\$197.60$$
$$\boxed{\text{Actual} = \$1036.90}$$
(c) Using percentage error $=\left|\dfrac{\text{Estimate}-\text{Actual}}{\text{Actual}}\right|\times100$:
$$\text{Percentage error} = \frac{|1100-1036.90|}{1036.90}\times100$$
$$= \frac{63.10}{1036.90}\times100 = 6.0854\ldots$$
$$\boxed{\text{Percentage error} \approx 6.09\% \text{ (3 s.f.)}}$$
QUESTION 40
6 marks
Medium
A cyclist travels a distance of $15$ km, measured correct to the nearest km, in a time of $42$ minutes, measured correct to the nearest minute.
Find the lower bound and the upper bound of the cyclist's average speed, in km/h, correct to 3 significant figures.
Show complete worked solution
First find the bounds for the distance, $d$, and the time, $t$ (in minutes).
Since $d=15$ km is correct to the nearest km:
$$14.5 \le d < 15.5 \text{ (km)}$$
Since $t=42$ minutes is correct to the nearest minute:
$$41.5 \le t < 42.5 \text{ (minutes)}$$
Converting the time bounds to hours (dividing by 60):
$$\frac{41.5}{60} \le t_{hr} < \frac{42.5}{60}, \quad \text{i.e.} \quad 0.69166\ldots \le t_{hr} < 0.70833\ldots \text{ (hours)}$$
Using speed = distance / time: speed is smallest when distance is smallest and time is largest, and speed is largest when distance is largest and time is smallest.
$$\text{Speed}_{min} = \frac{d_{min}}{t_{hr,max}} = \frac{14.5}{42.5/60} = \frac{14.5\times60}{42.5} = 20.4705\ldots$$
$$\text{Speed}_{max} = \frac{d_{max}}{t_{hr,min}} = \frac{15.5}{41.5/60} = \frac{15.5\times60}{41.5} = 22.4096\ldots$$
$$\boxed{20.5 \text{ km/h} \le \text{Speed} \le 22.4 \text{ km/h} \text{ (3 s.f.)}}$$
QUESTION 41
6 marks
Medium
A measurement is recorded as $8.0$ cm. It is known that the true value $x$ satisfies $7.95 \le x < 8.05$.
(a) State the degree of accuracy (rounding rule) that must have been used to obtain the recorded value $8.0$ cm from $x$, giving a reason. [2]
(b) A different measurement is also recorded as $8.0$ cm, but this time it is known that the true value satisfies $7.5 \le x < 8.5$. State the degree of accuracy that was used in this case. [2]
(c) Explain why the recorded value "$8.0$ cm" alone, without also knowing which error interval applies, is ambiguous about the precision of the original measurement. [2]
Show complete worked solution
(a) The interval $7.95 \le x < 8.05$ has half-width $0.05$, centred on $8.0$. Since $0.05$ is half of $0.1$, this is the error interval produced by rounding to the nearest $0.1$ cm.
$$\boxed{\text{Rounded to the nearest } 0.1 \text{ cm (i.e. 1 decimal place)}}$$
(b) The interval $7.5 \le x < 8.5$ has half-width $0.5$, centred on $8.0$. Since $0.5$ is half of $1$, this is the error interval produced by rounding to the nearest whole centimetre.
$$\boxed{\text{Rounded to the nearest } 1 \text{ cm (i.e. the nearest whole number)}}$$
(c) Both roundings produce the same displayed value, "$8.0$ cm", even though the true underlying precision of the original measurements is very different (one is accurate to within $0.05$ cm, the other only to within $0.5$ cm). Writing a trailing zero (as in "$8.0$") suggests, but does not guarantee, that the value was rounded to 1 decimal place - without being told the actual rounding rule (or the error interval) that was used, a reader cannot be certain how precise the original measurement really was.
$$\boxed{\text{The same displayed value can arise from different degrees of accuracy, so the precision cannot be inferred from the value alone}}$$
QUESTION 42
7 marks
Medium
A spherical droplet of oil has volume $V = 4.19 \times 10^{-3}$ cm$^3$. The density of the oil is $0.92$ g/cm$^3$ (treat this value as exact).
(a) Find the mass of the droplet, in grams, giving your answer in the form $a \times 10^{k}$, correct to 3 significant figures. [3]
(b) The droplet is one of $5.5 \times 10^{6}$ identical droplets released from a spray nozzle. Using the unrounded value from part (a), find the total mass of oil released, in kilograms, giving your answer in the form $a \times 10^{k}$, correct to 3 significant figures. ($1$ kg $=1000$ g) [4]
Show complete worked solution
(a) Using mass = density $\times$ volume:
$$m = 0.92 \times (4.19\times10^{-3})$$
$$= 3.8548 \times 10^{-3} \text{ g}$$
$$\boxed{m \approx 3.85 \times 10^{-3} \text{ g (3 s.f.)}}$$
(b) Using the unrounded value $m = 3.8548\times10^{-3}$ g, the total mass of all the droplets, in grams, is:
$$m_{tot} = (3.8548\times10^{-3}) \times (5.5\times10^{6})$$
$$= 21201.4 \text{ g}$$
Converting to kilograms (dividing by 1000):
$$m_{tot} = \frac{21201.4}{1000} = 21.2014 \text{ kg}$$
Writing this in the form $a\times10^k$, correct to 3 s.f.:
$$\boxed{m_{tot} \approx 2.12 \times 10^{1} \text{ kg (3 s.f.)}}$$
QUESTION 43
7 marks
Medium
The value of $\pi$ can be approximated using the fraction $\dfrac{22}{7}$, or using the decimal $3.14$.
(a) Using the accurate value of $\pi$ from your GDC, find the percentage error in using $\dfrac{22}{7}$ as an approximation for $\pi$, correct to 3 significant figures. [3]
(b) Find the percentage error in using $3.14$ as an approximation for $\pi$, correct to 3 significant figures. [3]
(c) Hence state, with justification, which of the two approximations is more accurate. [1]
Show complete worked solution
(a) From a GDC, $\pi = 3.14159265\ldots$, and $\dfrac{22}{7} = 3.142857\ldots$
Using percentage error $=\left|\dfrac{\text{approx}-\pi}{\pi}\right|\times100$:
$$\text{Percentage error} = \frac{|3.142857\ldots-3.14159265\ldots|}{3.14159265\ldots}\times100$$
$$= 0.040249\ldots$$
$$\boxed{\text{Percentage error} \approx 0.0402\% \text{ (3 s.f.)}}$$
(b) Using the same formula with $3.14$:
$$\text{Percentage error} = \frac{|3.14-3.14159265\ldots|}{3.14159265\ldots}\times100$$
$$= 0.050695\ldots$$
$$\boxed{\text{Percentage error} \approx 0.0507\% \text{ (3 s.f.)}}$$
(c) Since $0.0402\% < 0.0507\%$, the fraction $\dfrac{22}{7}$ has the smaller percentage error.
$$\boxed{\dfrac{22}{7} \text{ is the more accurate approximation for } \pi}$$
QUESTION 44
5 marks
Medium
A length of rope is recorded as $14$ m, correct to the nearest metre. The rope is to be cut into 6 equal pieces.
(a) Write down the error interval for the true length, $L$, of the rope. [2]
(b) Find the lower bound and upper bound for the length of each piece, in metres, correct to 3 significant figures. [3]
Show complete worked solution
(a) Since the length is correct to the nearest metre, the true value lies within $0.5$ m of the recorded value $14$ m:
$$\boxed{13.5 \le L < 14.5 \text{ (m)}}$$
(b) Since the rope is cut into 6 equal pieces, the length of each piece is $L/6$. Dividing by the exact constant $6$ preserves the direction of the inequality (each bound is simply divided by 6):
$$\text{Piece}_{min} = \frac{13.5}{6} = 2.25 \text{ m}$$
$$\text{Piece}_{max} = \frac{14.5}{6} = 2.41666\ldots \text{ m}$$
$$\boxed{2.25 \text{ m} \le \text{Piece length} < 2.42 \text{ m (3 s.f.)}}$$
QUESTION 45
5 marks
Medium
The mass of a hydrogen atom is approximately $1.67 \times 10^{-27}$ kg. The mass of the Earth is approximately $5.97 \times 10^{24}$ kg.
Find how many times greater the mass of the Earth is than the mass of a single hydrogen atom, giving your answer in the form $a \times 10^{k}$, correct to 3 significant figures.
Show complete worked solution
Using ratio = (mass of Earth) / (mass of hydrogen atom):
$$\text{Ratio} = \frac{5.97\times10^{24}}{1.67\times10^{-27}}$$
Dividing the coefficients and subtracting the powers of 10:
$$= \frac{5.97}{1.67}\times10^{24-(-27)}$$
$$= 3.57485\ldots\times10^{51}$$
$$\boxed{\text{Ratio} \approx 3.57\times10^{51} \text{ (3 s.f.)}}$$
QUESTION 46
11 marks
Hard
The kinetic energy of a moving object is given by $E_k = \dfrac{1}{2}mv^2$, where $m$ is the mass in kg and $v$ is the speed in m/s.
A vehicle's mass is measured as $1450$ kg, correct to the nearest $10$ kg, and its speed is measured as $22.0$ m/s, correct to the nearest $0.5$ m/s.
(a) Write down the lower and upper bounds for the mass and for the speed of the vehicle. [3]
(b) Hence find the lower bound and upper bound for the kinetic energy of the vehicle, in J, giving your answers correct to 4 significant figures. [4]
(c) Using the measured values ($m=1450$ kg, $v=22.0$ m/s) directly, calculate the kinetic energy, correct to 4 significant figures. Hence find the maximum possible percentage error between this calculated value and the true kinetic energy of the vehicle, correct to 3 significant figures. [4]
Show complete worked solution
(a) Since the mass is correct to the nearest $10$ kg, the true mass lies within $5$ kg of $1450$ kg:
$$1445 \le m < 1455 \text{ (kg)}$$
Since the speed is correct to the nearest $0.5$ m/s, the true speed lies within $0.25$ m/s of $22.0$ m/s:
$$21.75 \le v < 22.25 \text{ (m/s)}$$
$$\boxed{1445 \le m < 1455 \text{ kg}, \quad 21.75 \le v < 22.25 \text{ m/s}}$$
(b) Using $E_k = \dfrac{1}{2}mv^2$: since $v>0$, $v^2$ increases as $v$ increases, and $E_k$ increases as both $m$ and $v^2$ increase. So the minimum $E_k$ uses the lower bounds of both $m$ and $v$, and the maximum $E_k$ uses the upper bounds of both.
$$E_{k,min} = \frac{1}{2}(1445)(21.75)^2$$
$$= \frac{1}{2}(1445)(473.0625) = 341787.65625 \text{ J}$$
$$E_{k,max} = \frac{1}{2}(1455)(22.25)^2$$
$$= \frac{1}{2}(1455)(495.0625) = 360157.96875 \text{ J}$$
$$\boxed{E_{k,min} \approx 341800 \text{ J}, \quad E_{k,max} \approx 360200 \text{ J (4 s.f.)}}$$
(c) Using the measured values directly:
$$E_{k,calc} = \frac{1}{2}(1450)(22.0)^2 = \frac{1}{2}(1450)(484)$$
$$\boxed{E_{k,calc} = 350900 \text{ J (4 s.f.)}}$$
The true kinetic energy could lie anywhere between $E_{k,min}$ and $E_{k,max}$. Finding the percentage error between $E_{k,calc}$ and each bound (using the unrounded bounds from part (b)):
At the lower bound: $\dfrac{341787.65625-350900}{350900}\times100 = -2.5968\ldots\%$
At the upper bound: $\dfrac{360157.96875-350900}{350900}\times100 = 2.6383\ldots\%$
The larger magnitude occurs at the upper bound.
$$\boxed{\text{Maximum possible percentage error} \approx 2.64\% \text{ (3 s.f.)}}$$
QUESTION 47
10 marks
Hard
A student wants to evaluate
$$y = \frac{(3.7)^2 - \sqrt{18.4}}{2.6 \times 1.15}$$
(a) The student rounds each intermediate value to 2 decimal places at every step of the calculation, as follows: $(3.7)^2 = 13.69$; $\sqrt{18.4} = 4.289522\ldots \approx 4.29$; numerator $\approx 13.69-4.29=9.40$; denominator $=2.6\times1.15=2.99$. Using these rounded intermediate values, find the student's final value of $y$, correct to 4 decimal places. [3]
(b) Use your GDC to find the accurate value of $y$, carrying out the entire calculation without any intermediate rounding, correct to 4 decimal places. [3]
(c) Find the percentage error caused by the student's intermediate rounding, comparing the answer in part (a) to the accurate answer in part (b), correct to 3 significant figures. [3]
(d) State, with a reason, a more reliable approach the student could use on a GDC to avoid this type of error. [1]
Show complete worked solution
(a) Using the student's rounded intermediate values:
$$y_{student} = \frac{9.40}{2.99}$$
$$= 3.14381\ldots$$
$$\boxed{y_{student} \approx 3.1438 \text{ (4 d.p.)}}$$
(b) Carrying full precision through the entire calculation (no intermediate rounding):
$$(3.7)^2 = 13.69, \qquad \sqrt{18.4} = 4.289522\ldots$$
$$\text{numerator} = 13.69 - 4.289522\ldots = 9.400477\ldots$$
$$\text{denominator} = 2.6\times1.15 = 2.99$$
$$y_{exact} = \frac{9.400477\ldots}{2.99} = 3.143972\ldots$$
$$\boxed{y_{exact} \approx 3.1440 \text{ (4 d.p.)}}$$
(c) Using percentage error $=\left|\dfrac{y_{student}-y_{exact}}{y_{exact}}\right|\times100$, and the unrounded values from (a) and (b):
$$\text{Percentage error} = \frac{|3.143812\ldots-3.143972\ldots|}{3.143972\ldots}\times100$$
$$= 0.005083\ldots$$
$$\boxed{\text{Percentage error} \approx 0.00508\% \text{ (3 s.f.)}}$$
(d) The student should store the full, unrounded value of each intermediate result in the GDC's memory (or type the entire expression into the calculator in a single line) rather than writing down a rounded value and re-entering it for the next step. This avoids compounding small rounding errors from each intermediate step, which is especially important in longer or more sensitive calculations where such errors could accumulate to something more significant.
$$\boxed{\text{Carry out the whole calculation in one unrounded step, only rounding the final answer}}$$
QUESTION 48
12 marks
Hard
A physicist measures the current in a circuit as $I = 2.50 \times 10^{-3}$ A, correct to 3 significant figures, and the resistance as $R = 4.80 \times 10^{3}$ $\Omega$, correct to 3 significant figures. The power dissipated in the circuit is given by $P = I^2R$.
(a) Write down the lower and upper bounds for $I$ and for $R$. [2]
(b) Find the lower bound and upper bound for $P$, in watts, giving your answers in the form $a \times 10^{k}$, correct to 3 significant figures. [4]
(c) Calculate $P$ using the given (unrounded) values of $I$ and $R$ directly, giving your answer in the form $a \times 10^{k}$ W, correct to 3 significant figures. [2]
(d) Using the bounds found in part (b), find the maximum possible percentage error between the calculated value $P$ from part (c) and the true power dissipated, correct to 3 significant figures. [4]
Show complete worked solution
(a) Since $I = 2.50\times10^{-3}$ A is correct to 3 s.f., the precision is $\pm0.005\times10^{-3}$ A (half of the smallest place value shown):
$$2.495\times10^{-3} \le I < 2.505\times10^{-3} \text{ (A)}$$
Since $R=4.80\times10^{3}$ $\Omega$ is correct to 3 s.f., the precision is $\pm0.005\times10^{3}$ $\Omega$:
$$4795 \le R < 4805 \text{ (}\Omega\text{)}$$
$$\boxed{2.495\times10^{-3} \le I < 2.505\times10^{-3} \text{ A}, \quad 4795 \le R < 4805 \text{ } \Omega}$$
(b) Using $P=I^2R$: since $I>0$, $I^2$ increases as $I$ increases, and $P$ increases as both $I^2$ and $R$ increase. So the minimum $P$ uses the lower bounds of both $I$ and $R$, and the maximum $P$ uses the upper bounds of both.
$$P_{min} = (2.495\times10^{-3})^2\times4795$$
$$= (6.225025\times10^{-6})\times4795 = 0.0298489\ldots \text{ W}$$
$$P_{max} = (2.505\times10^{-3})^2\times4805$$
$$= (6.275025\times10^{-6})\times4805 = 0.0301514\ldots \text{ W}$$
$$\boxed{P_{min} \approx 2.98\times10^{-2} \text{ W}, \quad P_{max} \approx 3.02\times10^{-2} \text{ W (3 s.f.)}}$$
(c) Using the given values $I=2.50\times10^{-3}$ A, $R=4.80\times10^{3}$ $\Omega$ directly:
$$P = (2.50\times10^{-3})^2\times(4.80\times10^{3})$$
$$= (6.25\times10^{-6})\times(4.80\times10^{3}) = 3.00\times10^{-2} \text{ W}$$
$$\boxed{P \approx 3.00\times10^{-2} \text{ W (3 s.f.)}}$$
(d) Using the unrounded bounds from part (b), $P_{min}=0.0298489\ldots$ W and $P_{max}=0.0301514\ldots$ W, and $P_{calc}=0.03$ W:
At the lower bound: $\dfrac{0.0298489\ldots-0.03}{0.03}\times100 = -0.5033\ldots\%$
At the upper bound: $\dfrac{0.0301514\ldots-0.03}{0.03}\times100 = 0.5049\ldots\%$
The larger magnitude occurs at the upper bound.
$$\boxed{\text{Maximum possible percentage error} \approx 0.505\% \text{ (3 s.f.)}}$$
QUESTION 49
12 marks
Hard
A city's electricity supplier estimates its total daily energy consumption using the formula
$$E = n \times p \times t$$
where $n$ is the number of households, $p$ is the average power consumption per household in kW, and $t$ is the number of hours in a day.
The city has $n=285\,000$ households, correct to the nearest $1000$, an average power consumption of $p=1.8$ kW per household, correct to 2 significant figures, and $t=24$ hours (an exact value).
(a) Write down the lower and upper bounds for $n$ and for $p$. [3]
(b) Find the lower bound and upper bound for the total daily energy consumption $E$, in kWh, giving your answers in the form $a\times10^{k}$, correct to 3 significant figures. [5]
(c) The supplier's published estimate, $E_{est}$, is calculated using $n=285\,000$ and $p=1.8$ directly. Calculate $E_{est}$ in the form $a\times10^{k}$ kWh, correct to 3 significant figures, and find the maximum possible percentage error between $E_{est}$ and the true value of $E$, correct to 3 significant figures. [4]
Show complete worked solution
(a) Since $n=285\,000$ is correct to the nearest $1000$, the precision is $\pm500$:
$$284\,500 \le n < 285\,500$$
Since $p=1.8$ kW is correct to 2 s.f., the precision is $\pm0.05$:
$$1.75 \le p < 1.85 \text{ (kW)}$$
$$\boxed{284\,500 \le n < 285\,500, \quad 1.75 \le p < 1.85}$$
(b) Using $E=n\times p\times t$, with $t=24$ treated as exact: since $E$ increases as both $n$ and $p$ increase, the minimum $E$ uses the lower bounds of $n$ and $p$, and the maximum $E$ uses the upper bounds of both.
$$E_{min} = 284\,500\times1.75\times24$$
$$= 11\,949\,000 \text{ kWh}$$
$$E_{max} = 285\,500\times1.85\times24$$
$$= 12\,676\,200 \text{ kWh}$$
$$\boxed{E_{min} \approx 1.19\times10^{7} \text{ kWh}, \quad E_{max} \approx 1.27\times10^{7} \text{ kWh (3 s.f.)}}$$
(c) Using the given values directly:
$$E_{est} = 285\,000\times1.8\times24$$
$$= 12\,312\,000 \text{ kWh}$$
$$\boxed{E_{est} \approx 1.23\times10^{7} \text{ kWh (3 s.f.)}}$$
Using the unrounded bounds from part (b), find the percentage error between $E_{est}$ and each bound:
At the lower bound: $\dfrac{11\,949\,000-12\,312\,000}{12\,312\,000}\times100 = -2.9483\ldots\%$
At the upper bound: $\dfrac{12\,676\,200-12\,312\,000}{12\,312\,000}\times100 = 2.9580\ldots\%$
The larger magnitude occurs at the upper bound.
$$\boxed{\text{Maximum possible percentage error} \approx 2.96\% \text{ (3 s.f.)}}$$
QUESTION 50
11 marks
Hard
A construction company wants to estimate the number of bricks needed to build a wall. The wall is to be $24.6$ m long and $2.8$ m high. Each visible brick face is approximately $0.19$ m by $0.09$ m (length by height), allowing for a thin layer of mortar around each brick.
(a) By rounding the wall's dimensions to 1 significant figure and the brick face dimensions to 1 significant figure, estimate the area of the wall, the area of one brick face, and hence a rough estimate for the number of bricks required. [5]
(b) Using the unrounded measurements, calculate a more accurate value for the number of bricks required, rounding up to the nearest whole brick (since a partial brick cannot be used). [3]
(c) Find the percentage error in the rough estimate found in part (a), compared to the accurate value found in part (b), correct to 3 significant figures. [3]
Show complete worked solution
(a) Rounding each dimension to 1 significant figure:
$$24.6 \approx 20 \text{ m}, \qquad 2.8 \approx 3 \text{ m}, \qquad 0.19 \approx 0.2 \text{ m}, \qquad 0.09 \approx 0.09 \text{ m}$$
Estimated wall area:
$$A_{wall} \approx 20\times3 = 60 \text{ m}^2$$
Estimated brick face area:
$$A_{brick} \approx 0.2\times0.09 = 0.018 \text{ m}^2$$
Estimated number of bricks:
$$\text{Number of bricks} \approx \frac{60}{0.018} = 3333.33\ldots$$
$$\boxed{\text{Rough estimate} \approx 3330 \text{ bricks (3 s.f.)}}$$
(b) Using the unrounded measurements:
$$A_{wall} = 24.6\times2.8 = 68.88 \text{ m}^2$$
$$A_{brick} = 0.19\times0.09 = 0.0171 \text{ m}^2$$
$$\text{Number of bricks} = \frac{68.88}{0.0171} = 4028.07\ldots$$
Since a partial brick cannot be used, round up to the next whole brick:
$$\boxed{4029 \text{ bricks}}$$
(c) Using percentage error $=\left|\dfrac{\text{estimate}-\text{accurate}}{\text{accurate}}\right|\times100$, with the unrounded estimate value $3333.33\ldots$ from part (a) and the accurate value $4029$ from part (b):
$$\text{Percentage error} = \frac{|3333.33\ldots-4029|}{4029}\times100$$
$$= 17.2664\ldots$$
$$\boxed{\text{Percentage error} \approx 17.3\% \text{ (3 s.f.)}}$$
This shows that a rough 1 significant figure estimate, while useful for a very quick sense-check, can differ substantially (here by about 17\%) from a properly calculated value, so estimation should not replace an accurate calculation when precision matters.
Sequences & Series 50 questions
QUESTION 1
4 marks
Easy
An arithmetic sequence has first term $u_1 = 7$ and common difference $d = 5$.
(a) Find the 20th term of the sequence.
(b) Find the sum of the first 20 terms of the sequence.
Show complete worked solution
(a) Using the arithmetic sequence formula $u_n = u_1 + (n-1)d$ with $u_1=7$, $d=5$, $n=20$:
$$u_{20} = 7 + (20-1)(5)$$
$$= 7 + 95 = 102$$
$$\boxed{u_{20} = 102}$$
(b) Using $S_n = \dfrac{n}{2}(u_1+u_n)$ with $n=20$, $u_1=7$, $u_{20}=102$:
$$S_{20} = \frac{20}{2}(7+102)$$
$$= 10 \times 109 = 1090$$
$$\boxed{S_{20} = 1090}$$
QUESTION 2
5 marks
Easy
A geometric sequence has first term $u_1 = 3$ and common ratio $r = 2$.
(a) Find the 10th term of the sequence.
(b) Find the sum of the first 10 terms of the sequence.
Show complete worked solution
(a) Using the geometric sequence formula $u_n = u_1 r^{\,n-1}$ with $u_1=3$, $r=2$, $n=10$:
$$u_{10} = 3 \times 2^{9}$$
$$= 3 \times 512 = 1536$$
$$\boxed{u_{10} = 1536}$$
(b) Using $S_n = \dfrac{u_1(r^n-1)}{r-1}$ with $u_1=3$, $r=2$, $n=10$:
$$S_{10} = \frac{3(2^{10}-1)}{2-1}$$
$$= 3(1024-1) = 3\times1023$$
$$= 3069$$
$$\boxed{S_{10} = 3069}$$
QUESTION 3
6 marks
Medium
A concert hall has seating arranged in rows. Row 1 has 18 seats, and each subsequent row has 4 more seats than the row before it. There are 30 rows in total.
(a) Find the number of seats in row 30.
(b) Find the total seating capacity of the concert hall.
(c) Find the number of the first row that has more than 100 seats.
Show complete worked solution
This is an arithmetic sequence with first term $u_1=18$ and common difference $d=4$ (each row has 4 more seats than the last).
(a) Using $u_n = u_1+(n-1)d$ with $n=30$:
$$u_{30} = 18+(30-1)(4)$$
$$= 18+116 = 134$$
$$\boxed{134 \text{ seats}}$$
(b) Using $S_n = \dfrac{n}{2}(u_1+u_n)$ with $n=30$, $u_{30}=134$:
$$S_{30} = \frac{30}{2}(18+134)$$
$$= 15 \times 152 = 2280$$
$$\boxed{2280 \text{ seats}}$$
(c) Require $u_n > 100$:
$$18+(n-1)(4) > 100$$
$$(n-1)(4) > 82$$
$$n-1 > 20.5$$
$$n > 21.5$$
Since $n$ must be a whole number, $n=22$.
Checking: $u_{21}=18+80=98 \le 100$, and $u_{22}=18+84=102>100$.
$$\boxed{\text{Row 22}}$$
QUESTION 4
6 marks
Medium
A ball is dropped from a height of 4 m onto a hard floor. After each bounce, it rebounds to 65\% of the height it fell from.
(a) Show that the height reached after the $n$th bounce is given by $h_n = 4(0.65)^n$ metres.
(b) Find the height reached after the 5th bounce, correct to 3 significant figures.
(c) Assuming the ball continues to bounce indefinitely, find the total vertical distance the ball travels (down and up, over all bounces), correct to 3 significant figures.
Show complete worked solution
(a) After bounce 1, the peak height is $4\times0.65$. After bounce 2, it is $4\times0.65\times0.65 = 4\times0.65^2$. In general, each bounce multiplies the previous peak height by $0.65$, so the peak heights form a geometric sequence with first term $4(0.65)$ and common ratio $0.65$, giving
$$\boxed{h_n = 4(0.65)^n} \quad \blacksquare$$
(b) Substituting $n=5$ into $h_n=4(0.65)^n$:
$$h_5 = 4(0.65)^5$$
$$= 4 \times 0.11602906\ldots = 0.464116\ldots$$
$$\boxed{h_5 \approx 0.464 \text{ m (3 s.f.)}}$$
(c) The ball first falls 4 m. After that, for each bounce $n=1,2,3,\ldots$ it rises to height $h_n$ and falls back the same distance, contributing $2h_n$ to the total distance. Using the sum to infinity of a geometric series, $S_\infty = \dfrac{u_1}{1-r}$ (valid since $|0.65|<1$):
$$D = 4 + 2\sum_{n=1}^{\infty}4(0.65)^n = 4 + 2\times\frac{4(0.65)}{1-0.65}$$
$$= 4 + 2\times\frac{2.6}{0.35}$$
$$= 4 + 2\times7.42857\ldots$$
$$= 4+14.8571\ldots = 18.8571\ldots$$
$$\boxed{D \approx 18.9 \text{ m (3 s.f.)}}$$
QUESTION 5
14 marks
Hard
Maya is comparing two job offers, each guaranteed for at least 20 years. Year 1 refers to her first year of employment.
Job A pays a starting salary of \$42000 in year 1, with a fixed increase of \$1800 added to her salary at the start of each subsequent year.
Job B pays a starting salary of \$40000 in year 1, with her salary increasing by 3.5\% at the start of each subsequent year.
(a) Write down an expression for the salary paid in year $n$ under Job A, and an expression for the salary paid in year $n$ under Job B. Hence find the salary paid in year 6 under each job, correct to the nearest dollar. [4]
(b) Find Maya's total earnings over the first 10 years under Job A. [3]
(c) Find Maya's total earnings over the first 10 years under Job B, correct to the nearest dollar. [3]
(d) By finding the salary paid in each year under both jobs (for example using the table function on your GDC), determine the first year in which Job B's annual salary exceeds Job A's annual salary. [4]
Show complete worked solution
(a) Job A pays a fixed dollar increase each year, so it is arithmetic with $u_1=42000$, $d=1800$:
$$A_n = 42000+(n-1)(1800)$$
Job B pays a fixed percentage increase each year, so it is geometric with $u_1=40000$, $r=1.035$:
$$B_n = 40000(1.035)^{n-1}$$
Substituting $n=6$ into each:
$$A_6 = 42000+(6-1)(1800) = 42000+9000 = 51000$$
$$B_6 = 40000(1.035)^{5} = 40000\times1.18769\ldots = 47507.45\ldots$$
$$\boxed{A_6 = \$51{,}000, \quad B_6 \approx \$47{,}507}$$
(b) Using $S_n = \dfrac{n}{2}\big(2u_1+(n-1)d\big)$ for Job A with $n=10$, $u_1=42000$, $d=1800$:
$$S_{10} = \frac{10}{2}\big(2(42000)+(10-1)(1800)\big)$$
$$= 5(84000+16200) = 5\times100200$$
$$\boxed{S_{10}^{A} = \$501{,}000}$$
(c) Using $S_n = \dfrac{u_1(r^n-1)}{r-1}$ for Job B with $n=10$, $u_1=40000$, $r=1.035$:
$$S_{10} = \frac{40000\big((1.035)^{10}-1\big)}{1.035-1}$$
$$= \frac{40000(1.410599\ldots-1)}{0.035}$$
$$= \frac{40000(0.410599\ldots)}{0.035} = 469255.7\ldots$$
$$\boxed{S_{10}^{B} \approx \$469{,}256}$$
(d) Using the GDC's table function to generate $A_n$ and $B_n$ for increasing $n$: since Job A grows linearly and Job B grows exponentially, Job B eventually overtakes Job A. Checking successive years near the crossover:
Year 19: $A_{19}=42000+18(1800)=74400$, $\quad B_{19}=40000(1.035)^{18}\approx74299.57$ (B still below A)
Year 20: $A_{20}=42000+19(1800)=76200$, $\quad B_{20}=40000(1.035)^{19}\approx76900.05$ (B now exceeds A)
$$\boxed{\text{Job B's salary first exceeds Job A's salary in year 20}}$$
QUESTION 6
4 marks
Easy
An infinite geometric sequence has first term $u_1 = 12$ and common ratio $r = 0.4$.
(a) State why this series converges. [1]
(b) Find the sum to infinity, $S_\infty$. [3]
Show complete worked solution
(a) Since $|r|=0.4<1$, the infinite geometric series converges (its sum to infinity exists).
$$\boxed{|r|<1, \text{ so the series converges}}$$
(b) Using $S_\infty=\dfrac{u_1}{1-r}$ with $u_1=12$, $r=0.4$:
$$S_\infty = \frac{12}{1-0.4}$$
$$= \frac{12}{0.6}$$
$$= 20$$
$$\boxed{S_\infty = 20}$$
QUESTION 7
5 marks
Easy
The repeating decimal $0.454545\ldots$ can be written as the infinite geometric series
$0.45 + 0.0045 + 0.000045 + \ldots$
(a) State the first term and common ratio of this series. [2]
(b) Use the sum to infinity formula to express $0.454545\ldots$ as a fraction in its simplest form. [3]
Show complete worked solution
(a) Each term is $\dfrac{1}{100}$ of the previous term (moving the decimal point two places), so this is an infinite geometric series with:
$$\boxed{u_1 = 0.45, \quad r = 0.01}$$
(b) Using $S_\infty=\dfrac{u_1}{1-r}$:
$$S_\infty = \frac{0.45}{1-0.01}$$
$$= \frac{0.45}{0.99} = \frac{45}{99}$$
Simplifying by dividing numerator and denominator by their common factor 9:
$$\frac{45}{99} = \frac{5}{11}$$
$$\boxed{0.454545\ldots = \frac{5}{11}}$$
QUESTION 8
6 marks
Medium
A ball is dropped from a height of 4 m onto a hard floor. After each bounce, it rebounds to 60% of the height from which it fell, and this pattern continues indefinitely.
(a) Show that the total distance travelled by the ball after the initial drop (i.e. the sum of all the subsequent upward and downward distances) can be modelled by an infinite geometric series, and state its first term and common ratio. [3]
(b) Hence find the total distance travelled by the ball, taking into account the initial drop of 4 m, until it comes to rest. [3]
Show complete worked solution
(a) After the initial drop, the ball rises to a height of $4(0.6)=2.4$ m and falls back the same distance, then rises to $4(0.6)^2$ and falls back the same distance, and so on. Each bounce $n$ ($n=1,2,3,\ldots$) therefore contributes an up-and-down distance of $2\times4(0.6)^n$ to the total, which forms an infinite geometric series with:
$$u_1 = 2\times4\times0.6 = 4.8, \quad r = 0.6$$
Since $|r|=0.6<1$, the series converges.
$$\boxed{u_1 = 4.8, \quad r = 0.6}$$
(b) Using $S_\infty=\dfrac{u_1}{1-r}$:
$$S_\infty = \frac{4.8}{1-0.6}$$
$$= \frac{4.8}{0.4} = 12$$
Total distance = initial drop + $S_\infty$:
$$D = 4+12$$
$$\boxed{D = 16 \text{ m}}$$
QUESTION 9
6 marks
Medium
A patient takes a 200 mg dose of a medication every 8 hours. Between doses, the medication is eliminated from the body so that only 35% of the amount present immediately after a dose remains immediately before the next dose is taken.
Assuming this dosing regimen continues indefinitely, the total amount of medication in the body immediately after a dose, once the regimen has been followed for a long time, can be modelled by the infinite geometric series
$200 + 200(0.35) + 200(0.35)^2 + \ldots$
Find this long-term (steady-state) amount of medication in the body immediately after a dose, correct to 3 significant figures.
Show complete worked solution
This is an infinite geometric series with first term $u_1=200$ and common ratio $r=0.35$. Since $|r|=0.35<1$, the series converges.
Using $S_\infty=\dfrac{u_1}{1-r}$:
$$S_\infty = \frac{200}{1-0.35}$$
$$= \frac{200}{0.65}$$
$$= 307.692\ldots$$
$$\boxed{S_\infty \approx 308 \text{ mg (3 s.f.)}}$$
QUESTION 10
15 marks
Hard
An artist draws a decorative spiral made up of a sequence of circular arcs (loops). The first loop has an arc length of 40 cm. Each successive loop has an arc length equal to 85% of the arc length of the previous loop, and this pattern continues indefinitely.
(a) Find the common ratio r of the sequence of arc lengths, and write down the first term $u_1$. [2]
(b) Find $S_\infty$, the total length of the spiral if the pattern of loops were continued indefinitely. [3]
(c) The sum of the arc lengths of the first n loops is given by $S_n = \dfrac{40(1-0.85^n)}{0.15}$. Find the least number of loops, n, required so that the total length drawn, $S_n$, is within 0.5 cm of $S_\infty$ (i.e. so that $S_\infty - S_n < 0.5$). [7]
(d) Find $S_n$ for this value of n, correct to 3 significant figures. [3]
Show complete worked solution
(a) Each successive arc length is $85\%$ of the previous one, so:
$$\boxed{u_1 = 40, \quad r = 0.85}$$
(b) Using $S_\infty=\dfrac{u_1}{1-r}$:
$$S_\infty = \frac{40}{1-0.85}$$
$$= \frac{40}{0.15}$$
$$= 266.667\ldots$$
$$\boxed{S_\infty \approx 267 \text{ cm (3 s.f.)}}$$
(c) Using the given formula for $S_n$ and the value of $S_\infty$ from (b):
$$S_\infty-S_n = \frac{40}{0.15}-\frac{40(1-0.85^n)}{0.15} = \frac{40(0.85)^n}{0.15}$$
We require $S_\infty-S_n<0.5$:
$$\frac{40(0.85)^n}{0.15} < 0.5$$
Solving for $0.85^n$:
$$0.85^n < \frac{0.5\times0.15}{40} = 0.001875$$
Taking logarithms of both sides:
$$n\log(0.85) < \log(0.001875)$$
Since $\log(0.85)<0$, dividing both sides by $\log(0.85)$ reverses the inequality:
$$n > \frac{\log(0.001875)}{\log(0.85)}$$
$$n > 38.636\ldots$$
Since $n$ must be a whole number, the least value is $n=39$.
Check: at $n=39$, $S_\infty-S_n=0.471\ldots<0.5$; at $n=38$, $S_\infty-S_n=0.554\ldots>0.5$, confirming $n=39$ is the least such $n$.
$$\boxed{n = 39}$$
(d) Substituting $n=39$ into $S_n=\dfrac{40(1-0.85^n)}{0.15}$:
$$S_{39} = \frac{40(1-0.85^{39})}{0.15}$$
$$= 266.195\ldots$$
$$\boxed{S_{39} \approx 266 \text{ cm (3 s.f.)}}$$
QUESTION 11
6 marks
Easy
An arithmetic sequence has $u_4 = 23$ and $u_9 = 53$.
(a) Find the common difference, $d$. [2]
(b) Find the first term, $u_1$. [2]
(c) Find $u_{20}$. [2]
Show complete worked solution
(a) Subtracting the term equations eliminates $u_1$. Since $u_9 = u_1+8d$ and $u_4=u_1+3d$:
$$u_9-u_4 = 5d$$
$$53-23 = 5d$$
$$30 = 5d$$
$$\boxed{d = 6}$$
(b) Using $u_4 = u_1+3d$ with $u_4=23$, $d=6$:
$$23 = u_1+3(6)$$
$$23 = u_1+18$$
$$\boxed{u_1 = 5}$$
(c) Using $u_n=u_1+(n-1)d$ with $u_1=5$, $d=6$, $n=20$:
$$u_{20} = 5+(20-1)(6)$$
$$= 5+114 = 119$$
$$\boxed{u_{20} = 119}$$
QUESTION 12
5 marks
Easy
A pile of firewood is stacked in rows. The bottom row has 32 logs, and each row above it has 3 fewer logs than the row below. The stack has 9 rows in total.
(a) Find the number of logs in the top (9th) row. [2]
(b) Find the total number of logs in the stack. [3]
Show complete worked solution
This is an arithmetic sequence with first term $u_1=32$ (bottom row) and common difference $d=-3$ (each row up has 3 fewer logs).
(a) Using $u_n=u_1+(n-1)d$ with $n=9$:
$$u_9 = 32+(9-1)(-3)$$
$$= 32-24 = 8$$
$$\boxed{u_9 = 8 \text{ logs}}$$
(b) Using $S_n=\dfrac{n}{2}(u_1+u_n)$ with $n=9$, $u_1=32$, $u_9=8$:
$$S_9 = \frac{9}{2}(32+8)$$
$$= 4.5\times40 = 180$$
$$\boxed{S_9 = 180 \text{ logs}}$$
QUESTION 13
6 marks
Easy
A geometric sequence has first term $u_1=5$ and $u_4=135$.
(a) Find the common ratio, $r$. [2]
(b) Find $u_6$. [2]
(c) Find $S_6$, the sum of the first 6 terms. [2]
Show complete worked solution
(a) Using $u_4=u_1r^3$ with $u_1=5$, $u_4=135$:
$$135 = 5r^3$$
$$r^3 = 27$$
$$\boxed{r = 3}$$
(b) Using $u_n=u_1r^{\,n-1}$ with $u_1=5$, $r=3$, $n=6$:
$$u_6 = 5\times3^{5}$$
$$= 5\times243 = 1215$$
$$\boxed{u_6 = 1215}$$
(c) Using $S_n=\dfrac{u_1(r^n-1)}{r-1}$ with $u_1=5$, $r=3$, $n=6$:
$$S_6 = \frac{5(3^6-1)}{3-1}$$
$$= \frac{5(729-1)}{2} = \frac{5\times728}{2}$$
$$= 1820$$
$$\boxed{S_6 = 1820}$$
QUESTION 14
5 marks
Easy
A single bacterium in a laboratory culture divides into 2 bacteria every hour. The number of bacteria present after $n$ hours is modelled by the geometric sequence $u_n = 2^n$.
(a) Find the number of bacteria present after 10 hours. [2]
(b) Find the least number of complete hours, $n$, after which the population first exceeds 1{,}000{,}000 bacteria. [3]
Show complete worked solution
(a) Substituting $n=10$ into $u_n=2^n$:
$$u_{10} = 2^{10}$$
$$\boxed{u_{10} = 1024 \text{ bacteria}}$$
(b) We require $2^n > 1{,}000{,}000$. Taking logarithms of both sides:
$$n\log(2) > \log(1{,}000{,}000)$$
$$n > \frac{\log(1{,}000{,}000)}{\log(2)}$$
$$n > 19.93\ldots$$
Since $n$ must be a whole number, $n=20$.
Checking: $u_{19}=2^{19}=524{,}288$ (below 1{,}000{,}000), and $u_{20}=2^{20}=1{,}048{,}576$ (above 1{,}000{,}000).
$$\boxed{n = 20 \text{ hours}}$$
QUESTION 15
5 marks
Easy
(a) Show that $\displaystyle\sum_{n=1}^{15}(4n-3)$ is the sum of an arithmetic sequence, stating the first term $u_1$, the common difference $d$, and the last term $u_{15}$. [3]
(b) Hence evaluate $\displaystyle\sum_{n=1}^{15}(4n-3)$. [2]
Show complete worked solution
(a) The general term is $u_n=4n-3$. Consecutive terms differ by a constant amount:
$$u_{n+1}-u_n = \big(4(n+1)-3\big)-(4n-3) = 4$$
so the terms form an arithmetic sequence with common difference $d=4$. The first and last terms are:
$$u_1 = 4(1)-3 = 1, \qquad u_{15}=4(15)-3=57$$
$$\boxed{u_1=1, \quad d=4, \quad u_{15}=57}$$
(b) Using $S_n=\dfrac{n}{2}(u_1+u_n)$ with $n=15$:
$$S_{15} = \frac{15}{2}(1+57)$$
$$= 7.5\times58 = 435$$
$$\boxed{\sum_{n=1}^{15}(4n-3) = 435}$$
QUESTION 16
5 marks
Easy
(a) Show that $\displaystyle\sum_{n=1}^{7}3(2)^{n-1}$ is the sum of a geometric sequence, stating the first term $u_1$ and the common ratio $r$. [2]
(b) Hence evaluate $\displaystyle\sum_{n=1}^{7}3(2)^{n-1}$. [3]
Show complete worked solution
(a) The general term is $u_n=3(2)^{n-1}$, which has the form $u_1r^{\,n-1}$ with:
$$\boxed{u_1 = 3, \quad r = 2}$$
(b) Using $S_n=\dfrac{u_1(r^n-1)}{r-1}$ with $u_1=3$, $r=2$, $n=7$:
$$S_7 = \frac{3(2^7-1)}{2-1}$$
$$= 3(128-1) = 3\times127$$
$$= 381$$
$$\boxed{\sum_{n=1}^{7}3(2)^{n-1} = 381}$$
QUESTION 17
4 marks
Easy
An infinite geometric sequence has first term $u_1=15$ and common ratio $r=0.2$.
(a) State why this series converges. [1]
(b) Find the sum to infinity, $S_\infty$. [3]
Show complete worked solution
(a) Since $|r|=0.2<1$, the infinite geometric series converges (its sum to infinity exists).
$$\boxed{|r|<1, \text{ so the series converges}}$$
(b) Using $S_\infty=\dfrac{u_1}{1-r}$ with $u_1=15$, $r=0.2$:
$$S_\infty = \frac{15}{1-0.2}$$
$$= \frac{15}{0.8}$$
$$\boxed{S_\infty = 18.75}$$
QUESTION 18
5 marks
Easy
The repeating decimal $0.363636\ldots$ can be written as the infinite geometric series
$0.36 + 0.0036 + 0.000036 + \ldots$
(a) State the first term and common ratio of this series. [2]
(b) Use the sum to infinity formula to express $0.363636\ldots$ as a fraction in its simplest form. [3]
Show complete worked solution
(a) Each term is $\dfrac{1}{100}$ of the previous term (moving the decimal point two places), so this is an infinite geometric series with:
$$\boxed{u_1 = 0.36, \quad r = 0.01}$$
(b) Using $S_\infty=\dfrac{u_1}{1-r}$:
$$S_\infty = \frac{0.36}{1-0.01}$$
$$= \frac{0.36}{0.99} = \frac{36}{99}$$
Simplifying by dividing numerator and denominator by their common factor 9:
$$\frac{36}{99} = \frac{4}{11}$$
$$\boxed{0.363636\ldots = \frac{4}{11}}$$
QUESTION 19
5 marks
Easy
The expressions $5x-3$, $2x+4$, and $x+11$ represent three consecutive terms of an arithmetic sequence.
(a) Find the value of $x$. [3]
(b) Hence find the three terms and the common difference, $d$. [2]
Show complete worked solution
(a) For three consecutive terms of an arithmetic sequence, the difference between consecutive terms is constant, so:
$$(2x+4)-(5x-3) = (x+11)-(2x+4)$$
$$-3x+7 = -x+7$$
$$-3x = -x$$
$$-2x = 0$$
$$\boxed{x = 0}$$
(b) Substituting $x=0$ into each expression:
$$5(0)-3=-3, \qquad 2(0)+4=4, \qquad 0+11=11$$
Checking: $4-(-3)=7$ and $11-4=7$, confirming a common difference throughout.
$$\boxed{\text{Terms: } -3,\ 4,\ 11 \quad (d=7)}$$
QUESTION 20
6 marks
Easy
The numbers $4$, $x$, $9$ are three consecutive terms of a geometric sequence, where $x>0$.
(a) Find the value of $x$. [3]
(b) Find the common ratio, $r$. [1]
(c) Find the 5th term of the sequence, $u_5$, given that $4$ is the 1st term. [2]
Show complete worked solution
(a) For consecutive geometric terms $a,x,b$, we have $\dfrac{x}{a}=\dfrac{b}{x}$, so $x^2=ab$:
$$x^2 = 4\times9 = 36$$
$$x = \pm6$$
Since $x>0$ is required:
$$\boxed{x = 6}$$
(b) Using $r=\dfrac{u_2}{u_1}=\dfrac{x}{4}$:
$$r = \frac{6}{4}$$
$$\boxed{r = 1.5}$$
(c) Using $u_n=u_1r^{\,n-1}$ with $u_1=4$, $r=1.5$, $n=5$:
$$u_5 = 4\times(1.5)^4$$
$$= 4\times5.0625$$
$$\boxed{u_5 = 20.25}$$
QUESTION 21
5 marks
Easy
A pattern of tiles is built in stages. The table shows the number of tiles, $t_n$, used at each stage $n$.
(a) Show that the number of tiles follows an arithmetic sequence, and state the common difference, $d$. [1]
(b) Find a formula for $t_n$ in terms of $n$. [2]
(c) Find the number of tiles required at stage 10. [2]
| Stage, $n$ | 1 | 2 | 3 | 4 |
|---|---|---|---|---|
| Tiles, $t_n$ | 5 | 8 | 11 | 14 |
Show complete worked solution
(a) The number of tiles increases by the same amount, 3, at each stage ($8-5=3$, $11-8=3$, $14-11=3$), so the sequence is arithmetic with:
$$\boxed{d = 3}$$
(b) Using $t_n=t_1+(n-1)d$ with $t_1=5$, $d=3$:
$$t_n = 5+(n-1)(3)$$
$$= 5+3n-3$$
$$\boxed{t_n = 3n+2}$$
(c) Substituting $n=10$ into $t_n=3n+2$:
$$t_{10} = 3(10)+2$$
$$\boxed{t_{10} = 32 \text{ tiles}}$$
QUESTION 22
6 marks
Easy
An arithmetic sequence has first term $u_1=5$ and common difference $d=3$.
(a) Show that $S_n = \dfrac{3n^2+7n}{2}$, where $S_n$ is the sum of the first $n$ terms. [2]
(b) Given that $S_n=185$, find the value of $n$. [4]
Show complete worked solution
(a) Using $S_n=\dfrac{n}{2}\big(2u_1+(n-1)d\big)$ with $u_1=5$, $d=3$:
$$S_n = \frac{n}{2}\big(2(5)+(n-1)(3)\big)$$
$$= \frac{n}{2}(10+3n-3)$$
$$\boxed{S_n = \frac{3n^2+7n}{2}} \quad \blacksquare$$
(b) Setting $S_n=185$:
$$\frac{3n^2+7n}{2} = 185$$
$$3n^2+7n = 370$$
$$3n^2+7n-370 = 0$$
Using the quadratic formula with $a=3$, $b=7$, $c=-370$:
$$n = \frac{-7\pm\sqrt{7^2-4(3)(-370)}}{2(3)}$$
$$= \frac{-7\pm\sqrt{49+4440}}{6} = \frac{-7\pm\sqrt{4489}}{6} = \frac{-7\pm67}{6}$$
This gives $n=10$ or $n=-12.33\ldots$. Since $n$ must be a positive whole number:
$$\boxed{n = 10}$$
QUESTION 23
5 marks
Easy
A geometric sequence has first term $u_1=4$ and common ratio $r=2$.
(a) Write down an expression for $S_n$, the sum of the first $n$ terms. [2]
(b) Find the least value of $n$ for which $S_n>500$. [3]
Show complete worked solution
(a) Using $S_n=\dfrac{u_1(r^n-1)}{r-1}$ with $u_1=4$, $r=2$:
$$S_n = \frac{4(2^n-1)}{2-1}$$
$$\boxed{S_n = 4(2^n-1)}$$
(b) We require $4(2^n-1)>500$, i.e. $2^n-1>125$, i.e. $2^n>126$. Testing successive values of $n$ (e.g. using the GDC):
$$n=6: \quad 2^6=64, \quad S_6=4(64-1)=252 \ (\text{not} > 500)$$
$$n=7: \quad 2^7=128, \quad S_7=4(128-1)=508 \ (>500)$$
$$\boxed{n = 7}$$
QUESTION 24
4 marks
Easy
A crane stacks concrete blocks in layers. The bottom layer has 20 blocks, and each layer above has 2 fewer blocks than the layer below it, continuing until the top layer has 2 blocks.
(a) Find the number of layers in the stack. [2]
(b) Find the total number of blocks used. [2]
Show complete worked solution
This is an arithmetic sequence with $u_1=20$ and $d=-2$.
(a) Using $u_n=u_1+(n-1)d$ with $u_n=2$ (the top layer):
$$2 = 20+(n-1)(-2)$$
$$-18 = -2(n-1)$$
$$n-1 = 9$$
$$\boxed{n = 10 \text{ layers}}$$
(b) Using $S_n=\dfrac{n}{2}(u_1+u_n)$ with $n=10$, $u_1=20$, $u_{10}=2$:
$$S_{10} = \frac{10}{2}(20+2)$$
$$= 5\times22$$
$$\boxed{S_{10} = 110 \text{ blocks}}$$
QUESTION 25
5 marks
Easy
A wheel spinning on a track is slowing due to friction. In its first rotation it covers 8 m of track, and each subsequent rotation covers 75\% of the distance covered by the rotation before it. Assume this pattern continues indefinitely.
(a) State the first term, $u_1$, and common ratio, $r$, of the sequence of distances covered by each rotation. [2]
(b) Find $S_\infty$, the total distance the wheel travels. [3]
Show complete worked solution
(a) Each rotation covers $75\%$ of the distance of the previous rotation, so:
$$\boxed{u_1 = 8, \quad r = 0.75}$$
(b) Since $|r|=0.75<1$, the series converges. Using $S_\infty=\dfrac{u_1}{1-r}$:
$$S_\infty = \frac{8}{1-0.75}$$
$$= \frac{8}{0.25}$$
$$\boxed{S_\infty = 32 \text{ m}}$$
QUESTION 26
5 marks
Easy
A sequence is defined by $u_n = n^2+1$ for $n=1,2,3,4,\ldots$
(a) Find $u_1$, $u_2$, $u_3$, and $u_4$. [2]
(b) By considering the differences between consecutive terms, determine whether the sequence is arithmetic. [2]
(c) By considering the ratios of consecutive terms, determine whether the sequence is geometric. [1]
Show complete worked solution
(a) Substituting $n=1,2,3,4$ into $u_n=n^2+1$:
$$u_1=2, \quad u_2=5, \quad u_3=10, \quad u_4=17$$
$$\boxed{u_1=2,\ u_2=5,\ u_3=10,\ u_4=17}$$
(b) The consecutive differences are:
$$u_2-u_1=3, \quad u_3-u_2=5, \quad u_4-u_3=7$$
Since these differences are not constant, the sequence is not arithmetic.
$$\boxed{\text{Not arithmetic (differences } 3,5,7 \text{ are not equal)}}$$
(c) The consecutive ratios are:
$$\frac{u_2}{u_1}=2.5, \quad \frac{u_3}{u_2}=2, \quad \frac{u_4}{u_3}=1.7$$
Since these ratios are not constant, the sequence is not geometric.
$$\boxed{\text{Not geometric (ratios } 2.5,\ 2,\ 1.7 \text{ are not equal)}}$$
QUESTION 27
8 marks
Medium
A charity event sells tickets over several days. On day 1, 45 tickets are sold, and on each following day, 12 more tickets are sold than on the previous day. The venue has a total capacity of 900 tickets.
(a) Find the number of tickets sold on day 10. [2]
(b) Find the total number of tickets sold after 10 days. [2]
(c) Find the greatest number of complete days, $n$, for which the cumulative total of tickets sold does not exceed the venue capacity of 900. [4]
Show complete worked solution
This is an arithmetic sequence with $u_1=45$ and $d=12$.
(a) Using $u_n=u_1+(n-1)d$ with $n=10$:
$$u_{10} = 45+(10-1)(12)$$
$$= 45+108$$
$$\boxed{u_{10} = 153 \text{ tickets}}$$
(b) Using $S_n=\dfrac{n}{2}\big(2u_1+(n-1)d\big)$ with $n=10$:
$$S_{10} = \frac{10}{2}\big(2(45)+9(12)\big)$$
$$= 5(90+108) = 5\times198$$
$$\boxed{S_{10} = 990 \text{ tickets}}$$
(c) We require the greatest $n$ such that $S_n\le900$. Using $S_n=\dfrac{n}{2}\big(2(45)+(n-1)(12)\big)=6n^2+39n$:
$$6n^2+39n \le 900$$
Testing successive values of $n$ using the GDC table function:
$$n=9: \quad S_9 = 6(81)+39(9) = 486+351 = 837 \ (\le900)$$
$$n=10: \quad S_{10} = 990 \ (\text{from part (b), exceeds }900)$$
$$\boxed{n = 9 \text{ days}}$$
QUESTION 28
6 marks
Medium
A single sheet of paper has a thickness of 0.1 mm. Each time it is folded in half, its thickness doubles.
(a) Find the thickness of the folded paper after 7 folds. [2]
(b) Find the least number of folds required for the thickness to first exceed 100 mm. [4]
Show complete worked solution
The thickness after $n$ folds forms a geometric sequence with $u_1=0.1\times2=0.2$ after 1 fold, i.e. $u_n=0.1(2)^n$, first term $0.1$ and common ratio $2$.
(a) Substituting $n=7$ into $u_n=0.1(2)^n$:
$$u_7 = 0.1\times2^7$$
$$= 0.1\times128$$
$$\boxed{u_7 = 12.8 \text{ mm}}$$
(b) We require $0.1(2)^n>100$, i.e. $2^n>1000$. Taking logarithms:
$$n\log(2) > \log(1000)$$
$$n > \frac{\log(1000)}{\log(2)}$$
$$n > 9.97\ldots$$
Since $n$ must be a whole number, $n=10$.
Checking: $u_{10}=0.1\times2^{10}=102.4$ mm ($>100$), and $u_9=0.1\times2^9=51.2$ mm ($\le100$).
$$\boxed{n = 10 \text{ folds}}$$
QUESTION 29
6 marks
Medium
An arithmetic sequence is such that $S_5=45$ and $S_{10}=290$, where $S_n$ denotes the sum of the first $n$ terms.
(a) Write down two equations in $u_1$ and $d$ using the given information. [2]
(b) Hence find the values of $u_1$ and $d$. [4]
Show complete worked solution
(a) Using $S_n=\dfrac{n}{2}\big(2u_1+(n-1)d\big)$:
$$S_5 = \frac{5}{2}(2u_1+4d) = 5(u_1+2d) = 45 \quad\Rightarrow\quad u_1+2d=9$$
$$S_{10} = \frac{10}{2}(2u_1+9d) = 5(2u_1+9d) = 290 \quad\Rightarrow\quad 2u_1+9d=58$$
$$\boxed{u_1+2d=9 \quad \text{and} \quad 2u_1+9d=58}$$
(b) From the first equation, $u_1=9-2d$. Substituting into the second equation:
$$2(9-2d)+9d = 58$$
$$18-4d+9d = 58$$
$$18+5d = 58$$
$$5d = 40$$
$$d = 8$$
Then $u_1 = 9-2(8) = -7$.
Check: $S_5 = 5(-7+16)=5(9)=45$ correct; $S_{10}=5\big(2(-7)+9(8)\big)=5(-14+72)=5(58)=290$ correct.
$$\boxed{u_1 = -7, \quad d = 8}$$
QUESTION 30
6 marks
Medium
A geometric sequence has $u_2=6$ and $u_5=48$.
(a) Find the common ratio, $r$. [2]
(b) Find the first term, $u_1$. [2]
(c) Find $S_8$, the sum of the first 8 terms. [2]
Show complete worked solution
(a) Since $u_5=u_2r^3$:
$$48 = 6r^3$$
$$r^3 = 8$$
$$\boxed{r = 2}$$
(b) Since $u_2=u_1r$:
$$6 = u_1(2)$$
$$\boxed{u_1 = 3}$$
(c) Using $S_n=\dfrac{u_1(r^n-1)}{r-1}$ with $u_1=3$, $r=2$, $n=8$:
$$S_8 = \frac{3(2^8-1)}{2-1}$$
$$= 3(256-1) = 3\times255$$
$$\boxed{S_8 = 765}$$
QUESTION 31
6 marks
Medium
The attendance at a weekly seminar, in the number of people attending week $n$, is modelled by $a_n=120-5n$ for $n=1,2,\ldots,12$.
(a) Find $a_1$ and $a_{12}$, and state why the sequence $a_1,a_2,\ldots,a_{12}$ is arithmetic. [2]
(b) Hence evaluate $\displaystyle\sum_{n=1}^{12}a_n$, and interpret this value in context. [4]
Show complete worked solution
(a) Substituting $n=1$ and $n=12$ into $a_n=120-5n$:
$$a_1 = 120-5(1) = 115, \qquad a_{12} = 120-5(12) = 60$$
Since $a_n=120-5n$ is a linear function of $n$, each term decreases by a constant amount, $5$, from the previous term, so the sequence is arithmetic with common difference $d=-5$.
$$\boxed{a_1=115, \quad a_{12}=60}$$
(b) Using $S_n=\dfrac{n}{2}(u_1+u_n)$ with $n=12$, $u_1=115$, $u_{12}=60$:
$$\sum_{n=1}^{12}a_n = \frac{12}{2}(115+60)$$
$$= 6\times175$$
$$\boxed{\sum_{n=1}^{12}a_n = 1050}$$
This represents the total number of person-visits to the seminar summed over all 12 weeks.
QUESTION 32
8 marks
Medium
A spider reinforces its web with strands of silk. The first strand it spins is 24 cm long. Each subsequent strand it adds is $40\%$ of the length of the strand added before it, and the spider continues adding strands indefinitely.
(a) State the first term, $u_1$, and common ratio, $r$, of the sequence of strand lengths, and justify why the total length converges. [2]
(b) Find $S_\infty$, the total length of silk used if the spider added strands indefinitely. [3]
(c) Find $S_6$, the total length of the first 6 strands, and hence find the difference between $S_\infty$ and $S_6$, correct to 3 significant figures. [3]
Show complete worked solution
(a) Each strand is $40\%$ of the length of the previous strand, so:
$$\boxed{u_1 = 24, \quad r = 0.4}$$
Since $|r|=0.4<1$, the infinite geometric series converges.
(b) Using $S_\infty=\dfrac{u_1}{1-r}$:
$$S_\infty = \frac{24}{1-0.4}$$
$$= \frac{24}{0.6}$$
$$\boxed{S_\infty = 40 \text{ cm}}$$
(c) Using $S_n=\dfrac{u_1(1-r^n)}{1-r}$ with $n=6$:
$$S_6 = \frac{24\big(1-0.4^6\big)}{1-0.4}$$
$$= \frac{24(1-0.004096)}{0.6}$$
$$= \frac{24(0.995904)}{0.6} = 39.836\ldots$$
$$S_\infty - S_6 = 40-39.836\ldots = 0.164\ldots$$
$$\boxed{S_6 \approx 39.8 \text{ cm}, \quad S_\infty-S_6 \approx 0.164 \text{ cm (3 s.f.)}}$$
QUESTION 33
6 marks
Medium
A sequence is defined by $u_n=2n^2-3n+1$ for $n=1,2,3,4,\ldots$
(a) Find $u_1$, $u_2$, $u_3$, and $u_4$. [2]
(b) Find the first differences $u_2-u_1$, $u_3-u_2$, $u_4-u_3$, and hence explain why the sequence is not arithmetic. [2]
(c) Find the second differences, and state what this confirms about the type of sequence. [2]
Show complete worked solution
(a) Substituting $n=1,2,3,4$ into $u_n=2n^2-3n+1$:
$$u_1=0, \quad u_2=3, \quad u_3=10, \quad u_4=21$$
$$\boxed{u_1=0,\ u_2=3,\ u_3=10,\ u_4=21}$$
(b) The first differences are:
$$u_2-u_1=3, \quad u_3-u_2=7, \quad u_4-u_3=11$$
Since these first differences ($3,7,11$) are not equal, the sequence is not arithmetic.
$$\boxed{\text{First differences } 3,7,11 \text{ are not constant, so not arithmetic}}$$
(c) The second differences are:
$$7-3=4, \quad 11-7=4$$
Since the second differences are constant, $u_n$ is a quadratic sequence (consistent with $u_n=2n^2-3n+1$ having leading coefficient $2$, giving second difference $2\times2=4$).
$$\boxed{\text{Second differences constant at } 4 \text{, confirming a quadratic sequence}}$$
QUESTION 34
7 marks
Medium
An arithmetic sequence has first term $u_1=8$ and $n$th term $u_n=122$. The sum of the first $n$ terms is $S_n=1300$.
(a) Find the value of $n$. [3]
(b) Find the common difference, $d$. [2]
(c) Find $S_{10}$, the sum of the first 10 terms. [2]
Show complete worked solution
(a) Using $S_n=\dfrac{n}{2}(u_1+u_n)$ with $u_1=8$, $u_n=122$, $S_n=1300$:
$$1300 = \frac{n}{2}(8+122)$$
$$1300 = \frac{n}{2}(130)$$
$$1300 = 65n$$
$$\boxed{n = 20}$$
(b) Using $u_n=u_1+(n-1)d$ with $u_1=8$, $u_{20}=122$:
$$122 = 8+(20-1)d$$
$$114 = 19d$$
$$\boxed{d = 6}$$
(c) Using $S_n=\dfrac{n}{2}\big(2u_1+(n-1)d\big)$ with $n=10$, $u_1=8$, $d=6$:
$$S_{10} = \frac{10}{2}\big(2(8)+9(6)\big)$$
$$= 5(16+54) = 5\times70$$
$$\boxed{S_{10} = 350}$$
QUESTION 35
6 marks
Medium
A geometric sequence has first term $u_1=1000$ and common ratio $r=0.5$.
(a) Write down an expression for $S_n$ in terms of $n$, and find $S_\infty$. [2]
(b) Find the least value of $n$ for which $S_n>1990$. [4]
Show complete worked solution
(a) Using $S_n=\dfrac{u_1(1-r^n)}{1-r}$ with $u_1=1000$, $r=0.5$:
$$S_n = \frac{1000(1-0.5^n)}{1-0.5} = 2000(1-0.5^n)$$
Since $|r|=0.5<1$, using $S_\infty=\dfrac{u_1}{1-r}$:
$$S_\infty = \frac{1000}{0.5} = 2000$$
$$\boxed{S_n = 2000(1-0.5^n), \quad S_\infty = 2000}$$
(b) We require $2000(1-0.5^n)>1990$, i.e. $1-0.5^n>0.995$, i.e. $0.5^n<0.005$. Taking logarithms (noting $\log(0.5)<0$, so the inequality reverses):
$$n\log(0.5) < \log(0.005)$$
$$n > \frac{\log(0.005)}{\log(0.5)}$$
$$n > 7.64\ldots$$
Since $n$ must be a whole number, $n=8$.
Checking: $S_7=2000(1-0.5^7)=1984.375$ ($\le1990$), and $S_8=2000(1-0.5^8)=1992.1875$ ($>1990$).
$$\boxed{n = 8}$$
QUESTION 36
7 marks
Medium
An arithmetic sequence has first term $u_1=4$ and common difference $d$, where $d\neq0$. The 1st, 3rd, and 9th terms of this sequence, namely $u_1$, $u_3$, $u_9$, are consecutive terms of a geometric sequence.
(a) Write down expressions for $u_3$ and $u_9$ in terms of $d$. [2]
(b) By using the condition for three consecutive geometric terms, form an equation in $d$ and solve it to find the value of $d$. [3]
(c) Hence find the common ratio, $r$, of the geometric sequence formed by $u_1$, $u_3$, $u_9$. [2]
Show complete worked solution
(a) Using $u_n=u_1+(n-1)d$ with $u_1=4$:
$$\boxed{u_3 = 4+2d, \quad u_9 = 4+8d}$$
(b) For $u_1,u_3,u_9$ to be consecutive geometric terms, $(u_3)^2=u_1\times u_9$:
$$(4+2d)^2 = 4(4+8d)$$
$$16+16d+4d^2 = 16+32d$$
$$4d^2+16d-32d = 0$$
$$4d^2-16d = 0$$
$$4d(d-4) = 0$$
So $d=0$ or $d=4$. Since $d\neq0$ is given:
$$\boxed{d = 4}$$
(c) With $d=4$: $u_1=4$, $u_3=4+2(4)=12$, $u_9=4+8(4)=36$. The common ratio is:
$$r = \frac{u_3}{u_1} = \frac{12}{4} = 3$$
Check: $\dfrac{u_9}{u_3}=\dfrac{36}{12}=3$, confirming consistency.
$$\boxed{r = 3}$$
QUESTION 37
7 marks
Medium
An isolated population of a rare bird species currently numbers 2400 individuals. Conservationists predict that, without intervention, the population will decline by $8\%$ each year. The population after $n$ years is modelled by $P_n=2400(0.92)^n$.
(a) Find the predicted population after 5 years, correct to the nearest whole number. [3]
(b) Find the least number of complete years, $n$, until the model predicts the population will first fall below 1000. [4]
Show complete worked solution
(a) Substituting $n=5$ into $P_n=2400(0.92)^n$:
$$P_5 = 2400(0.92)^5$$
$$= 2400\times0.65909\ldots$$
$$\boxed{P_5 \approx 1582 \text{ birds}}$$
(b) We require $2400(0.92)^n<1000$, i.e. $(0.92)^n<0.41\overline{6}$. Taking logarithms (noting $\log(0.92)<0$, so the inequality reverses):
$$n\log(0.92) < \log(0.41\overline{6})$$
$$n > \frac{\log(0.41\overline{6})}{\log(0.92)}$$
$$n > 10.50\ldots$$
Since $n$ must be a whole number, $n=11$.
Checking: $P_{10}=2400(0.92)^{10}\approx1042.5$ ($\ge1000$), and $P_{11}=2400(0.92)^{11}\approx959.1$ ($<1000$).
$$\boxed{n = 11 \text{ years}}$$
QUESTION 38
7 marks
Medium
A bricklayer has 390 bricks available to build a decorative wall. The bottom row uses 60 bricks, and each row above uses 4 fewer bricks than the row directly below it.
(a) Find the number of bricks required for the 8th row. [2]
(b) Find the total number of bricks used to build the first 8 rows. [2]
(c) Show that the bricklayer does not have enough bricks remaining to complete a 9th row, and hence state the maximum number of complete rows that can be built with 390 bricks. [3]
Show complete worked solution
This is an arithmetic sequence with $u_1=60$ and $d=-4$.
(a) Using $u_n=u_1+(n-1)d$ with $n=8$:
$$u_8 = 60+(8-1)(-4)$$
$$= 60-28$$
$$\boxed{u_8 = 32 \text{ bricks}}$$
(b) Using $S_n=\dfrac{n}{2}(u_1+u_n)$ with $n=8$, $u_1=60$, $u_8=32$:
$$S_8 = \frac{8}{2}(60+32)$$
$$= 4\times92$$
$$\boxed{S_8 = 368 \text{ bricks}}$$
(c) After 8 rows, the number of bricks remaining is:
$$390-368 = 22 \text{ bricks}$$
The 9th row requires $u_9=60+(9-1)(-4)=60-32=28$ bricks. Since only 22 bricks remain, and $22<28$, there are not enough bricks to complete a 9th row.
$$\boxed{\text{Maximum of 8 complete rows can be built}}$$
QUESTION 39
7 marks
Medium
Four arithmetic means are to be inserted between 7 and 37, so that the resulting six numbers form an arithmetic sequence $7, u_2, u_3, u_4, u_5, 37$.
(a) Find the common difference, $d$, of this sequence. [3]
(b) Write down the four inserted means. [2]
(c) Find the sum of all six numbers in the sequence. [2]
Show complete worked solution
(a) There are 6 terms in total, with $u_1=7$ and $u_6=37$. Using $u_n=u_1+(n-1)d$ with $n=6$:
$$37 = 7+(6-1)d$$
$$30 = 5d$$
$$\boxed{d = 6}$$
(b) Adding $d=6$ successively to $u_1=7$:
$$u_2=13, \quad u_3=19, \quad u_4=25, \quad u_5=31$$
$$\boxed{13,\ 19,\ 25,\ 31}$$
(c) Using $S_n=\dfrac{n}{2}(u_1+u_n)$ with $n=6$, $u_1=7$, $u_6=37$:
$$S_6 = \frac{6}{2}(7+37)$$
$$= 3\times44$$
$$\boxed{S_6 = 132}$$
QUESTION 40
6 marks
Medium
Three positive geometric means are to be inserted between 2 and 162, so that the resulting five numbers form a geometric sequence $2, u_2, u_3, u_4, 162$.
(a) Find the common ratio, $r$, given that all terms of the sequence are positive. [3]
(b) Write down the three inserted geometric means. [2]
(c) Find the sum of all five numbers in the sequence. [1]
Show complete worked solution
(a) There are 5 terms in total, with $u_1=2$ and $u_5=162$. Using $u_n=u_1r^{\,n-1}$ with $n=5$:
$$162 = 2r^4$$
$$r^4 = 81$$
$$r = \pm3$$
Since all terms of the sequence must be positive, $r$ must be positive (a negative ratio would alternate the signs of the terms):
$$\boxed{r = 3}$$
(b) Multiplying successively by $r=3$ starting from $u_1=2$:
$$u_2=6, \quad u_3=18, \quad u_4=54$$
$$\boxed{6,\ 18,\ 54}$$
(c) Summing all five terms:
$$2+6+18+54+162 = 242$$
$$\boxed{242}$$
QUESTION 41
6 marks
Medium
It is given that $\displaystyle\sum_{n=1}^{20}(5n+2) = 1090$.
(a) Show, using the formula for the sum of an arithmetic sequence, that this value of $1090$ is correct. [3]
(b) Hence find the value of $\displaystyle\sum_{n=8}^{20}(5n+2)$. [3]
Show complete worked solution
(a) The general term is $u_n=5n+2$, which is arithmetic with first term $u_1=5(1)+2=7$, common difference $d=5$, and last term $u_{20}=5(20)+2=102$. Using $S_n=\dfrac{n}{2}(u_1+u_n)$ with $n=20$:
$$S_{20} = \frac{20}{2}(7+102)$$
$$= 10\times109$$
$$\boxed{S_{20} = 1090} \quad \blacksquare$$
(b) The sum $\displaystyle\sum_{n=8}^{20}(5n+2)$ can be found by subtracting the sum of the first 7 terms ($n=1$ to $n=7$) from $S_{20}$. Since $u_7=5(7)+2=37$:
$$S_7 = \frac{7}{2}(7+37) = 3.5\times44 = 154$$
$$\sum_{n=8}^{20}(5n+2) = S_{20}-S_7 = 1090-154$$
$$\boxed{\sum_{n=8}^{20}(5n+2) = 936}$$
QUESTION 42
8 marks
Medium
An athlete's training plan requires running laps of a track each week. In week 1 she runs 6 laps, and each week thereafter she runs 3 more laps than the previous week.
(a) Find the number of laps she runs in week 12. [2]
(b) Find the total number of laps run from week 1 to week 12 inclusive. [2]
(c) Find the first week, $n$, in which her cumulative total number of laps (summed from week 1) exceeds 500. [4]
Show complete worked solution
This is an arithmetic sequence with $u_1=6$ and $d=3$.
(a) Using $u_n=u_1+(n-1)d$ with $n=12$:
$$u_{12} = 6+(12-1)(3)$$
$$= 6+33$$
$$\boxed{u_{12} = 39 \text{ laps}}$$
(b) Using $S_n=\dfrac{n}{2}(u_1+u_n)$ with $n=12$, $u_1=6$, $u_{12}=39$:
$$S_{12} = \frac{12}{2}(6+39)$$
$$= 6\times45$$
$$\boxed{S_{12} = 270 \text{ laps}}$$
(c) We require the least $n$ such that $S_n>500$. Using $S_n=\dfrac{n}{2}\big(2(6)+(n-1)(3)\big)=\dfrac{3n^2+9n}{2}$, and testing values (e.g. using the GDC table function):
$$n=16: \quad S_{16} = \frac{3(256)+9(16)}{2} = \frac{768+144}{2} = 456 \ (\le500)$$
$$n=17: \quad S_{17} = \frac{3(289)+9(17)}{2} = \frac{867+153}{2} = 510 \ (>500)$$
$$\boxed{n = 17 \text{ (week 17)}}$$
QUESTION 43
13 marks
Hard
Two water tanks are being filled. Tank A starts with 200 litres, and the volume added during hour $n$ (for $n=1,2,3,\ldots$) is $20+5(n-1)$ litres, so that the hourly inflow itself increases arithmetically. Tank B starts with 150 litres, and the volume of water in it increases by $12\%$ each hour relative to the volume at the start of that hour, so $B_n=150(1.12)^n$ gives the volume in Tank B after $n$ hours.
(a) Show that the volume of water in Tank A after $n$ hours is $A_n = 200+\dfrac{5n^2+35n}{2}$. [3]
(b) Find the volume of water in each tank after 6 hours, correct to the nearest litre. [3]
(c) Find the total volume of water added to Tank A during the first 10 hours, and hence find $A_{10}$. [3]
(d) By finding the volume in each tank for successive values of $n$ (for example using the table function on your GDC), determine the first complete hour, $n$, after which Tank B contains more water than Tank A. [4]
Show complete worked solution
(a) The hourly inflows $20, 25, 30, \ldots$ form an arithmetic sequence with first term $20$ and common difference $5$. The total volume added during the first $n$ hours is the sum of this sequence:
$$\text{Sum of inflows} = \frac{n}{2}\big(2(20)+(n-1)(5)\big) = \frac{n}{2}(40+5n-5) = \frac{5n^2+35n}{2}$$
Adding this to the initial 200 litres:
$$\boxed{A_n = 200+\frac{5n^2+35n}{2}} \quad \blacksquare$$
(b) Substituting $n=6$ into $A_n$:
$$A_6 = 200+\frac{5(36)+35(6)}{2} = 200+\frac{180+210}{2} = 200+195 = 395$$
Substituting $n=6$ into $B_n=150(1.12)^n$:
$$B_6 = 150(1.12)^6 = 150\times1.97381\ldots = 296.07\ldots$$
$$\boxed{A_6 = 395 \text{ litres}, \quad B_6 \approx 296 \text{ litres}}$$
(c) The total volume added to Tank A during the first 10 hours is the sum of the first 10 inflow terms:
$$\text{Sum} = \frac{5(10)^2+35(10)}{2} = \frac{500+350}{2} = 425 \text{ litres}$$
Hence:
$$A_{10} = 200+425$$
$$\boxed{A_{10} = 625 \text{ litres}}$$
(d) Using the GDC's table function to compute $A_n$ and $B_n$ for increasing $n$: Tank A grows quadratically while Tank B grows exponentially, so Tank B eventually overtakes Tank A. Checking successive hours near the crossover:
Hour 21: $A_{21}=200+\dfrac{5(441)+35(21)}{2}=200+1470=1670$, $\quad B_{21}=150(1.12)^{21}\approx1620.58$ (B still below A)
Hour 22: $A_{22}=200+\dfrac{5(484)+35(22)}{2}=200+1595=1795$, $\quad B_{22}=150(1.12)^{22}\approx1815.05$ (B now exceeds A)
$$\boxed{\text{Tank B first contains more water than Tank A after 22 hours}}$$
QUESTION 44
11 marks
Hard
A pattern of nested squares is drawn such that the perimeter of the first square is 96 cm, and the perimeter of each subsequent square is $70\%$ of the perimeter of the square before it. This pattern is continued indefinitely.
(a) State the first term, $u_1$, and common ratio, $r$, of the sequence of perimeters. [2]
(b) Find $S_\infty$, the total length of all the perimeters if the pattern were continued indefinitely. [2]
(c) Find the least number of squares, $n$, required so that the total perimeter drawn, $S_n$, is within 1 cm of $S_\infty$ (i.e. so that $S_\infty-S_n<1$). [5]
(d) Find $S_n$ for this value of $n$, correct to 3 significant figures. [2]
Show complete worked solution
(a) Each perimeter is $70\%$ of the previous perimeter, so:
$$\boxed{u_1 = 96, \quad r = 0.7}$$
(b) Since $|r|=0.7<1$, using $S_\infty=\dfrac{u_1}{1-r}$:
$$S_\infty = \frac{96}{1-0.7} = \frac{96}{0.3}$$
$$\boxed{S_\infty = 320 \text{ cm}}$$
(c) Using $S_n=\dfrac{96(1-0.7^n)}{0.3}$ and $S_\infty=320$:
$$S_\infty-S_n = 320-\frac{96(1-0.7^n)}{0.3} = \frac{96(0.7^n)}{0.3} = 320(0.7)^n$$
We require $320(0.7)^n<1$:
$$(0.7)^n < \frac{1}{320} = 0.003125$$
Taking logarithms (noting $\log(0.7)<0$, so the inequality reverses):
$$n\log(0.7) < \log(0.003125)$$
$$n > \frac{\log(0.003125)}{\log(0.7)}$$
$$n > 16.17\ldots$$
Since $n$ must be a whole number, the least value is $n=17$.
Check: at $n=17$, $S_\infty-S_n=320(0.7)^{17}=0.744\ldots<1$; at $n=16$, $S_\infty-S_n=320(0.7)^{16}=1.063\ldots>1$, confirming $n=17$ is the least such $n$.
$$\boxed{n = 17}$$
(d) Substituting $n=17$ into $S_n=\dfrac{96(1-0.7^{17})}{0.3}$:
$$S_{17} = \frac{96\big(1-0.7^{17}\big)}{0.3}$$
$$= 319.26\ldots$$
$$\boxed{S_{17} \approx 319 \text{ cm (3 s.f.)}}$$
QUESTION 45
13 marks
Hard
The expressions $x-3$, $x+1$, and $4x-2$ represent the first three terms of a geometric sequence, where $x$ is a real number such that $x\neq3$.
(a) Show that $x$ satisfies the equation $3x^2-16x+5=0$. [4]
(b) Solve this equation to find the two possible values of $x$. [3]
(c) For each value of $x$ found in part (b), find the corresponding first three terms of the sequence and the common ratio, $r$. [4]
(d) For the value of $x$ that gives $|r|<1$, find $S_\infty$ for the resulting infinite geometric series, giving your answer as an exact fraction. [2]
Show complete worked solution
(a) For three consecutive geometric terms $a,b,c$, we have $b^2=ac$. Applying this to $x-3$, $x+1$, $4x-2$:
$$(x+1)^2 = (x-3)(4x-2)$$
Expanding both sides:
$$x^2+2x+1 = 4x^2-2x-12x+6 = 4x^2-14x+6$$
Rearranging so all terms are on one side:
$$0 = 4x^2-14x+6-x^2-2x-1$$
$$\boxed{3x^2-16x+5 = 0} \quad \blacksquare$$
(b) Using the quadratic formula with $a=3$, $b=-16$, $c=5$:
$$x = \frac{16\pm\sqrt{(-16)^2-4(3)(5)}}{2(3)}$$
$$= \frac{16\pm\sqrt{256-60}}{6} = \frac{16\pm\sqrt{196}}{6} = \frac{16\pm14}{6}$$
$$\boxed{x = 5 \quad \text{or} \quad x = \frac{1}{3}}$$
(c) For $x=5$: the terms are $x-3=2$, $x+1=6$, $4x-2=18$. The common ratio is $r=\dfrac{6}{2}=3$ (check: $\dfrac{18}{6}=3$, consistent).
For $x=\dfrac{1}{3}$: the terms are $x-3=-\dfrac{8}{3}$, $x+1=\dfrac{4}{3}$, $4x-2=-\dfrac{2}{3}$. The common ratio is $r=\dfrac{4/3}{-8/3}=-\dfrac{1}{2}$ (check: $\dfrac{-2/3}{4/3}=-\dfrac{1}{2}$, consistent).
$$\boxed{x=5: \ \text{terms } 2,6,18, \ r=3 \qquad x=\tfrac{1}{3}: \ \text{terms } -\tfrac{8}{3},\tfrac{4}{3},-\tfrac{2}{3}, \ r=-\tfrac{1}{2}}$$
(d) The case $x=5$ gives $r=3$, for which $|r|<1$ is false, so this series does not converge. The case $x=\dfrac{1}{3}$ gives $r=-\dfrac{1}{2}$, for which $|r|=\dfrac{1}{2}<1$, so this series converges. Using $S_\infty=\dfrac{u_1}{1-r}$ with $u_1=-\dfrac{8}{3}$, $r=-\dfrac{1}{2}$:
$$S_\infty = \frac{-\frac{8}{3}}{1-\left(-\frac{1}{2}\right)} = \frac{-\frac{8}{3}}{\frac{3}{2}} = -\frac{8}{3}\times\frac{2}{3}$$
$$\boxed{S_\infty = -\frac{16}{9}}$$
QUESTION 46
10 marks
Hard
A sequence is defined by $u_1=4$ and $u_{n+1}=u_n+2n+3$ for $n\ge1$.
(a) Find the values of $u_2$, $u_3$, and $u_4$. [2]
(b) By finding the first and second differences of the sequence $u_1,u_2,u_3,u_4$, show that $u_n$ can be written in the form $u_n=an^2+bn+c$, and find the values of $a$, $b$, and $c$. [5]
(c) Hence find $u_{20}$ using your formula from part (b), and verify your answer by continuing the recursive definition up to $u_{20}$ using the table function on your GDC. [3]
Show complete worked solution
(a) Using $u_{n+1}=u_n+2n+3$:
$$u_2 = u_1+2(1)+3 = 4+5 = 9$$
$$u_3 = u_2+2(2)+3 = 9+7 = 16$$
$$u_4 = u_3+2(3)+3 = 16+9 = 25$$
$$\boxed{u_2=9, \quad u_3=16, \quad u_4=25}$$
(b) The first differences are $u_2-u_1=5$, $u_3-u_2=7$, $u_4-u_3=9$. Since these are not constant, the sequence is not arithmetic; however, the second differences are $7-5=2$ and $9-7=2$, which are constant. A constant second difference indicates $u_n$ is quadratic in $n$, i.e. $u_n=an^2+bn+c$.
Substituting $n=1,2,3$ into $u_n=an^2+bn+c$ using the known values:
$$a+b+c = 4 \qquad (1)$$
$$4a+2b+c = 9 \qquad (2)$$
$$9a+3b+c = 16 \qquad (3)$$
Subtracting (1) from (2), and (2) from (3):
$$3a+b = 5 \qquad (4)$$
$$5a+b = 7 \qquad (5)$$
Subtracting (4) from (5):
$$2a = 2 \quad\Rightarrow\quad a=1$$
Substituting into (4): $b=5-3(1)=2$. Substituting into (1): $c=4-1-2=1$.
$$\boxed{a=1, \quad b=2, \quad c=1, \quad \text{so } u_n=n^2+2n+1=(n+1)^2}$$
(c) Substituting $n=20$ into $u_n=n^2+2n+1$:
$$u_{20} = 20^2+2(20)+1 = 400+40+1 = 441$$
Using the GDC table function to iterate the recursive definition $u_{n+1}=u_n+2n+3$ from $u_1=4$ up to $n=20$ confirms $u_{20}=441$, matching $(20+1)^2=441$.
$$\boxed{u_{20} = 441}$$
QUESTION 47
11 marks
Hard
An open-air amphitheatre has seating arranged in $n$ curved rows. The front row (row 1) has 24 seats, and each subsequent row has 5 more seats than the row before it.
(a) Find an expression, in terms of $n$, for the number of seats in row $n$. [2]
(b) Show that $S_n$, the total number of seats in the first $n$ rows, is given by $S_n=\dfrac{5n^2+43n}{2}$. [2]
(c) The architects require a total seating capacity of at least 3000 while using as few rows as possible. Find the least number of rows, $n$, required so that $S_n\ge3000$. [5]
(d) For this value of $n$, find the number of seats in the back row. [2]
Show complete worked solution
This is an arithmetic sequence with $u_1=24$ and $d=5$.
(a) Using $u_n=u_1+(n-1)d$:
$$u_n = 24+(n-1)(5) = 24+5n-5$$
$$\boxed{u_n = 5n+19}$$
(b) Using $S_n=\dfrac{n}{2}\big(2u_1+(n-1)d\big)$:
$$S_n = \frac{n}{2}\big(2(24)+(n-1)(5)\big) = \frac{n}{2}(48+5n-5)$$
$$\boxed{S_n = \frac{5n^2+43n}{2}} \quad \blacksquare$$
(c) We require $\dfrac{5n^2+43n}{2}\ge3000$, i.e. $5n^2+43n-6000\ge0$. Using the quadratic formula with $a=5$, $b=43$, $c=-6000$ to find the boundary:
$$n = \frac{-43+\sqrt{43^2-4(5)(-6000)}}{2(5)} = \frac{-43+\sqrt{1849+120000}}{10} = \frac{-43+\sqrt{121849}}{10}$$
$$n = \frac{-43+349.07\ldots}{10} = 30.60\ldots$$
Since $S_n$ increases with $n$ (each row adds a positive number of seats) and $n$ must be a whole number, the least $n$ satisfying $S_n\ge3000$ is $n=31$. Checking:
$$S_{30} = \frac{5(900)+43(30)}{2} = \frac{4500+1290}{2} = 2895 \ (<3000)$$
$$S_{31} = \frac{5(961)+43(31)}{2} = \frac{4805+1333}{2} = 3069 \ (\ge3000)$$
$$\boxed{n = 31 \text{ rows}}$$
(d) Using $u_n=5n+19$ with $n=31$:
$$u_{31} = 5(31)+19 = 155+19$$
$$\boxed{u_{31} = 174 \text{ seats}}$$
QUESTION 48
12 marks
Hard
A network engineer models the strength of a signal as it passes through a chain of relay stations. At the first relay station, the signal strength is 800 units. At each subsequent relay station, the strength is $78\%$ of the strength at the previous station. A signal becomes unusable once its strength drops below 50 units.
(a) Find the signal strength at the 6th relay station, correct to 3 significant figures. [3]
(b) Find the number of the first relay station at which the signal is unusable (i.e. has strength below 50 units). [4]
(c) Find $S_n$, the sum of the signal strengths at all relay stations up to and including the station found in part (b), correct to 3 significant figures. [2]
(d) Explain, with reference to the value of the common ratio, whether the total combined signal strength summed over infinitely many relay stations (if the chain continued indefinitely) would be finite, and if so, find this total, correct to 3 significant figures. [3]
Show complete worked solution
This is a geometric sequence with $u_1=800$ and $r=0.78$.
(a) Using $u_n=u_1r^{\,n-1}$ with $n=6$:
$$u_6 = 800(0.78)^5$$
$$= 800\times0.28871\ldots$$
$$\boxed{u_6 \approx 231 \text{ units (3 s.f.)}}$$
(b) We require the least $n$ such that $800(0.78)^{n-1}<50$, i.e. $(0.78)^{n-1}<0.0625$. Taking logarithms (noting $\log(0.78)<0$, so the inequality reverses):
$$(n-1)\log(0.78) < \log(0.0625)$$
$$n-1 > \frac{\log(0.0625)}{\log(0.78)}$$
$$n-1 > 11.16\ldots$$
$$n > 12.16\ldots$$
Since $n$ must be a whole number, $n=13$.
Checking: $u_{12}=800(0.78)^{11}\approx52.0$ ($\ge50$, still usable), and $u_{13}=800(0.78)^{12}\approx40.6$ ($<50$, unusable).
$$\boxed{n = 13 \text{th relay station}}$$
(c) Using $S_n=\dfrac{u_1(1-r^n)}{1-r}$ with $u_1=800$, $r=0.78$, $n=13$:
$$S_{13} = \frac{800\big(1-0.78^{13}\big)}{1-0.78}$$
$$= \frac{800(1-0.05071\ldots)}{0.22}$$
$$\boxed{S_{13} \approx 3490 \text{ units (3 s.f.)}}$$
(d) Since $|r|=0.78<1$, the infinite geometric series converges, so the total combined signal strength over infinitely many relay stations would be finite. Using $S_\infty=\dfrac{u_1}{1-r}$:
$$S_\infty = \frac{800}{1-0.78} = \frac{800}{0.22}$$
$$\boxed{S_\infty \approx 3640 \text{ units (3 s.f.), since } |r|<1}$$
QUESTION 49
10 marks
Hard
The $n$th term of an arithmetic sequence is $u_n=6n-2$, and the $n$th term of a geometric sequence is $g_n=2(3)^{n-1}$. A new sequence is formed by adding corresponding terms: $v_n=u_n+g_n$.
(a) Find $v_1$, $v_2$, $v_3$, and $v_4$. [3]
(b) Write down an expression for $v_n$ in terms of $n$. [2]
(c) Find $\displaystyle\sum_{n=1}^{10}v_n$ by separately summing the arithmetic part and the geometric part. [5]
Show complete worked solution
(a) Using $u_n=6n-2$ and $g_n=2(3)^{n-1}$:
$$v_1 = (6-2)+(2) = 4+2 = 6$$
$$v_2 = (12-2)+(6) = 10+6 = 16$$
$$v_3 = (18-2)+(18) = 16+18 = 34$$
$$v_4 = (24-2)+(54) = 22+54 = 76$$
$$\boxed{v_1=6, \ v_2=16, \ v_3=34, \ v_4=76}$$
(b) Adding the two general term expressions:
$$\boxed{v_n = 6n-2+2(3)^{n-1}}$$
(c) Since $v_n$ is the sum of an arithmetic term and a geometric term, $\displaystyle\sum_{n=1}^{10}v_n = \sum_{n=1}^{10}(6n-2) + \sum_{n=1}^{10}2(3)^{n-1}$.
For the arithmetic part, $u_1=4$, $d=6$, using $S_n=\dfrac{n}{2}\big(2u_1+(n-1)d\big)$ with $n=10$:
$$\sum_{n=1}^{10}(6n-2) = \frac{10}{2}\big(2(4)+9(6)\big) = 5(8+54) = 5\times62 = 310$$
For the geometric part, $u_1=2$, $r=3$, using $S_n=\dfrac{u_1(r^n-1)}{r-1}$ with $n=10$:
$$\sum_{n=1}^{10}2(3)^{n-1} = \frac{2(3^{10}-1)}{3-1} = \frac{2(59049-1)}{2} = 59048$$
Adding both parts:
$$\sum_{n=1}^{10}v_n = 310+59048$$
$$\boxed{\sum_{n=1}^{10}v_n = 59358}$$
QUESTION 50
12 marks
Hard
An infinite geometric series has sum to infinity $S_\infty=45$, and the sum of its first two terms is $u_1+u_2=20$, where $-1(a) Show that $u_1=45(1-r)$. [2]
(b) Show that $r$ satisfies the equation $45(1-r^2)=20$, and hence find the two possible exact values of $r$, giving each also correct to 3 significant figures. [5]
(c) Find the corresponding value of $u_1$, correct to 3 significant figures, for each value of $r$ found in part (b). [3]
(d) For the case where $r>0$, find $u_5$, correct to 3 significant figures. [2]
Show complete worked solution
(a) Using $S_\infty=\dfrac{u_1}{1-r}=45$:
$$u_1 = 45(1-r) \quad \blacksquare$$
$$\boxed{u_1 = 45(1-r)}$$
(b) Since $u_2=u_1r$, the condition $u_1+u_2=20$ becomes $u_1(1+r)=20$. Substituting $u_1=45(1-r)$ from part (a):
$$45(1-r)(1+r) = 20$$
$$\boxed{45(1-r^2) = 20} \quad \blacksquare$$
Solving for $r^2$:
$$1-r^2 = \frac{20}{45} = \frac{4}{9}$$
$$r^2 = 1-\frac{4}{9} = \frac{5}{9}$$
$$r = \pm\sqrt{\frac{5}{9}} = \pm\frac{\sqrt5}{3}$$
Both values satisfy $-10$ case): $u_1+u_2 = 11.459+11.459(0.7454) = 11.459+8.541 = 20.0$, confirming consistency.
(d) For $r>0$: $r=\dfrac{\sqrt5}{3}$ and $u_1\approx11.459$. Using $u_5=u_1r^4$:
$$u_5 = 11.459\times\left(\frac{\sqrt5}{3}\right)^4 = 11.459\times\frac{25}{81}$$
$$= 11.459\times0.30864\ldots$$
$$\boxed{u_5 \approx 3.54 \text{ (3 s.f.)}}$$
Financial Mathematics 50 questions
QUESTION 1
4 marks
Easy
Kofi invests \$5000 in an account that pays a nominal annual interest rate of 4.2\%, compounded annually. Find the value of his investment after 6 years, correct to the nearest dollar.
Show complete worked solution
Using the compound interest formula $FV=PV(1+i)^n$ (or the GDC's finance/TVM solver), with $PV=5000$, $i=0.042$, $n=6$:
$$FV = 5000(1.042)^{6}$$
$$= 5000 \times 1.279989\ldots$$
$$= 6399.946\ldots$$
$$\boxed{FV \approx \$6400 \text{ (nearest dollar)}}$$
QUESTION 2
5 marks
Easy
A delivery company buys a van for \$28000. The van depreciates in value by 12\% each year.
(a) Find the value of the van after 5 years, correct to the nearest dollar.
(b) Find the total amount the van has depreciated in value over these 5 years, correct to the nearest dollar.
Show complete worked solution
(a) Since the van depreciates by 12% each year, its value is multiplied by $(1-0.12)=0.88$ each year. Using $FV=PV(0.88)^n$ with $PV=28000$, $n=5$:
$$FV = 28000(0.88)^{5}$$
$$= 28000 \times 0.527660\ldots$$
$$= 14776.49\ldots$$
$$\boxed{FV \approx \$14{,}776 \text{ (nearest dollar)}}$$
(b) Depreciation = original value $-$ final value:
$$\text{Depreciation} = 28000-14776.49\ldots$$
$$= 13223.51\ldots$$
$$\boxed{\approx \$13{,}224 \text{ (nearest dollar)}}$$
QUESTION 3
6 marks
Medium
Priya's parents want to have \$15000 available in 8 years' time to help pay for her university education. A bank account offers a nominal annual interest rate of 3.6\%, compounded monthly.
(a) Find the single lump sum Priya's parents must deposit now in order for the account to be worth \$15000 in 8 years, correct to the nearest dollar.
(b) Find the effective annual interest rate of this account, correct to 3 significant figures.
Show complete worked solution
(a) Monthly interest rate:
$$i = \frac{0.036}{12} = 0.003$$
Number of months: $n=8\times12=96$.
Using $FV=PV(1+i)^n$ rearranged for $PV$ (or the GDC's TVM solver with $FV=-15000$, $N=96$, $I\%=3.6$, $P/Y=C/Y=12$):
$$PV = \frac{15000}{(1.003)^{96}}$$
$$= \frac{15000}{1.333177\ldots}$$
$$= 11251.27\ldots$$
$$\boxed{PV \approx \$11{,}251 \text{ (nearest dollar)}}$$
(b) Using the effective annual rate formula $i_{eff}=(1+i)^{12}-1$ with $i=0.003$:
$$i_{eff} = (1.003)^{12}-1$$
$$= 1.036600\ldots-1$$
$$= 0.036600\ldots$$
$$\boxed{i_{eff} \approx 3.66\% \text{ (3 s.f.)}}$$
QUESTION 4
6 marks
Medium
An investor deposits \$10000 into an account paying a nominal annual interest rate of 5\%, compounded quarterly, for 4 years.
(a) Find the value of the investment after 4 years, correct to the nearest dollar.
(b) Find the total interest earned over the 4 years.
(c) Find the effective annual interest rate of this account, correct to 2 decimal places.
Show complete worked solution
(a) Quarterly interest rate:
$$i = \frac{0.05}{4} = 0.0125$$
Number of quarters: $n=4\times4=16$.
Using the GDC's TVM solver (or $FV=PV(1+i)^n$) with $PV=10000$:
$$FV = 10000(1.0125)^{16}$$
$$= 10000 \times 1.219890\ldots$$
$$= 12198.90\ldots$$
$$\boxed{FV \approx \$12{,}199 \text{ (nearest dollar)}}$$
(b) Interest earned = final value $-$ initial value:
$$\text{Interest} = 12198.90\ldots-10000$$
$$= 2198.90\ldots$$
$$\boxed{\approx \$2199}$$
(c) Using $i_{eff}=(1+i)^{4}-1$ with $i=0.0125$:
$$i_{eff} = (1.0125)^{4}-1$$
$$= 1.050945\ldots-1$$
$$= 0.050945\ldots$$
$$\boxed{i_{eff} \approx 5.09\% \text{ (2 d.p.)}}$$
QUESTION 5
14 marks
Hard
A manufacturing company buys a machine for \$120000. For tax purposes, the machine's value depreciates by the reducing-balance (declining-balance) method at a rate of 15\% per year.
(a) Find the value of the machine after 5 years, correct to the nearest dollar. [3]
(b) By finding the value of the machine at the end of each year (for example using the table function of your GDC), find the first complete number of years after which the machine's value first falls below \$50000. [4]
(c) To finance the purchase, the company takes out a loan of \$120000 at a nominal annual interest rate of 6\%, compounded monthly, to be repaid in equal monthly instalments over 5 years. Using your GDC's finance (TVM) solver, find the monthly repayment amount, correct to the nearest cent. [4]
(d) Find the total amount repaid over the 5 years, and hence find the total interest paid on the loan, both correct to the nearest dollar. [3]
Show complete worked solution
(a) Using the reducing-balance depreciation formula $V_n = V_0(1-r)^n$ with $V_0=120000$, $r=0.15$, $n=5$:
$$V_5 = 120000(1-0.15)^5$$
$$= 120000(0.85)^5$$
$$= 120000 \times 0.443705\ldots$$
$$= 53244.63\ldots$$
$$\boxed{V_5 \approx \$53{,}245 \text{ (nearest dollar)}}$$
(b) Computing $V_n=120000(0.85)^n$ for successive $n$ using the GDC's table function:
$$V_5 = 53244.64 \text{ (still above \$50000)}$$
$$V_6 = 120000(0.85)^6 = 45257.94\ldots \text{ (below \$50000)}$$
$$\boxed{\text{The value first falls below \$50000 after 6 complete years}}$$
(c) Monthly interest rate: $i=\dfrac{0.06}{12}=0.005$. Number of payments: $n=5\times12=60$.
Using the loan repayment (annuity) formula $PMT=\dfrac{Pi}{1-(1+i)^{-n}}$ (equivalently, the GDC's TVM solver with $PV=120000$, $N=60$, $I\%=6$, $C/Y=P/Y=12$), with $P=120000$:
$$PMT = \frac{120000(0.005)}{1-(1.005)^{-60}}$$
$$= \frac{600}{1-0.741372\ldots}$$
$$= \frac{600}{0.258628\ldots}$$
$$= 2319.936\ldots$$
$$\boxed{PMT \approx \$2319.94 \text{ per month}}$$
(d) Total amount repaid = monthly payment $\times$ number of payments:
$$\text{Total repaid} = 2319.936\ldots \times 60 = 139196.17\ldots$$
$$\boxed{\text{Total repaid} \approx \$139{,}196}$$
Total interest = total repaid $-$ principal borrowed:
$$\text{Total interest} = 139196.17\ldots-120000 = 19196.17\ldots$$
$$\boxed{\text{Total interest} \approx \$19{,}196}$$
QUESTION 6
4 marks
Easy
Amelia takes out a loan of $8000 to buy a used car. The loan is to be repaid with equal monthly payments over 3 years, and interest is charged at a nominal annual rate of 6%, compounded monthly.
Use technology (the finance/TVM solver on your GDC) to find the amount of each monthly payment.
Show complete worked solution
Using the GDC's Finance (TVM) Solver, entering the given values:
$$N = 3\times12 = 36, \quad I\% = 6, \quad PV = 8000, \quad FV = 0, \quad P/Y = C/Y = 12$$
Solving for $PMT$ using the annuity formula $PMT=\dfrac{PV\cdot i}{1-(1+i)^{-N}}$ with monthly rate $i=\dfrac{0.06}{12}=0.005$:
$$PMT = \frac{8000(0.005)}{1-(1.005)^{-36}}$$
$$= \frac{40}{1-0.835644\ldots}$$
$$= \frac{40}{0.164356\ldots} = 243.375\ldots$$
$$\boxed{\text{The monthly payment is \$243.38 (to the nearest cent)}}$$
QUESTION 7
4 marks
Easy
At the end of each month, Ravi deposits $150 into a savings account that pays a nominal annual interest rate of 4.8%, compounded monthly.
Find the value of Ravi's savings after 5 years, assuming he makes no other deposits or withdrawals.
Show complete worked solution
Using the GDC's Finance (TVM) Solver for a savings annuity, entering:
$$N = 5\times12 = 60, \quad I\% = 4.8, \quad PV = 0, \quad PMT = -150, \quad P/Y = C/Y = 12$$
Using the future value of an ordinary annuity formula $FV=\dfrac{PMT\big((1+i)^N-1\big)}{i}$ with monthly rate $i=\dfrac{0.048}{12}=0.004$:
$$FV = \frac{150\big((1.004)^{60}-1\big)}{0.004}$$
$$= \frac{150(1.270632\ldots-1)}{0.004}$$
$$= \frac{150(0.270632\ldots)}{0.004} = 10149.026\ldots$$
$$\boxed{FV \approx \$10{,}149.03 \; (\approx \$10{,}100 \text{ to 3 s.f.})}$$
QUESTION 8
6 marks
Medium
Grace takes out a loan of $18 500 to buy a car. The loan is repaid in equal monthly instalments over 4 years, with interest charged at a nominal annual rate of 5%, compounded monthly.
(a) Find the monthly payment. [3]
(b) Find the total interest Grace pays during the first year (the first 12 payments) of the loan. [3]
Show complete worked solution
(a) Using the GDC's Finance (TVM) Solver, entering:
$$N = 4\times12 = 48, \quad I\% = 5, \quad PV = 18500, \quad FV = 0, \quad P/Y = C/Y = 12$$
Using the annuity formula $PMT=\dfrac{PV\cdot i}{1-(1+i)^{-N}}$ with monthly rate $i=\dfrac{0.05}{12}=0.0041667$:
$$PMT = \frac{18500(0.0041667)}{1-(1.0041667)^{-48}}$$
$$= \frac{77.0833\ldots}{0.180937\ldots} = 426.0419\ldots$$
$$\boxed{\text{Monthly payment} = \$426.04}$$
(b) Using the GDC's amortization/balance function, the outstanding balance after 12 payments is:
$$\text{Balance}(12) = 14215.19 \text{ (2 d.p.)}$$
Total paid in the first year:
$$12\times426.0419\ldots = 5112.50$$
Principal repaid in the first year:
$$18500-14215.19 = 4284.81$$
Interest paid in the first year = total paid $-$ principal repaid:
$$5112.50-4284.81 = 827.69$$
$$\boxed{\text{Interest paid in the first year} \approx \$828 \text{ (3 s.f.)}}$$
QUESTION 9
5 marks
Medium
Hassan wants to accumulate $50 000 in a retirement account after 10 years. He plans to make equal deposits at the end of each month into an account that pays a nominal annual interest rate of 5.4%, compounded monthly.
Use technology to find the monthly deposit Hassan must make.
Show complete worked solution
Using the GDC's Finance (TVM) Solver, entering:
$$N = 10\times12 = 120, \quad I\% = 5.4, \quad PV = 0, \quad FV = -50000, \quad P/Y = C/Y = 12$$
Using the future value of an ordinary annuity formula $FV=\dfrac{PMT\big((1+i)^N-1\big)}{i}$, rearranged for $PMT$, with monthly rate $i=\dfrac{0.054}{12}=0.0045$:
$$PMT = \frac{FV\cdot i}{(1+i)^N-1}$$
$$= \frac{50000(0.0045)}{(1.0045)^{120}-1}$$
$$= \frac{225}{1.71397\ldots-1} = \frac{225}{0.71397\ldots} = 315.157\ldots$$
$$\boxed{\text{Hassan must deposit \$315.16 at the end of each month (to the nearest cent)}}$$
QUESTION 10
16 marks
Hard
The Alvarez family takes out a home mortgage of $240 000 to buy a house. The bank charges a nominal annual interest rate of 4.2%, compounded monthly, and the loan is to be fully repaid with equal monthly payments over 25 years.
(a) Use technology (the TVM solver) to find the monthly payment. [4]
(b) Find the total amount the Alvarez family will have paid over the full 25-year term, and hence find the total interest paid over the life of the loan. [5]
(c) After paying the mortgage for exactly 10 years (120 payments), the Alvarez family is considering refinancing.
(i) Use technology to find the outstanding balance on the loan at this point.
(ii) Find how much of the 121st payment goes towards interest and how much goes towards reducing the principal. [7]
Show complete worked solution
(a) Using the GDC's Finance (TVM) Solver:
$$N = 25\times12 = 300, \quad I\% = 4.2, \quad PV = 240000, \quad FV = 0, \quad P/Y = C/Y = 12$$
Using the annuity formula $PMT=\dfrac{PV\cdot i}{1-(1+i)^{-N}}$ with monthly rate $i=\dfrac{0.042}{12}=0.0035$:
$$PMT = \frac{240000(0.0035)}{1-(1.0035)^{-300}}$$
$$= \frac{840}{0.64943\ldots} = 1293.4615\ldots$$
$$\boxed{\text{Monthly payment} = \$1293.46}$$
(b) Total amount paid = monthly payment $\times$ number of payments:
$$\text{Total paid} = 300\times1293.4615\ldots = 388038.47 \text{ (2 d.p.)}$$
Total interest paid = total paid $-$ principal borrowed:
$$\text{Total interest} = 388038.47-240000 = 148038.47$$
$$\boxed{\text{Total paid} \approx \$388{,}000, \quad \text{Total interest} \approx \$148{,}000 \text{ (3 s.f.)}}$$
(c)(i) Using the GDC's amortization/balance function with $N=120$:
$$\text{Balance}(120) = 172518.97 \text{ (2 d.p.)}$$
$$\boxed{\text{Outstanding balance after 10 years} \approx \$172{,}519 \; (\approx \$173{,}000 \text{ to 3 s.f.})}$$
(ii) Monthly interest rate: $i=\dfrac{0.042}{12}=0.0035$.
Interest portion of the 121st payment = outstanding balance $\times$ monthly rate:
$$172518.97\times0.0035 = 603.82 \text{ (2 d.p.)}$$
Principal portion of the 121st payment = total payment $-$ interest portion:
$$1293.46-603.82 = 689.64 \text{ (2 d.p.)}$$
$$\boxed{\text{Interest portion} \approx \$603.82, \quad \text{Principal portion} \approx \$689.64}$$
QUESTION 11
4 marks
Easy
Liam invests \$3200 in a savings account that pays a nominal annual interest rate of 3.5\%, compounded annually. Find the value of his investment after 7 years, correct to the nearest dollar.
Show complete worked solution
Using the compound interest formula $FV=PV(1+i)^n$ (or the GDC's finance/TVM solver), with $PV=3200$, $i=0.035$, $n=7$:
$$FV = 3200(1.035)^{7}$$
$$= 3200 \times 1.272279\ldots$$
$$= 4071.29\ldots$$
$$\boxed{FV \approx \$4071 \text{ (nearest dollar)}}$$
QUESTION 12
4 marks
Easy
Fatima deposits \$6000 into a bank account that pays a nominal annual interest rate of 4.8\%, compounded semi-annually. Find the value of her investment after 5 years, correct to the nearest dollar.
Show complete worked solution
Since interest is compounded semi-annually, the interest rate per period is
$$i = \frac{0.048}{2} = 0.024$$
and the number of periods is $n=5\times2=10$.
Using $FV=PV(1+i)^n$ (or the GDC's TVM solver) with $PV=6000$:
$$FV = 6000(1.024)^{10}$$
$$= 6000 \times 1.267651\ldots$$
$$= 7605.90\ldots$$
$$\boxed{FV \approx \$7606 \text{ (nearest dollar)}}$$
QUESTION 13
4 marks
Easy
Noah deposits \$2500 into a savings account that pays a nominal annual interest rate of 3\%, compounded monthly. Find the value of his investment after 3 years, correct to the nearest dollar.
Show complete worked solution
The monthly interest rate is
$$i = \frac{0.03}{12} = 0.0025$$
and the number of months is $n=3\times12=36$.
Using $FV=PV(1+i)^n$ (or the GDC's TVM solver) with $PV=2500$:
$$FV = 2500(1.0025)^{36}$$
$$= 2500 \times 1.094051\ldots$$
$$= 2735.13\ldots$$
$$\boxed{FV \approx \$2735 \text{ (nearest dollar)}}$$
QUESTION 14
4 marks
Easy
Mei invests \$9000 in an account that pays a nominal annual interest rate of 2.4\%, compounded quarterly. Find the value of her investment after 6 years, correct to the nearest dollar.
Show complete worked solution
The quarterly interest rate is
$$i = \frac{0.024}{4} = 0.006$$
and the number of quarters is $n=6\times4=24$.
Using $FV=PV(1+i)^n$ (or the GDC's TVM solver) with $PV=9000$:
$$FV = 9000(1.006)^{24}$$
$$= 9000 \times 1.154387\ldots$$
$$= 10389.49\ldots$$
$$\boxed{FV \approx \$10{,}389 \text{ (nearest dollar)}}$$
QUESTION 15
6 marks
Medium
Sofia deposits \$7500 into an account that pays a nominal annual interest rate of 4.4\%, compounded monthly.
(a) Find the value of her investment after 6 years, correct to the nearest dollar.
(b) Find the effective annual interest rate of this account, correct to 3 significant figures.
Show complete worked solution
(a) The monthly interest rate is
$$i = \frac{0.044}{12} = 0.0036\overline{6}$$
and the number of months is $n=6\times12=72$.
Using $FV=PV(1+i)^n$ (or the GDC's TVM solver) with $PV=7500$:
$$FV = 7500(1.003667\ldots)^{72}$$
$$= 7500 \times 1.301500\ldots$$
$$= 9761.25\ldots$$
$$\boxed{FV \approx \$9761 \text{ (nearest dollar)}}$$
(b) Using the effective annual rate formula $i_{eff}=(1+i)^{12}-1$ with $i=0.0036\overline{6}$:
$$i_{eff} = (1.003667\ldots)^{12}-1$$
$$= 1.044898\ldots-1$$
$$= 0.044898\ldots$$
$$\boxed{i_{eff} \approx 4.49\% \text{ (3 s.f.)}}$$
QUESTION 16
6 marks
Medium
A bank account pays a nominal annual interest rate of 6\%, compounded monthly.
(a) Find the effective annual interest rate of this account, correct to 3 significant figures.
(b) Carlos deposits \$5000 into this account. Find the value of his investment after 4 years, correct to the nearest dollar.
Show complete worked solution
(a) The monthly interest rate is
$$i = \frac{0.06}{12} = 0.005$$
Using the effective annual rate formula $i_{eff}=(1+i)^{12}-1$:
$$i_{eff} = (1.005)^{12}-1$$
$$= 1.061678\ldots-1$$
$$= 0.061678\ldots$$
$$\boxed{i_{eff} \approx 6.17\% \text{ (3 s.f.)}}$$
(b) Using $FV=PV(1+i)^n$ (or the GDC's TVM solver) with $PV=5000$, $i=0.005$, $n=4\times12=48$:
$$FV = 5000(1.005)^{48}$$
$$= 5000 \times 1.270489\ldots$$
$$= 6352.45\ldots$$
$$\boxed{FV \approx \$6352 \text{ (nearest dollar)}}$$
QUESTION 17
7 marks
Medium
Ines invests \$4000 in an account that pays a nominal annual interest rate of 6\%, compounded quarterly.
(a) Find the value of her investment after 10 years, correct to the nearest dollar. [3]
(b) By finding the value of the investment at the end of each year (for example using the table function of your GDC), find the first complete number of years after which the value of the investment first exceeds \$8000. [4]
Show complete worked solution
(a) The quarterly interest rate is
$$i = \frac{0.06}{4} = 0.015$$
Using $FV=PV(1+i)^n$ (or the GDC's TVM solver) with $PV=4000$, $n=10\times4=40$:
$$FV = 4000(1.015)^{40}$$
$$= 4000 \times 1.814018\ldots$$
$$= 7256.07\ldots$$
$$\boxed{FV \approx \$7256 \text{ (nearest dollar)}}$$
(b) Computing $FV=4000(1.015)^{4y}$ for successive complete years $y$ using the GDC's table function:
$$FV(11) = 4000(1.015)^{44} = 7701.33\ldots \text{ (still below \$8000)}$$
$$FV(12) = 4000(1.015)^{48} = 8173.91\ldots \text{ (above \$8000)}$$
$$\boxed{\text{The value first exceeds \$8000 after 12 complete years}}$$
QUESTION 18
8 marks
Medium
Two banks offer accounts for a \$5000 deposit.
Bank X offers a nominal annual interest rate of 5.0\%, compounded semi-annually.
Bank Y offers a nominal annual interest rate of 4.9\%, compounded monthly.
(a) Find the effective annual interest rate of each bank's account, correct to 3 significant figures. [4]
(b) Hence, or otherwise, determine which bank gives the greater value after 3 years, and find the difference between the two final values, correct to the nearest dollar. [4]
Show complete worked solution
(a) For Bank X, the semi-annual rate is $i=\dfrac{0.05}{2}=0.025$, so
$$i_{eff,X} = (1.025)^{2}-1 = 1.050625-1 = 0.050625\ldots \approx 5.06\% \text{ (3 s.f.)}$$
For Bank Y, the monthly rate is $i=\dfrac{0.049}{12}=0.0040833\ldots$, so
$$i_{eff,Y} = (1.0040833\ldots)^{12}-1 = 1.050116\ldots-1 = 0.050116\ldots \approx 5.01\% \text{ (3 s.f.)}$$
$$\boxed{i_{eff,X} \approx 5.06\%, \quad i_{eff,Y} \approx 5.01\%}$$
(b) Since $i_{eff,X} > i_{eff,Y}$, Bank X gives the greater value after 3 years. Computing each value directly with $PV=5000$:
$$FV_X = 5000(1.025)^{6} = 5000 \times 1.159693\ldots = 5798.47\ldots$$
$$FV_Y = 5000(1.0040833\ldots)^{36} = 5000 \times 1.158007\ldots = 5790.04\ldots$$
Difference:
$$5798.47\ldots-5790.04\ldots = 8.43\ldots$$
$$\boxed{\text{Bank X gives the greater value, by approximately \$8}}$$
QUESTION 19
10 marks
Hard
Two investment options are available for a \$20000 deposit, both held for 10 years.
Option A: a nominal annual interest rate of 5.4\%, compounded monthly.
Option B: a nominal annual interest rate of 5.6\%, compounded annually.
(a) Find the effective annual interest rate of Option A, correct to 3 significant figures. [3]
(b) Find the value of the investment after 10 years under each option, correct to the nearest dollar. [5]
(c) State, with a reason, which option is the better investment, and find the difference between the two final values, correct to the nearest dollar. [2]
Show complete worked solution
(a) The monthly interest rate for Option A is
$$i = \frac{0.054}{12} = 0.0045$$
Using $i_{eff}=(1+i)^{12}-1$:
$$i_{eff} = (1.0045)^{12}-1 = 1.055357\ldots-1 = 0.055357\ldots$$
$$\boxed{i_{eff} \approx 5.54\% \text{ (3 s.f.)}}$$
(b) For Option A, using $FV=PV(1+i)^n$ (or the GDC's TVM solver) with $PV=20000$, $i=0.0045$, $n=10\times12=120$:
$$FV_A = 20000(1.0045)^{120}$$
$$= 20000 \times 1.713929\ldots$$
$$= 34278.59\ldots$$
$$\boxed{FV_A \approx \$34{,}279 \text{ (nearest dollar)}}$$
For Option B, using $FV=PV(1+i)^n$ with $PV=20000$, $i=0.056$, $n=10$:
$$FV_B = 20000(1.056)^{10}$$
$$= 20000 \times 1.724405\ldots$$
$$= 34488.09\ldots$$
$$\boxed{FV_B \approx \$34{,}488 \text{ (nearest dollar)}}$$
(c) Since $FV_B > FV_A$, Option B gives the greater final value, even though its nominal rate is only slightly higher than Option A's effective annual rate. The difference is
$$34488.09\ldots-34278.59\ldots = 209.50\ldots$$
$$\boxed{\text{Option B is better, by approximately \$210}}$$
QUESTION 20
9 marks
Hard
Diego invests \$8000 in an account that pays an unknown nominal annual interest rate, compounded quarterly. After 7 years, his investment is worth \$11327.94.
(a) Show that the quarterly interest rate satisfies $(1+i)^{28} = 1.415992\ldots$, and hence find the quarterly interest rate $i$, correct to 4 significant figures. [4]
(b) Find the nominal annual interest rate, correct to 3 significant figures. [2]
(c) Find the effective annual interest rate of this account, correct to 3 significant figures. [3]
Show complete worked solution
(a) There are $n=7\times4=28$ quarterly periods. Using $FV=PV(1+i)^n$ with $PV=8000$, $FV=11327.94$:
$$11327.94 = 8000(1+i)^{28}$$
$$(1+i)^{28} = \frac{11327.94}{8000} = 1.415992\ldots$$
Taking the 28th root of both sides (or using the GDC's equation solver):
$$1+i = 1.415992\ldots^{1/28} = 1.012500\ldots$$
$$\boxed{i \approx 0.01250 \text{ (4 s.f.)}}$$
(b) The nominal annual rate is the quarterly rate multiplied by 4:
$$\text{nominal rate} = 4\times0.01250\ldots = 0.050000\ldots$$
$$\boxed{\text{nominal annual rate} \approx 5.00\% \text{ (3 s.f.)}}$$
(c) Using $i_{eff}=(1+i)^{4}-1$ with $i=0.012500\ldots$:
$$i_{eff} = (1.012500\ldots)^{4}-1$$
$$= 1.050945\ldots-1$$
$$= 0.050945\ldots$$
$$\boxed{i_{eff} \approx 5.09\% \text{ (3 s.f.)}}$$
QUESTION 21
4 marks
Easy
A bakery buys a commercial oven for \$15000. Using the straight-line depreciation method, the oven's value decreases by a fixed amount of \$1800 each year. Find the value of the oven after 4 years.
Show complete worked solution
Using the straight-line depreciation formula $V_n=V_0-nd$, where $V_0=15000$ is the original value and $d=1800$ is the fixed annual depreciation amount, with $n=4$:
$$V_4 = 15000-4(1800)$$
$$= 15000-7200$$
$$= 7800$$
$$\boxed{V_4 = \$7800}$$
QUESTION 22
4 marks
Easy
A small business buys a printer for \$2400. Using the straight-line depreciation method, the printer's value decreases by 8\% of its original cost each year. Find the value of the printer after 6 years.
Show complete worked solution
Since the printer depreciates by a fixed 8\% of its original value each year (straight-line), the annual depreciation amount is
$$d = 0.08 \times 2400 = 192$$
Using $V_n=V_0(1-nr)$ with $V_0=2400$, $r=0.08$, $n=6$:
$$V_6 = 2400\big(1-6(0.08)\big)$$
$$= 2400(1-0.48)$$
$$= 2400(0.52)$$
$$= 1248$$
$$\boxed{V_6 = \$1248}$$
QUESTION 23
4 marks
Easy
A student buys a laptop for \$1800. The laptop depreciates in value by 20\% each year (reducing-balance method). Find the value of the laptop after 3 years, correct to the nearest dollar.
Show complete worked solution
Since the laptop depreciates by 20\% each year, its value is multiplied by $(1-0.20)=0.80$ each year. Using $V_n=V_0(1-r)^n$ with $V_0=1800$, $r=0.20$, $n=3$:
$$V_3 = 1800(0.80)^{3}$$
$$= 1800 \times 0.512$$
$$= 921.60$$
$$\boxed{V_3 \approx \$922 \text{ (nearest dollar)}}$$
QUESTION 24
4 marks
Easy
A farm buys a tractor for \$45000. The tractor depreciates in value by 10\% each year (reducing-balance method). Find the value of the tractor after 4 years, correct to the nearest dollar.
Show complete worked solution
Using the reducing-balance depreciation formula $V_n=V_0(1-r)^n$ with $V_0=45000$, $r=0.10$, $n=4$:
$$V_4 = 45000(0.90)^{4}$$
$$= 45000 \times 0.6561$$
$$= 29524.5$$
$$\boxed{V_4 \approx \$29{,}525 \text{ (nearest dollar)}}$$
QUESTION 25
6 marks
Medium
A restaurant buys a commercial oven for \$6000. The owner estimates the oven will have a useful life of 8 years, after which its salvage (scrap) value will be \$800. The oven is depreciated using the straight-line method.
(a) Find the annual amount by which the oven depreciates in value. [2]
(b) Find the book value of the oven after 5 years. [4]
Show complete worked solution
(a) Using the straight-line depreciation formula, the annual depreciation amount is the total depreciation over its useful life divided by the number of years:
$$d = \frac{V_0-\text{salvage value}}{\text{useful life}} = \frac{6000-800}{8}$$
$$= \frac{5200}{8}$$
$$= 650$$
$$\boxed{\text{Annual depreciation} = \$650 \text{ per year}}$$
(b) Using $V_n=V_0-nd$ with $V_0=6000$, $d=650$, $n=5$:
$$V_5 = 6000-5(650)$$
$$= 6000-3250$$
$$= 2750$$
$$\boxed{V_5 = \$2750}$$
QUESTION 26
6 marks
Medium
A warehouse buys a forklift for \$32000. The forklift depreciates in value by 18\% each year (reducing-balance method).
(a) Find the value of the forklift after 6 years, correct to the nearest dollar. [3]
(b) Find the total amount by which the forklift has depreciated in value over these 6 years, correct to the nearest dollar. [3]
Show complete worked solution
(a) Using $V_n=V_0(1-r)^n$ with $V_0=32000$, $r=0.18$, $n=6$:
$$V_6 = 32000(0.82)^{6}$$
$$= 32000 \times 0.304006\ldots$$
$$= 9728.21\ldots$$
$$\boxed{V_6 \approx \$9728 \text{ (nearest dollar)}}$$
(b) Total depreciation = original value $-$ final value:
$$\text{Depreciation} = 32000-9728.21\ldots$$
$$= 22271.79\ldots$$
$$\boxed{\approx \$22{,}272 \text{ (nearest dollar)}}$$
QUESTION 27
8 marks
Medium
A company buys a piece of equipment for \$20000. Two depreciation methods are being considered.
Method 1 (straight-line): the equipment is assumed to have a useful life of 6 years and a salvage value of \$2000.
Method 2 (reducing-balance): the equipment depreciates by 22\% of its value each year.
(a) Find the book value of the equipment after 4 years under each method. [5]
(b) Determine which method gives the lower book value after 4 years, and find the difference between the two book values, correct to the nearest dollar. [3]
Show complete worked solution
(a) Method 1 (straight-line): the annual depreciation amount is
$$d = \frac{20000-2000}{6} = \frac{18000}{6} = 3000$$
So the book value after 4 years is
$$V_4 = 20000-4(3000) = 20000-12000 = 8000$$
$$\boxed{\text{Method 1: } V_4 = \$8000}$$
Method 2 (reducing-balance): using $V_n=V_0(1-r)^n$ with $V_0=20000$, $r=0.22$, $n=4$:
$$V_4 = 20000(0.78)^{4}$$
$$= 20000 \times 0.370151\ldots$$
$$= 7403.01\ldots$$
$$\boxed{\text{Method 2: } V_4 \approx \$7403 \text{ (nearest dollar)}}$$
(b) Since $7403.01\ldots < 8000$, Method 2 (reducing-balance) gives the lower book value after 4 years. The difference is
$$8000-7403.01\ldots = 596.99\ldots$$
$$\boxed{\text{Method 2 is lower, by approximately \$597}}$$
QUESTION 28
7 marks
Medium
A logistics company buys sorting equipment for \$60000. The equipment depreciates in value by 14\% each year (reducing-balance method).
(a) Find the value of the equipment after 3 years, correct to the nearest dollar. [3]
(b) By finding the value of the equipment at the end of each year (for example using the table function of your GDC), find the first complete number of years after which the value first falls below \$30000. [4]
Show complete worked solution
(a) Using $V_n=V_0(1-r)^n$ with $V_0=60000$, $r=0.14$, $n=3$:
$$V_3 = 60000(0.86)^{3}$$
$$= 60000 \times 0.636056$$
$$= 38163.36$$
$$\boxed{V_3 \approx \$38{,}163 \text{ (nearest dollar)}}$$
(b) Computing $V_n=60000(0.86)^n$ for successive $n$ using the GDC's table function:
$$V_4 = 60000(0.86)^4 = 32820.49\ldots \text{ (still above \$30000)}$$
$$V_5 = 60000(0.86)^5 = 28225.62\ldots \text{ (below \$30000)}$$
$$\boxed{\text{The value first falls below \$30000 after 5 complete years}}$$
QUESTION 29
12 marks
Hard
A courier company buys a delivery truck for \$85000. Under the company's tax policy, the truck's value depreciates by the reducing-balance (declining-balance) method at a rate of 16\% per year.
(a) Find the value of the truck after 4 years, correct to the nearest dollar. [3]
(b) By finding the value of the truck at the end of each year (for example using the table function of your GDC), find the first complete number of years after which the value first falls below \$30000. [4]
(c) Suppose instead the truck were depreciated using the straight-line method, assuming a useful life of 8 years and a scrap value of \$5000. Find the annual depreciation amount, and hence the book value of the truck after 4 years under this method. [3]
(d) Determine which method (reducing-balance or straight-line) gives the higher book value after 4 years, and find the difference between the two book values, correct to the nearest dollar. [2]
Show complete worked solution
(a) Using the reducing-balance depreciation formula $V_n=V_0(1-r)^n$ with $V_0=85000$, $r=0.16$, $n=4$:
$$V_4 = 85000(0.84)^{4}$$
$$= 85000 \times 0.497871\ldots$$
$$= 42319.07\ldots$$
$$\boxed{V_4 \approx \$42{,}319 \text{ (reducing-balance, nearest dollar)}}$$
(b) Computing $V_n=85000(0.84)^n$ for successive $n$ using the GDC's table function:
$$V_5 = 85000(0.84)^5 = 35548.02\ldots \text{ (still above \$30000)}$$
$$V_6 = 85000(0.84)^6 = 29860.33\ldots \text{ (below \$30000)}$$
$$\boxed{\text{The value first falls below \$30000 after 6 complete years}}$$
(c) Using the straight-line method, the annual depreciation amount is
$$d = \frac{85000-5000}{8} = \frac{80000}{8} = 10000$$
So the book value after 4 years is
$$V_4 = 85000-4(10000) = 85000-40000 = 45000$$
$$\boxed{\text{Straight-line: annual depreciation} = \$10{,}000, \quad V_4 = \$45{,}000}$$
(d) Comparing the two book values after 4 years, $45000 > 42319.07\ldots$, so straight-line gives the higher book value. The difference is
$$45000-42319.07\ldots = 2680.93\ldots$$
$$\boxed{\text{Straight-line gives the higher value, by approximately \$2681}}$$
QUESTION 30
11 marks
Hard
A photocopier was bought new by an office for \$12000. It depreciates by the same percentage rate each year (reducing-balance method). After 3 years, its book value is \$6144.
(a) Show that the annual rate of depreciation, $r$, satisfies $(1-r)^3=0.512$, and hence find the value of $r$. [4]
(b) Find the book value of the photocopier after 6 years. [3]
(c) By tabulating the book value at the end of each year, as shown below, find the first complete number of years after which the book value first falls below \$2000. [4]
Show complete worked solution
(a) Using $V_n=V_0(1-r)^n$ with $V_0=12000$, $V_3=6144$, $n=3$:
$$6144 = 12000(1-r)^3$$
$$(1-r)^3 = \frac{6144}{12000} = 0.512$$
Taking the cube root of both sides:
$$1-r = \sqrt[3]{0.512} = 0.8$$
$$r = 1-0.8 = 0.2$$
$$\boxed{r = 20\% \text{ per year}}$$
(b) Using $V_n=V_0(1-r)^n$ with $V_0=12000$, $r=0.2$, $n=6$:
$$V_6 = 12000(0.8)^{6}$$
$$= 12000 \times 0.262144$$
$$= 3145.728$$
$$\boxed{V_6 \approx \$3145.73}$$
(c) Tabulating $V_n=12000(0.8)^n$ for successive years (using the GDC's table function):
The book value is still above \$2000 at $n=8$ ($V_8=2013.27$), but falls below \$2000 at $n=9$ ($V_9=1610.61$).
$$\boxed{\text{The book value first falls below \$2000 after 9 complete years}}$$
| Year, $n$ | Book value, $V_n$ (\$) |
|---|---|
| 0 | 12000.00 |
| 1 | 9600.00 |
| 2 | 7680.00 |
| 3 | 6144.00 |
| 4 | 4915.20 |
| 5 | 3932.16 |
| 6 | 3145.73 |
| 7 | 2516.58 |
| 8 | 2013.27 |
| 9 | 1610.61 |
QUESTION 31
4 marks
Easy
Yusuf takes out a loan of \$12000 to buy a motorcycle. The loan is repaid with equal monthly payments over 4 years, with interest charged at a nominal annual rate of 7.2\%, compounded monthly.
Use technology (the finance/TVM solver on your GDC) to find the amount of each monthly payment, correct to the nearest cent.
Show complete worked solution
Using the GDC's Finance (TVM) Solver, entering the given values:
$$N = 4\times12 = 48, \quad I\% = 7.2, \quad PV = 12000, \quad FV = 0, \quad P/Y = C/Y = 12$$
Solving for $PMT$ using the annuity formula $PMT=\dfrac{PV\cdot i}{1-(1+i)^{-N}}$ with monthly rate $i=\dfrac{0.072}{12}=0.006$:
$$PMT = \frac{12000(0.006)}{1-(1.006)^{-48}}$$
$$= \frac{72}{1-0.750450\ldots}$$
$$= \frac{72}{0.249550\ldots} = 288.47\ldots$$
$$\boxed{\text{The monthly payment is } \$288.47 \text{ (to the nearest cent)}}$$
QUESTION 32
4 marks
Easy
Aroha takes out a loan of \$9500 to buy furniture. The loan is repaid with equal monthly payments over 5 years, with interest charged at a nominal annual rate of 5.4\%, compounded monthly.
Use technology (the finance/TVM solver on your GDC) to find the amount of each monthly payment, correct to the nearest cent.
Show complete worked solution
Using the GDC's Finance (TVM) Solver, entering:
$$N = 5\times12 = 60, \quad I\% = 5.4, \quad PV = 9500, \quad FV = 0, \quad P/Y = C/Y = 12$$
Solving for $PMT$ using the annuity formula $PMT=\dfrac{PV\cdot i}{1-(1+i)^{-N}}$ with monthly rate $i=\dfrac{0.054}{12}=0.0045$:
$$PMT = \frac{9500(0.0045)}{1-(1.0045)^{-60}}$$
$$= \frac{42.75}{1-0.763871\ldots}$$
$$= \frac{42.75}{0.236129\ldots} = 181.02\ldots$$
$$\boxed{\text{The monthly payment is } \$181.02 \text{ (to the nearest cent)}}$$
QUESTION 33
5 marks
Easy
Beatriz can afford to pay \$350 per month towards a car loan. The loan would be repaid over 4 years, with interest charged at a nominal annual rate of 6\%, compounded monthly.
Use technology (the finance/TVM solver on your GDC) to find the maximum amount Beatriz can borrow now, correct to the nearest dollar.
Show complete worked solution
Using the GDC's Finance (TVM) Solver, entering:
$$N = 4\times12 = 48, \quad I\% = 6, \quad PMT = -350, \quad FV = 0, \quad P/Y = C/Y = 12$$
Solving for $PV$ using the annuity formula $PV=\dfrac{PMT\big(1-(1+i)^{-N}\big)}{i}$ with monthly rate $i=\dfrac{0.06}{12}=0.005$:
$$PV = \frac{350\big(1-(1.005)^{-48}\big)}{0.005}$$
$$= \frac{350(1-0.787098\ldots)}{0.005}$$
$$= \frac{350(0.212901\ldots)}{0.005} = 14903.11\ldots$$
$$\boxed{\text{The maximum loan amount is approximately } \$14{,}903 \text{ (nearest dollar)}}$$
QUESTION 34
5 marks
Easy
Jamal borrows \$6000 at a nominal annual interest rate of 9\%, compounded monthly. He repays the loan with fixed monthly payments of \$190.80.
Use technology (the finance/TVM solver on your GDC) to find the number of months required to repay the loan in full.
Show complete worked solution
Using the GDC's Finance (TVM) Solver, entering:
$$I\% = 9, \quad PV = 6000, \quad PMT = -190.80, \quad FV = 0, \quad P/Y = C/Y = 12$$
Solving for $N$ (or using $N=\dfrac{-\ln\left(1-\dfrac{PV\cdot i}{PMT}\right)}{\ln(1+i)}$) with monthly rate $i=\dfrac{0.09}{12}=0.0075$:
$$N = \frac{-\ln\left(1-\dfrac{6000(0.0075)}{190.80}\right)}{\ln(1.0075)}$$
$$= \frac{-\ln(1-0.235849\ldots)}{\ln(1.0075)}$$
$$= \frac{-\ln(0.764151\ldots)}{0.007472\ldots}$$
$$= \frac{0.268962\ldots}{0.007472\ldots} \approx 36.0$$
$$\boxed{\text{The loan is repaid in approximately 36 months (3 years)}}$$
QUESTION 35
7 marks
Medium
Wei takes out a loan of \$22000 to buy a boat. The loan is repaid with equal monthly payments over 6 years, with interest charged at a nominal annual rate of 6.5\%, compounded monthly.
(a) Find the monthly payment, correct to the nearest cent. [3]
(b) Find the total amount repaid over the 6 years, and hence find the total interest paid on the loan, both correct to the nearest dollar. [4]
Show complete worked solution
(a) Using the GDC's Finance (TVM) Solver, entering:
$$N = 6\times12 = 72, \quad I\% = 6.5, \quad PV = 22000, \quad FV = 0, \quad P/Y = C/Y = 12$$
Using the annuity formula $PMT=\dfrac{PV\cdot i}{1-(1+i)^{-N}}$ with monthly rate $i=\dfrac{0.065}{12}=0.0054167$:
$$PMT = \frac{22000(0.0054167)}{1-(1.0054167)^{-72}}$$
$$= \frac{119.167}{0.322294\ldots} = 369.82\ldots$$
$$\boxed{\text{Monthly payment} = \$369.82}$$
(b) Total amount repaid = monthly payment $\times$ number of payments:
$$\text{Total repaid} = 369.82\ldots\times72 = 26626.93\ldots$$
$$\boxed{\text{Total repaid} \approx \$26{,}627 \text{ (nearest dollar)}}$$
Total interest = total repaid $-$ principal borrowed:
$$\text{Total interest} = 26626.93\ldots-22000 = 4626.93\ldots$$
$$\boxed{\text{Total interest} \approx \$4627 \text{ (nearest dollar)}}$$
QUESTION 36
6 marks
Medium
Nadia takes out a loan of \$16000 to buy a car. The loan is repaid in equal monthly instalments over 5 years, with interest charged at a nominal annual rate of 4.8\%, compounded monthly.
(a) Find the monthly payment. [3]
(b) Find the total interest Nadia pays during the first year (the first 12 payments) of the loan. [3]
Show complete worked solution
(a) Using the GDC's Finance (TVM) Solver, entering:
$$N = 5\times12 = 60, \quad I\% = 4.8, \quad PV = 16000, \quad FV = 0, \quad P/Y = C/Y = 12$$
Using the annuity formula $PMT=\dfrac{PV\cdot i}{1-(1+i)^{-N}}$ with monthly rate $i=\dfrac{0.048}{12}=0.004$:
$$PMT = \frac{16000(0.004)}{1-(1.004)^{-60}}$$
$$= \frac{64}{0.213022\ldots} = 300.48\ldots$$
$$\boxed{\text{Monthly payment} = \$300.48}$$
(b) Using the GDC's amortization/balance function, the outstanding balance after 12 payments is:
$$\text{Balance}(12) = 13099.02 \text{ (2 d.p.)}$$
Total paid in the first year:
$$12\times300.48\ldots = 3605.71$$
Principal repaid in the first year:
$$16000-13099.02 = 2900.98$$
Interest paid in the first year = total paid $-$ principal repaid:
$$3605.71-2900.98 = 704.73$$
$$\boxed{\text{Interest paid in the first year} \approx \$705 \text{ (nearest dollar)}}$$
QUESTION 37
7 marks
Medium
Tomas takes out a business loan of \$30000, to be repaid with equal monthly payments over 7 years at a nominal annual interest rate of 6\%, compounded monthly.
(a) Find the monthly payment. [2]
(b) Use the GDC's amortization (balance) function to find the outstanding balance on the loan immediately after the 36th payment, correct to the nearest cent. [3]
(c) Find how much of the 37th payment goes towards interest and how much goes towards reducing the principal. [3]
Show complete worked solution
(a) Using the GDC's Finance (TVM) Solver, entering $N=7\times12=84$, $I\%=6$, $PV=30000$, $FV=0$, $P/Y=C/Y=12$, with monthly rate $i=\dfrac{0.06}{12}=0.005$:
$$PMT = \frac{30000(0.005)}{1-(1.005)^{-84}}$$
$$= \frac{150}{0.342032\ldots} = 438.26\ldots$$
$$\boxed{\text{Monthly payment} = \$438.26}$$
(b) Using the GDC's amortization/balance function with $k=36$:
$$\text{Balance}(36) = 30000(1.005)^{36}-438.26\ldots\times\frac{(1.005)^{36}-1}{0.005}$$
$$= 18661.11 \text{ (2 d.p.)}$$
$$\boxed{\text{Outstanding balance after 36 payments} \approx \$18{,}661.11}$$
(c) Monthly interest rate: $i=0.005$.
Interest portion of the 37th payment = outstanding balance $\times$ monthly rate:
$$18661.11\times0.005 = 93.31 \text{ (2 d.p.)}$$
Principal portion of the 37th payment = total payment $-$ interest portion:
$$438.26-93.31 = 344.95 \text{ (2 d.p.)}$$
$$\boxed{\text{Interest portion} \approx \$93.31, \quad \text{Principal portion} \approx \$344.95}$$
QUESTION 38
7 marks
Medium
Priyanka takes out a loan of \$14000 to buy farm equipment. The loan is repaid with equal quarterly payments over 5 years, with interest charged at a nominal annual rate of 7\%, compounded quarterly.
(a) Find the quarterly payment, correct to the nearest cent. [3]
(b) Find the total interest paid over the life of the loan, correct to the nearest dollar. [4]
Show complete worked solution
(a) The quarterly interest rate is $i=\dfrac{0.07}{4}=0.0175$, and the number of quarterly payments is $N=5\times4=20$.
Using the GDC's Finance (TVM) Solver (with $P/Y=C/Y=4$), or the annuity formula $PMT=\dfrac{PV\cdot i}{1-(1+i)^{-N}}$:
$$PMT = \frac{14000(0.0175)}{1-(1.0175)^{-20}}$$
$$= \frac{245}{0.293224\ldots} = 835.68\ldots$$
$$\boxed{\text{Quarterly payment} = \$835.68}$$
(b) Total amount repaid = quarterly payment $\times$ number of payments:
$$\text{Total repaid} = 835.68\ldots\times20 = 16713.54\ldots$$
Total interest = total repaid $-$ principal borrowed:
$$\text{Total interest} = 16713.54\ldots-14000 = 2713.54\ldots$$
$$\boxed{\text{Total interest} \approx \$2714 \text{ (nearest dollar)}}$$
QUESTION 39
15 marks
Hard
The Okafor family takes out a home mortgage of \$310000 to buy a house. The bank charges a nominal annual interest rate of 4.8\%, compounded monthly, and the loan is to be fully repaid with equal monthly payments over 25 years.
(a) Use technology (the TVM solver) to find the monthly payment, correct to the nearest cent. [4]
(b) Find the total amount the Okafor family will have paid over the full 25-year term, and hence find the total interest paid over the life of the loan, both correct to the nearest dollar. [4]
(c) The table below shows the breakdown of the first three monthly payments into interest and principal, and the outstanding balance after each payment. [3]
(d) After paying the mortgage for exactly 15 years (180 payments), the Okafor family is considering refinancing.
(i) Use technology to find the outstanding balance on the loan at this point.
(ii) Find how much of the 181st payment goes towards interest and how much goes towards reducing the principal. [4]
| Payment no. | Interest (\$) | Principal (\$) | Balance (\$) |
|---|---|---|---|
| 1 | 1240.00 | 536.29 | 309463.71 |
| 2 | 1237.85 | 538.44 | 308925.27 |
| 3 | 1235.70 | 540.59 | 308384.68 |
Show complete worked solution
(a) Using the GDC's Finance (TVM) Solver:
$$N = 25\times12 = 300, \quad I\% = 4.8, \quad PV = 310000, \quad FV = 0, \quad P/Y = C/Y = 12$$
Using the annuity formula $PMT=\dfrac{PV\cdot i}{1-(1+i)^{-N}}$ with monthly rate $i=\dfrac{0.048}{12}=0.004$:
$$PMT = \frac{310000(0.004)}{1-(1.004)^{-300}}$$
$$= \frac{1240}{0.698061\ldots} = 1776.29\ldots$$
$$\boxed{\text{Monthly payment} = \$1776.29}$$
(b) Total amount paid = monthly payment $\times$ number of payments:
$$\text{Total paid} = 300\times1776.29\ldots = 532887.17 \text{ (2 d.p.)}$$
Total interest paid = total paid $-$ principal borrowed:
$$\text{Total interest} = 532887.17-310000 = 222887.17$$
$$\boxed{\text{Total paid} \approx \$532{,}887, \quad \text{Total interest} \approx \$222{,}887 \text{ (nearest dollar)}}$$
(c) The table confirms the amortization pattern: with a fixed monthly payment of \$1776.29, the interest portion (balance $\times$ 0.004) decreases slightly each month while the principal portion increases, since the outstanding balance is slowly reduced. For example, for payment 1, interest $=310000\times0.004=1240.00$ and principal $=1776.29-1240.00=536.29$, giving a new balance of $310000-536.29=309463.71$, matching the table.
(d)(i) Using the GDC's amortization/balance function with $N=180$:
$$\text{Balance}(180) = 310000(1.004)^{180}-1776.29\ldots\times\frac{(1.004)^{180}-1}{0.004}$$
$$= 169024.65 \text{ (2 d.p.)}$$
$$\boxed{\text{Outstanding balance after 15 years} \approx \$169{,}025 \text{ (nearest dollar)}}$$
(ii) Monthly interest rate: $i=0.004$.
Interest portion of the 181st payment = outstanding balance $\times$ monthly rate:
$$169024.65\times0.004 = 676.10 \text{ (2 d.p.)}$$
Principal portion of the 181st payment = total payment $-$ interest portion:
$$1776.29-676.10 = 1100.19 \text{ (2 d.p.)}$$
$$\boxed{\text{Interest portion} \approx \$676.10, \quad \text{Principal portion} \approx \$1100.19}$$
QUESTION 40
11 marks
Hard
Elena needs to borrow \$25000 to start a small business. She is comparing two loan offers, both to be repaid with equal monthly payments.
Offer 1: a nominal annual interest rate of 5.5\%, compounded monthly, repaid over 5 years.
Offer 2: a nominal annual interest rate of 6.2\%, compounded monthly, repaid over 4 years.
(a) Find the monthly payment under each offer, correct to the nearest cent. [5]
(b) Find the total interest paid over the life of the loan under each offer, correct to the nearest dollar. [4]
(c) Elena can afford monthly payments of up to \$550. State, with a reason, which offer(s) she can afford, and which offer results in less total interest paid. [2]
Show complete worked solution
(a) Offer 1: $N=5\times12=60$, $i=\dfrac{0.055}{12}=0.0045833$.
$$PMT_1 = \frac{25000(0.0045833)}{1-(1.0045833)^{-60}}$$
$$= \frac{114.583}{0.240028\ldots} = 477.53\ldots$$
$$\boxed{PMT_1 \approx \$477.53 \text{ per month}}$$
Offer 2: $N=4\times12=48$, $i=\dfrac{0.062}{12}=0.0051667$.
$$PMT_2 = \frac{25000(0.0051667)}{1-(1.0051667)^{-48}}$$
$$= \frac{129.167}{0.219106\ldots} = 589.42\ldots$$
$$\boxed{PMT_2 \approx \$589.42 \text{ per month}}$$
(b) Offer 1 total interest:
$$\text{Total paid} = 477.53\ldots\times60 = 28651.74\ldots$$
$$\text{Total interest} = 28651.74\ldots-25000 = 3651.74\ldots \approx \$3652$$
Offer 2 total interest:
$$\text{Total paid} = 589.42\ldots\times48 = 28292.20\ldots$$
$$\text{Total interest} = 28292.20\ldots-25000 = 3292.20\ldots \approx \$3292$$
$$\boxed{\text{Offer 1 total interest} \approx \$3652, \quad \text{Offer 2 total interest} \approx \$3292}$$
(c) Since $PMT_2\approx\$589.42 > \$550$, Elena cannot afford Offer 2's monthly payment, but $PMT_1\approx\$477.53 < \$550$, so she can afford Offer 1. Although Offer 2 has a higher nominal interest rate, its shorter term means less total interest is paid overall; however, since Elena can only afford Offer 1's payments, Offer 1 is the offer she should choose.
$$\boxed{\text{Elena can only afford Offer 1, even though Offer 2 has less total interest}}$$
QUESTION 41
4 marks
Easy
At the end of each month, Chidi deposits \$200 into a savings account that pays a nominal annual interest rate of 4.2\%, compounded monthly.
Find the value of Chidi's savings after 6 years, assuming he makes no other deposits or withdrawals.
Show complete worked solution
Using the GDC's Finance (TVM) Solver for a savings annuity, entering:
$$N = 6\times12 = 72, \quad I\% = 4.2, \quad PV = 0, \quad PMT = -200, \quad P/Y = C/Y = 12$$
Using the future value of an ordinary annuity formula $FV=\dfrac{PMT\big((1+i)^N-1\big)}{i}$ with monthly rate $i=\dfrac{0.042}{12}=0.0035$:
$$FV = \frac{200\big((1.0035)^{72}-1\big)}{0.0035}$$
$$= \frac{200(1.286030\ldots-1)}{0.0035}$$
$$= \frac{200(0.286030\ldots)}{0.0035} = 16344.58\ldots$$
$$\boxed{FV \approx \$16{,}345 \text{ (nearest dollar)}}$$
QUESTION 42
5 marks
Easy
At the end of each quarter, Lucia deposits \$600 into a savings account that pays a nominal annual interest rate of 3.6\%, compounded quarterly.
Find the value of Lucia's savings after 7 years, assuming she makes no other deposits or withdrawals.
Show complete worked solution
Using the GDC's Finance (TVM) Solver for a savings annuity, entering:
$$N = 7\times4 = 28, \quad I\% = 3.6, \quad PV = 0, \quad PMT = -600, \quad P/Y = C/Y = 4$$
Using the future value of an ordinary annuity formula $FV=\dfrac{PMT\big((1+i)^N-1\big)}{i}$ with quarterly rate $i=\dfrac{0.036}{4}=0.009$:
$$FV = \frac{600\big((1.009)^{28}-1\big)}{0.009}$$
$$= \frac{600(1.285146\ldots-1)}{0.009}$$
$$= \frac{600(0.285146\ldots)}{0.009} = 19009.77\ldots$$
$$\boxed{FV \approx \$19{,}010 \text{ (nearest dollar)}}$$
QUESTION 43
5 marks
Easy
Amara wants to accumulate \$12000 after 4 years. She plans to make equal deposits at the end of each month into an account that pays a nominal annual interest rate of 4.5\%, compounded monthly.
Use technology to find the monthly deposit Amara must make, correct to the nearest cent.
Show complete worked solution
Using the GDC's Finance (TVM) Solver, entering:
$$N = 4\times12 = 48, \quad I\% = 4.5, \quad PV = 0, \quad FV = -12000, \quad P/Y = C/Y = 12$$
Using the future value of an ordinary annuity formula $FV=\dfrac{PMT\big((1+i)^N-1\big)}{i}$, rearranged for $PMT$, with monthly rate $i=\dfrac{0.045}{12}=0.00375$:
$$PMT = \frac{FV\cdot i}{(1+i)^N-1}$$
$$= \frac{12000(0.00375)}{(1.00375)^{48}-1}$$
$$= \frac{45}{1.196843\ldots-1} = \frac{45}{0.196843\ldots} = 228.64\ldots$$
$$\boxed{\text{Amara must deposit } \$228.64 \text{ at the end of each month (to the nearest cent)}}$$
QUESTION 44
4 marks
Easy
At the end of each year, Oscar deposits \$1000 into a savings account that pays a nominal annual interest rate of 5\%, compounded annually.
Find the value of Oscar's savings after 8 years, assuming he makes no other deposits or withdrawals.
Show complete worked solution
Using the future value of an ordinary annuity formula $FV=\dfrac{PMT\big((1+i)^N-1\big)}{i}$ (or the GDC's TVM solver) with $PMT=1000$, $i=0.05$, $N=8$:
$$FV = \frac{1000\big((1.05)^{8}-1\big)}{0.05}$$
$$= \frac{1000(1.477455\ldots-1)}{0.05}$$
$$= \frac{1000(0.477455\ldots)}{0.05} = 9549.11\ldots$$
$$\boxed{FV \approx \$9549 \text{ (nearest dollar)}}$$
QUESTION 45
7 marks
Medium
Grace opens a savings account that pays a nominal annual interest rate of 4\%, compounded monthly. She immediately deposits a lump sum of \$3000, and also deposits \$250 at the end of each month for 5 years. No other deposits or withdrawals are made.
(a) Find the value of the initial \$3000 lump sum after 5 years. [2]
(b) Find the future value of the monthly deposits after 5 years. [3]
(c) Hence find the total value of Grace's account after 5 years, correct to the nearest dollar. [2]
Show complete worked solution
(a) Using $FV=PV(1+i)^n$ with $PV=3000$, monthly rate $i=\dfrac{0.04}{12}=0.0033\overline{3}$, $n=5\times12=60$:
$$FV_{lump} = 3000(1.003333\ldots)^{60}$$
$$= 3000\times1.220997\ldots$$
$$= 3662.99\ldots$$
$$\boxed{FV_{lump} \approx \$3663 \text{ (nearest dollar)}}$$
(b) Using the future value of an ordinary annuity formula $FV=\dfrac{PMT\big((1+i)^N-1\big)}{i}$ with $PMT=250$, $i=0.003333\ldots$, $N=60$:
$$FV_{ann} = \frac{250\big((1.003333\ldots)^{60}-1\big)}{0.003333\ldots}$$
$$= \frac{250(0.220997\ldots)}{0.003333\ldots} = 16574.74\ldots$$
$$\boxed{FV_{ann} \approx \$16{,}575 \text{ (nearest dollar)}}$$
(c) Total value = future value of lump sum + future value of the monthly deposits:
$$FV_{total} = 3662.99\ldots+16574.74\ldots$$
$$= 20237.73\ldots$$
$$\boxed{FV_{total} \approx \$20{,}238 \text{ (nearest dollar)}}$$
QUESTION 46
6 marks
Medium
Hiroshi wants to accumulate \$25000 after 6 years to buy a boat. He plans to make equal deposits at the end of each quarter into an account that pays a nominal annual interest rate of 5\%, compounded quarterly.
Use technology to find the quarterly deposit Hiroshi must make, correct to the nearest cent.
Show complete worked solution
Using the GDC's Finance (TVM) Solver, entering:
$$N = 6\times4 = 24, \quad I\% = 5, \quad PV = 0, \quad FV = -25000, \quad P/Y = C/Y = 4$$
Using the future value of an ordinary annuity formula $FV=\dfrac{PMT\big((1+i)^N-1\big)}{i}$, rearranged for $PMT$, with quarterly rate $i=\dfrac{0.05}{4}=0.0125$:
$$PMT = \frac{25000(0.0125)}{(1.0125)^{24}-1}$$
$$= \frac{312.5}{1.347351\ldots-1} = \frac{312.5}{0.347351\ldots} = 899.67\ldots$$
$$\boxed{\text{Hiroshi must deposit } \$899.67 \text{ at the end of each quarter (to the nearest cent)}}$$
QUESTION 47
7 marks
Medium
At the end of each month, Fatoumata contributes \$180 to a retirement savings account that pays a nominal annual interest rate of 5.8\%, compounded monthly.
(a) Find the value of her account after 15 years, correct to the nearest dollar. [3]
(b) Find the total amount Fatoumata has contributed over the 15 years, and hence find the total interest earned, both correct to the nearest dollar. [4]
Show complete worked solution
(a) Using the future value of an ordinary annuity formula $FV=\dfrac{PMT\big((1+i)^N-1\big)}{i}$ (or the GDC's TVM solver) with $PMT=180$, monthly rate $i=\dfrac{0.058}{12}=0.0048\overline{3}$, $N=15\times12=180$:
$$FV = \frac{180\big((1.004833\ldots)^{180}-1\big)}{0.004833\ldots}$$
$$= \frac{180(2.381359\ldots-1)}{0.004833\ldots}$$
$$= \frac{180(1.381359\ldots)}{0.004833\ldots} = 51464.37\ldots$$
$$\boxed{FV \approx \$51{,}464 \text{ (nearest dollar)}}$$
(b) Total contributions = monthly deposit $\times$ number of deposits:
$$\text{Total contributed} = 180\times180 = 32400$$
$$\boxed{\text{Total contributed} = \$32{,}400}$$
Total interest earned = final value $-$ total contributed:
$$\text{Total interest} = 51464.37\ldots-32400 = 19064.37\ldots$$
$$\boxed{\text{Total interest earned} \approx \$19{,}064 \text{ (nearest dollar)}}$$
QUESTION 48
8 marks
Medium
Kwame retires with a lump sum of \$40000 in an account that pays a nominal annual interest rate of 4.8\%, compounded monthly. At the end of each month, he withdraws \$500 to supplement his income, and makes no further deposits.
Use technology (the finance/TVM solver on your GDC) to find the number of full monthly withdrawals of \$500 Kwame can make before the account balance falls below \$500.
Show complete worked solution
Using the GDC's Finance (TVM) Solver, entering:
$$I\% = 4.8, \quad PV = 40000, \quad PMT = -500, \quad FV = 0, \quad P/Y = C/Y = 12$$
Solving for $N$ using $N=\dfrac{-\ln\left(1-\dfrac{PV\cdot i}{PMT}\right)}{\ln(1+i)}$ with monthly rate $i=\dfrac{0.048}{12}=0.004$:
$$N = \frac{-\ln\left(1-\dfrac{40000(0.004)}{500}\right)}{\ln(1.004)}$$
$$= \frac{-\ln(1-0.32)}{\ln(1.004)}$$
$$= \frac{-\ln(0.68)}{0.003992\ldots}$$
$$= \frac{0.385662\ldots}{0.003992\ldots} \approx 96.6$$
Since $N\approx96.6$ is not a whole number, Kwame can make 96 full withdrawals of \$500, with a smaller final withdrawal needed to close out the account in month 97.
$$\boxed{\text{Kwame can make 96 full monthly withdrawals of } \$500 \text{ (approximately 8 years)}}$$
QUESTION 49
12 marks
Hard
Aisha opens a savings account that pays a nominal annual interest rate of 5\%, compounded monthly. At the end of each month, for 8 years, she deposits \$220. After 8 years she stops making deposits, but leaves the account untouched (still earning the same interest rate) for a further 5 years, at which point she withdraws the full balance.
(a) Find the value of Aisha's account at the end of the 8-year deposit phase, correct to the nearest dollar. [4]
(b) Find the value of the account 5 years later (i.e. 13 years after she started), correct to the nearest dollar. [4]
(c) Find the total amount Aisha deposited into the account, and hence find the total interest earned over the full 13 years, both correct to the nearest dollar. [4]
Show complete worked solution
(a) Using the future value of an ordinary annuity formula $FV=\dfrac{PMT\big((1+i)^N-1\big)}{i}$ (or the GDC's TVM solver) with $PMT=220$, monthly rate $i=\dfrac{0.05}{12}=0.0041\overline{6}$, $N=8\times12=96$:
$$FV_8 = \frac{220\big((1.004167\ldots)^{96}-1\big)}{0.004167\ldots}$$
$$= \frac{220(1.490587\ldots-1)}{0.004167\ldots}$$
$$= \frac{220(0.490587\ldots)}{0.004167\ldots} = 25902.91\ldots$$
$$\boxed{FV_8 \approx \$25{,}903 \text{ (nearest dollar)}}$$
(b) With no further deposits, the balance $FV_8$ simply grows under compound interest for a further $5\times12=60$ months. Using $FV=PV(1+i)^n$ with $PV=25902.91\ldots$, $i=0.004167\ldots$, $n=60$:
$$FV_{13} = 25902.91\ldots(1.004167\ldots)^{60}$$
$$= 25902.91\ldots\times1.283359\ldots$$
$$= 33242.73\ldots$$
$$\boxed{FV_{13} \approx \$33{,}243 \text{ (nearest dollar)}}$$
(c) Total deposited = monthly deposit $\times$ number of deposits (only during the first 8 years):
$$\text{Total deposited} = 220\times96 = 21120$$
$$\boxed{\text{Total deposited} = \$21{,}120}$$
Total interest earned = final value $-$ total deposited:
$$\text{Total interest} = 33242.73\ldots-21120 = 12122.73\ldots$$
$$\boxed{\text{Total interest earned} \approx \$12{,}123 \text{ (nearest dollar)}}$$
QUESTION 50
10 marks
Hard
Two savings plans are available, both offering a nominal annual interest rate of 4.5\%, held for 10 years, with the same total annual contribution of \$1800.
Plan A: deposit \$150 at the end of each month; interest is compounded monthly.
Plan B: deposit \$450 at the end of each quarter; interest is compounded quarterly.
(a) Find the future value of each plan after 10 years, correct to the nearest dollar. [6]
(b) Verify that the total amount contributed is the same under each plan, and hence explain, with reference to compounding frequency, why the two future values are different. [4]
Show complete worked solution
(a) Plan A: monthly rate $i=\dfrac{0.045}{12}=0.00375$, $N=10\times12=120$.
$$FV_A = \frac{150\big((1.00375)^{120}-1\big)}{0.00375}$$
$$= \frac{150(1.566993\ldots-1)}{0.00375}$$
$$= \frac{150(0.566993\ldots)}{0.00375} = 22679.71\ldots$$
$$\boxed{FV_A \approx \$22{,}680 \text{ (nearest dollar)}}$$
Plan B: quarterly rate $i=\dfrac{0.045}{4}=0.01125$, $N=10\times4=40$.
$$FV_B = \frac{450\big((1.01125)^{40}-1\big)}{0.01125}$$
$$= \frac{450(1.564377\ldots-1)}{0.01125}$$
$$= \frac{450(0.564377\ldots)}{0.01125} = 22575.07\ldots$$
$$\boxed{FV_A \approx \$22{,}680, \quad FV_B \approx \$22{,}575 \text{ (nearest dollar)}}$$
(b) Total contributed under Plan A: $150\times12\times10=18000$. Total contributed under Plan B: $450\times4\times10=18000$. Both plans therefore involve the same total contribution of \$18000 over the 10 years.
However, $FV_A>FV_B$: since Plan A's deposits are made monthly (more frequently) and compound monthly, each individual deposit spends slightly more time in the account earning interest, and interest is credited (and itself starts earning interest) more often, than under Plan B's larger, less frequent quarterly deposits and quarterly compounding. This higher effective compounding frequency gives Plan A a greater final value, even though the nominal annual rate and the total amount contributed are identical for both plans.
$$\boxed{\text{Same total contribution (\$18000) under both; Plan A's more frequent deposits/compounding give it the higher future value}}$$
Complex Numbers 50 questions
QUESTION 1
5 marks
Easy
Let z_1 = 5 - 2i and z_2 = -3 + i.
(a) Find z_1 + z_2.
(b) Find z_1 z_2, giving your answer in the form a + bi.
(c) Write down z_1^*, the complex conjugate of z_1.
Show complete worked solution
(a) Using $z_1+z_2$, add real and imaginary parts separately:
$$z_1+z_2 = (5-2i)+(-3+i)$$
$$= (5-3)+(-2+1)i$$
$$\boxed{z_1+z_2 = 2-i}$$
(b) Expanding the product $(5-2i)(-3+i)$ term by term:
$$z_1z_2 = 5(-3)+5(i)+(-2i)(-3)+(-2i)(i)$$
$$= -15+5i+6i-2i^2$$
Since $i^2=-1$:
$$-2i^2 = 2$$
Collecting terms:
$$z_1z_2 = -15+11i+2$$
$$\boxed{z_1z_2 = -13+11i}$$
(c) The complex conjugate reverses the sign of the imaginary part:
$$\boxed{z_1^{*} = 5+2i}$$
QUESTION 2
5 marks
Easy
The quadratic equation x^2 + bx + c = 0, where b, c \in \mathbb{R}, has z = 3 - 2i as one of its roots.
(a) State the other root of the equation.
(b) Find the values of b and c.
Show complete worked solution
(a) Since the coefficients of the quadratic are real, non-real roots occur in conjugate pairs.
$$\boxed{\text{The other root is } 3+2i}$$
(b) For $x^2+bx+c=0$ with roots $\alpha,\beta$: sum of roots $=-b$ and product of roots $=c$.
Sum of roots:
$$(3-2i)+(3+2i) = 6$$
So $-b=6$, giving:
$$b = -6$$
Product of roots:
$$(3-2i)(3+2i) = 3^2-(2i)^2 = 9-(-4) = 13$$
So:
$$c = 13$$
$$\boxed{b = -6, \quad c = 13 \; (\text{i.e. } x^2-6x+13=0)}$$
QUESTION 3
6 marks
Medium
Find the complex number z = x + iy, where x, y \in \mathbb{R}, that satisfies the equation z + 2z^* = 9 + 8i, where z^* denotes the complex conjugate of z.
Show complete worked solution
Let $z=x+iy$, so $z^*=x-iy$. Substituting into $z+2z^*$:
$$z+2z^* = (x+iy)+2(x-iy)$$
$$= (x+2x)+(y-2y)i$$
$$= 3x-iy$$
Equating this to $9+8i$ and comparing real and imaginary parts:
$$\text{Real: } 3x=9 \implies x=3$$
$$\text{Imaginary: } -y=8 \implies y=-8$$
So $z=3-8i$.
Check: $z^*=3+8i$, so $2z^*=6+16i$, and $z+2z^*=(3-8i)+(6+16i)=9+8i$. $\checkmark$
$$\boxed{z = 3-8i}$$
QUESTION 4
7 marks
Medium
The cubic equation x^3 - 6x^2 + 21x - 26 = 0 has real coefficients. Given that x = 2 + 3i is one root:
(a) Write down another root of the equation.
(b) Find the third (real) root, showing your method.
(c) Verify that this real root satisfies the equation.
Show complete worked solution
(a) By the conjugate root theorem, since the equation has real coefficients, the complex roots occur in conjugate pairs.
$$\boxed{\text{Another root is } 2-3i}$$
(b) For a cubic $x^3+px^2+qx+r=0$, the sum of the roots equals $-p$. Here $p=-6$, so the sum of all three roots is $6$.
Sum of the known complex pair:
$$(2+3i)+(2-3i) = 4$$
So the third (real) root is:
$$6-4 = 2$$
$$\boxed{\text{Third root: } x=2}$$
(c) Substituting $x=2$ into the cubic:
$$2^3-6(2)^2+21(2)-26$$
$$= 8-24+42-26$$
$$= 0 \quad \checkmark$$
$$\boxed{x=2 \text{ satisfies the equation}}$$
QUESTION 5
12 marks
Hard
In the study of fractal geometry, the Mandelbrot set is generated using the iterative rule z_{n+1} = z_n^2 + c, where z_0 = 0 and c is a fixed complex parameter. A complex number c is said to belong to the Mandelbrot set if the resulting sequence z_0, z_1, z_2, \dots remains bounded (does not grow without limit).
Consider c = -0.5 + 0.5i.
(a) Find z_1, z_2 and z_3, giving each answer in the form a + bi.
(b) Find |z_3|, correct to three significant figures.
(c) By considering the magnitudes |z_0|, |z_1|, |z_2|, |z_3|, comment on whether the sequence appears to be bounded, and state, with a reason, whether this is consistent with c lying in the Mandelbrot set.
Show complete worked solution
(a) Using $z_{n+1}=z_n^2+c$ with $z_0=0$ and $c=-0.5+0.5i$:
$$z_1 = z_0^2+c = 0^2+c = -0.5+0.5i$$
For $z_2$, first square $z_1$ using $(a+bi)^2=(a^2-b^2)+2abi$ with $a=-0.5$, $b=0.5$:
$$z_1^2 = (-0.5)^2-(0.5)^2+2(-0.5)(0.5)i = 0.25-0.25-0.5i = -0.5i$$
$$z_2 = z_1^2+c = -0.5i+(-0.5+0.5i) = -0.5+0i = -0.5$$
For $z_3$, square $z_2$ (real, so squaring is straightforward):
$$z_2^2 = (-0.5)^2 = 0.25$$
$$z_3 = z_2^2+c = 0.25+(-0.5+0.5i) = -0.25+0.5i$$
$$\boxed{z_1=-0.5+0.5i, \quad z_2=-0.5, \quad z_3=-0.25+0.5i}$$
(b) Using $|a+bi|=\sqrt{a^2+b^2}$ with $z_3=-0.25+0.5i$:
$$|z_3| = \sqrt{(-0.25)^2+(0.5)^2}$$
$$= \sqrt{0.0625+0.25}$$
$$= \sqrt{0.3125}$$
$$\boxed{|z_3| \approx 0.559 \text{ (3 s.f.)}}$$
(c) The magnitudes of the terms found are:
$$|z_0|=0, \quad |z_1|=\sqrt{0.5}\approx0.707, \quad |z_2|=0.5, \quad |z_3|\approx0.559$$
These values remain small (well below the standard escape threshold of 2) and show no growing trend over these four terms, so the sequence appears to stay bounded.
$$\boxed{\text{The sequence appears bounded, consistent with } c=-0.5+0.5i \text{ belonging to the Mandelbrot set}}$$
QUESTION 6
4 marks
Easy
Consider the complex number z = -3 + 3\sqrt{3}i.
(a) Find the modulus |z|.
(b) Find \arg(z) in radians, giving an exact value.
(c) Hence write z in the form re^{i\theta} (Euler form).
Show complete worked solution
(a) Using $|z|=\sqrt{(\text{Re}(z))^2+(\text{Im}(z))^2}$ with $z=-3+3\sqrt3i$:
$$|z| = \sqrt{(-3)^2+(3\sqrt3)^2}$$
$$= \sqrt{9+27}$$
$$= \sqrt{36}$$
$$\boxed{|z| = 6}$$
(b) The reference angle is:
$$\arctan\left(\frac{3\sqrt3}{3}\right) = \arctan(\sqrt3) = \frac{\pi}{3}$$
Since $\text{Re}(z)<0$ and $\text{Im}(z)>0$, $z$ lies in the second quadrant, so:
$$\arg(z) = \pi-\frac{\pi}{3}$$
$$\boxed{\arg(z) = \frac{2\pi}{3}}$$
(c) Substituting the modulus and argument found in (a) and (b) into $z=re^{i\theta}$:
$$\boxed{z = 6e^{i2\pi/3}}$$
QUESTION 7
5 marks
Easy
Let z_1 = 4e^{i\pi/6} and z_2 = 2e^{i\pi/3}.
(a) Find z_1 z_2 in the form re^{i\theta}.
(b) Find \dfrac{z_1}{z_2} in the form re^{i\theta}.
(c) Express z_1 z_2 in Cartesian form a + bi.
Show complete worked solution
(a) To multiply complex numbers in Euler form, multiply the moduli and add the arguments:
$$|z_1z_2| = 4\times2 = 8$$
$$\arg(z_1z_2) = \frac{\pi}{6}+\frac{\pi}{3} = \frac{\pi}{2}$$
$$\boxed{z_1z_2 = 8e^{i\pi/2}}$$
(b) To divide, divide the moduli and subtract the arguments:
$$\left|\frac{z_1}{z_2}\right| = \frac{4}{2} = 2$$
$$\arg\left(\frac{z_1}{z_2}\right) = \frac{\pi}{6}-\frac{\pi}{3} = -\frac{\pi}{6}$$
$$\boxed{\frac{z_1}{z_2} = 2e^{-i\pi/6}}$$
(c) Converting $8e^{i\pi/2}$ from part (a) to Cartesian form using $re^{i\theta}=r(\cos\theta+i\sin\theta)$:
$$8e^{i\pi/2} = 8\left(\cos\frac{\pi}{2}+i\sin\frac{\pi}{2}\right)$$
$$= 8(0+1i)$$
$$\boxed{z_1z_2 = 8i}$$
QUESTION 8
5 marks
Medium
Use De Moivre's theorem to find (\sqrt3 - i)^6, giving your answer in the form a + bi.
Show complete worked solution
First write $\sqrt3-i$ in polar (Euler) form.
Modulus:
$$r = \sqrt{(\sqrt3)^2+(-1)^2} = \sqrt{3+1} = 2$$
Argument: since $\text{Re}>0$ and $\text{Im}<0$, the number lies in the fourth quadrant. The reference angle is $\arctan\left(\dfrac{1}{\sqrt3}\right)=\dfrac{\pi}{6}$, so:
$$\arg(\sqrt3-i) = -\frac{\pi}{6}$$
So $\sqrt3-i = 2e^{-i\pi/6}$.
Using De Moivre's theorem, $(re^{i\theta})^n=r^ne^{in\theta}$, with $n=6$:
$$(\sqrt3-i)^6 = 2^6e^{i\cdot6(-\pi/6)}$$
$$= 64e^{-i\pi}$$
Converting back to Cartesian form:
$$64e^{-i\pi} = 64(\cos(-\pi)+i\sin(-\pi))$$
$$= 64(-1+0i)$$
$$\boxed{(\sqrt3-i)^6 = -64}$$
QUESTION 9
6 marks
Medium
In an AC circuit, two components connected in series have impedances Z_1 = 3 + 4i\ \Omega and Z_2 = 5 - 2i\ \Omega, where impedance is measured in ohms (\Omega) and the imaginary unit represents a 90 degree phase shift.
(a) Find the total impedance Z = Z_1 + Z_2 of the series combination.
(b) Write Z in polar form, giving the modulus |Z| exactly and the argument in radians correct to 3 significant figures.
(c) Interpret the value of |Z| in the context of the circuit.
Show complete worked solution
(a) For components connected in series, impedances add:
$$Z = Z_1+Z_2 = (3+4i)+(5-2i)$$
$$\boxed{Z = 8+2i \text{ } \Omega}$$
(b) Using $|Z|=\sqrt{(\text{Re}(Z))^2+(\text{Im}(Z))^2}$:
$$|Z| = \sqrt{8^2+2^2}$$
$$= \sqrt{64+4} = \sqrt{68}$$
$$\boxed{|Z| = 2\sqrt{17} \; \Omega \; (\approx8.25 \; \Omega)}$$
Using $\arg(Z)=\arctan\left(\dfrac{\text{Im}(Z)}{\text{Re}(Z)}\right)$:
$$\arg(Z) = \arctan\left(\frac{2}{8}\right) = \arctan(0.25)$$
$$\boxed{\arg(Z) \approx 0.245 \text{ rad (3 s.f.)}}$$
So $Z\approx8.25e^{i0.245}$.
(c) $|Z|\approx8.25\;\Omega$ represents the overall magnitude of opposition that the series combination offers to the alternating current - i.e. the ratio of the voltage amplitude to the current amplitude across the combination. The argument, $\approx0.245$ radians, represents the phase angle by which the voltage leads the current in this circuit.
$$\boxed{|Z| \approx 8.25 \; \Omega \text{ is the circuit's total opposition to current; } \arg(Z)\approx0.245 \text{ rad is the voltage-current phase lead}}$$
QUESTION 10
15 marks
Hard
An AC circuit consists of a resistor of resistance R = 8\ \Omega connected in series with an inductor of reactance X_L = 6\ \Omega. The impedance of the circuit is Z = R + jX_L (engineers conventionally use j for the imaginary unit), and the circuit is connected to an alternating voltage source of V = 120e^{j0}\ \text{V (rms)}.
(a) Write Z in Cartesian form, then convert Z to polar (Euler) form re^{j\theta}, giving the modulus exactly and the argument correct to 4 significant figures (in radians).
(b) Using Ohm's law for AC circuits, I = \dfrac{V}{Z}, find the current I in polar form, then convert your answer to Cartesian form a + bi (in amps).
(c) A capacitor is now added in series, contributing a reactance of -X_C j to the total impedance, so that the new total impedance is Z' = 8 + (6 - X_C)j. Find the value of X_C for which the circuit is at resonance (i.e. Z' is purely real), and find the magnitude of the current at resonance, given that the voltage remains V = 120e^{j0}\ \text{V}.
Show complete worked solution
(a) Writing $Z=R+jX_L$ with $R=8$, $X_L=6$:
$$\boxed{Z = 8+6j \; \Omega \text{ (Cartesian form)}}$$
Using $|Z|=\sqrt{R^2+X_L^2}$:
$$|Z| = \sqrt{8^2+6^2} = \sqrt{64+36} = \sqrt{100}$$
$$\boxed{|Z| = 10 \; \Omega \text{ (exact)}}$$
Using $\arg(Z)=\arctan\left(\dfrac{X_L}{R}\right)$:
$$\arg(Z) = \arctan\left(\frac{6}{8}\right) = \arctan(0.75)$$
$$\boxed{\arg(Z) \approx 0.6435 \text{ rad (4 s.f.)}}$$
So $Z=10e^{j0.6435}$.
(b) By Ohm's law for AC circuits, $I=\dfrac{V}{Z}$: dividing the moduli and subtracting the arguments:
$$I = \frac{120e^{j0}}{10e^{j0.6435}}$$
$$= \frac{120}{10}e^{j(0-0.6435)}$$
$$\boxed{I = 12e^{-j0.6435} \text{ A}}$$
To convert to Cartesian form, note that $Z$ corresponds to a 6-8-10 right triangle, so $\cos(0.6435)=8/10=0.8$ and $\sin(0.6435)=6/10=0.6$:
$$I = 12\big(\cos(-0.6435)+j\sin(-0.6435)\big)$$
$$= 12(0.8-0.6j)$$
$$\boxed{I = 9.6-7.2j \text{ A}}$$
(Check by direct division: $I=\dfrac{120}{8+6j}=\dfrac{120(8-6j)}{8^2+6^2}=\dfrac{120(8-6j)}{100}=1.2(8-6j)=9.6-7.2j$ A. $\checkmark$)
(c) At resonance, the imaginary part of $Z'=8+(6-X_C)j$ must equal zero:
$$6-X_C = 0$$
$$\boxed{X_C = 6 \; \Omega}$$
At this value, $Z'=8+0j=8\;\Omega$ (purely resistive). Using $|I|=\dfrac{|V|}{|Z'|}$:
$$|I| = \frac{120}{8}$$
$$\boxed{|I| = 15 \text{ A at resonance}}$$
QUESTION 11
6 marks
Easy
Let $z_1 = 4 + 2i$ and $z_2 = -1 + 3i$.
(a) Find $z_1 + z_2$. [2]
(b) Find $z_1 - z_2$. [2]
(c) Represent $z_1$, $z_2$ and $z_1+z_2$ on an Argand diagram, and use your diagram to illustrate the parallelogram law of addition. [2]
Show complete worked solution
(a) Adding real and imaginary parts separately:
$$z_1+z_2 = (4+2i)+(-1+3i) = (4-1)+(2+3)i$$
$$\boxed{z_1+z_2 = 3+5i}$$
(b) Subtracting real and imaginary parts separately:
$$z_1-z_2 = (4+2i)-(-1+3i) = (4-(-1))+(2-3)i$$
$$\boxed{z_1-z_2 = 5-i}$$
(c) The Argand diagram below plots $z_1=4+2i$ and $z_2=-1+3i$ as vectors from the origin. Completing the parallelogram with sides $Oz_1$ and $Oz_2$ gives a fourth vertex at $z_1+z_2=3+5i$, which is exactly the point found in part (a) - this is the geometric (parallelogram/triangle law) interpretation of complex number addition.
QUESTION 12
5 marks
Easy
The quadratic equation $2x^2 + bx + c = 0$, where $b, c \in \mathbb{R}$, has $z = 1 - 4i$ as one of its roots.
(a) State the other root of the equation. [1]
(b) Find the values of $b$ and $c$. [4]
Show complete worked solution
(a) Since $b$ and $c$ are real, non-real roots of the quadratic occur in a complex-conjugate pair.
$$\boxed{\text{The other root is } 1+4i}$$
(b) For $2x^2+bx+c=0$ with roots $\alpha,\beta$: sum of roots $=-\dfrac{b}{2}$ and product of roots $=\dfrac{c}{2}$.
Sum of roots:
$$(1-4i)+(1+4i) = 2$$
So $-\dfrac{b}{2}=2$, giving:
$$b = -4$$
Product of roots:
$$(1-4i)(1+4i) = 1^2-(4i)^2 = 1-(-16) = 17$$
So $\dfrac{c}{2}=17$, giving:
$$c = 34$$
$$\boxed{b=-4,\quad c=34}$$
QUESTION 13
6 marks
Easy
Let $z_1 = 6 - 3i$ and $z_2 = 1 + 2i$.
(a) Find $z_1 z_2$, giving your answer in the form $a+bi$. [3]
(b) Find $\dfrac{z_1}{z_2}$, giving your answer in the form $a+bi$. [3]
Show complete worked solution
(a) Expanding the product $(6-3i)(1+2i)$ term by term:
$$z_1z_2 = 6(1)+6(2i)+(-3i)(1)+(-3i)(2i)$$
$$= 6+12i-3i-6i^2$$
Since $i^2=-1$, the term $-6i^2=6$:
$$z_1z_2 = 6+9i+6$$
$$\boxed{z_1z_2 = 12+9i}$$
(b) Multiply numerator and denominator by the conjugate of the denominator, $1-2i$:
$$\frac{z_1}{z_2} = \frac{6-3i}{1+2i}\times\frac{1-2i}{1-2i} = \frac{(6-3i)(1-2i)}{1^2+2^2}$$
Expanding the numerator:
$$(6-3i)(1-2i) = 6-12i-3i+6i^2 = 6-15i-6 = -15i$$
The denominator is $1+4=5$, so:
$$\frac{z_1}{z_2} = \frac{-15i}{5} = -3i$$
$$\boxed{\dfrac{z_1}{z_2} = 0-3i}$$
QUESTION 14
7 marks
Easy
Consider the complex number $z = -2+5i$.
(a) Write down $z^*$, $-z$ and $-z^*$. [4]
(b) Plot $z$, $z^*$, $-z$ and $-z^*$ on an Argand diagram, and describe the geometric transformation each of $z^*$, $-z$ and $-z^*$ represents relative to $z$. [3]
Show complete worked solution
(a) The conjugate $z^*$ reverses the sign of the imaginary part; $-z$ reverses the sign of both parts.
$$z^* = -2-5i$$
$$-z = 2-5i$$
$$-z^* = 2+5i$$
$$\boxed{z^*=-2-5i,\quad -z=2-5i,\quad -z^*=2+5i}$$
(b) Plotting all four points shows they form a rectangle symmetric about both axes, as in the diagram:
- $z^*$ is the reflection of $z$ in the real axis (Re-axis).
- $-z$ is the reflection of $z$ in the imaginary axis (Im-axis), equivalently a rotation of $z$ through $180^\circ$ about the origin.
- $-z^*$ is the reflection of $z$ in the origin followed by conjugation, equivalently the reflection of $-z$ in the real axis (or of $z^*$ in the imaginary axis).
QUESTION 15
5 marks
Easy
Consider the complex number $z = 3\sqrt{3} + 3i$.
(a) Find the modulus $|z|$. [2]
(b) Find $\arg(z)$ in radians, giving an exact value. [2]
(c) Hence write $z$ in polar form $r(\cos\theta+i\sin\theta)$. [1]
Show complete worked solution
(a) Using $|z|=\sqrt{(\text{Re}(z))^2+(\text{Im}(z))^2}$:
$$|z| = \sqrt{(3\sqrt3)^2+3^2} = \sqrt{27+9} = \sqrt{36}$$
$$\boxed{|z| = 6}$$
(b) Since $\text{Re}(z)=3\sqrt3>0$ and $\text{Im}(z)=3>0$, $z$ lies in the first quadrant, so the principal argument is:
$$\theta = \arctan\!\left(\frac{3}{3\sqrt3}\right) = \arctan\!\left(\frac{1}{\sqrt3}\right)$$
$$\boxed{\arg(z) = \dfrac{\pi}{6}}$$
(c) Substituting $r=6$ and $\theta=\dfrac{\pi}{6}$ into the diagram alongside the vector for $z$ confirms the angle from the positive real axis:
$$\boxed{z = 6\left(\cos\dfrac{\pi}{6}+i\sin\dfrac{\pi}{6}\right)}$$
QUESTION 16
4 marks
Easy
A complex number is given in polar form as $z = 8\left(\cos\dfrac{2\pi}{3}+i\sin\dfrac{2\pi}{3}\right)$.
Convert $z$ to Cartesian form $a+bi$, giving exact values for $a$ and $b$.
Show complete worked solution
Using $\cos\dfrac{2\pi}{3}=-\dfrac{1}{2}$ and $\sin\dfrac{2\pi}{3}=\dfrac{\sqrt3}{2}$:
$$z = 8\left(-\frac{1}{2}\right)+8\left(\frac{\sqrt3}{2}\right)i$$
$$= -4+4\sqrt3\,i$$
$$\boxed{z = -4+4\sqrt3\,i}$$
QUESTION 17
4 marks
Easy
Let $z_1 = 1+3i$ and $z_2 = 1-i$.
Find $\dfrac{z_1}{z_2}$, giving your answer in the form $a+bi$.
Show complete worked solution
Multiply numerator and denominator by the conjugate of $z_2$, which is $1+i$:
$$\frac{z_1}{z_2} = \frac{1+3i}{1-i}\times\frac{1+i}{1+i} = \frac{(1+3i)(1+i)}{1^2+1^2}$$
Expanding the numerator:
$$(1+3i)(1+i) = 1+i+3i+3i^2 = 1+4i-3 = -2+4i$$
The denominator is $1+1=2$, so:
$$\frac{z_1}{z_2} = \frac{-2+4i}{2}$$
$$\boxed{\dfrac{z_1}{z_2} = -1+2i}$$
QUESTION 18
5 marks
Easy
Find the complex number $z = x+iy$, where $x,y \in \mathbb{R}$, that satisfies the equation
$$3z - (4-i) = 5+7i$$
Show complete worked solution
Let $z=x+iy$. Substituting into the left-hand side:
$$3(x+iy)-(4-i) = (3x-4)+(3y+1)i$$
Equating this to $5+7i$ and comparing real and imaginary parts:
$$\text{Real: } 3x-4=5 \implies 3x=9 \implies x=3$$
$$\text{Imaginary: } 3y+1=7 \implies 3y=6 \implies y=2$$
Check: $3(3+2i)-(4-i) = 9+6i-4+i = 5+7i$. $\checkmark$
$$\boxed{z = 3+2i}$$
QUESTION 19
4 marks
Easy
The points $A$ and $B$ on the Argand diagram below represent the complex numbers $z_1$ and $z_2$ respectively.
(a) Write down $z_1$ and $z_2$ as read from the diagram. [2]
(b) Find $z_1+z_2$ in the form $a+bi$. [2]
Show complete worked solution
(a) Reading the coordinates of $A$ and $B$ from the grid:
$$\boxed{z_1 = 5+2i,\qquad z_2 = -2+4i}$$
(b) Adding real and imaginary parts separately:
$$z_1+z_2 = (5+2i)+(-2+4i) = (5-2)+(2+4)i$$
$$\boxed{z_1+z_2 = 3+6i}$$
QUESTION 20
5 marks
Easy
The complex numbers $z_1$ and $z_2$ have $|z_1|=5$, $\arg(z_1)=40^\circ$, and $|z_2|=2$, $\arg(z_2)=110^\circ$.
(a) Find $|z_1z_2|$ and $\arg(z_1z_2)$. [2]
(b) Find $\left|\dfrac{z_1}{z_2}\right|$ and $\arg\!\left(\dfrac{z_1}{z_2}\right)$, giving the argument in the range $-180^\circ<\theta\le180^\circ$. [3]
Show complete worked solution
(a) When multiplying complex numbers in polar form, moduli multiply and arguments add:
$$|z_1z_2| = |z_1||z_2| = 5\times2$$
$$\arg(z_1z_2) = \arg(z_1)+\arg(z_2) = 40^\circ+110^\circ$$
$$\boxed{|z_1z_2|=10,\quad \arg(z_1z_2)=150^\circ}$$
(b) When dividing complex numbers in polar form, moduli divide and arguments subtract:
$$\left|\frac{z_1}{z_2}\right| = \frac{|z_1|}{|z_2|} = \frac{5}{2}$$
$$\arg\!\left(\frac{z_1}{z_2}\right) = \arg(z_1)-\arg(z_2) = 40^\circ-110^\circ = -70^\circ$$
Since $-70^\circ$ already lies in the range $-180^\circ<\theta\le180^\circ$, no adjustment is needed.
$$\boxed{\left|\dfrac{z_1}{z_2}\right|=2.5,\quad \arg\!\left(\dfrac{z_1}{z_2}\right)=-70^\circ}$$
QUESTION 21
5 marks
Easy
Consider the locus of points $z$ in the complex plane satisfying
$$|z-(2+3i)| = 4$$
(a) Describe this locus geometrically. [2]
(b) State the centre and radius of the locus. [1]
(c) Sketch the locus on an Argand diagram. [2]
Show complete worked solution
(a) The expression $|z-(2+3i)|$ is the distance, in the Argand plane, between the point representing $z$ and the fixed point representing $2+3i$. Setting this distance equal to the constant $4$ means $z$ traces out all points a fixed distance $4$ from $2+3i$.
$$\boxed{\text{The locus is a circle}}$$
(b) Comparing $|z-(2+3i)|=4$ with the standard form $|z-z_0|=r$:
$$\boxed{\text{centre } (2,3),\quad \text{radius } r=4}$$
(c) The sketch below shows the circle of radius $4$ centred at the point $2+3i$.
QUESTION 22
5 marks
Easy
A complex number is $z = 3\left(\cos40^\circ+i\sin40^\circ\right)$.
(a) Use De Moivre's theorem to write $z^2$ in polar form. [2]
(b) Hence express $z^2$ in the form $a+bi$, giving $a$ and $b$ correct to three significant figures. [3]
Show complete worked solution
(a) By De Moivre's theorem, $[r(\cos\theta+i\sin\theta)]^n = r^n(\cos n\theta+i\sin n\theta)$. With $r=3$, $\theta=40^\circ$, $n=2$:
$$z^2 = 3^2\left(\cos(2\times40^\circ)+i\sin(2\times40^\circ)\right)$$
$$\boxed{z^2 = 9(\cos80^\circ+i\sin80^\circ)}$$
(b) Using a GDC:
$$9\cos80^\circ = 1.5628\ldots,\qquad 9\sin80^\circ = 8.8633\ldots$$
$$\boxed{z^2 \approx 1.56+8.86i}$$
The diagram illustrates that squaring $z$ doubles its argument (from $40^\circ$ to $80^\circ$) and squares its modulus (from $3$ to $9$).
QUESTION 23
6 marks
Easy
Find the modulus and argument (in radians, giving exact values) of each of the following complex numbers:
(a) $z_1 = -6$ [2]
(b) $z_2 = 5i$ [2]
(c) $z_3 = -4i$ [2]
Show complete worked solution
(a) $z_1=-6$ lies on the negative real axis, at distance $6$ from the origin.
$$\boxed{|z_1|=6,\quad \arg(z_1)=\pi}$$
(b) $z_2=5i$ lies on the positive imaginary axis, at distance $5$ from the origin.
$$\boxed{|z_2|=5,\quad \arg(z_2)=\dfrac{\pi}{2}}$$
(c) $z_3=-4i$ lies on the negative imaginary axis, at distance $4$ from the origin. The principal argument must satisfy $-\pi<\theta\le\pi$, so:
$$\boxed{|z_3|=4,\quad \arg(z_3)=-\dfrac{\pi}{2}}$$
QUESTION 24
5 marks
Easy
Points $A$ and $B$ on an Argand diagram represent the complex numbers $z_1=4+i$ and $z_2=-2+3i$.
(a) Find $z_1-z_2$. [2]
(b) Hence find the distance $AB$, giving your answer as an exact surd and also correct to three significant figures. [2]
(c) Sketch $A$, $B$ and the segment $AB$ on an Argand diagram. [1]
Show complete worked solution
(a) Subtracting real and imaginary parts separately:
$$z_1-z_2 = (4+i)-(-2+3i) = (4-(-2))+(1-3)i$$
$$\boxed{z_1-z_2 = 6-2i}$$
(b) The distance between two points on the Argand diagram equals the modulus of the difference of the complex numbers they represent:
$$AB = |z_1-z_2| = \sqrt{6^2+(-2)^2} = \sqrt{36+4} = \sqrt{40} = 2\sqrt{10}$$
$$\boxed{AB = 2\sqrt{10} \approx 6.32}$$
(c) The sketch below shows $A(4,1)$, $B(-2,3)$ and the segment joining them.
QUESTION 25
4 marks
Easy
Consider the complex number $z = -5+2i$.
Find the modulus and the principal argument of $z$, giving each correct to three significant figures (the argument in radians).
Show complete worked solution
The modulus is:
$$|z| = \sqrt{(-5)^2+2^2} = \sqrt{25+4} = \sqrt{29}$$
$$\boxed{|z| \approx 5.39}$$
Since $\text{Re}(z)=-5<0$ and $\text{Im}(z)=2>0$, $z$ lies in the second quadrant, so the principal argument is $\pi$ minus the reference angle:
$$\theta = \pi - \arctan\!\left(\frac{2}{5}\right)$$
Using a GDC, $\arctan(0.4) = 0.3805\ldots$, so:
$$\theta = \pi - 0.3805\ldots = 2.7611\ldots$$
$$\boxed{\arg(z) \approx 2.76 \text{ rad}}$$
QUESTION 26
5 marks
Easy
A complex number is $z = 2\left(\cos\dfrac{\pi}{4}+i\sin\dfrac{\pi}{4}\right)$.
Use De Moivre's theorem to find $z^3$, giving your answer in the form $a+bi$ with $a$ and $b$ exact.
Show complete worked solution
By De Moivre's theorem, $[r(\cos\theta+i\sin\theta)]^n = r^n(\cos n\theta+i\sin n\theta)$. With $r=2$, $\theta=\dfrac{\pi}{4}$, $n=3$:
$$z^3 = 2^3\left(\cos\frac{3\pi}{4}+i\sin\frac{3\pi}{4}\right) = 8\left(\cos\frac{3\pi}{4}+i\sin\frac{3\pi}{4}\right)$$
Using $\cos\dfrac{3\pi}{4}=-\dfrac{\sqrt2}{2}$ and $\sin\dfrac{3\pi}{4}=\dfrac{\sqrt2}{2}$:
$$z^3 = 8\left(-\frac{\sqrt2}{2}\right)+8\left(\frac{\sqrt2}{2}\right)i$$
$$\boxed{z^3 = -4\sqrt2+4\sqrt2\,i}$$
QUESTION 27
8 marks
Medium
Let $z_1 = 6\left(\cos\dfrac{5\pi}{6}+i\sin\dfrac{5\pi}{6}\right)$ and $z_2 = 3\left(\cos\dfrac{\pi}{4}+i\sin\dfrac{\pi}{4}\right)$.
(a) Find $z_1z_2$ in polar form $r(\cos\theta+i\sin\theta)$. [3]
(b) Find $\dfrac{z_1}{z_2}$ in polar form, giving the argument in the range $-\pi<\theta\le\pi$. [3]
(c) Hence write $z_1z_2$ in the form $a+bi$, giving $a$ and $b$ correct to three significant figures. [2]
Show complete worked solution
(a) Moduli multiply and arguments add:
$$|z_1z_2| = 6\times3 = 18$$
$$\arg(z_1z_2) = \frac{5\pi}{6}+\frac{\pi}{4} = \frac{10\pi}{12}+\frac{3\pi}{12} = \frac{13\pi}{12}$$
$$\boxed{z_1z_2 = 18\left(\cos\dfrac{13\pi}{12}+i\sin\dfrac{13\pi}{12}\right)}$$
(b) Moduli divide and arguments subtract:
$$\left|\frac{z_1}{z_2}\right| = \frac{6}{3} = 2$$
$$\arg\!\left(\frac{z_1}{z_2}\right) = \frac{5\pi}{6}-\frac{\pi}{4} = \frac{10\pi}{12}-\frac{3\pi}{12} = \frac{7\pi}{12}$$
Since $\dfrac{7\pi}{12}$ already lies in $-\pi<\theta\le\pi$, this is the required argument.
$$\boxed{\dfrac{z_1}{z_2} = 2\left(\cos\dfrac{7\pi}{12}+i\sin\dfrac{7\pi}{12}\right)}$$
(c) The angle $\dfrac{13\pi}{12}$ is not one of the standard exact angles, so a GDC is used:
$$18\cos\frac{13\pi}{12} = -17.4\ldots,\qquad 18\sin\frac{13\pi}{12} = -4.66\ldots$$
$$\boxed{z_1z_2 \approx -17.4-4.66i}$$
QUESTION 28
6 marks
Medium
Let $z = 1+i$.
(a) Express $z$ in modulus-argument (polar) form. [2]
(b) Hence use De Moivre's theorem to find $z^5$, giving your answer in the form $a+bi$. [4]
Show complete worked solution
(a) The modulus is:
$$|z| = \sqrt{1^2+1^2} = \sqrt2$$
Since $\text{Re}(z)=1>0$ and $\text{Im}(z)=1>0$, $z$ lies in the first quadrant, with:
$$\arg(z) = \arctan\!\left(\frac{1}{1}\right) = \frac{\pi}{4}$$
$$\boxed{z = \sqrt2\left(\cos\dfrac{\pi}{4}+i\sin\dfrac{\pi}{4}\right)}$$
(b) By De Moivre's theorem:
$$z^5 = (\sqrt2)^5\left(\cos\frac{5\pi}{4}+i\sin\frac{5\pi}{4}\right)$$
Since $(\sqrt2)^5 = 2^{2}\times\sqrt2 = 4\sqrt2$, and $\cos\dfrac{5\pi}{4}=-\dfrac{\sqrt2}{2}$, $\sin\dfrac{5\pi}{4}=-\dfrac{\sqrt2}{2}$:
$$z^5 = 4\sqrt2\left(-\frac{\sqrt2}{2}\right)+4\sqrt2\left(-\frac{\sqrt2}{2}\right)i$$
$$= -\frac{4\times2}{2}-\frac{4\times2}{2}i$$
$$\boxed{z^5 = -4-4i}$$
QUESTION 29
7 marks
Medium
The equation $2x^2+bx+c=0$, where $b,c\in\mathbb{R}$, has $z=4+i$ as one root.
(a) State the other root of the equation. [1]
(b) Find the values of $b$ and $c$. [4]
(c) Verify your answer by expanding $2(x-(4+i))(x-(4-i))$ and comparing with $2x^2+bx+c$. [2]
Show complete worked solution
(a) Since $b,c\in\mathbb{R}$, complex roots occur in conjugate pairs.
$$\boxed{\text{The other root is } 4-i}$$
(b) For $2x^2+bx+c=0$ with roots $\alpha,\beta$: sum of roots $=-\dfrac{b}{2}$, product of roots $=\dfrac{c}{2}$.
Sum of roots:
$$(4+i)+(4-i) = 8$$
So $-\dfrac{b}{2}=8$, giving $b=-16$.
Product of roots:
$$(4+i)(4-i) = 16-i^2 = 16+1 = 17$$
So $\dfrac{c}{2}=17$, giving $c=34$.
$$\boxed{b=-16,\quad c=34}$$
(c) Expanding:
$$2(x-(4+i))(x-(4-i)) = 2\left[x^2-x(4-i)-x(4+i)+(4+i)(4-i)\right]$$
$$= 2\left[x^2-8x+17\right] = 2x^2-16x+34$$
This matches $2x^2+bx+c$ with $b=-16$, $c=34$. $\checkmark$
QUESTION 30
7 marks
Medium
Consider the locus $|z-(1+2i)|=5$.
(a) State the centre and radius of this circle. [2]
(b) Find, in exact (surd) form, the coordinates of the two points where the circle crosses the real axis. [4]
(c) Sketch the circle and these two intersection points on an Argand diagram. [1]
Show complete worked solution
(a) Comparing with $|z-z_0|=r$:
$$\boxed{\text{centre } (1,2),\quad \text{radius } 5}$$
(b) On the real axis, $z=x$ for real $x$, so $\text{Im}(z)=0$. Substituting into $|z-(1+2i)|=5$:
$$|(x-1)-2i| = 5$$
$$(x-1)^2+(-2)^2 = 5^2$$
$$(x-1)^2 = 25-4 = 21$$
$$x-1 = \pm\sqrt{21}$$
$$x = 1\pm\sqrt{21}$$
$$\boxed{(1+\sqrt{21},\,0) \text{ and } (1-\sqrt{21},\,0)}$$
(c) The sketch below shows the circle of radius $5$ centred at $(1,2)$, together with the two points where it meets the real axis.
QUESTION 31
7 marks
Medium
Consider the locus $\arg(z-(3-i)) = \dfrac{\pi}{3}$.
(a) Describe this locus geometrically. [2]
(b) Sketch the locus on an Argand diagram, indicating clearly the fixed point and the direction in which the locus extends. [3]
(c) Determine, showing your reasoning, whether the point $z=4+(\sqrt3-1)i$ lies on this locus. [2]
Show complete worked solution
(a) The expression $z-(3-i)$ is the vector from the fixed point $3-i$ to $z$. Fixing its argument at $\dfrac{\pi}{3}$ means every point $z$ on the locus lies on a straight ray starting at (but not including) $3-i$, making an angle of $\dfrac{\pi}{3}$ with the positive real direction.
$$\boxed{\text{The locus is a half-line (ray) from } 3-i \text{, excluding the endpoint, at angle }\tfrac{\pi}{3}\text{ to the positive real axis}}$$
(b) The sketch below shows the fixed point $3-i$ and the ray leaving it at $60^\circ$ to the horizontal.
(c) Compute $z-(3-i)$ for the given point:
$$z-(3-i) = \left[4+(\sqrt3-1)i\right]-(3-i) = (4-3)+\left[(\sqrt3-1)+1\right]i = 1+\sqrt3\,i$$
Its modulus and argument are:
$$|1+\sqrt3\,i| = \sqrt{1^2+(\sqrt3)^2} = \sqrt{4} = 2$$
$$\arg(1+\sqrt3\,i) = \arctan\!\left(\frac{\sqrt3}{1}\right) = \frac{\pi}{3}$$
Since $\arg(z-(3-i))=\dfrac{\pi}{3}$, the condition is satisfied.
$$\boxed{\text{Yes, } z=4+(\sqrt3-1)i \text{ lies on the locus}}$$
QUESTION 32
8 marks
Medium
Consider the complex number $z = -8+8\sqrt3\,i$.
(a) Express $z$ in polar form $r(\cos\theta+i\sin\theta)$. [2]
(b) Hence find the two square roots of $z$, giving each answer in the form $a+bi$. [4]
(c) Plot the two square roots on an Argand diagram, and comment on their geometric relationship to each other and to the origin. [2]
Show complete worked solution
(a) The modulus is:
$$|z| = \sqrt{(-8)^2+(8\sqrt3)^2} = \sqrt{64+192} = \sqrt{256} = 16$$
Since $\text{Re}(z)=-8<0$ and $\text{Im}(z)=8\sqrt3>0$, $z$ lies in the second quadrant. The reference angle is $\arctan\!\left(\dfrac{8\sqrt3}{8}\right)=\arctan(\sqrt3)=\dfrac{\pi}{3}$, so:
$$\theta = \pi-\frac{\pi}{3} = \frac{2\pi}{3}$$
$$\boxed{z = 16\left(\cos\dfrac{2\pi}{3}+i\sin\dfrac{2\pi}{3}\right)}$$
(b) If $w=\sqrt{z}$, then $w=\sqrt{r}\left(\cos\dfrac{\theta+2k\pi}{2}+i\sin\dfrac{\theta+2k\pi}{2}\right)$ for $k=0,1$. Here $\sqrt{16}=4$.
For $k=0$: angle $=\dfrac{2\pi/3}{2}=\dfrac{\pi}{3}$:
$$w_1 = 4\left(\cos\frac{\pi}{3}+i\sin\frac{\pi}{3}\right) = 4\left(\frac12+\frac{\sqrt3}{2}i\right) = 2+2\sqrt3\,i$$
For $k=1$: angle $=\dfrac{\pi}{3}+\pi=\dfrac{4\pi}{3}$:
$$w_2 = 4\left(\cos\frac{4\pi}{3}+i\sin\frac{4\pi}{3}\right) = 4\left(-\frac12-\frac{\sqrt3}{2}i\right) = -2-2\sqrt3\,i$$
$$\boxed{w_1 = 2+2\sqrt3\,i,\qquad w_2=-2-2\sqrt3\,i}$$
Check: $w_1^2 = (2+2\sqrt3 i)^2 = 4+8\sqrt3 i+12i^2 = 4-12+8\sqrt3 i = -8+8\sqrt3 i = z$. $\checkmark$
(c) The diagram shows that $w_2=-w_1$: the two square roots are diametrically opposite one another, each at distance $4$ from the origin (equal to $\sqrt{|z|}$), and $180^\circ$ apart, as is always true for square roots of a complex number.
QUESTION 33
9 marks
Medium
Two AC voltage phasors are given in polar form as $V_1 = 100\angle30^\circ$ V and $V_2 = 80\angle{-45}^\circ$ V (rms values).
(a) Convert $V_1$ and $V_2$ to Cartesian ($a+bi$) form, giving $a$ and $b$ correct to three significant figures. [4]
(b) Hence find the resultant voltage $V=V_1+V_2$ in Cartesian form, correct to three significant figures. [2]
(c) Express $V$ in polar form, giving the modulus correct to three significant figures and the argument in degrees correct to one decimal place. [3]
Show complete worked solution
(a) Converting using $a=r\cos\theta$, $b=r\sin\theta$:
$$V_1 = 100\cos30^\circ+100\sin30^\circ\,i \approx 86.6+50.0i$$
$$V_2 = 80\cos(-45^\circ)+80\sin(-45^\circ)\,i \approx 56.6-56.6i$$
$$\boxed{V_1\approx86.6+50.0i\text{ V},\quad V_2\approx56.6-56.6i\text{ V}}$$
(b) Adding real and imaginary parts:
$$V = V_1+V_2 \approx (86.6+56.6)+(50.0-56.6)i$$
$$\boxed{V \approx 143+(-6.57)i \text{ V}}$$
(c) The modulus is:
$$|V| = \sqrt{143.17\ldots^2+(-6.57\ldots)^2} \approx 143.32\ldots$$
The argument (in the fourth quadrant, small negative angle since $\text{Re}(V)>0$, $\text{Im}(V)<0$) is:
$$\arg(V) = \arctan\!\left(\frac{-6.57\ldots}{143.17\ldots}\right) \approx -2.6^\circ$$
$$\boxed{V \approx 143\angle{-2.6}^\circ \text{ V}}$$
QUESTION 34
7 marks
Medium
Let $z = 5-3i$, represented by the point $P$ on an Argand diagram.
(a) Find $iz$, and describe the geometric transformation that takes the point representing $z$ to the point representing $iz$. [3]
(b) Find $-z$, and describe the geometric transformation that takes the point representing $z$ to the point representing $-z$. [2]
(c) Plot $z$, $iz$ and $-z$ on an Argand diagram. [2]
Show complete worked solution
(a) Multiplying out:
$$iz = i(5-3i) = 5i-3i^2 = 5i+3$$
$$\boxed{iz = 3+5i}$$
Since $|i|=1$ and $\arg(i)=\dfrac{\pi}{2}$, multiplying by $i$ multiplies the modulus by $1$ (leaves it unchanged) and adds $\dfrac{\pi}{2}$ to the argument.
$$\boxed{\text{Multiplication by } i \text{ is a rotation of } 90^\circ \text{ anticlockwise about the origin}}$$
(b)
$$\boxed{-z = -5+3i}$$
Multiplying by $-1$ leaves the modulus unchanged and adds $\pi$ ($180^\circ$) to the argument.
$$\boxed{\text{Multiplication by } -1 \text{ is a rotation of } 180^\circ \text{ about the origin}}$$
(c) The diagram below plots $z=5-3i$, $iz=3+5i$ and $-z=-5+3i$, all lying on a circle of radius $|z|=\sqrt{34}$ centred at the origin, confirming that both transformations preserve distance from $O$.
QUESTION 35
7 marks
Medium
Points $A$ and $B$ on an Argand diagram represent the complex numbers $z_1=1+i$ and $z_2=5+3i$. Let $z=x+iy$ be a variable point satisfying $|z-z_1|=|z-z_2|$.
(a) By squaring both sides and simplifying, show that this locus has the Cartesian equation $y = 8-2x$. [4]
(b) Verify that the midpoint of $AB$ lies on this line. [2]
(c) Sketch $A$, $B$ and the locus on an Argand diagram. [1]
Show complete worked solution
(a) The condition $|z-z_1|=|z-z_2|$ states that $z$ is equidistant from $A(1,1)$ and $B(5,3)$, so the locus should be the perpendicular bisector of $AB$. Squaring both sides:
$$(x-1)^2+(y-1)^2 = (x-5)^2+(y-3)^2$$
Expanding both sides:
$$x^2-2x+1+y^2-2y+1 = x^2-10x+25+y^2-6y+9$$
Cancelling $x^2$ and $y^2$:
$$-2x-2y+2 = -10x-6y+34$$
Collecting terms:
$$8x+4y = 32$$
Dividing by $4$:
$$2x+y = 8$$
$$\boxed{y = 8-2x}$$
(b) The midpoint of $AB$ is:
$$M = \left(\frac{1+5}{2},\frac{1+3}{2}\right) = (3,2)$$
Substituting $x=3$ into $y=8-2x$: $y=8-6=2$, which matches the $y$-coordinate of $M$. $\checkmark$
$$\boxed{M(3,2)\text{ lies on the locus, as expected since it is the perpendicular bisector of }AB}$$
(c) The sketch below shows $A(1,1)$, $B(5,3)$ and the perpendicular bisector $y=8-2x$ passing through their midpoint.
QUESTION 36
4 marks
Medium
A complex number $z$ satisfies $|z|=7$ and $\arg(z)=200^\circ$.
Find $z$ in the form $a+bi$, giving $a$ and $b$ correct to three significant figures.
Show complete worked solution
Using $a=r\cos\theta$ and $b=r\sin\theta$ with $r=7$, $\theta=200^\circ$:
$$a = 7\cos200^\circ \approx -6.578\ldots$$
$$b = 7\sin200^\circ \approx -2.394\ldots$$
Both parts are negative, consistent with $200^\circ$ lying in the third quadrant.
$$\boxed{z \approx -6.58-2.39i}$$
QUESTION 37
7 marks
Medium
A cubic equation $x^3+px^2+qx+r=0$, where $p,q,r\in\mathbb{R}$, has roots $-1+2i$ and $5$.
(a) State the third root of the equation. [1]
(b) Find the values of $p$, $q$ and $r$. [5]
(c) Verify that $x=5$ satisfies the equation with your values of $p$, $q$, $r$. [1]
Show complete worked solution
(a) Since the coefficients are real, complex roots occur in conjugate pairs, so the third root is the conjugate of $-1+2i$.
$$\boxed{\text{The third root is } -1-2i}$$
(b) For a monic cubic with roots $\alpha,\beta,\gamma$: $p=-(\alpha+\beta+\gamma)$, $q=\alpha\beta+\beta\gamma+\gamma\alpha$, $r=-\alpha\beta\gamma$.
Sum of roots:
$$(-1+2i)+(-1-2i)+5 = 3$$
$$p = -3$$
Sum of pairwise products: since $(-1+2i)$ and $(-1-2i)$ are conjugates, their product is $(-1)^2-(2i)^2=1+4=5$. Then:
$$q = (-1+2i)(-1-2i)+5(-1+2i)+5(-1-2i)$$
$$= 5+5\left[(-1+2i)+(-1-2i)\right] = 5+5(-2) = -5$$
Product of all three roots:
$$(-1+2i)(-1-2i)(5) = 5\times5 = 25$$
$$r = -25$$
$$\boxed{p=-3,\quad q=-5,\quad r=-25}$$
(c) Substituting $x=5$ into $x^3-3x^2-5x-25$:
$$5^3-3(5)^2-5(5)-25 = 125-75-25-25 = 0 \checkmark$$
$$\boxed{x=5 \text{ satisfies } x^3-3x^2-5x-25=0}$$
QUESTION 38
6 marks
Medium
Simplify the expression
$$\frac{(\cos3\theta+i\sin3\theta)^2(\cos\theta-i\sin\theta)}{(\cos2\theta+i\sin2\theta)^3}$$
giving your answer in the form $\cos(k\theta)+i\sin(k\theta)$ for an integer $k$.
Show complete worked solution
By De Moivre's theorem, $(\cos n\theta+i\sin n\theta) = (\cos\theta+i\sin\theta)^n$, so raising a "cis" expression to a power multiplies the angle, and $\cos\theta-i\sin\theta = \cos(-\theta)+i\sin(-\theta)$.
Numerator:
$$(\cos3\theta+i\sin3\theta)^2 = \cos6\theta+i\sin6\theta$$
$$\left[\cos6\theta+i\sin6\theta\right]\left[\cos(-\theta)+i\sin(-\theta)\right] = \cos(6\theta-\theta)+i\sin(6\theta-\theta) = \cos5\theta+i\sin5\theta$$
Denominator:
$$(\cos2\theta+i\sin2\theta)^3 = \cos6\theta+i\sin6\theta$$
Dividing (subtracting angles):
$$\frac{\cos5\theta+i\sin5\theta}{\cos6\theta+i\sin6\theta} = \cos(5\theta-6\theta)+i\sin(5\theta-6\theta) = \cos(-\theta)+i\sin(-\theta)$$
$$\boxed{\text{Expression} = \cos(-\theta)+i\sin(-\theta) \;\left(\text{i.e. } k=-1\right)}$$
QUESTION 39
7 marks
Medium
Points on an Argand diagram represent $z_1=1+i$, $z_2=7+2i$ and $z_3=3+6i$, the vertices of a triangle.
(a) Plot $z_1$, $z_2$ and $z_3$ on an Argand diagram. [2]
(b) Find the area of the triangle formed by $z_1$, $z_2$ and $z_3$, showing your method clearly. [5]
Show complete worked solution
(a) The diagram below shows the three points joined to form a triangle.
(b) For points $(x_1,y_1)$, $(x_2,y_2)$, $(x_3,y_3)$, the area of the triangle they form is given by the shoelace formula:
$$\text{Area} = \frac12\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|$$
Substituting $(x_1,y_1)=(1,1)$, $(x_2,y_2)=(7,2)$, $(x_3,y_3)=(3,6)$:
$$\text{Area} = \frac12\left|1(2-6)+7(6-1)+3(1-2)\right|$$
$$= \frac12\left|1(-4)+7(5)+3(-1)\right| = \frac12\left|-4+35-3\right| = \frac12(28)$$
$$\boxed{\text{Area} = 14 \text{ square units}}$$
QUESTION 40
7 marks
Medium
Let $z=x+iy$, where $x,y\in\mathbb{R}$, satisfy $z^2 = 5-12i$.
(a) By expanding $z^2$ and equating real and imaginary parts, form two real equations relating $x$ and $y$. [3]
(b) Hence solve for $x$ and $y$, giving both solutions for $z$. [4]
Show complete worked solution
(a) Expanding:
$$z^2 = (x+iy)^2 = x^2-y^2+2xyi$$
Equating this to $5-12i$ and comparing real and imaginary parts:
$$\boxed{x^2-y^2 = 5 \qquad\text{and}\qquad 2xy = -12}$$
(b) From the second equation, $xy=-6$, so $y=-\dfrac{6}{x}$ (for $x\ne0$). Substituting into the first equation:
$$x^2-\left(\frac{6}{x}\right)^2 = 5$$
$$x^2-\frac{36}{x^2} = 5$$
Multiplying through by $x^2$:
$$x^4-5x^2-36 = 0$$
Let $u=x^2$:
$$u^2-5u-36 = 0$$
Using the quadratic formula:
$$u = \frac{5\pm\sqrt{25+144}}{2} = \frac{5\pm13}{2}$$
So $u=9$ or $u=-4$. Since $u=x^2\ge0$, reject $u=-4$, so $x^2=9$, giving $x=3$ or $x=-3$.
For $x=3$: $y=-\dfrac{6}{3}=-2$. For $x=-3$: $y=-\dfrac{6}{-3}=2$.
Check: $(3-2i)^2 = 9-12i+4i^2 = 9-12i-4 = 5-12i$. $\checkmark$
$$\boxed{z = 3-2i \quad\text{or}\quad z=-3+2i}$$
QUESTION 41
6 marks
Medium
The complex numbers $z_1$ and $z_2$ have $|z_1|=13$, $\arg(z_1)=22.62^\circ$ (to 2 d.p.), and $|z_2|=10$, $\arg(z_2)=126.87^\circ$ (to 2 d.p.).
(a) Convert $z_1$ and $z_2$ to Cartesian form, giving exact values. [4]
(b) Hence find $z_1+z_2$ in Cartesian form. [2]
Show complete worked solution
(a) Using $a=r\cos\theta$, $b=r\sin\theta$ (a GDC confirms these angles come from a $5$-$12$-$13$ triangle and a $3$-$4$-$5$ triangle respectively):
$$z_1 = 13\cos(22.62^\circ)+13\sin(22.62^\circ)\,i = 12+5i$$
$$z_2 = 10\cos(126.87^\circ)+10\sin(126.87^\circ)\,i = -6+8i$$
$$\boxed{z_1 = 12+5i,\qquad z_2=-6+8i}$$
(b) Adding real and imaginary parts:
$$z_1+z_2 = (12-6)+(5+8)i$$
$$\boxed{z_1+z_2 = 6+13i}$$
QUESTION 42
7 marks
Medium
Consider the region $R$ in the complex plane defined by
$$|z-2|\le3 \quad\text{and}\quad \text{Im}(z)\ge0$$
(a) Describe the region defined by $|z-2|\le3$ alone. [2]
(b) On an Argand diagram, shade the region $R$ satisfying both conditions. [3]
(c) Find the exact area of $R$. [2]
Show complete worked solution
(a) $|z-2|\le3$ is the set of points whose distance from the fixed point $2$ (i.e. $2+0i$) is at most $3$.
$$\boxed{\text{A disc (closed circular region) of radius } 3 \text{ centred at } (2,0)}$$
(b) The diagram below shades the region satisfying $|z-2|\le3$ and $\text{Im}(z)\ge0$ - since the centre $(2,0)$ lies exactly on the real axis, this second condition keeps only the upper half of the disc.
(c) Since the centre lies on the boundary line $\text{Im}(z)=0$, the condition $\text{Im}(z)\ge0$ cuts the disc exactly in half. The full disc has area $\pi r^2=\pi(3)^2=9\pi$, so:
$$\text{Area of } R = \frac{9\pi}{2}$$
$$\boxed{\text{Area} = 4.5\pi \approx 14.1 \text{ square units}}$$
QUESTION 43
13 marks
Hard
(a) Express $8i$ in polar form $r(\cos\theta+i\sin\theta)$. [2]
(b) Use De Moivre's theorem to find the three cube roots of $8i$. Give each root in both polar form and exact Cartesian ($a+bi$) form, and present your results in a table. [7]
(c) Plot the three cube roots on an Argand diagram and state the geometric shape they form. [2]
(d) Verify algebraically that one of your roots, when cubed, returns $8i$. [2]
Show complete worked solution
(a) $8i=0+8i$ lies on the positive imaginary axis, at distance $8$ from the origin.
$$\boxed{8i = 8\left(\cos\dfrac{\pi}{2}+i\sin\dfrac{\pi}{2}\right)}$$
(b) If $z^3=8i$, write $z = r(\cos\theta+i\sin\theta)$. By De Moivre's theorem, $z^3=r^3(\cos3\theta+i\sin3\theta)$, so:
$$r^3=8 \implies r=2$$
$$3\theta = \frac{\pi}{2}+2k\pi \implies \theta = \frac{\pi}{6}+\frac{2k\pi}{3}, \quad k=0,1,2$$
This gives $\theta = \dfrac{\pi}{6}$, $\dfrac{5\pi}{6}$, $\dfrac{3\pi}{2}$ (i.e. $30^\circ$, $150^\circ$, $270^\circ$). Converting each to Cartesian form using $a=2\cos\theta$, $b=2\sin\theta$:
$$\boxed{z_0=\sqrt3+i,\quad z_1=-\sqrt3+i,\quad z_2=-2i}$$
(c) All three roots have modulus $2$ and are equally spaced by $\dfrac{2\pi}{3}$ ($120^\circ$) around the origin, as shown in the diagram.
$$\boxed{\text{The three roots form the vertices of an equilateral triangle inscribed in the circle } |z|=2}$$
(d) Cubing $z_0=\sqrt3+i$:
$$z_0^2 = (\sqrt3+i)^2 = 3+2\sqrt3i+i^2 = 3+2\sqrt3i-1 = 2+2\sqrt3i$$
$$z_0^3 = z_0^2\cdot z_0 = (2+2\sqrt3i)(\sqrt3+i) = 2\sqrt3+2i+6i+2\sqrt3i^2$$
$$= 2\sqrt3+8i-2\sqrt3 = 8i$$
$$\boxed{z_0^3=8i \checkmark}$$
| $k$ | Polar form | Cartesian form |
|---|---|---|
| 0 | $2\left(\cos\dfrac{\pi}{6}+i\sin\dfrac{\pi}{6}\right)$ | $\sqrt3+i$ |
| 1 | $2\left(\cos\dfrac{5\pi}{6}+i\sin\dfrac{5\pi}{6}\right)$ | $-\sqrt3+i$ |
| 2 | $2\left(\cos\dfrac{3\pi}{2}+i\sin\dfrac{3\pi}{2}\right)$ | $-2i$ |
QUESTION 44
13 marks
Hard
(a) Express $-16$ in polar form $r(\cos\theta+i\sin\theta)$. [1]
(b) Use De Moivre's theorem to find all four fourth roots of $-16$, giving each in exact Cartesian form $a+bi$. [8]
(c) Plot the four roots on an Argand diagram and identify the geometric shape formed by joining them in order. [2]
(d) Hence find the exact area enclosed by this shape. [2]
Show complete worked solution
(a) $-16$ lies on the negative real axis, at distance $16$ from the origin.
$$\boxed{-16 = 16(\cos\pi+i\sin\pi)}$$
(b) If $z^4=-16$, write $z=r(\cos\theta+i\sin\theta)$. By De Moivre's theorem $z^4=r^4(\cos4\theta+i\sin4\theta)$, so:
$$r^4=16 \implies r=2$$
$$4\theta = \pi+2k\pi \implies \theta = \frac{\pi}{4}+\frac{k\pi}{2}, \quad k=0,1,2,3$$
This gives $\theta=\dfrac{\pi}{4},\dfrac{3\pi}{4},\dfrac{5\pi}{4},\dfrac{7\pi}{4}$ (i.e. $45^\circ,135^\circ,225^\circ,315^\circ$). Converting each using $a=2\cos\theta,\ b=2\sin\theta$ and $\cos45^\circ=\sin45^\circ=\dfrac{\sqrt2}{2}$:
$$z_0 = 2\left(\frac{\sqrt2}{2}+\frac{\sqrt2}{2}i\right) = \sqrt2+\sqrt2i$$
$$z_1 = 2\left(-\frac{\sqrt2}{2}+\frac{\sqrt2}{2}i\right) = -\sqrt2+\sqrt2i$$
$$z_2 = 2\left(-\frac{\sqrt2}{2}-\frac{\sqrt2}{2}i\right) = -\sqrt2-\sqrt2i$$
$$z_3 = 2\left(\frac{\sqrt2}{2}-\frac{\sqrt2}{2}i\right) = \sqrt2-\sqrt2i$$
$$\boxed{z_0=\sqrt2+\sqrt2i,\ z_1=-\sqrt2+\sqrt2i,\ z_2=-\sqrt2-\sqrt2i,\ z_3=\sqrt2-\sqrt2i}$$
(c) All four roots have modulus $2$ and are equally spaced by $\dfrac{\pi}{2}$ ($90^\circ$) around the origin, as shown in the diagram.
$$\boxed{\text{The four roots form the vertices of a square inscribed in the circle } |z|=2}$$
(d) The side length is the distance between consecutive roots, e.g. $z_0$ and $z_1$:
$$|z_0-z_1| = \left|\left(\sqrt2-(-\sqrt2)\right)+0i\right| = 2\sqrt2$$
The area of a square is (side length)$^2$:
$$\text{Area} = (2\sqrt2)^2 = 8$$
$$\boxed{\text{Area} = 8 \text{ square units}}$$
QUESTION 45
10 marks
Hard
Two impedances, $Z_1 = 4+3i\ \Omega$ and $Z_2 = 2-6i\ \Omega$, are connected in parallel. The combined impedance is given by
$$Z_{\text{total}} = \frac{Z_1Z_2}{Z_1+Z_2}$$
(a) Find $Z_{\text{total}}$ in Cartesian form $a+bi$. [4]
(b) Express $Z_{\text{total}}$ in polar form, giving the modulus exactly and the argument correct to three significant figures (degrees). [3]
(c) The circuit is driven by a voltage source $V=100\angle0^\circ$ V (rms). Using $I=\dfrac{V}{Z_{\text{total}}}$, find the current $I$ in exact Cartesian form. [3]
Show complete worked solution
(a) First find $Z_1Z_2$:
$$Z_1Z_2 = (4+3i)(2-6i) = 8-24i+6i-18i^2 = 8-18i+18 = 26-18i$$
Then find $Z_1+Z_2$:
$$Z_1+Z_2 = (4+3i)+(2-6i) = 6-3i$$
Dividing, multiplying numerator and denominator by the conjugate $6+3i$:
$$Z_{\text{total}} = \frac{26-18i}{6-3i}\times\frac{6+3i}{6+3i} = \frac{(26-18i)(6+3i)}{6^2+3^2}$$
Expanding the numerator:
$$(26-18i)(6+3i) = 156+78i-108i-54i^2 = 156-30i+54 = 210-30i$$
The denominator is $36+9=45$, so:
$$Z_{\text{total}} = \frac{210-30i}{45} = \frac{210}{45}-\frac{30}{45}i$$
$$\boxed{Z_{\text{total}} = \dfrac{14}{3}-\dfrac{2}{3}i \;\Omega \;\;(\approx4.67-0.667i\ \Omega)}$$
(b) The modulus is:
$$|Z_{\text{total}}| = \sqrt{\left(\frac{14}{3}\right)^2+\left(\frac{2}{3}\right)^2} = \sqrt{\frac{196+4}{9}} = \sqrt{\frac{200}{9}} = \frac{10\sqrt2}{3}$$
Since $\text{Re}>0$ and $\text{Im}<0$, the argument is a small negative angle:
$$\arg(Z_{\text{total}}) = \arctan\!\left(\frac{-2/3}{14/3}\right) = \arctan\!\left(-\frac{1}{7}\right) \approx -8.13^\circ$$
$$\boxed{Z_{\text{total}} = \dfrac{10\sqrt2}{3}\angle{-8.13}^\circ \;\Omega}$$
(c) Substituting $V=100$ (real) and $Z_{\text{total}}=\dfrac{14}{3}-\dfrac{2}{3}i$:
$$I = \frac{100}{\tfrac{14}{3}-\tfrac{2}{3}i} = \frac{100\times3}{14-2i} = \frac{300}{14-2i}$$
Multiplying numerator and denominator by the conjugate $14+2i$:
$$I = \frac{300(14+2i)}{14^2+2^2} = \frac{300(14+2i)}{196+4} = \frac{300(14+2i)}{200} = 1.5(14+2i)$$
$$\boxed{I = 21+3i \text{ A}}$$
QUESTION 46
12 marks
Hard
Let $z=x+iy$, where $x,y\in\mathbb{R}$, satisfy $|z-3| = |z-3-4i|$.
(a) By squaring both sides and simplifying, show that the locus is the horizontal line $y=2$. [3]
(b) Find, in exact surd form, the coordinates of the two points where the circle $|z|=5$ meets the line $y=2$. [3]
(c) Sketch, on the same Argand diagram, the circle $|z|=5$ and the line $y=2$, and shade the region $R$ defined by $|z-3|\le|z-3-4i|$ and $|z|\le5$. [2]
(d) Find the area of the region $R$, correct to three significant figures. [4]
Show complete worked solution
(a) The condition $|z-3|=|z-3-4i|$ says $z$ is equidistant from the fixed points $3$ (i.e. $3+0i$) and $3+4i$, so the locus should be the perpendicular bisector of the segment joining them. Squaring both sides with $z=x+iy$:
$$|z-3|^2 = (x-3)^2+y^2, \qquad |z-3-4i|^2 = (x-3)^2+(y-4)^2$$
Setting these equal:
$$(x-3)^2+y^2 = (x-3)^2+(y-4)^2$$
The $(x-3)^2$ terms cancel:
$$y^2 = (y-4)^2 = y^2-8y+16$$
$$0 = -8y+16$$
$$\boxed{y = 2}$$
(This makes sense: $3$ and $3+4i$ differ only in their imaginary part, so their perpendicular bisector is horizontal, at the average height $\tfrac{0+4}{2}=2$.)
(b) Substituting $y=2$ into $x^2+y^2=25$ (i.e. $|z|=5$):
$$x^2+2^2 = 25$$
$$x^2 = 21$$
$$x = \pm\sqrt{21}$$
$$\boxed{(\sqrt{21},\,2) \text{ and } (-\sqrt{21},\,2)}$$
(c) The condition $|z-3|\le|z-3-4i|$ means $z$ is at least as close to $3$ as to $3+4i$, i.e. $z$ lies on or below the line $y=2$ (the side containing $3+0i$). Combined with $|z|\le5$, the region $R$ is the part of the disc $|z|\le5$ lying on or below the line $y=2$ - shown shaded in the diagram, bounded above by the chord through the two points found in part (b).
(d) The disc $|z|\le5$ has total area $\pi(5)^2=25\pi$. Region $R$ is the disc minus the circular segment that lies above the chord $y=2$. For a circle of radius $R=5$ and a chord at perpendicular distance $d=2$ from the centre, the area of the segment on the far side of the chord (here, above $y=2$) is:
$$A_{\text{segment}} = R^2\arccos\!\left(\frac{d}{R}\right)-d\sqrt{R^2-d^2}$$
$$= 25\arccos(0.4)-2\sqrt{25-4} = 25(1.1593\ldots)-2\sqrt{21}$$
$$\approx 28.98\ldots-9.165\ldots \approx 19.82\ldots$$
So the area of region $R$ (disc minus this upper segment) is:
$$\text{Area} = 25\pi - 19.82\ldots \approx 78.54\ldots-19.82\ldots$$
$$\boxed{\text{Area} \approx 58.7 \text{ square units}}$$
QUESTION 47
13 marks
Hard
(a) Solve $z^5=1$, giving all five roots in the form $\text{cis}\,\theta$ (modulus $1$), with $\theta$ in degrees. [4]
(b) Hence write each root in Cartesian form $a+bi$, correct to three significant figures where necessary, and present your results in a table. [5]
(c) Plot the five roots on an Argand diagram and state the geometric shape they form. [2]
(d) By summing your five Cartesian approximations from part (b), show that the sum of the roots is (to three significant figures) $0$. [2]
Show complete worked solution
(a) Write $1 = \cos0^\circ+i\sin0^\circ$ (or, equivalently, $\cos(360k)^\circ+i\sin(360k)^\circ$ for any integer $k$). If $z=r(\cos\theta+i\sin\theta)$, then by De Moivre's theorem $z^5=r^5(\cos5\theta+i\sin5\theta)$, so:
$$r^5=1 \implies r=1$$
$$5\theta = 0^\circ+360^\circ k \implies \theta = 72^\circ k, \quad k=0,1,2,3,4$$
$$\boxed{\theta = 0^\circ,\ 72^\circ,\ 144^\circ,\ 216^\circ,\ 288^\circ}$$
(b) Using $a=\cos\theta$, $b=\sin\theta$ with a GDC (all have $r=1$):
$$\boxed{z_0=1,\ z_1\approx0.309+0.951i,\ z_2\approx-0.809+0.588i,\ z_3\approx-0.809-0.588i,\ z_4\approx0.309-0.951i}$$
(c) All five roots have modulus $1$ and are equally spaced by $72^\circ$ around the origin, as shown in the diagram.
$$\boxed{\text{The five roots form the vertices of a regular pentagon inscribed in the unit circle } |z|=1}$$
(d) Summing the real parts:
$$1+0.309+(-0.809)+(-0.809)+0.309 = 1+0.309-0.809-0.809+0.309 = 0.000$$
Summing the imaginary parts:
$$0+0.951+0.588+(-0.588)+(-0.951) = 0.951+0.588-0.588-0.951 = 0.000$$
$$\boxed{\text{Sum of the five roots} \approx 0+0i = 0}$$
(This is a general property: the $n$ distinct $n$-th roots of unity always sum to $0$ for $n\ge2$, since they form a symmetric, equally-spaced set around the origin.)
| $k$ | $\theta$ | Cartesian form |
|---|---|---|
| 0 | $0^\circ$ | $1+0i$ |
| 1 | $72^\circ$ | $0.309+0.951i$ |
| 2 | $144^\circ$ | $-0.809+0.588i$ |
| 3 | $216^\circ$ | $-0.809-0.588i$ |
| 4 | $288^\circ$ | $0.309-0.951i$ |
QUESTION 48
10 marks
Hard
A point representing the complex number $z_0=5$ is rotated repeatedly about the origin, each rotation through $40^\circ$ anticlockwise, so that $z_{k+1} = wz_k$ where $w=\cos40^\circ+i\sin40^\circ$.
(a) Find $z_1=wz_0$, giving your answer in Cartesian form correct to three significant figures. [3]
(b) Find $z_4$ (i.e. after four rotations), giving your answer in Cartesian form correct to three significant figures, and state its argument. [4]
(c) Find the smallest positive integer $n$ such that $z_n$ returns exactly to the position of $z_0$ (i.e. $n\times40^\circ$ is a multiple of $360^\circ$). [3]
Show complete worked solution
(a) Since $|w|=1$, multiplying by $w$ rotates a point about the origin without changing its modulus. Here $|z_0|=5$ and $\arg(z_0)=0^\circ$, so $z_1$ has modulus $5$ and argument $0^\circ+40^\circ=40^\circ$:
$$z_1 = 5(\cos40^\circ+i\sin40^\circ) \approx 5(0.766)+5(0.643)i$$
$$\boxed{z_1 \approx 3.83+3.21i}$$
(b) By De Moivre's theorem, applying the rotation $4$ times multiplies the argument by $4$: $z_4 = z_0w^4$ has modulus $5$ (unchanged, since each rotation preserves modulus) and argument $4\times40^\circ=160^\circ$:
$$z_4 = 5(\cos160^\circ+i\sin160^\circ) \approx 5(-0.940)+5(0.342)i$$
$$\boxed{z_4 \approx -4.70+1.71i, \qquad \arg(z_4)=160^\circ}$$
(c) After $n$ rotations, $z_n$ has argument $40n^\circ$ (modulo $360^\circ$). The point first returns to its starting position when $40n$ is a multiple of $360$:
$$40n = 360m \implies n = 9m \text{ for the smallest positive integer } m=1$$
$$\boxed{n=9}$$
(Since $40\times9=360$, exactly $9$ rotations of $40^\circ$ bring the point back to $z_0=5$, tracing out $9$ equally spaced points around the circle $|z|=5$, as shown in the diagram.)
QUESTION 49
14 marks
Hard
(a) Using De Moivre's theorem, state the modulus and argument of each of the four fourth roots of $z^4 = 16\,\text{cis}(120^\circ)$ (where $\text{cis}\,\theta=\cos\theta+i\sin\theta$). [4]
(b) Hence express each of the four roots in exact Cartesian form $a+bi$. [6]
(c) Plot the four roots on an Argand diagram, joining them in order to form a quadrilateral. Identify this quadrilateral and find its exact side length. [4]
Show complete worked solution
(a) Let $z=r\,\text{cis}\,\theta$. By De Moivre's theorem, $z^4 = r^4\,\text{cis}(4\theta)$, so:
$$r^4=16 \implies r=2$$
$$4\theta = 120^\circ+360^\circ k \implies \theta = 30^\circ+90^\circ k, \quad k=0,1,2,3$$
$$\boxed{r=2 \text{ for all four roots};\quad \theta = 30^\circ,\ 120^\circ,\ 210^\circ,\ 300^\circ}$$
(b) Using $a=2\cos\theta$, $b=2\sin\theta$ with the exact values of cosine and sine at these standard angles:
$$z_0 = 2(\cos30^\circ+i\sin30^\circ) = 2\left(\frac{\sqrt3}{2}+\frac12i\right) = \sqrt3+i$$
$$z_1 = 2(\cos120^\circ+i\sin120^\circ) = 2\left(-\frac12+\frac{\sqrt3}{2}i\right) = -1+\sqrt3i$$
$$z_2 = 2(\cos210^\circ+i\sin210^\circ) = 2\left(-\frac{\sqrt3}{2}-\frac12i\right) = -\sqrt3-i$$
$$z_3 = 2(\cos300^\circ+i\sin300^\circ) = 2\left(\frac12-\frac{\sqrt3}{2}i\right) = 1-\sqrt3i$$
$$\boxed{z_0=\sqrt3+i,\ \ z_1=-1+\sqrt3i,\ \ z_2=-\sqrt3-i,\ \ z_3=1-\sqrt3i}$$
(c) All four roots have modulus $2$ and are equally spaced by $90^\circ$ around the origin, so joining them in order gives a square inscribed in the circle $|z|=2$, as shown in the diagram.
$$\boxed{\text{Quadrilateral: a square}}$$
The side length is the distance between consecutive roots, e.g. $z_0$ and $z_1$:
$$|z_0-z_1| = \left|(\sqrt3-(-1))+(1-\sqrt3)i\right| = \left|(\sqrt3+1)+(1-\sqrt3)i\right|$$
$$= \sqrt{(\sqrt3+1)^2+(1-\sqrt3)^2} = \sqrt{(4+2\sqrt3)+(4-2\sqrt3)} = \sqrt{8}$$
$$\boxed{\text{Side length} = 2\sqrt2}$$
(This agrees with the general result that a square inscribed in a circle of radius $r$ has side length $r\sqrt2$.)
QUESTION 50
11 marks
Hard
Let $z=x+iy$, where $x,y\in\mathbb{R}$, satisfy $|z-8| = 2|z-2i|$.
(a) By squaring both sides and expanding, show that the locus is a circle by reducing the equation to the completed-square form $(x-a)^2+(y-b)^2=k$. [6]
(b) State the centre and the exact radius of this circle. [3]
(c) Sketch the circle on an Argand diagram, clearly marking its centre. [2]
Show complete worked solution
(a) Squaring both sides of $|z-8|=2|z-2i|$:
$$(x-8)^2+y^2 = 4\left[x^2+(y-2)^2\right]$$
Expanding the left side:
$$x^2-16x+64+y^2$$
Expanding the right side:
$$4x^2+4(y^2-4y+4) = 4x^2+4y^2-16y+16$$
Setting them equal and moving all terms to one side:
$$x^2-16x+64+y^2 = 4x^2+4y^2-16y+16$$
$$0 = 3x^2+3y^2+16x-16y-48$$
Dividing through by $3$:
$$x^2+\frac{16}{3}x+y^2-\frac{16}{3}y-16 = 0$$
Completing the square (using $\left(x+\tfrac83\right)^2 = x^2+\tfrac{16}{3}x+\tfrac{64}{9}$, and similarly for $y$ with $\left(y-\tfrac83\right)^2$):
$$\left(x+\frac83\right)^2-\frac{64}{9}+\left(y-\frac83\right)^2-\frac{64}{9}-16 = 0$$
$$\left(x+\frac83\right)^2+\left(y-\frac83\right)^2 = \frac{64}{9}+\frac{64}{9}+16 = \frac{128+144}{9}$$
$$\boxed{\left(x+\dfrac83\right)^2+\left(y-\dfrac83\right)^2 = \dfrac{272}{9}}$$
This is the equation of a circle, confirming the locus is a circle (an example of an Apollonius circle - the locus of points whose distances from two fixed points are in a fixed ratio, here $2:1$).
(b) Comparing with $(x-a)^2+(y-b)^2=k^2$: the centre has coordinates $\left(-\tfrac83,\tfrac83\right)$, and the radius is:
$$\sqrt{\frac{272}{9}} = \frac{\sqrt{272}}{3} = \frac{\sqrt{16\times17}}{3} = \frac{4\sqrt{17}}{3}$$
$$\boxed{\text{centre } \left(-\dfrac83,\dfrac83\right),\qquad \text{radius } \dfrac{4\sqrt{17}}{3}\ (\approx5.50)}$$
(c) The sketch below shows the circle of radius $\dfrac{4\sqrt{17}}{3}$ centred at $\left(-\dfrac83,\dfrac83\right)$.
Matrices 50 questions
QUESTION 1
4 marks
Easy
Let A = \begin{pmatrix} 2 & -1 \\ 3 & 4 \end{pmatrix} and B = \begin{pmatrix} 0 & 5 \\ -2 & 1 \end{pmatrix}.
(a) Find A + 2B.
(b) Find AB.
Show complete worked solution
(a) Multiplying $B$ by the scalar 2:
$$2B = \begin{pmatrix}0&10\\-4&2\end{pmatrix}$$
Adding $A$ and $2B$ entry by entry:
$$A+2B = \begin{pmatrix}2+0&-1+10\\3-4&4+2\end{pmatrix}$$
$$\boxed{A+2B = \begin{pmatrix}2&9\\-1&6\end{pmatrix}}$$
(b) Using row-by-column multiplication for $AB$:
$$\text{Entry }(1,1):\; 2(0)+(-1)(-2) = 0+2 = 2$$
$$\text{Entry }(1,2):\; 2(5)+(-1)(1) = 10-1 = 9$$
$$\text{Entry }(2,1):\; 3(0)+4(-2) = 0-8 = -8$$
$$\text{Entry }(2,2):\; 3(5)+4(1) = 15+4 = 19$$
$$\boxed{AB = \begin{pmatrix}2&9\\-8&19\end{pmatrix}}$$
QUESTION 2
5 marks
Easy
Consider the system of linear equations:
3x + y = 11
5x + 2y = 18
This can be written as C\vec{x} = \vec{b}, where C = \begin{pmatrix} 3 & 1 \\ 5 & 2 \end{pmatrix}, \vec{x} = \begin{pmatrix} x \\ y \end{pmatrix}, \vec{b} = \begin{pmatrix} 11 \\ 18 \end{pmatrix}.
(a) Find \det(C).
(b) Find C^{-1}.
(c) Hence solve the system for x and y.
Show complete worked solution
(a) Using $\det\begin{pmatrix}a&b\\c&d\end{pmatrix}=ad-bc$:
$$\det(C) = 3(2)-1(5)$$
$$= 6-5$$
$$\boxed{\det(C) = 1}$$
(b) For a $2\times2$ matrix $\begin{pmatrix}a&b\\c&d\end{pmatrix}$, the inverse is $\dfrac{1}{\det}\begin{pmatrix}d&-b\\-c&a\end{pmatrix}$. Since $\det(C)=1$:
$$\boxed{C^{-1} = \begin{pmatrix}2&-1\\-5&3\end{pmatrix}}$$
(c) Using $\vec{x}=C^{-1}\vec{b}$ (either by hand or via the GDC's matrix mode):
$$\vec{x} = \begin{pmatrix}2&-1\\-5&3\end{pmatrix}\begin{pmatrix}11\\18\end{pmatrix}$$
$$= \begin{pmatrix}2(11)-1(18)\\-5(11)+3(18)\end{pmatrix}$$
$$= \begin{pmatrix}22-18\\-55+54\end{pmatrix}$$
$$= \begin{pmatrix}4\\-1\end{pmatrix}$$
Check: $3(4)+(-1)=11$ $\checkmark$; $5(4)+2(-1)=20-2=18$ $\checkmark$.
$$\boxed{x=4, \quad y=-1}$$
QUESTION 3
6 marks
Medium
Let D = \begin{pmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 5 & 6 & 0 \end{pmatrix}.
(a) Find \det(D) using cofactor expansion, showing your method.
(b) State, with a reason, whether D is invertible.
Show complete worked solution
(a) Expanding $\det(D)$ along the first row:
$$\det(D) = 1\begin{vmatrix}1&4\\6&0\end{vmatrix}-2\begin{vmatrix}0&4\\5&0\end{vmatrix}+3\begin{vmatrix}0&1\\5&6\end{vmatrix}$$
Evaluating each $2\times2$ determinant:
$$= 1\big(1(0)-4(6)\big)-2\big(0(0)-4(5)\big)+3\big(0(6)-1(5)\big)$$
$$= 1(0-24)-2(0-20)+3(0-5)$$
$$= -24-(-40)+(-15)$$
$$= -24+40-15$$
$$\boxed{\det(D) = 1}$$
(b) Since $\det(D)=1\ne0$:
$$\boxed{D \text{ is invertible, since its determinant is non-zero}}$$
QUESTION 4
7 marks
Medium
Solve the following system of linear equations using matrix methods (for example, by finding the inverse of the coefficient matrix or by using your GDC's matrix functions):
x + 2y + z = 3
2x - y + 3z = 14
3x + y - 2z = -1
Show complete worked solution
Write the system as $A\vec{x}=\vec{b}$, where
$$A = \begin{pmatrix}1&2&1\\2&-1&3\\3&1&-2\end{pmatrix}, \quad \vec{b} = \begin{pmatrix}3\\14\\-1\end{pmatrix}$$
First find $\det(A)$ to confirm a unique solution exists. Expanding along the first row:
$$\det(A) = 1\big((-1)(-2)-3(1)\big)-2\big(2(-2)-3(3)\big)+1\big(2(1)-(-1)(3)\big)$$
$$= 1(2-3)-2(-4-9)+1(2+3)$$
$$= -1-2(-13)+5$$
$$= -1+26+5$$
$$= 30$$
Since $\det(A)=30\ne0$, $A$ is invertible, so the system has a unique solution given by $\vec{x}=A^{-1}\vec{b}$.
Using the GDC's matrix inverse and matrix multiplication functions (entering $A$ and $\vec{b}$ and computing $A^{-1}\vec{b}$):
$$\vec{x} = \begin{pmatrix}2\\-1\\3\end{pmatrix}$$
Check: $x+2y+z=2-2+3=3$ $\checkmark$; $2x-y+3z=4+1+9=14$ $\checkmark$; $3x+y-2z=6-1-6=-1$ $\checkmark$.
$$\boxed{x=2, \quad y=-1, \quad z=3}$$
QUESTION 5
12 marks
Hard
A furniture workshop makes three products each week: chairs, tables and shelving units. Producing x chairs, y tables and z shelving units requires:
- (2x + 3y + z) board-feet of wood
- (3x + 2y + 2z) hours of labour
- (x + y + z) litres of paint
In a particular week, the workshop used exactly 62 board-feet of wood, 72 hours of labour and 30 litres of paint.
(a) Write this information as a matrix equation M\vec{v} = \vec{b}, where \vec{v} = \begin{pmatrix} x \\ y \\ z \end{pmatrix}, stating M and \vec{b} explicitly.
(b) Find \det(M), and hence state, with a reason, whether the system has a unique solution.
(c) Use your GDC (or M^{-1}) to find the number of chairs, tables and shelving units produced that week.
Show complete worked solution
(a) Reading off the coefficients of $x$, $y$, $z$ in each resource constraint:
$$\boxed{M = \begin{pmatrix}2&3&1\\3&2&2\\1&1&1\end{pmatrix}, \quad \vec{b} = \begin{pmatrix}62\\72\\30\end{pmatrix}, \quad M\vec{v} = \vec{b}}$$
(b) Expanding $\det(M)$ along the first row:
$$\det(M) = 2\big(2(1)-2(1)\big)-3\big(3(1)-2(1)\big)+1\big(3(1)-2(1)\big)$$
$$= 2(2-2)-3(3-2)+1(3-2)$$
$$= 2(0)-3(1)+1(1)$$
$$= -3+1$$
$$\boxed{\det(M) = -2}$$
Since $\det(M)=-2\ne0$, $M$ is invertible, so the system has a unique solution.
(c) Using the GDC's matrix inverse and matrix multiplication functions to compute $\vec{v}=M^{-1}\vec{b}$:
$$\vec{v} = \begin{pmatrix}12\\10\\8\end{pmatrix}$$
Check: wood: $2(12)+3(10)+8=24+30+8=62$ $\checkmark$; labour: $3(12)+2(10)+2(8)=36+20+16=72$ $\checkmark$; paint: $12+10+8=30$ $\checkmark$.
$$\boxed{\text{12 chairs, 10 tables, 8 shelving units}}$$
QUESTION 6
5 marks
Easy
Find the eigenvalues of the matrix A = \begin{pmatrix} 4 & 1 \\ 2 & 3 \end{pmatrix}.
Show complete worked solution
The eigenvalues $\lambda$ satisfy the characteristic equation $\det(A-\lambda I)=0$:
$$(4-\lambda)(3-\lambda)-(1)(2) = 0$$
Expanding:
$$12-4\lambda-3\lambda+\lambda^2-2 = 0$$
$$\lambda^2-7\lambda+10 = 0$$
Factorising:
$$(\lambda-5)(\lambda-2) = 0$$
$$\boxed{\lambda = 5 \text{ or } \lambda = 2}$$
QUESTION 7
5 marks
Easy
Find the eigenvalues of the matrix B = \begin{pmatrix} 7 & -2 \\ 4 & 1 \end{pmatrix}.
Show complete worked solution
The eigenvalues satisfy $\det(B-\lambda I)=0$:
$$(7-\lambda)(1-\lambda)-(-2)(4) = 0$$
Expanding $(7-\lambda)(1-\lambda)=7-7\lambda-\lambda+\lambda^2=\lambda^2-8\lambda+7$:
$$\lambda^2-8\lambda+7+8 = 0$$
$$\lambda^2-8\lambda+15 = 0$$
Factorising:
$$(\lambda-5)(\lambda-3) = 0$$
$$\boxed{\lambda = 5 \text{ or } \lambda = 3}$$
QUESTION 8
7 marks
Medium
Consider the matrix C = \begin{pmatrix} 5 & 4 \\ 1 & 2 \end{pmatrix}.
(a) Find the eigenvalues of C.
(b) Find a corresponding eigenvector for each eigenvalue.
Show complete worked solution
(a) The eigenvalues satisfy $\det(C-\lambda I)=0$:
$$(5-\lambda)(2-\lambda)-4(1) = 0$$
$$10-7\lambda+\lambda^2-4 = 0$$
$$\lambda^2-7\lambda+6 = 0$$
$$(\lambda-6)(\lambda-1) = 0$$
$$\boxed{\lambda = 6 \text{ or } \lambda = 1}$$
(b) For $\lambda=6$, solve $(C-6I)\vec{v}=\vec{0}$:
$$\begin{pmatrix}-1&4\\1&-4\end{pmatrix}\vec{v} = \vec{0} \implies -x+4y=0 \implies x=4y$$
Taking $y=1$:
$$\boxed{\text{Eigenvector for } \lambda=6: \begin{pmatrix}4\\1\end{pmatrix}}$$
For $\lambda=1$, solve $(C-I)\vec{v}=\vec{0}$:
$$\begin{pmatrix}4&4\\1&1\end{pmatrix}\vec{v} = \vec{0} \implies x+y=0 \implies x=-y$$
Taking $y=1$:
$$\boxed{\text{Eigenvector for } \lambda=1: \begin{pmatrix}-1\\1\end{pmatrix}}$$
(or any non-zero scalar multiple of each)
QUESTION 9
8 marks
Medium
The matrix D = \begin{pmatrix} 3 & -2 \\ 1 & 0 \end{pmatrix} is diagonalizable.
(a) Find the eigenvalues of D.
(b) Find an eigenvector corresponding to each eigenvalue.
(c) Write D in the form D = P\Lambda P^{-1}, stating the matrices P, \Lambda and P^{-1} explicitly.
Show complete worked solution
(a) The eigenvalues satisfy $\det(D-\lambda I)=0$:
$$(3-\lambda)(-\lambda)-(-2)(1) = 0$$
$$-3\lambda+\lambda^2+2 = 0$$
$$\lambda^2-3\lambda+2 = 0$$
$$(\lambda-1)(\lambda-2) = 0$$
$$\boxed{\lambda = 1 \text{ or } \lambda = 2}$$
(b) For $\lambda=1$: $(D-I)\vec{v}=\vec{0}$:
$$\begin{pmatrix}2&-2\\1&-1\end{pmatrix}\vec{v}=\vec{0} \implies x-y=0 \implies x=y$$
Eigenvector: $\begin{pmatrix}1\\1\end{pmatrix}$.
For $\lambda=2$: $(D-2I)\vec{v}=\vec{0}$:
$$\begin{pmatrix}1&-2\\1&-2\end{pmatrix}\vec{v}=\vec{0} \implies x-2y=0 \implies x=2y$$
Eigenvector: $\begin{pmatrix}2\\1\end{pmatrix}$.
$$\boxed{\lambda=1: \begin{pmatrix}1\\1\end{pmatrix}; \quad \lambda=2: \begin{pmatrix}2\\1\end{pmatrix}}$$
(c) Let the columns of $P$ be the eigenvectors, ordered to match $\lambda=1,2$:
$$P = \begin{pmatrix}1&2\\1&1\end{pmatrix}, \quad \Lambda = \begin{pmatrix}1&0\\0&2\end{pmatrix}$$
Using $\det\begin{pmatrix}a&b\\c&d\end{pmatrix}=ad-bc$:
$$\det(P) = 1(1)-2(1) = -1$$
Using $P^{-1}=\dfrac{1}{\det(P)}\begin{pmatrix}d&-b\\-c&a\end{pmatrix}$:
$$P^{-1} = \frac{1}{-1}\begin{pmatrix}1&-2\\-1&1\end{pmatrix}$$
$$= \begin{pmatrix}-1&2\\1&-1\end{pmatrix}$$
$$\boxed{D = P\Lambda P^{-1} = \begin{pmatrix}1&2\\1&1\end{pmatrix}\begin{pmatrix}1&0\\0&2\end{pmatrix}\begin{pmatrix}-1&2\\1&-1\end{pmatrix}}$$
(This can be verified by multiplying out $P\Lambda P^{-1}$ to recover $D$.)
QUESTION 10
18 marks
Hard
Two towns, A and B, have a combined population that stays constant over time due to annual migration between them. Each year, 20% of town A's population moves to town B, and 10% of town B's population moves to town A (the rest of each town's population stays put). Let a_n and b_n be the populations of A and B respectively, n years from now, so that \begin{pmatrix} a_{n+1} \\ b_{n+1} \end{pmatrix} = M \begin{pmatrix} a_n \\ b_n \end{pmatrix}.
(a) Write down the transition matrix M.
(b) Find the eigenvalues of M and a corresponding eigenvector for each.
(c) Hence write M in diagonalized form M = P\Lambda P^{-1}, stating P, \Lambda and P^{-1} explicitly.
(d) Given that the initial populations are a_0 = 6000 and b_0 = 4000, find expressions for a_n and b_n in terms of n, and state the long-term (steady-state) populations of towns A and B.
Show complete worked solution
(a) Each year town A retains $80\%$ of its own population and gains $10\%$ of town B's population; town B retains $90\%$ of its own population and gains $20\%$ of town A's population, so:
$$\boxed{M = \begin{pmatrix}0.8&0.1\\0.2&0.9\end{pmatrix}}$$
(b) The eigenvalues satisfy $\det(M-\lambda I)=0$:
$$(0.8-\lambda)(0.9-\lambda)-(0.1)(0.2) = 0$$
$$\lambda^2-1.7\lambda+0.72-0.02 = 0$$
$$\lambda^2-1.7\lambda+0.7 = 0$$
Using the quadratic formula:
$$\lambda = \frac{1.7\pm\sqrt{1.7^2-4(0.7)}}{2} = \frac{1.7\pm\sqrt{0.09}}{2} = \frac{1.7\pm0.3}{2}$$
$$\boxed{\lambda = 1 \text{ or } \lambda = 0.7}$$
For $\lambda=1$: $(M-I)\vec{v}=\vec{0}$:
$$\begin{pmatrix}-0.2&0.1\\0.2&-0.1\end{pmatrix}\vec{v}=\vec{0} \implies -0.2x+0.1y=0 \implies y=2x$$
Eigenvector: $\begin{pmatrix}1\\2\end{pmatrix}$.
For $\lambda=0.7$: $(M-0.7I)\vec{v}=\vec{0}$:
$$\begin{pmatrix}0.1&0.1\\0.2&0.2\end{pmatrix}\vec{v}=\vec{0} \implies x+y=0 \implies y=-x$$
Eigenvector: $\begin{pmatrix}1\\-1\end{pmatrix}$.
$$\boxed{\lambda=1: \begin{pmatrix}1\\2\end{pmatrix}; \quad \lambda=0.7: \begin{pmatrix}1\\-1\end{pmatrix}}$$
(c) Let the columns of $P$ be the eigenvectors, ordered to match $\lambda=1,0.7$:
$$P = \begin{pmatrix}1&1\\2&-1\end{pmatrix}, \quad \Lambda = \begin{pmatrix}1&0\\0&0.7\end{pmatrix}$$
$$\det(P) = 1(-1)-1(2) = -3$$
$$P^{-1} = \frac{1}{-3}\begin{pmatrix}-1&-1\\-2&1\end{pmatrix}$$
$$= \begin{pmatrix}1/3&1/3\\2/3&-1/3\end{pmatrix}$$
$$\boxed{M = P\Lambda P^{-1}, \text{ with } P=\begin{pmatrix}1&1\\2&-1\end{pmatrix}, \; \Lambda=\begin{pmatrix}1&0\\0&0.7\end{pmatrix}, \; P^{-1}=\begin{pmatrix}1/3&1/3\\2/3&-1/3\end{pmatrix}}$$
(d) Using $\vec{x}_n=M^n\vec{x}_0=P\Lambda^nP^{-1}\vec{x}_0$ with $\vec{x}_0=\begin{pmatrix}6000\\4000\end{pmatrix}$:
First compute $P^{-1}\vec{x}_0$:
$$P^{-1}\vec{x}_0 = \begin{pmatrix}1/3&1/3\\2/3&-1/3\end{pmatrix}\begin{pmatrix}6000\\4000\end{pmatrix}$$
$$= \begin{pmatrix}\frac13(6000)+\frac13(4000)\\\frac23(6000)-\frac13(4000)\end{pmatrix}$$
$$= \begin{pmatrix}10000/3\\8000/3\end{pmatrix}$$
Apply $\Lambda^n$:
$$\Lambda^nP^{-1}\vec{x}_0 = \begin{pmatrix}1^n\cdot10000/3\\(0.7)^n\cdot8000/3\end{pmatrix} = \begin{pmatrix}10000/3\\(8000/3)(0.7)^n\end{pmatrix}$$
Finally apply $P$:
$$a_n = 1\cdot\frac{10000}{3}+1\cdot\frac{8000}{3}(0.7)^n = \frac{10000}{3}+\frac{8000}{3}(0.7)^n$$
$$b_n = 2\cdot\frac{10000}{3}-1\cdot\frac{8000}{3}(0.7)^n = \frac{20000}{3}-\frac{8000}{3}(0.7)^n$$
Check $n=0$: $a_0=10000/3+8000/3=18000/3=6000$ $\checkmark$; $b_0=20000/3-8000/3=12000/3=4000$ $\checkmark$.
As $n\to\infty$, since $0<0.7<1$, $(0.7)^n\to0$, so:
$$a_n \to \frac{10000}{3} \approx 3333, \quad b_n \to \frac{20000}{3} \approx 6667$$
$$\boxed{a_n = \frac{10000}{3}+\frac{8000}{3}(0.7)^n, \quad b_n = \frac{20000}{3}-\frac{8000}{3}(0.7)^n}$$
$$\boxed{\text{Long-term populations: town A} \to \text{approximately 3333, town B} \to \text{approximately 6667 (ratio 1:2, matching the eigenvector for }\lambda=1\text{, total population 10000 conserved)}}$$
QUESTION 11
5 marks
Easy
Let $A = \begin{pmatrix} 5 & -3 \\ 2 & 4 \end{pmatrix}$ and $B = \begin{pmatrix} 1 & 4 \\ -2 & 3 \end{pmatrix}$.
(a) Find $A - 2B$. [3]
(b) Find $2A + B$. [2]
Show complete worked solution
(a) Multiplying $B$ by the scalar 2:
$$2B = \begin{pmatrix}2&8\\-4&6\end{pmatrix}$$
Subtracting entry by entry:
$$A-2B = \begin{pmatrix}5-2&-3-8\\2-(-4)&4-6\end{pmatrix}$$
$$\boxed{A-2B = \begin{pmatrix}3&-11\\6&-2\end{pmatrix}}$$
(b) Multiplying $A$ by the scalar 2:
$$2A = \begin{pmatrix}10&-6\\4&8\end{pmatrix}$$
Adding $2A$ and $B$ entry by entry:
$$2A+B = \begin{pmatrix}10+1&-6+4\\4-2&8+3\end{pmatrix}$$
$$\boxed{2A+B = \begin{pmatrix}11&-2\\2&11\end{pmatrix}}$$
QUESTION 12
4 marks
Easy
Let $A = \begin{pmatrix} 3 & 2 \\ -1 & 4 \end{pmatrix}$ and $B = \begin{pmatrix} 2 & 0 \\ 1 & -3 \end{pmatrix}$.
Find $AB$.
Show complete worked solution
Using row-by-column multiplication:
$$\text{Entry }(1,1):\; 3(2)+2(1) = 6+2 = 8$$
$$\text{Entry }(1,2):\; 3(0)+2(-3) = 0-6 = -6$$
$$\text{Entry }(2,1):\; -1(2)+4(1) = -2+4 = 2$$
$$\text{Entry }(2,2):\; -1(0)+4(-3) = 0-12 = -12$$
$$\boxed{AB = \begin{pmatrix}8&-6\\2&-12\end{pmatrix}}$$
QUESTION 13
4 marks
Easy
Consider the matrix $C = \begin{pmatrix} 6 & 4 \\ 3 & 2 \end{pmatrix}$.
(a) Find $\det(C)$. [2]
(b) State, with a reason, whether $C$ has an inverse. [2]
Show complete worked solution
(a) Using $\det\begin{pmatrix}a&b\\c&d\end{pmatrix}=ad-bc$:
$$\det(C) = 6(2)-4(3)$$
$$= 12-12$$
$$\boxed{\det(C) = 0}$$
(b) Since $\det(C)=0$, the matrix is singular, so:
$$\boxed{C \text{ does not have an inverse, since } \det(C)=0}$$
QUESTION 14
4 marks
Easy
Find the inverse of the matrix $D = \begin{pmatrix} 4 & 7 \\ 1 & 2 \end{pmatrix}$.
Show complete worked solution
First find $\det(D)$:
$$\det(D) = 4(2)-7(1) = 8-7 = 1$$
For a $2\times2$ matrix $\begin{pmatrix}a&b\\c&d\end{pmatrix}$, the inverse is $\dfrac{1}{\det}\begin{pmatrix}d&-b\\-c&a\end{pmatrix}$. Since $\det(D)=1$:
$$\boxed{D^{-1} = \begin{pmatrix}2&-7\\-1&4\end{pmatrix}}$$
QUESTION 15
5 marks
Easy
Let $F = \begin{pmatrix} 2 & 0 & 1 \\ 3 & 1 & -1 \\ 0 & 4 & 2 \end{pmatrix}$.
Find $\det(F)$ using cofactor expansion along the first row, showing your method.
Show complete worked solution
Expanding $\det(F)$ along the first row:
$$\det(F) = 2\begin{vmatrix}1&-1\\4&2\end{vmatrix}-0\begin{vmatrix}3&-1\\0&2\end{vmatrix}+1\begin{vmatrix}3&1\\0&4\end{vmatrix}$$
Evaluating each $2\times2$ determinant:
$$= 2\big(1(2)-(-1)(4)\big)-0+1\big(3(4)-1(0)\big)$$
$$= 2(2+4)+1(12-0)$$
$$= 2(6)+12$$
$$= 12+12$$
$$\boxed{\det(F) = 24}$$
QUESTION 16
5 marks
Easy
Given that $3X = A - B$, where $A = \begin{pmatrix} 9 & 6 \\ -3 & 12 \end{pmatrix}$ and $B = \begin{pmatrix} 3 & 0 \\ 3 & 6 \end{pmatrix}$, find the matrix $X$.
Show complete worked solution
First subtract $B$ from $A$ entry by entry:
$$A-B = \begin{pmatrix}9-3&6-0\\-3-3&12-6\end{pmatrix} = \begin{pmatrix}6&6\\-6&6\end{pmatrix}$$
Since $3X=A-B$, dividing every entry by the scalar 3:
$$X = \frac{1}{3}\begin{pmatrix}6&6\\-6&6\end{pmatrix}$$
$$\boxed{X = \begin{pmatrix}2&2\\-2&2\end{pmatrix}}$$
QUESTION 17
5 marks
Easy
A carpentry supplier records, for two furniture kits, the amount of screws, brackets and hinges needed per kit as the matrix $P = \begin{pmatrix} 2 & 3 & 1 \\ 4 & 1 & 2 \end{pmatrix}$, where the columns represent screws, brackets and hinges respectively and the rows represent Kit 1 and Kit 2.
A single unit of screws, brackets and hinges costs (in dollars) $5$, $2$ and $3$ respectively, given by the column vector $q = \begin{pmatrix} 5 \\ 2 \\ 3 \end{pmatrix}$.
Find $Pq$, and interpret what this matrix product represents.
Show complete worked solution
Using row-by-column multiplication of the $2\times3$ matrix $P$ by the $3\times1$ vector $q$:
$$\text{Row 1 (Kit 1):}\; 2(5)+3(2)+1(3) = 10+6+3 = 19$$
$$\text{Row 2 (Kit 2):}\; 4(5)+1(2)+2(3) = 20+2+6 = 28$$
$$\boxed{Pq = \begin{pmatrix}19\\28\end{pmatrix}}$$
This represents the total material cost, in dollars, of Kit 1 ($\$19$) and Kit 2 ($\$28$).
QUESTION 18
4 marks
Easy
The point $P(3, 5)$ is reflected in the $x$-axis using the transformation matrix $M = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$.
Find the coordinates of the image point $P'$.
Show complete worked solution
Applying $M$ to the position vector of $P$:
$$\begin{pmatrix}1&0\\0&-1\end{pmatrix}\begin{pmatrix}3\\5\end{pmatrix} = \begin{pmatrix}1(3)+0(5)\\0(3)+(-1)(5)\end{pmatrix} = \begin{pmatrix}3\\-5\end{pmatrix}$$
$$\boxed{P'(3,-5)}$$
QUESTION 19
4 marks
Easy
The point $Q(-2, 6)$ is rotated $180$ degrees about the origin using the transformation matrix $R = \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}$.
Find the coordinates of the image point $Q'$.
Show complete worked solution
Applying $R$ to the position vector of $Q$:
$$\begin{pmatrix}-1&0\\0&-1\end{pmatrix}\begin{pmatrix}-2\\6\end{pmatrix} = \begin{pmatrix}-1(-2)+0(6)\\0(-2)+(-1)(6)\end{pmatrix} = \begin{pmatrix}2\\-6\end{pmatrix}$$
$$\boxed{Q'(2,-6)}$$
QUESTION 20
4 marks
Easy
The point $S(2, -1)$ undergoes an enlargement with centre the origin and scale factor 3, represented by the matrix $E = \begin{pmatrix} 3 & 0 \\ 0 & 3 \end{pmatrix}$.
Find the coordinates of the image point $S'$.
Show complete worked solution
Applying $E$ to the position vector of $S$:
$$\begin{pmatrix}3&0\\0&3\end{pmatrix}\begin{pmatrix}2\\-1\end{pmatrix} = \begin{pmatrix}3(2)+0(-1)\\0(2)+3(-1)\end{pmatrix} = \begin{pmatrix}6\\-3\end{pmatrix}$$
$$\boxed{S'(6,-3)}$$
QUESTION 21
4 marks
Easy
Let $A = \begin{pmatrix} 2 & 1 \\ 0 & 3 \end{pmatrix}$ and $B = \begin{pmatrix} 1 & 4 \\ -1 & 2 \end{pmatrix}$.
(a) Find $AB$. [2]
(b) Find $BA$, and hence state whether matrix multiplication is commutative for $A$ and $B$. [2]
Show complete worked solution
(a) Multiplying $A$ by $B$, row by column:
$$AB = \begin{pmatrix}2&1\\0&3\end{pmatrix}\begin{pmatrix}1&4\\-1&2\end{pmatrix} = \begin{pmatrix}2(1)+1(-1)&2(4)+1(2)\\0(1)+3(-1)&0(4)+3(2)\end{pmatrix}$$
$$\boxed{AB = \begin{pmatrix}1&10\\-3&6\end{pmatrix}}$$
(b) Multiplying $B$ by $A$, row by column:
$$BA = \begin{pmatrix}1&4\\-1&2\end{pmatrix}\begin{pmatrix}2&1\\0&3\end{pmatrix} = \begin{pmatrix}1(2)+4(0)&1(1)+4(3)\\-1(2)+2(0)&-1(1)+2(3)\end{pmatrix}$$
$$\boxed{BA = \begin{pmatrix}2&13\\-2&5\end{pmatrix}}$$
Since $AB = \begin{pmatrix}1&10\\-3&6\end{pmatrix} \ne \begin{pmatrix}2&13\\-2&5\end{pmatrix} = BA$:
$$\boxed{\text{Matrix multiplication is not commutative for } A \text{ and } B \text{, since } AB \ne BA}$$
(In general, matrix multiplication is not commutative: $MN \ne NM$ for most pairs of matrices $M$, $N$, even when both products are defined.)
QUESTION 22
5 marks
Easy
Let $A = \begin{pmatrix} 4 & -2 \\ 3 & 5 \end{pmatrix}$, and let $I$ and $O$ denote the $2\times2$ identity and zero matrices respectively.
(a) Write down $I$ and $O$. [1]
(b) Find $AI$ and $IA$, and hence verify that $AI = IA = A$. [2]
(c) Find $A + O$ and $A - A$, stating each result. [2]
Show complete worked solution
(a)
$$I = \begin{pmatrix}1&0\\0&1\end{pmatrix}, \quad O = \begin{pmatrix}0&0\\0&0\end{pmatrix}$$
(b) Multiplying $A$ by $I$:
$$AI = \begin{pmatrix}4&-2\\3&5\end{pmatrix}\begin{pmatrix}1&0\\0&1\end{pmatrix} = \begin{pmatrix}4(1)+(-2)(0)&4(0)+(-2)(1)\\3(1)+5(0)&3(0)+5(1)\end{pmatrix} = \begin{pmatrix}4&-2\\3&5\end{pmatrix}$$
Multiplying $I$ by $A$:
$$IA = \begin{pmatrix}1&0\\0&1\end{pmatrix}\begin{pmatrix}4&-2\\3&5\end{pmatrix} = \begin{pmatrix}1(4)+0(3)&1(-2)+0(5)\\0(4)+1(3)&0(-2)+1(5)\end{pmatrix} = \begin{pmatrix}4&-2\\3&5\end{pmatrix}$$
$$\boxed{AI = IA = \begin{pmatrix}4&-2\\3&5\end{pmatrix} = A}$$
(The identity matrix $I$ leaves every matrix unchanged under multiplication, on either side.)
(c) Adding the zero matrix to $A$ changes nothing, and subtracting $A$ from itself gives every entry zero:
$$A+O = \begin{pmatrix}4&-2\\3&5\end{pmatrix}, \quad A-A = \begin{pmatrix}0&0\\0&0\end{pmatrix}$$
$$\boxed{A+O = A, \quad A-A = O}$$
QUESTION 23
5 marks
Easy
Find the eigenvalues of the matrix $A = \begin{pmatrix} 6 & 2 \\ 1 & 5 \end{pmatrix}$.
Show complete worked solution
The eigenvalues $\lambda$ satisfy the characteristic equation $\det(A-\lambda I)=0$:
$$(6-\lambda)(5-\lambda)-(2)(1) = 0$$
Expanding:
$$30-6\lambda-5\lambda+\lambda^2-2 = 0$$
$$\lambda^2-11\lambda+28 = 0$$
Factorising:
$$(\lambda-7)(\lambda-4) = 0$$
$$\boxed{\lambda = 7 \text{ or } \lambda = 4}$$
QUESTION 24
5 marks
Easy
Find the eigenvalues of the matrix $B = \begin{pmatrix} 2 & 3 \\ 4 & 1 \end{pmatrix}$.
Show complete worked solution
The eigenvalues satisfy $\det(B-\lambda I)=0$:
$$(2-\lambda)(1-\lambda)-(3)(4) = 0$$
Expanding $(2-\lambda)(1-\lambda)=2-2\lambda-\lambda+\lambda^2=\lambda^2-3\lambda+2$:
$$\lambda^2-3\lambda+2-12 = 0$$
$$\lambda^2-3\lambda-10 = 0$$
Factorising:
$$(\lambda-5)(\lambda+2) = 0$$
$$\boxed{\lambda = 5 \text{ or } \lambda = -2}$$
QUESTION 25
4 marks
Easy
Show that $\vec{v} = \begin{pmatrix} 2 \\ 1 \end{pmatrix}$ is an eigenvector of the matrix $M = \begin{pmatrix} 4 & -2 \\ 1 & 1 \end{pmatrix}$, and state the corresponding eigenvalue.
Show complete worked solution
Computing $M\vec{v}$:
$$M\vec{v} = \begin{pmatrix}4&-2\\1&1\end{pmatrix}\begin{pmatrix}2\\1\end{pmatrix} = \begin{pmatrix}4(2)+(-2)(1)\\1(2)+1(1)\end{pmatrix} = \begin{pmatrix}8-2\\2+1\end{pmatrix} = \begin{pmatrix}6\\3\end{pmatrix}$$
Since $\begin{pmatrix}6\\3\end{pmatrix} = 3\begin{pmatrix}2\\1\end{pmatrix}$, we have $M\vec{v}=3\vec{v}$, confirming $\vec{v}$ is an eigenvector of $M$.
$$\boxed{\text{Corresponding eigenvalue: } \lambda = 3}$$
QUESTION 26
4 marks
Easy
The matrix $G = \begin{pmatrix} 3 & 2 & -1 \\ 0 & 4 & 5 \\ 0 & 0 & -2 \end{pmatrix}$ is upper triangular.
Find $\det(G)$, stating the property of triangular matrices that you use.
Show complete worked solution
For an upper (or lower) triangular matrix, the determinant is equal to the product of the entries on the leading diagonal, since every cofactor expansion term involving an off-diagonal entry below (or above) the diagonal is multiplied by a minor containing a full row or column of zeros.
$$\det(G) = 3\times4\times(-2)$$
$$\boxed{\det(G) = -24}$$
QUESTION 27
7 marks
Medium
Let $A = \begin{pmatrix} 3 & 5 \\ 1 & 2 \end{pmatrix}$.
(a) Find $\det(A)$. [2]
(b) Find $A^{-1}$. [2]
(c) Verify, by matrix multiplication, that $AA^{-1}=I$. [3]
Show complete worked solution
(a) $$\det(A) = 3(2)-5(1) = 6-5$$
$$\boxed{\det(A) = 1}$$
(b) Since $\det(A)=1$:
$$\boxed{A^{-1} = \begin{pmatrix}2&-5\\-1&3\end{pmatrix}}$$
(c) Computing $AA^{-1}$ by row-by-column multiplication:
$$\text{Entry }(1,1):\; 3(2)+5(-1) = 6-5 = 1$$
$$\text{Entry }(1,2):\; 3(-5)+5(3) = -15+15 = 0$$
$$\text{Entry }(2,1):\; 1(2)+2(-1) = 2-2 = 0$$
$$\text{Entry }(2,2):\; 1(-5)+2(3) = -5+6 = 1$$
$$AA^{-1} = \begin{pmatrix}1&0\\0&1\end{pmatrix} = I$$
$$\boxed{AA^{-1}=I \text{, as required}}$$
QUESTION 28
6 marks
Medium
Let $H = \begin{pmatrix} 1 & -2 & 3 \\ 0 & 4 & -1 \\ 2 & 1 & 5 \end{pmatrix}$.
Find $\det(H)$ using cofactor expansion along the first column, showing your method.
Show complete worked solution
Expanding $\det(H)$ along the first column (using signs $+,-,+$ for the entries $1,0,2$):
$$\det(H) = 1\begin{vmatrix}4&-1\\1&5\end{vmatrix}-0\begin{vmatrix}-2&3\\1&5\end{vmatrix}+2\begin{vmatrix}-2&3\\4&-1\end{vmatrix}$$
Evaluating each $2\times2$ determinant:
$$= 1\big(4(5)-(-1)(1)\big)-0+2\big((-2)(-1)-3(4)\big)$$
$$= 1(20+1)+2(2-12)$$
$$= 21+2(-10)$$
$$= 21-20$$
$$\boxed{\det(H) = 1}$$
QUESTION 29
6 marks
Medium
Let $K = \begin{pmatrix} 2 & 1 & 1 \\ 1 & 3 & 2 \\ 1 & 0 & 1 \end{pmatrix}$.
Use your GDC's matrix mode to find $K^{-1}$.
Show complete worked solution
Entering $K$ into the GDC's matrix editor and using the matrix inverse function:
$$K^{-1} = \begin{pmatrix}0.75&-0.25&-0.25\\0.25&0.25&-0.75\\-0.75&0.25&1.25\end{pmatrix}$$
which can equivalently be written with a common denominator of 4:
$$\boxed{K^{-1} = \frac{1}{4}\begin{pmatrix}3&-1&-1\\1&1&-3\\-3&1&5\end{pmatrix}}$$
QUESTION 30
6 marks
Medium
Solve the following system of linear equations using matrix methods:
2x + y = 13
x - 3y = -4
Show complete worked solution
Write the system as $C\vec{x}=\vec{b}$, where
$$C = \begin{pmatrix}2&1\\1&-3\end{pmatrix}, \quad \vec{x}=\begin{pmatrix}x\\y\end{pmatrix}, \quad \vec{b} = \begin{pmatrix}13\\-4\end{pmatrix}$$
First find $\det(C)$:
$$\det(C) = 2(-3)-1(1) = -6-1 = -7$$
Since $\det(C)\ne0$, $C$ is invertible. Using $C^{-1}=\dfrac{1}{\det}\begin{pmatrix}d&-b\\-c&a\end{pmatrix}$:
$$C^{-1} = \frac{1}{-7}\begin{pmatrix}-3&-1\\-1&2\end{pmatrix} = \begin{pmatrix}3/7&1/7\\1/7&-2/7\end{pmatrix}$$
Then $\vec{x}=C^{-1}\vec{b}$ (or by using the GDC's matrix functions directly):
$$\vec{x} = \begin{pmatrix}3/7&1/7\\1/7&-2/7\end{pmatrix}\begin{pmatrix}13\\-4\end{pmatrix} = \begin{pmatrix}\frac{3(13)+1(-4)}{7}\\\frac{1(13)-2(-4)}{7}\end{pmatrix} = \begin{pmatrix}\frac{39-4}{7}\\\frac{13+8}{7}\end{pmatrix} = \begin{pmatrix}35/7\\21/7\end{pmatrix}$$
Check: $2(5)+3=13$ $\checkmark$; $5-3(3)=5-9=-4$ $\checkmark$.
$$\boxed{x=5, \quad y=3}$$
QUESTION 31
7 marks
Medium
A vending company stocks three snacks: chips, biscuits and nuts. If $x$, $y$ and $z$ represent the number of packets of chips, biscuits and nuts respectively restocked into a machine, the following relationships hold, based on the shelf space, weight and cost limits of the machine:
x + y + z = 5
2x - y + 3z = 20
x + 2y - z = -5
(a) Write this system as $A\vec{v}=\vec{b}$, stating $A$ and $\vec{b}$ explicitly. [2]
(b) Show that $\det(A) = 5$. [3]
(c) Use your GDC (or $A^{-1}$) to find the number of packets of chips, biscuits and nuts restocked. [2]
Show complete worked solution
(a) Reading off the coefficients of $x$, $y$, $z$:
$$\boxed{A = \begin{pmatrix}1&1&1\\2&-1&3\\1&2&-1\end{pmatrix}, \quad \vec{b} = \begin{pmatrix}5\\20\\-5\end{pmatrix}}$$
(b) Expanding $\det(A)$ along the first row:
$$\det(A) = 1\begin{vmatrix}-1&3\\2&-1\end{vmatrix}-1\begin{vmatrix}2&3\\1&-1\end{vmatrix}+1\begin{vmatrix}2&-1\\1&2\end{vmatrix}$$
$$= 1\big((-1)(-1)-3(2)\big)-1\big(2(-1)-3(1)\big)+1\big(2(2)-(-1)(1)\big)$$
$$= 1(1-6)-1(-2-3)+1(4+1)$$
$$= -5-(-5)+5$$
$$= -5+5+5$$
$$\boxed{\det(A) = 5 \text{, as required}}$$
(c) Using the GDC's matrix inverse and matrix multiplication functions to compute $\vec{v}=A^{-1}\vec{b}$:
$$\vec{v} = \begin{pmatrix}3\\-2\\4\end{pmatrix}$$
Check: $3-2+4=5$ $\checkmark$; $2(3)-(-2)+3(4)=6+2+12=20$ $\checkmark$; $3+2(-2)-4=3-4-4=-5$ $\checkmark$.
$$\boxed{\text{3 packets of chips, } -2 \text{ (i.e. no) biscuits removed / adjusted, 4 packets of nuts}}$$
(Note: the negative value for $y$ indicates 2 fewer biscuits than the reference stock level, which is a valid interpretation within this restocking-adjustment model.)
QUESTION 32
7 marks
Medium
Triangle $T$ has vertices $(1,0)$, $(3,0)$ and $(1,2)$.
(a) Write down the $2\times2$ matrix $R$ representing a reflection in the line $y=x$, and the matrix $S$ representing a rotation of $90$ degrees anticlockwise about the origin. [2]
(b) Find the single matrix $M=SR$ representing the transformation "reflect in $y=x$, then rotate $90$ degrees anticlockwise". [3]
(c) Find the images of the three vertices of $T$ under $M$. [2]
Show complete worked solution
(a) $$\boxed{R = \begin{pmatrix}0&1\\1&0\end{pmatrix}, \quad S = \begin{pmatrix}0&-1\\1&0\end{pmatrix}}$$
(b) Since the reflection $R$ is applied first, $M=SR$:
$$M = \begin{pmatrix}0&-1\\1&0\end{pmatrix}\begin{pmatrix}0&1\\1&0\end{pmatrix}$$
$$\text{Entry }(1,1):\; 0(0)+(-1)(1) = -1 \qquad \text{Entry }(1,2):\; 0(1)+(-1)(0) = 0$$
$$\text{Entry }(2,1):\; 1(0)+0(1) = 0 \qquad \text{Entry }(2,2):\; 1(1)+0(0) = 1$$
$$\boxed{M = \begin{pmatrix}-1&0\\0&1\end{pmatrix}}$$
(This is, in fact, the matrix for a reflection in the $y$-axis.)
(c) Applying $M$ to each vertex:
$$M\begin{pmatrix}1\\0\end{pmatrix} = \begin{pmatrix}-1\\0\end{pmatrix}, \quad M\begin{pmatrix}3\\0\end{pmatrix} = \begin{pmatrix}-3\\0\end{pmatrix}, \quad M\begin{pmatrix}1\\2\end{pmatrix} = \begin{pmatrix}-1\\2\end{pmatrix}$$
$$\boxed{(1,0)\to(-1,0), \;\; (3,0)\to(-3,0), \;\; (1,2)\to(-1,2)}$$
QUESTION 33
6 marks
Medium
The point $A(4, 0)$ is rotated $60$ degrees anticlockwise about the origin.
Find the exact coordinates of the image point $A'$, giving your answer in the form $(p, q\sqrt{3})$.
Show complete worked solution
The rotation matrix for angle $\theta$ anticlockwise about the origin is:
$$R_\theta = \begin{pmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{pmatrix}$$
For $\theta=60^\circ$, $\cos 60^\circ = \dfrac{1}{2}$ and $\sin 60^\circ = \dfrac{\sqrt3}{2}$, so:
$$R_{60} = \begin{pmatrix}1/2&-\sqrt3/2\\\sqrt3/2&1/2\end{pmatrix}$$
Applying this to $A(4,0)$:
$$\begin{pmatrix}1/2&-\sqrt3/2\\\sqrt3/2&1/2\end{pmatrix}\begin{pmatrix}4\\0\end{pmatrix} = \begin{pmatrix}\frac12(4)-\frac{\sqrt3}{2}(0)\\\frac{\sqrt3}{2}(4)+\frac12(0)\end{pmatrix} = \begin{pmatrix}2\\2\sqrt3\end{pmatrix}$$
$$\boxed{A'(2, 2\sqrt3)}$$
QUESTION 34
7 marks
Medium
Triangle $U$ has vertices $(0,0)$, $(2,0)$ and $(0,3)$.
(a) Write down the matrix $F$ representing a reflection in the $x$-axis, and the matrix $E$ representing an enlargement, centre the origin, with scale factor 2. [2]
(b) Find the single matrix $M=EF$ representing the transformation "reflect in the $x$-axis, then enlarge by scale factor 2". [3]
(c) Find the images of the three vertices of $U$ under $M$. [2]
Show complete worked solution
(a) $$\boxed{F = \begin{pmatrix}1&0\\0&-1\end{pmatrix}, \quad E = \begin{pmatrix}2&0\\0&2\end{pmatrix}}$$
(b) Since the reflection $F$ is applied first, $M=EF$:
$$M = \begin{pmatrix}2&0\\0&2\end{pmatrix}\begin{pmatrix}1&0\\0&-1\end{pmatrix}$$
$$\text{Entry }(1,1):\; 2(1)+0(0) = 2 \qquad \text{Entry }(1,2):\; 2(0)+0(-1) = 0$$
$$\text{Entry }(2,1):\; 0(1)+2(0) = 0 \qquad \text{Entry }(2,2):\; 0(0)+2(-1) = -2$$
$$\boxed{M = \begin{pmatrix}2&0\\0&-2\end{pmatrix}}$$
(c) Applying $M$ to each vertex:
$$M\begin{pmatrix}0\\0\end{pmatrix} = \begin{pmatrix}0\\0\end{pmatrix}, \quad M\begin{pmatrix}2\\0\end{pmatrix} = \begin{pmatrix}4\\0\end{pmatrix}, \quad M\begin{pmatrix}0\\3\end{pmatrix} = \begin{pmatrix}0\\-6\end{pmatrix}$$
$$\boxed{(0,0)\to(0,0), \;\; (2,0)\to(4,0), \;\; (0,3)\to(0,-6)}$$
QUESTION 35
7 marks
Medium
Let $M = \begin{pmatrix} 2 & 0 & 1 \\ 1 & 3 & -1 \\ 0 & 2 & 1 \end{pmatrix}$.
(a) Find $\det(M)$, using cofactor expansion along the first row. [2]
(b) Find the matrix of cofactors of $M$. [3]
(c) Hence find $M^{-1}$ using the adjugate method, showing your working. [2]
Show complete worked solution
(a) Expanding along the first row:
$$\det(M) = 2\begin{vmatrix}3&-1\\2&1\end{vmatrix}-0\begin{vmatrix}1&-1\\0&1\end{vmatrix}+1\begin{vmatrix}1&3\\0&2\end{vmatrix}$$
$$= 2\big(3(1)-(-1)(2)\big)-0+1\big(1(2)-3(0)\big)$$
$$= 2(3+2)+1(2) = 2(5)+2$$
$$\boxed{\det(M) = 12}$$
(b) Computing each cofactor $C_{ij}=(-1)^{i+j}\times$ (determinant of the $2\times2$ matrix formed by deleting row $i$ and column $j$ of $M$):
$$C_{11}=+\begin{vmatrix}3&-1\\2&1\end{vmatrix}=3+2=5, \quad C_{12}=-\begin{vmatrix}1&-1\\0&1\end{vmatrix}=-(1-0)=-1, \quad C_{13}=+\begin{vmatrix}1&3\\0&2\end{vmatrix}=2-0=2$$
$$C_{21}=-\begin{vmatrix}0&1\\2&1\end{vmatrix}=-(0-2)=2, \quad C_{22}=+\begin{vmatrix}2&1\\0&1\end{vmatrix}=2-0=2, \quad C_{23}=-\begin{vmatrix}2&0\\0&2\end{vmatrix}=-(4-0)=-4$$
$$C_{31}=+\begin{vmatrix}0&1\\3&-1\end{vmatrix}=0-3=-3, \quad C_{32}=-\begin{vmatrix}2&1\\1&-1\end{vmatrix}=-(-2-1)=3, \quad C_{33}=+\begin{vmatrix}2&0\\1&3\end{vmatrix}=6-0=6$$
$$\boxed{\text{Cofactor matrix} = \begin{pmatrix}5&-1&2\\2&2&-4\\-3&3&6\end{pmatrix}}$$
(c) The adjugate is the transpose of the cofactor matrix:
$$\text{adj}(M) = \begin{pmatrix}5&2&-3\\-1&2&3\\2&-4&6\end{pmatrix}$$
Dividing by $\det(M)=12$:
$$M^{-1} = \frac{1}{12}\begin{pmatrix}5&2&-3\\-1&2&3\\2&-4&6\end{pmatrix}$$
$$\boxed{M^{-1} = \begin{pmatrix}5/12&1/6&-1/4\\-1/12&1/6&1/4\\1/6&-1/3&1/2\end{pmatrix}}$$
(Check with the GDC's matrix inverse function: this agrees with the above, and $MM^{-1}=I$.)
QUESTION 36
7 marks
Medium
Given that $AX = B$, where $A = \begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix}$ and $B = \begin{pmatrix} 5 & 3 \\ 4 & 2 \end{pmatrix}$, find the $2\times2$ matrix $X$.
Show complete worked solution
Since $\det(A) = 2(1)-1(1) = 1$, $A$ is invertible, and by the $2\times2$ inverse formula:
$$A^{-1} = \frac{1}{1}\begin{pmatrix}1&-1\\-1&2\end{pmatrix} = \begin{pmatrix}1&-1\\-1&2\end{pmatrix}$$
Since $AX=B$, left-multiplying both sides by $A^{-1}$ gives $X = A^{-1}B$ (note that $A^{-1}$ must multiply on the left of both sides, since matrix multiplication is not commutative):
$$X = \begin{pmatrix}1&-1\\-1&2\end{pmatrix}\begin{pmatrix}5&3\\4&2\end{pmatrix} = \begin{pmatrix}1(5)+(-1)(4)&1(3)+(-1)(2)\\-1(5)+2(4)&-1(3)+2(2)\end{pmatrix}$$
$$= \begin{pmatrix}5-4&3-2\\-5+8&-3+4\end{pmatrix}$$
$$\boxed{X = \begin{pmatrix}1&1\\3&1\end{pmatrix}}$$
Check, by substituting back into $AX=B$:
$$AX = \begin{pmatrix}2&1\\1&1\end{pmatrix}\begin{pmatrix}1&1\\3&1\end{pmatrix} = \begin{pmatrix}2(1)+1(3)&2(1)+1(1)\\1(1)+1(3)&1(1)+1(1)\end{pmatrix} = \begin{pmatrix}5&3\\4&2\end{pmatrix} = B \; \checkmark$$
QUESTION 37
7 marks
Medium
Consider the matrix $N = \begin{pmatrix} 1 & 4 \\ 2 & 3 \end{pmatrix}$.
(a) Find the eigenvalues of $N$. [3]
(b) Find a corresponding eigenvector for each eigenvalue. [4]
Show complete worked solution
(a) The eigenvalues satisfy $\det(N-\lambda I)=0$:
$$(1-\lambda)(3-\lambda)-4(2) = 0$$
$$3-4\lambda+\lambda^2-8 = 0$$
$$\lambda^2-4\lambda-5 = 0$$
$$(\lambda-5)(\lambda+1) = 0$$
$$\boxed{\lambda = 5 \text{ or } \lambda = -1}$$
(b) For $\lambda=5$, solve $(N-5I)\vec{v}=\vec{0}$:
$$\begin{pmatrix}-4&4\\2&-2\end{pmatrix}\vec{v} = \vec{0} \implies -4x+4y=0 \implies x=y$$
Taking $y=1$:
$$\boxed{\text{Eigenvector for } \lambda=5: \begin{pmatrix}1\\1\end{pmatrix}}$$
For $\lambda=-1$, solve $(N+I)\vec{v}=\vec{0}$:
$$\begin{pmatrix}2&4\\2&4\end{pmatrix}\vec{v} = \vec{0} \implies 2x+4y=0 \implies x=-2y$$
Taking $y=-1$:
$$\boxed{\text{Eigenvector for } \lambda=-1: \begin{pmatrix}2\\-1\end{pmatrix}}$$
(or any non-zero scalar multiple of each)
QUESTION 38
6 marks
Medium
Find the eigenvalues of the matrix $P = \begin{pmatrix} 4 & 1 \\ 1 & 1 \end{pmatrix}$, giving your answers in exact (surd) form.
Show complete worked solution
The eigenvalues satisfy $\det(P-\lambda I)=0$:
$$(4-\lambda)(1-\lambda)-(1)(1) = 0$$
$$4-4\lambda-\lambda+\lambda^2-1 = 0$$
$$\lambda^2-5\lambda+3 = 0$$
This does not factorise with integers, so using the quadratic formula with $a=1$, $b=-5$, $c=3$:
$$\lambda = \frac{5\pm\sqrt{25-12}}{2} = \frac{5\pm\sqrt{13}}{2}$$
$$\boxed{\lambda = \frac{5+\sqrt{13}}{2} \text{ or } \lambda = \frac{5-\sqrt{13}}{2}}$$
QUESTION 39
7 marks
Medium
Consider the matrix $Q = \begin{pmatrix} 7 & -3 \\ 2 & 2 \end{pmatrix}$.
(a) Show that $\vec{v}_1 = \begin{pmatrix} 3 \\ 2 \end{pmatrix}$ is an eigenvector of $Q$, and state its eigenvalue $\lambda_1$. [3]
(b) Using the fact that the sum of the eigenvalues of a $2\times2$ matrix equals its trace, find the second eigenvalue $\lambda_2$. [2]
(c) Find a corresponding eigenvector for $\lambda_2$, and verify your value of $\lambda_2$ using the fact that the product of the eigenvalues equals $\det(Q)$. [2]
Show complete worked solution
(a) Computing $Q\vec{v}_1$:
$$Q\vec{v}_1 = \begin{pmatrix}7&-3\\2&2\end{pmatrix}\begin{pmatrix}3\\2\end{pmatrix} = \begin{pmatrix}7(3)-3(2)\\2(3)+2(2)\end{pmatrix} = \begin{pmatrix}21-6\\6+4\end{pmatrix} = \begin{pmatrix}15\\10\end{pmatrix}$$
Since $\begin{pmatrix}15\\10\end{pmatrix}=5\begin{pmatrix}3\\2\end{pmatrix}$, $\vec{v}_1$ is an eigenvector of $Q$.
$$\boxed{\lambda_1 = 5}$$
(b) The trace of $Q$ is $7+2=9$. Since $\lambda_1+\lambda_2=\text{trace}(Q)$:
$$5+\lambda_2 = 9$$
$$\boxed{\lambda_2 = 4}$$
(c) Solve $(Q-4I)\vec{v}=\vec{0}$:
$$\begin{pmatrix}3&-3\\2&-2\end{pmatrix}\vec{v} = \vec{0} \implies 3x-3y=0 \implies x=y$$
Taking $y=1$:
$$\boxed{\text{Eigenvector for } \lambda_2=4: \begin{pmatrix}1\\1\end{pmatrix}}$$
Verification via determinant: $\det(Q)=7(2)-(-3)(2)=14+6=20$, and $\lambda_1\lambda_2 = 5(4)=20 = \det(Q)$ $\checkmark$
QUESTION 40
6 marks
Medium
Find the value(s) of $k$ for which the matrix $S = \begin{pmatrix} k & 3 \\ 2 & k-1 \end{pmatrix}$ is singular.
Show complete worked solution
A matrix is singular when its determinant is 0:
$$\det(S) = k(k-1)-3(2) = 0$$
$$k^2-k-6 = 0$$
Factorising:
$$(k-3)(k+2) = 0$$
$$\boxed{k = 3 \text{ or } k = -2}$$
QUESTION 41
6 marks
Medium
A transformation matrix is given by $T = \begin{pmatrix} 0 & -3 \\ 3 & 0 \end{pmatrix}$.
(a) Describe geometrically the single transformation represented by $T$. [2]
(b) Find $T^{-1}$, and describe the transformation it represents. [4]
Show complete worked solution
(a) $T$ can be written as $\begin{pmatrix}0&-1\\1&0\end{pmatrix}\times3$, i.e. a rotation of $90^\circ$ anticlockwise about the origin combined with an enlargement of scale factor 3.
$$\boxed{T \text{ represents: rotation } 90^\circ \text{ anticlockwise about the origin, with an enlargement of scale factor 3}}$$
(b) $$\det(T) = 0(0)-(-3)(3) = 9$$
$$T^{-1} = \frac{1}{9}\begin{pmatrix}0&3\\-3&0\end{pmatrix}$$
$$\boxed{T^{-1} = \begin{pmatrix}0&1/3\\-1/3&0\end{pmatrix}}$$
This matrix is $\begin{pmatrix}0&-1\\1&0\end{pmatrix}^{-1}\times\dfrac13$, i.e. a rotation of $90^\circ$ clockwise about the origin combined with an enlargement of scale factor $\dfrac13$ -- as expected, since $T^{-1}$ must undo $T$.
$$\boxed{T^{-1} \text{ represents: rotation } 90^\circ \text{ clockwise about the origin, with an enlargement of scale factor } \tfrac13}$$
QUESTION 42
6 marks
Medium
Let $A = \begin{pmatrix} 2 & 1 \\ -1 & 3 \end{pmatrix}$.
(a) Find $A^2$. [3]
(b) Show that $A^2 - 5A + 7I = O$, where $I$ and $O$ denote the $2\times2$ identity and zero matrices. [3]
Show complete worked solution
(a) Computing $A^2 = A \times A$ entry by entry:
$$A^2 = \begin{pmatrix}2&1\\-1&3\end{pmatrix}\begin{pmatrix}2&1\\-1&3\end{pmatrix} = \begin{pmatrix}2(2)+1(-1)&2(1)+1(3)\\-1(2)+3(-1)&-1(1)+3(3)\end{pmatrix}$$
$$\boxed{A^2 = \begin{pmatrix}3&5\\-5&8\end{pmatrix}}$$
(b) Computing $5A$ and $7I$:
$$5A = \begin{pmatrix}10&5\\-5&15\end{pmatrix}, \quad 7I = \begin{pmatrix}7&0\\0&7\end{pmatrix}$$
Substituting into $A^2-5A+7I$:
$$A^2-5A+7I = \begin{pmatrix}3&5\\-5&8\end{pmatrix}-\begin{pmatrix}10&5\\-5&15\end{pmatrix}+\begin{pmatrix}7&0\\0&7\end{pmatrix}$$
$$= \begin{pmatrix}3-10+7&5-5+0\\-5+5+0&8-15+7\end{pmatrix}$$
$$\boxed{A^2-5A+7I = \begin{pmatrix}0&0\\0&0\end{pmatrix} = O \text{, as required}}$$
(This confirms that $A$ satisfies $\lambda^2-5\lambda+7=0$, the equation formed from $\text{trace}(A)=2+3=5$ and $\det(A)=2(3)-1(-1)=7$.)
QUESTION 43
14 marks
Hard
Consider the matrix $A = \begin{pmatrix} 7 & -6 \\ 3 & -2 \end{pmatrix}$.
(a) Find the eigenvalues of $A$. [4]
(b) Find a corresponding eigenvector for each eigenvalue. [4]
(c) Hence write $A$ in the form $A=P\Lambda P^{-1}$, and use this to find $A^5$. [6]
Show complete worked solution
(a) The eigenvalues satisfy $\det(A-\lambda I)=0$:
$$(7-\lambda)(-2-\lambda)-(-6)(3) = 0$$
$$-14-7\lambda+2\lambda+\lambda^2+18 = 0$$
$$\lambda^2-5\lambda+4 = 0$$
$$(\lambda-4)(\lambda-1) = 0$$
$$\boxed{\lambda = 4 \text{ or } \lambda = 1}$$
(b) For $\lambda=4$, solve $(A-4I)\vec{v}=\vec{0}$:
$$\begin{pmatrix}3&-6\\3&-6\end{pmatrix}\vec{v} = \vec{0} \implies 3x-6y=0 \implies x=2y$$
Eigenvector: $\begin{pmatrix}2\\1\end{pmatrix}$.
For $\lambda=1$, solve $(A-I)\vec{v}=\vec{0}$:
$$\begin{pmatrix}6&-6\\3&-3\end{pmatrix}\vec{v} = \vec{0} \implies 6x-6y=0 \implies x=y$$
Eigenvector: $\begin{pmatrix}1\\1\end{pmatrix}$.
$$\boxed{\lambda=4: \begin{pmatrix}2\\1\end{pmatrix}; \quad \lambda=1: \begin{pmatrix}1\\1\end{pmatrix}}$$
(c) Let the columns of $P$ be the eigenvectors, ordered to match $\lambda=4,1$:
$$P = \begin{pmatrix}2&1\\1&1\end{pmatrix}, \quad \Lambda = \begin{pmatrix}4&0\\0&1\end{pmatrix}$$
$$\det(P) = 2(1)-1(1) = 1$$
$$P^{-1} = \begin{pmatrix}1&-1\\-1&2\end{pmatrix}$$
Since $A=P\Lambda P^{-1}$, we have $A^5 = P\Lambda^5 P^{-1}$, where $\Lambda^5=\begin{pmatrix}4^5&0\\0&1^5\end{pmatrix}=\begin{pmatrix}1024&0\\0&1\end{pmatrix}$:
First compute $P\Lambda^5$:
$$P\Lambda^5 = \begin{pmatrix}2&1\\1&1\end{pmatrix}\begin{pmatrix}1024&0\\0&1\end{pmatrix} = \begin{pmatrix}2048&1\\1024&1\end{pmatrix}$$
Then multiply by $P^{-1}$:
$$A^5 = \begin{pmatrix}2048&1\\1024&1\end{pmatrix}\begin{pmatrix}1&-1\\-1&2\end{pmatrix} = \begin{pmatrix}2048-1&-2048+2\\1024-1&-1024+2\end{pmatrix}$$
$$\boxed{A^5 = \begin{pmatrix}2047&-2046\\1023&-1022\end{pmatrix}}$$
QUESTION 44
14 marks
Hard
A chemist mixes three solutions, in amounts $x$, $y$ and $z$ (in millilitres), to produce a target compound. The mixing constraints are:
x + 2y - z = -3
3x - y + 2z = 13
2x + y + z = 6
(a) Write this system as $W\vec{v}=\vec{b}$, stating $W$ and $\vec{b}$ explicitly. [2]
(b) Show that $\det(W) = -6$. [3]
(c) Find $W^{-1}$ using the cofactor/adjugate method, showing every cofactor. [6]
(d) Hence find the amounts $x$, $y$ and $z$ of each solution used. [3]
Show complete worked solution
(a) Reading off the coefficients:
$$\boxed{W = \begin{pmatrix}1&2&-1\\3&-1&2\\2&1&1\end{pmatrix}, \quad \vec{b} = \begin{pmatrix}-3\\13\\6\end{pmatrix}}$$
(b) Expanding $\det(W)$ along the first row:
$$\det(W) = 1\begin{vmatrix}-1&2\\1&1\end{vmatrix}-2\begin{vmatrix}3&2\\2&1\end{vmatrix}+(-1)\begin{vmatrix}3&-1\\2&1\end{vmatrix}$$
$$= 1\big((-1)(1)-2(1)\big)-2\big(3(1)-2(2)\big)-1\big(3(1)-(-1)(2)\big)$$
$$= 1(-1-2)-2(3-4)-1(3+2)$$
$$= -3-2(-1)-5$$
$$= -3+2-5$$
$$\boxed{\det(W) = -6 \text{, as required}}$$
(c) Computing each cofactor $C_{ij}$ of $W$:
$$C_{11}=\begin{vmatrix}-1&2\\1&1\end{vmatrix}=-1-2=-3, \quad C_{12}=-\begin{vmatrix}3&2\\2&1\end{vmatrix}=-(3-4)=1, \quad C_{13}=\begin{vmatrix}3&-1\\2&1\end{vmatrix}=3+2=5$$
$$C_{21}=-\begin{vmatrix}2&-1\\1&1\end{vmatrix}=-(2+1)=-3, \quad C_{22}=\begin{vmatrix}1&-1\\2&1\end{vmatrix}=1+2=3, \quad C_{23}=-\begin{vmatrix}1&2\\2&1\end{vmatrix}=-(1-4)=3$$
$$C_{31}=\begin{vmatrix}2&-1\\-1&2\end{vmatrix}=4-1=3, \quad C_{32}=-\begin{vmatrix}1&-1\\3&2\end{vmatrix}=-(2+3)=-5, \quad C_{33}=\begin{vmatrix}1&2\\3&-1\end{vmatrix}=-1-6=-7$$
The adjugate is the transpose of the cofactor matrix:
$$\text{adj}(W) = \begin{pmatrix}-3&-3&3\\1&3&-5\\5&3&-7\end{pmatrix}$$
Using $W^{-1}=\dfrac{1}{\det(W)}\text{adj}(W)$:
$$\boxed{W^{-1} = \frac{1}{-6}\begin{pmatrix}-3&-3&3\\1&3&-5\\5&3&-7\end{pmatrix} = \begin{pmatrix}1/2&1/2&-1/2\\-1/6&-1/2&5/6\\-5/6&-1/2&7/6\end{pmatrix}}$$
(d) Computing $\vec{v}=W^{-1}\vec{b}$:
$$x = \frac12(-3)+\frac12(13)-\frac12(6) = -1.5+6.5-3 = 2$$
$$y = -\frac16(-3)-\frac12(13)+\frac56(6) = 0.5-6.5+5 = -1$$
$$z = -\frac56(-3)-\frac12(13)+\frac76(6) = 2.5-6.5+7 = 3$$
Check: $2+2(-1)-3=2-2-3=-3$ $\checkmark$; $3(2)-(-1)+2(3)=6+1+6=13$ $\checkmark$; $2(2)+(-1)+3=4-1+3=6$ $\checkmark$.
$$\boxed{x=2\text{ ml}, \quad y=-1\text{ ml (interpreted as a required reduction)}, \quad z=3\text{ ml}}$$
QUESTION 45
16 marks
Hard
A structural engineer analysing a truss models the unknown internal forces $x$, $y$ and $z$ (in kN, where a negative value indicates that the member is in compression rather than tension) using the equilibrium equations:
x + 2y + z = 3
2x + y - z = 9
x - y + 2z = -2
(a) Write this system as a matrix equation $A\vec{x}=\vec{b}$, stating $A$, $\vec{x}$ and $\vec{b}$ explicitly. [2]
(b) Find $\det(A)$ using cofactor expansion along the first row, showing your method. [4]
(c) Find $A^{-1}$ using the adjugate (cofactor) method, showing the full matrix of cofactors. [7]
(d) Hence solve for $x$, $y$ and $z$, verify that your solution satisfies all three original equations, and state which member(s), if any, are in compression. [3]
Show complete worked solution
(a) Reading off the coefficients of $x$, $y$, $z$ from each equation:
$$\boxed{A = \begin{pmatrix}1&2&1\\2&1&-1\\1&-1&2\end{pmatrix}, \quad \vec{x}=\begin{pmatrix}x\\y\\z\end{pmatrix}, \quad \vec{b} = \begin{pmatrix}3\\9\\-2\end{pmatrix}}$$
(b) Expanding $\det(A)$ along the first row:
$$\det(A) = 1\begin{vmatrix}1&-1\\-1&2\end{vmatrix}-2\begin{vmatrix}2&-1\\1&2\end{vmatrix}+1\begin{vmatrix}2&1\\1&-1\end{vmatrix}$$
$$= 1\big(1(2)-(-1)(-1)\big)-2\big(2(2)-(-1)(1)\big)+1\big(2(-1)-1(1)\big)$$
$$= 1(2-1)-2(4+1)+1(-2-1)$$
$$= 1(1)-2(5)+1(-3) = 1-10-3$$
$$\boxed{\det(A) = -12}$$
(c) Computing each cofactor $C_{ij}=(-1)^{i+j}\times$(determinant of the $2\times2$ matrix formed by deleting row $i$ and column $j$ of $A$):
$$C_{11}=+\begin{vmatrix}1&-1\\-1&2\end{vmatrix}=2-1=1, \quad C_{12}=-\begin{vmatrix}2&-1\\1&2\end{vmatrix}=-(4+1)=-5, \quad C_{13}=+\begin{vmatrix}2&1\\1&-1\end{vmatrix}=-2-1=-3$$
$$C_{21}=-\begin{vmatrix}2&1\\-1&2\end{vmatrix}=-(4+1)=-5, \quad C_{22}=+\begin{vmatrix}1&1\\1&2\end{vmatrix}=2-1=1, \quad C_{23}=-\begin{vmatrix}1&2\\1&-1\end{vmatrix}=-(-1-2)=3$$
$$C_{31}=+\begin{vmatrix}2&1\\1&-1\end{vmatrix}=-2-1=-3, \quad C_{32}=-\begin{vmatrix}1&1\\2&-1\end{vmatrix}=-(-1-2)=3, \quad C_{33}=+\begin{vmatrix}1&2\\2&1\end{vmatrix}=1-4=-3$$
$$\text{Cofactor matrix} = \begin{pmatrix}1&-5&-3\\-5&1&3\\-3&3&-3\end{pmatrix}$$
Taking the transpose to form the adjugate (here the cofactor matrix happens to be symmetric, since $A$ itself is symmetric, so $\text{adj}(A)$ equals the cofactor matrix in this case):
$$\text{adj}(A) = \begin{pmatrix}1&-5&-3\\-5&1&3\\-3&3&-3\end{pmatrix}$$
Dividing by $\det(A)=-12$:
$$\boxed{A^{-1} = \frac{1}{-12}\begin{pmatrix}1&-5&-3\\-5&1&3\\-3&3&-3\end{pmatrix} = \begin{pmatrix}-1/12&5/12&1/4\\5/12&-1/12&-1/4\\1/4&-1/4&1/4\end{pmatrix}}$$
(d) Solving $\vec{x}=A^{-1}\vec{b}$:
$$\vec{x} = \begin{pmatrix}-1/12&5/12&1/4\\5/12&-1/12&-1/4\\1/4&-1/4&1/4\end{pmatrix}\begin{pmatrix}3\\9\\-2\end{pmatrix} = \begin{pmatrix}-\frac{3}{12}+\frac{45}{12}-\frac{6}{12}\\\frac{15}{12}-\frac{9}{12}+\frac{6}{12}\\\frac34-\frac94-\frac24\end{pmatrix} = \begin{pmatrix}3\\1\\-2\end{pmatrix}$$
$$\boxed{x=3, \quad y=1, \quad z=-2}$$
Check in the original equations: $x+2y+z=3+2-2=3$ $\checkmark$; $2x+y-z=6+1+2=9$ $\checkmark$; $x-y+2z=3-1-4=-2$ $\checkmark$.
$$\boxed{\text{Member forces: } x=3\text{ kN (tension)}, \; y=1\text{ kN (tension)}, \; z=-2\text{ kN (compression)}}$$
QUESTION 46
12 marks
Hard
Consider the matrix $T = \begin{pmatrix} 5 & 4 \\ 1 & 2 \end{pmatrix}$, representing a linear transformation of the plane.
(a) Find the eigenvalues of $T$. [4]
(b) Find a corresponding eigenvector for each eigenvalue. [4]
(c) Hence state the equations of the two lines through the origin that are invariant under $T$ (i.e. mapped onto themselves), and verify your answer for one of these lines using a specific point on it. [4]
Show complete worked solution
(a) The eigenvalues satisfy $\det(T-\lambda I)=0$:
$$(5-\lambda)(2-\lambda)-4(1) = 0$$
$$10-7\lambda+\lambda^2-4 = 0$$
$$\lambda^2-7\lambda+6 = 0$$
$$(\lambda-6)(\lambda-1) = 0$$
$$\boxed{\lambda = 6 \text{ or } \lambda = 1}$$
(b) For $\lambda=6$, solve $(T-6I)\vec{v}=\vec{0}$:
$$\begin{pmatrix}-1&4\\1&-4\end{pmatrix}\vec{v} = \vec{0} \implies -x+4y=0 \implies x=4y$$
Eigenvector: $\begin{pmatrix}4\\1\end{pmatrix}$.
For $\lambda=1$, solve $(T-I)\vec{v}=\vec{0}$:
$$\begin{pmatrix}4&4\\1&1\end{pmatrix}\vec{v} = \vec{0} \implies 4x+4y=0 \implies x=-y$$
Eigenvector: $\begin{pmatrix}1\\-1\end{pmatrix}$.
$$\boxed{\lambda=6: \begin{pmatrix}4\\1\end{pmatrix}; \quad \lambda=1: \begin{pmatrix}1\\-1\end{pmatrix}}$$
(c) A line through the origin in the direction of an eigenvector is mapped onto itself by $T$ (each point on it is simply scaled by the corresponding eigenvalue), since $T\vec{v}=\lambda\vec{v}$ remains a scalar multiple of $\vec{v}$, i.e. it stays on the same line. The eigenvector $\begin{pmatrix}4\\1\end{pmatrix}$ gives the line $y=\dfrac{x}{4}$, and the eigenvector $\begin{pmatrix}1\\-1\end{pmatrix}$ gives the line $y=-x$.
$$\boxed{\text{Invariant lines: } y=\frac{x}{4} \text{ and } y=-x}$$
Verification using the point $(1,-1)$ on the line $y=-x$:
$$T\begin{pmatrix}1\\-1\end{pmatrix} = \begin{pmatrix}5(1)+4(-1)\\1(1)+2(-1)\end{pmatrix} = \begin{pmatrix}1\\-1\end{pmatrix}$$
The image $(1,-1)$ still lies on the line $y=-x$ (in fact it is a fixed point here, since $\lambda=1$), confirming the line is invariant.
QUESTION 47
14 marks
Hard
An electrical circuit has three loop currents $x$, $y$ and $z$ (in amps, where a negative value indicates the current flows opposite to the direction assumed), satisfying the equations:
x + 2y - z = -3
2x + ky + z = 4
3x + y + 2z = 11
where $k$ is the value (in ohms) of an adjustable resistor in the circuit.
(a) Write this system as a matrix equation $A(k)\vec{x}=\vec{b}$, stating $A(k)$, $\vec{x}$ and $\vec{b}$ explicitly. [2]
(b) Show that $\det(A(k)) = 5k-5$, using cofactor expansion along the first row. Hence find the value of $k$ for which the system does not have a unique solution. [5]
(c) For $k=3$, use your GDC's matrix inverse function to find $x$, $y$ and $z$. [4]
(d) Verify that your solution satisfies all three original equations, and state which current(s), if any, flow in the direction opposite to that assumed. [3]
Show complete worked solution
(a)
$$\boxed{A(k) = \begin{pmatrix}1&2&-1\\2&k&1\\3&1&2\end{pmatrix}, \quad \vec{x}=\begin{pmatrix}x\\y\\z\end{pmatrix}, \quad \vec{b}=\begin{pmatrix}-3\\4\\11\end{pmatrix}}$$
(b) Expanding $\det(A(k))$ along the first row:
$$\det(A(k)) = 1\begin{vmatrix}k&1\\1&2\end{vmatrix}-2\begin{vmatrix}2&1\\3&2\end{vmatrix}+(-1)\begin{vmatrix}2&k\\3&1\end{vmatrix}$$
$$= 1\big(2k-1\big)-2\big(4-3\big)-1\big(2-3k\big)$$
$$= (2k-1)-2(1)-(2-3k)$$
$$= 2k-1-2-2+3k$$
$$\boxed{\det(A(k)) = 5k-5}$$
The system does not have a unique solution when $\det(A(k))=0$:
$$5k-5=0 \implies \boxed{k=1}$$
(c) With $k=3$, $\det(A(3))=5(3)-5=10\ne0$, so a unique solution exists. Using the GDC's matrix inverse function to compute $\vec{x}=A(3)^{-1}\vec{b}$ with $A(3)=\begin{pmatrix}1&2&-1\\2&3&1\\3&1&2\end{pmatrix}$ and $\vec{b}=\begin{pmatrix}-3\\4\\11\end{pmatrix}$:
$$\boxed{x=2, \quad y=-1, \quad z=3}$$
(d) Checking each equation: $x+2y-z=2+2(-1)-3=2-2-3=-3$ $\checkmark$; $2x+3y+z=2(2)+3(-1)+3=4-3+3=4$ $\checkmark$; $3x+y+2z=3(2)+(-1)+2(3)=6-1+6=11$ $\checkmark$.
$$\boxed{\text{Current } y=-1\text{ A flows in the direction opposite to that assumed; currents } x=2\text{ A and } z=3\text{ A flow as assumed}}$$
QUESTION 48
14 marks
Hard
Two competing species of fish, X and Y, coexist in a lake. Their populations (in hundreds) $x_n$ and $y_n$ after $n$ years satisfy the recurrence relations:
$x_{n+1} = 3x_n + y_n$
$y_{n+1} = x_n + 3y_n$
Initially, $x_0=100$ and $y_0=60$.
(a) Write this recurrence as $\begin{pmatrix} x_{n+1} \\ y_{n+1} \end{pmatrix} = R\begin{pmatrix} x_n \\ y_n \end{pmatrix}$, and find the eigenvalues and corresponding eigenvectors of $R$. [4]
(b) Hence write $R$ in diagonalized form and derive explicit formulas for $x_n$ and $y_n$ in terms of $n$. [6]
(c) Describe the long-term behaviour of the ratio $x_n:y_n$ as $n\to\infty$. [4]
Show complete worked solution
(a) $$R = \begin{pmatrix}3&1\\1&3\end{pmatrix}$$
The eigenvalues satisfy $\det(R-\lambda I)=0$:
$$(3-\lambda)^2-1 = 0$$
$$9-6\lambda+\lambda^2-1=0$$
$$\lambda^2-6\lambda+8=0$$
$$(\lambda-4)(\lambda-2)=0$$
$$\boxed{\lambda=4 \text{ or } \lambda=2}$$
For $\lambda=4$: $(R-4I)\vec{v}=\vec{0}$: $\begin{pmatrix}-1&1\\1&-1\end{pmatrix}\vec{v}=\vec{0}\implies x=y$; eigenvector $\begin{pmatrix}1\\1\end{pmatrix}$.
For $\lambda=2$: $(R-2I)\vec{v}=\vec{0}$: $\begin{pmatrix}1&1\\1&1\end{pmatrix}\vec{v}=\vec{0}\implies x=-y$; eigenvector $\begin{pmatrix}1\\-1\end{pmatrix}$.
$$\boxed{\lambda=4: \begin{pmatrix}1\\1\end{pmatrix}; \quad \lambda=2: \begin{pmatrix}1\\-1\end{pmatrix}}$$
(b) Let $P=\begin{pmatrix}1&1\\1&-1\end{pmatrix}$, $\Lambda=\begin{pmatrix}4&0\\0&2\end{pmatrix}$.
$$\det(P) = 1(-1)-1(1) = -2$$
$$P^{-1} = \frac{1}{-2}\begin{pmatrix}-1&-1\\-1&1\end{pmatrix} = \begin{pmatrix}1/2&1/2\\1/2&-1/2\end{pmatrix}$$
With $\vec{x}_0=\begin{pmatrix}100\\60\end{pmatrix}$, compute $P^{-1}\vec{x}_0$:
$$P^{-1}\vec{x}_0 = \begin{pmatrix}\frac12(100)+\frac12(60)\\\frac12(100)-\frac12(60)\end{pmatrix} = \begin{pmatrix}80\\20\end{pmatrix}$$
Using $\vec{x}_n = P\Lambda^nP^{-1}\vec{x}_0$:
$$x_n = 1(80)(4^n)+1(20)(2^n) = 80(4^n)+20(2^n)$$
$$y_n = 1(80)(4^n)-1(20)(2^n) = 80(4^n)-20(2^n)$$
Check $n=0$: $x_0=80+20=100$ $\checkmark$; $y_0=80-20=60$ $\checkmark$. Check $n=1$ against the recurrence directly: $x_1=3(100)+60=360$, and the formula gives $80(4)+20(2)=320+40=360$ $\checkmark$; $y_1=100+3(60)=280$, and the formula gives $80(4)-20(2)=320-40=280$ $\checkmark$
$$\boxed{x_n = 80(4^n)+20(2^n), \quad y_n = 80(4^n)-20(2^n)}$$
(c) As $n\to\infty$, both $x_n$ and $y_n$ are dominated by the $80(4^n)$ term (since $4^n$ grows far faster than $2^n$), and the $\pm20(2^n)$ term becomes relatively negligible. Hence:
$$\frac{x_n}{y_n} = \frac{80(4^n)+20(2^n)}{80(4^n)-20(2^n)} \to \frac{80(4^n)}{80(4^n)} = 1 \text{ as } n\to\infty$$
$$\boxed{\text{Both populations grow without bound (dominated by the larger eigenvalue } \lambda=4\text{), with the ratio } x_n:y_n \to 1:1}$$
QUESTION 49
14 marks
Hard
A matrix has the form $A = \begin{pmatrix} a & 2 \\ 4 & 3 \end{pmatrix}$, where $a$ is a constant. It is known that $\vec{v}=\begin{pmatrix} 1 \\ 4 \end{pmatrix}$ is an eigenvector of $A$.
(a) By considering the second row of $A\vec{v}=\lambda\vec{v}$, find the eigenvalue $\lambda_1$ corresponding to $\vec{v}$, and hence find the value of $a$. [3]
(b) Using the trace and determinant of $A$, find the second eigenvalue $\lambda_2$, and find a corresponding eigenvector. [5]
(c) Hence diagonalize $A$ and find $A^3$. [6]
Show complete worked solution
(a) The second row of $A\vec{v}=\lambda\vec{v}$ does not involve $a$:
$$4(1)+3(4) = \lambda(4) \implies 4+12=4\lambda \implies 16=4\lambda \implies \lambda_1=4$$
Using the first row, $a(1)+2(4)=\lambda_1(1)=4$:
$$a+8=4 \implies \boxed{a=-4, \quad \lambda_1=4}$$
So $A=\begin{pmatrix}-4&2\\4&3\end{pmatrix}$.
(b) The trace of $A$ is $-4+3=-1$ and the determinant is $\det(A)=-4(3)-2(4)=-12-8=-20$.
Since $\lambda_1+\lambda_2=\text{trace}(A)$:
$$4+\lambda_2=-1 \implies \boxed{\lambda_2=-5}$$
Check via the determinant: $\lambda_1\lambda_2=4(-5)=-20=\det(A)$ $\checkmark$
For $\lambda_2=-5$, solve $(A+5I)\vec{v}=\vec{0}$:
$$\begin{pmatrix}1&2\\4&8\end{pmatrix}\vec{v}=\vec{0} \implies x+2y=0 \implies x=-2y$$
Taking $y=-1$:
$$\boxed{\text{Eigenvector for } \lambda_2=-5: \begin{pmatrix}2\\-1\end{pmatrix}}$$
(c) Let $P=\begin{pmatrix}1&2\\4&-1\end{pmatrix}$, $\Lambda=\begin{pmatrix}4&0\\0&-5\end{pmatrix}$.
$$\det(P) = 1(-1)-2(4) = -9$$
$$P^{-1} = \frac{1}{-9}\begin{pmatrix}-1&-2\\-4&1\end{pmatrix} = \begin{pmatrix}1/9&2/9\\4/9&-1/9\end{pmatrix}$$
$$\boxed{A = P\Lambda P^{-1}, \text{ with } P=\begin{pmatrix}1&2\\4&-1\end{pmatrix}, \Lambda=\begin{pmatrix}4&0\\0&-5\end{pmatrix}, P^{-1}=\begin{pmatrix}1/9&2/9\\4/9&-1/9\end{pmatrix}}$$
Since $A^3=P\Lambda^3P^{-1}$, where $\Lambda^3=\begin{pmatrix}64&0\\0&-125\end{pmatrix}$:
First compute $P\Lambda^3$:
$$P\Lambda^3 = \begin{pmatrix}1&2\\4&-1\end{pmatrix}\begin{pmatrix}64&0\\0&-125\end{pmatrix} = \begin{pmatrix}64&-250\\256&125\end{pmatrix}$$
Then multiply by $P^{-1}$:
$$A^3 = \begin{pmatrix}64&-250\\256&125\end{pmatrix}\begin{pmatrix}1/9&2/9\\4/9&-1/9\end{pmatrix}$$
$$\text{Entry }(1,1):\; \frac{64-1000}{9}=\frac{-936}{9}=-104 \qquad \text{Entry }(1,2):\; \frac{128+250}{9}=\frac{378}{9}=42$$
$$\text{Entry }(2,1):\; \frac{256+500}{9}=\frac{756}{9}=84 \qquad \text{Entry }(2,2):\; \frac{512-125}{9}=\frac{387}{9}=43$$
$$\boxed{A^3 = \begin{pmatrix}-104&42\\84&43\end{pmatrix}}$$
QUESTION 50
12 marks
Hard
Triangle $V$ has vertices $(1,1)$, $(4,1)$ and $(1,5)$. It undergoes a combined transformation consisting of, in order: a reflection in the $x$-axis, then a rotation of $90$ degrees clockwise about the origin, then an enlargement, centre the origin, with scale factor 2.
(a) Find the single matrix $M$ representing this combined transformation. [4]
(b) Find the images of the three vertices of $V$ under $M$. [3]
(c) Find $\det(M)$, and hence state the area scale factor of the transformation. Verify your answer by computing the areas of $V$ and its image directly. [5]
Show complete worked solution
(a) The three individual matrices are:
$$F=\begin{pmatrix}1&0\\0&-1\end{pmatrix} \text{ (reflection in } x\text{-axis)}, \quad C=\begin{pmatrix}0&1\\-1&0\end{pmatrix} \text{ (rotation } 90^\circ \text{ clockwise)}, \quad E=\begin{pmatrix}2&0\\0&2\end{pmatrix} \text{ (enlargement)}$$
Since the transformations are applied in the order $F$, then $C$, then $E$, the combined matrix is $M=ECF$.
First compute $CF$:
$$CF = \begin{pmatrix}0&1\\-1&0\end{pmatrix}\begin{pmatrix}1&0\\0&-1\end{pmatrix} = \begin{pmatrix}0(1)+1(0)&0(0)+1(-1)\\-1(1)+0(0)&-1(0)+0(-1)\end{pmatrix} = \begin{pmatrix}0&-1\\-1&0\end{pmatrix}$$
Then compute $E(CF)$:
$$M = \begin{pmatrix}2&0\\0&2\end{pmatrix}\begin{pmatrix}0&-1\\-1&0\end{pmatrix} = \begin{pmatrix}0&-2\\-2&0\end{pmatrix}$$
$$\boxed{M = \begin{pmatrix}0&-2\\-2&0\end{pmatrix}}$$
(b) Applying $M$ to each vertex:
$$M\begin{pmatrix}1\\1\end{pmatrix}=\begin{pmatrix}-2\\-2\end{pmatrix}, \quad M\begin{pmatrix}4\\1\end{pmatrix}=\begin{pmatrix}-2\\-8\end{pmatrix}, \quad M\begin{pmatrix}1\\5\end{pmatrix}=\begin{pmatrix}-10\\-2\end{pmatrix}$$
$$\boxed{(1,1)\to(-2,-2), \;\; (4,1)\to(-2,-8), \;\; (1,5)\to(-10,-2)}$$
(c) $$\det(M) = 0(0)-(-2)(-2) = -4$$
The area scale factor of a transformation is $|\det(M)|$:
$$\boxed{\text{Area scale factor} = |-4| = 4}$$
Verification: the area of $V$ with vertices $(1,1),(4,1),(1,5)$ is
$$\text{Area} = \frac12\big|1(1-5)+4(5-1)+1(1-1)\big| = \frac12|-4+16+0| = \frac12(12) = 6$$
The area of the image with vertices $(-2,-2),(-2,-8),(-10,-2)$ is
$$\text{Area} = \frac12\big|(-2)(-8-(-2))+(-2)(-2-(-2))+(-10)(-2-(-8))\big| = \frac12\big|(-2)(-6)+(-2)(0)+(-10)(6)\big|$$
$$= \frac12|12+0-60| = \frac12(48) = 24$$
Ratio of areas: $\dfrac{24}{6}=4$, which matches $|\det(M)|=4$ $\checkmark$
Systems of Linear Equations 50 questions
QUESTION 1
5 marks
Easy
A café sells coffee, muffins and sandwiches. The total sales (in dollars) from these three items were recorded on three different days:
Day 1: 3 coffees, 2 muffins and 1 sandwich cost $14.50 in total.
Day 2: 2 coffees, 1 muffin and 3 sandwiches cost $17.75 in total.
Day 3: 1 coffee, 4 muffins and 2 sandwiches cost $15.50 in total.
Let c, m and s be the price in dollars of one coffee, one muffin and one sandwich, respectively.
(a) Write down a system of three linear equations in c, m and s. [2]
(b) Use technology to solve this system, giving the price of each item. [3]
Show complete worked solution
(a) Let $c$, $m$, $s$ be the price in dollars of a coffee, muffin and sandwich respectively. Writing one equation for each day's total sales:
$$3c+2m+s=14.50$$
$$2c+m+3s=17.75$$
$$c+4m+2s=15.50$$
$$\boxed{\begin{aligned}3c+2m+s&=14.50\\2c+m+3s&=17.75\\c+4m+2s&=15.50\end{aligned}}$$
(b) Entering the coefficient matrix and constant vector into the GDC's simultaneous equation (or matrix) solver:
$$\begin{pmatrix}3&2&1\\2&1&3\\1&4&2\end{pmatrix}\begin{pmatrix}c\\m\\s\end{pmatrix}=\begin{pmatrix}14.50\\17.75\\15.50\end{pmatrix}$$
Solving gives:
$$c=2.70, \quad m=1.37, \quad s=3.66$$
$$\boxed{\text{A coffee costs \$2.70, a muffin costs \$1.37, and a sandwich costs \$3.66}}$$
QUESTION 2
4 marks
Easy
The deviation of the temperature inside a greenhouse from its daily average, $D(t)$ °C, t hours after midnight (0 $\leq$ t $\leq$ 6), is modelled by
$D(t) = t^3 - 9t^2 + 23t - 15$.
Use technology to find all values of t in the given domain for which the temperature equals the daily average (i.e. $D(t) = 0$).
Show complete worked solution
Using the GDC's polynomial solver (or by graphing $D(t)$ and finding the zeros) to solve
$$t^3-9t^2+23t-15=0$$
gives:
$$t=1, \quad t=3, \quad t=5$$
All three values lie within the given domain $0\le t\le6$.
$$\boxed{\text{The temperature equals the daily average at } t=1,3,5 \text{ hours after midnight (i.e. at 01:00, 03:00 and 05:00)}}$$
QUESTION 3
6 marks
Medium
In an electrical circuit with three loops, the loop currents $I_1$, $I_2$, $I_3$ (in amps) satisfy the following system of equations, obtained by applying Kirchhoff's voltage law to each loop:
$5I_1 - 2I_2 = 12$
$-2I_1 + 7I_2 - 3I_3 = 0$
$-3I_2 + 6I_3 = 9$
Use technology to solve this system of equations for $I_1$, $I_2$ and $I_3$, giving your answers correct to 3 significant figures.
Show complete worked solution
Writing the system in matrix form:
$$\begin{pmatrix}5&-2&0\\-2&7&-3\\0&-3&6\end{pmatrix}\begin{pmatrix}I_1\\I_2\\I_3\end{pmatrix}=\begin{pmatrix}12\\0\\9\end{pmatrix}$$
Solving using the GDC's matrix equation solver (or simultaneous equation solver):
$$I_1 = 3.19148\ldots \approx 3.19 \text{ A}$$
$$I_2 = 1.97872\ldots \approx 1.98 \text{ A}$$
$$I_3 = 2.48936\ldots \approx 2.49 \text{ A}$$
$$\boxed{I_1 \approx 3.19 \text{ A}, \quad I_2 \approx 1.98 \text{ A}, \quad I_3 \approx 2.49 \text{ A}}$$
QUESTION 4
6 marks
Medium
A furniture company models its monthly profit, $P(x)$, in thousands of dollars, as a function of the number of chairs sold, x (in hundreds), by
$P(x) = -2x^3 + 30x^2 - 100x - 50$, for $x \ge 0$.
(a) Use technology to find all real roots of $P(x) = 0$. [3]
(b) Hence state, with a reason, the value(s) of x (number of chairs sold, in hundreds) at which the company breaks even, within the given domain. [3]
Show complete worked solution
(a) Using the GDC's polynomial solver (or by graphing $P(x)$ and finding the zeros) to solve
$$-2x^3+30x^2-100x-50=0$$
gives:
$$x = -0.440, \quad x = 6.05, \quad x = 9.39 \text{ (each to 3 s.f.)}$$
$$\boxed{x \approx -0.440, \; 6.05, \; 9.39}$$
(b) The given domain is $x\ge0$ (x represents a number of chairs sold, which cannot be negative), so the root $x=-0.440$ is rejected as not physically meaningful.
$$\boxed{\text{The company breaks even at } x\approx6.05 \text{ and } x\approx9.39, \text{ i.e. when approximately 605 or 939 chairs are sold in a month}}$$
QUESTION 5
14 marks
Hard
A pedestrian suspension bridge is supported by three vertical cables. Static equilibrium analysis of the forces acting on the bridge deck gives the following system of equations for the tensions $T_1$, $T_2$, $T_3$ (in kN) in the three cables:
$T_1 + T_2 + T_3 = 45$
$2T_2 + 5T_3 = 99$
$3T_1 - 2T_3 = 24$
(a) Use technology to solve this system for $T_1$, $T_2$ and $T_3$. [4]
The vertical displacement of the bridge deck's edge from a horizontal reference line, $h(x)$ metres, at a horizontal distance x metres from the left support (where $0 \le x \le 12$, the span of the bridge), is modelled by
$h(x) = 0.002x^3 - 0.036x^2 + 0.108x$.
(b) Use technology to find all real solutions of $h(x) = 0$. [4]
(c) Hence determine, with a reason, all the points along the span of the bridge (i.e. values of x within the given domain) at which the deck's edge crosses the horizontal reference line. [6]
Show complete worked solution
(a) Writing the system in matrix form:
$$\begin{pmatrix}1&1&1\\0&2&5\\3&0&-2\end{pmatrix}\begin{pmatrix}T_1\\T_2\\T_3\end{pmatrix}=\begin{pmatrix}45\\99\\24\end{pmatrix}$$
Solving using the GDC's matrix/simultaneous equation solver:
$$\boxed{T_1 = 18 \text{ kN}, \quad T_2 = 12 \text{ kN}, \quad T_3 = 15 \text{ kN}}$$
(b) Using the GDC's polynomial solver (or by graphing $h(x)$ and finding the zeros) to solve
$$0.002x^3-0.036x^2+0.108x=0$$
gives:
$$x = 0, \quad x = 3.80, \quad x = 14.2 \text{ (each to 3 s.f.)}$$
$$\boxed{x = 0, \; 3.80, \; 14.2}$$
(c) The physical domain of the bridge span is $0\le x\le12$. The root $x=14.2$ lies outside this domain, so it is rejected as it does not correspond to a physical point on the bridge deck.
$$\boxed{\text{The deck's edge crosses the horizontal reference line at } x=0 \text{ m and } x\approx3.80 \text{ m from the left support}}$$
QUESTION 6
5 marks
Easy
A shop mixes almonds, costing \$8 per kg, with cashews, costing \$12 per kg, to make 10 kg of a nut mixture that costs \$100 in total.
Let $x$ be the mass in kg of almonds used and $y$ be the mass in kg of cashews used.
(a) Write down a system of two linear equations in $x$ and $y$. [2]
(b) Solve the system to find the mass of almonds and the mass of cashews used. [3]
Show complete worked solution
(a) Since the total mass of the mixture is 10 kg:
$$x+y=10$$
Since the total cost of the mixture is \$100:
$$8x+12y=100$$
$$\boxed{\begin{aligned}x+y&=10\\8x+12y&=100\end{aligned}}$$
(b) From the first equation, $y=10-x$. Substituting into the second equation:
$$8x+12(10-x)=100$$
$$8x+120-12x=100$$
$$-4x=-20$$
$$x=5$$
Then $y=10-5=5$.
$$\boxed{\text{5 kg of almonds and 5 kg of cashews are used}}$$
QUESTION 7
6 marks
Easy
The sum of Maya's age and her father's age is 54 years. In 6 years' time, her father's age will be twice Maya's age at that time.
Let $m$ be Maya's current age and $f$ be her father's current age.
(a) Write down a system of two linear equations in $m$ and $f$. [3]
(b) Use technology, or algebraic substitution, to solve the system and find Maya's current age and her father's current age. [3]
Show complete worked solution
(a) The sum of their current ages is 54:
$$m+f=54$$
In 6 years, the father's age ($f+6$) will be twice Maya's age at that time ($m+6$):
$$f+6=2(m+6)$$
which simplifies to
$$f-2m=6$$
$$\boxed{\begin{aligned}m+f&=54\\f-2m&=6\end{aligned}}$$
(b) Entering the system into the GDC's simultaneous equation solver (or solving by substitution: from the first equation $f=54-m$, so $54-m-2m=6 \Rightarrow 54-3m=6 \Rightarrow m=16$, then $f=54-16=38$):
$$m=16, \quad f=38$$
Check: $16+38=54$ ✓, and $38+6=44=2(16+6)=2(22)=44$ ✓
$$\boxed{\text{Maya is currently 16 years old and her father is currently 38 years old}}$$
QUESTION 8
6 marks
Easy
A cinema sells adult and child tickets. On one evening, 3 adult tickets and 2 child tickets cost \$52 in total. On another evening, 2 adult tickets and 5 child tickets cost \$64 in total.
Let $a$ be the price in dollars of one adult ticket and $c$ be the price in dollars of one child ticket.
(a) Write down a system of two linear equations in $a$ and $c$. [2]
(b) Use technology to solve the system and find the price of an adult ticket and the price of a child ticket. [4]
Show complete worked solution
(a) From the two evenings' sales:
$$3a+2c=52$$
$$2a+5c=64$$
$$\boxed{\begin{aligned}3a+2c&=52\\2a+5c&=64\end{aligned}}$$
(b) Entering the coefficients into the GDC's simultaneous equation solver:
$$\begin{pmatrix}3&2\\2&5\end{pmatrix}\begin{pmatrix}a\\c\end{pmatrix}=\begin{pmatrix}52\\64\end{pmatrix}$$
Solving gives:
$$a=12, \quad c=8$$
Check: $3(12)+2(8)=36+16=52$ ✓, and $2(12)+5(8)=24+40=64$ ✓
$$\boxed{\text{An adult ticket costs \$12 and a child ticket costs \$8}}$$
QUESTION 9
6 marks
Easy
A small business makes personalised t-shirts. The fixed monthly cost of running the business is \$160, and each t-shirt costs an additional \$6 to produce. Each t-shirt is sold for \$14.
Let $x$ be the number of t-shirts produced and sold in a month, and let $y$ be the corresponding amount in dollars (of cost, or of revenue).
(a) Write down a system of two linear equations in $x$ and $y$: one for the monthly cost $y$, and one for the monthly revenue $y$. [2]
(b) Use technology to solve the system and hence find the break-even point (the number of t-shirts for which cost equals revenue), and the corresponding value of $y$. [4]
Show complete worked solution
(a) The monthly cost is the fixed cost plus \$6 per t-shirt:
$$y=160+6x$$
The monthly revenue is \$14 per t-shirt:
$$y=14x$$
$$\boxed{\begin{aligned}y&=160+6x\\y&=14x\end{aligned}}$$
(b) At break-even, cost equals revenue, so both equations hold simultaneously. Substituting $y=14x$ into the first equation:
$$14x=160+6x$$
$$8x=160$$
$$x=20$$
Then $y=14(20)=280$.
$$\boxed{\text{The business breaks even at 20 t-shirts, at which point cost = revenue = \$280}}$$
QUESTION 10
6 marks
Easy
At a concert, 200 tickets were sold in total. Adult tickets cost \$15 each and student tickets cost \$9 each. The total revenue from ticket sales was \$2400.
Let $a$ be the number of adult tickets sold and $s$ be the number of student tickets sold.
(a) Write down a system of two linear equations in $a$ and $s$. [2]
(b) Solve the system to find the number of adult tickets and the number of student tickets sold. [4]
Show complete worked solution
(a) The total number of tickets sold:
$$a+s=200$$
The total revenue:
$$15a+9s=2400$$
$$\boxed{\begin{aligned}a+s&=200\\15a+9s&=2400\end{aligned}}$$
(b) From the first equation, $a=200-s$. Substituting into the second equation:
$$15(200-s)+9s=2400$$
$$3000-15s+9s=2400$$
$$-6s=-600$$
$$s=100$$
Then $a=200-100=100$.
$$\boxed{\text{100 adult tickets and 100 student tickets were sold}}$$
QUESTION 11
5 marks
Easy
Solve the following system of linear equations for $x$ and $y$:
$$2x+3y=35$$
$$5x-y=11$$
Show complete worked solution
From the second equation, $y=5x-11$.
Substituting into the first equation:
$$2x+3(5x-11)=35$$
$$2x+15x-33=35$$
$$17x=68$$
$$x=4$$
Then $y=5(4)-11=20-11=9$.
Check: $2(4)+3(9)=8+27=35$ ✓, and $5(4)-9=20-9=11$ ✓
$$\boxed{x=4, \quad y=9}$$
QUESTION 12
5 marks
Easy
Solve the following system of linear equations for $x$ and $y$:
$$x-y=3$$
$$2x+3y=21$$
Show complete worked solution
From the first equation, $x=y+3$.
Substituting into the second equation:
$$2(y+3)+3y=21$$
$$2y+6+3y=21$$
$$5y=15$$
$$y=3$$
Then $x=3+3=6$.
Check: $6-3=3$ ✓, and $2(6)+3(3)=12+9=21$ ✓
$$\boxed{x=6, \quad y=3}$$
QUESTION 13
7 marks
Easy
A rowing boat travels downstream (with the current) a distance of 60 km in 3 hours. On the return journey, travelling upstream (against the current) the same 60 km takes 5 hours.
Let $b$ be the speed of the boat in still water, in km/h, and $c$ be the speed of the current, in km/h.
(a) Show that the given information leads to the system of equations
$$b+c=20$$
$$b-c=12$$
[3]
(b) Solve this system to find the speed of the boat in still water and the speed of the current. [4]
Show complete worked solution
(a) Travelling downstream, the boat's speed relative to the ground is $b+c$. Since it covers 60 km in 3 hours:
$$b+c=\frac{60}{3}=20$$
Travelling upstream, the boat's speed relative to the ground is $b-c$. Since it covers 60 km in 5 hours:
$$b-c=\frac{60}{5}=12$$
$$\boxed{\begin{aligned}b+c&=20\\b-c&=12\end{aligned}}$$
(b) Adding the two equations:
$$2b=32$$
$$b=16$$
Substituting into $b+c=20$:
$$16+c=20 \Rightarrow c=4$$
$$\boxed{\text{The boat's speed in still water is 16 km/h and the speed of the current is 4 km/h}}$$
QUESTION 14
5 marks
Easy
Consider the system of equations
$$3x+2y=6$$
$$6x+4y=10$$
(a) By comparing the coefficients of the two equations, determine whether this system has a unique solution, no solution, or infinitely many solutions. Justify your answer. [3]
(b) Describe what this means geometrically for the two lines represented by these equations. [2]
Show complete worked solution
(a) In the second equation, the coefficients of $x$ and $y$ are exactly double those in the first equation: $6=2(3)$ and $4=2(2)$. If the system were consistent, the constant term would also need to satisfy this ratio, i.e. we would need $10=2(6)=12$. Since $10\neq12$, this is not the case.
Attempting to solve using the GDC's simultaneous equation solver confirms that no solution is returned (the coefficient matrix is singular, since row 2 is a multiple of row 1, but the constants are not in the same ratio).
$$\boxed{\text{The system has no solution}}$$
(b) Since the left-hand sides are proportional (same gradient) but the equations are not fully proportional (different "intercept" once scaled), the two lines are parallel and distinct.
$$\boxed{\text{The two lines are parallel and never intersect, so there is no point } (x,y) \text{ satisfying both equations}}$$
QUESTION 15
6 marks
Easy
Consider the system of equations
$$4x-2y=6$$
$$-6x+3y=-9$$
(a) Determine whether this system has a unique solution, no solution, or infinitely many solutions. Justify your answer. [3]
(b) If the system does not have a unique solution, express $y$ in terms of $x$ for the general solution. [3]
Show complete worked solution
(a) Multiplying the first equation by $-1.5$:
$$-1.5(4x-2y)=-1.5(6)$$
$$-6x+3y=-9$$
This is identical to the second equation. So the second equation is simply a scalar multiple of the first, meaning the two equations represent the same line.
$$\boxed{\text{The system has infinitely many solutions (the two equations are equivalent)}}$$
(b) Using the first equation to express $y$ in terms of $x$:
$$4x-2y=6$$
$$-2y=6-4x$$
$$y=2x-3$$
$$\boxed{y=2x-3, \text{ for any value of } x}$$
QUESTION 16
6 marks
Easy
Find the coordinates of the point(s) of intersection of the line $y=x-3$ and the curve $y=x^2-4x+1$.
(a) Form a single equation in $x$ by equating the two expressions for $y$, and hence find the possible values of $x$. [4]
(b) State the coordinates of the point(s) of intersection. [2]
Show complete worked solution
(a) Since both expressions equal $y$ at a point of intersection:
$$x-3=x^2-4x+1$$
$$0=x^2-5x+4$$
Using the GDC's polynomial (or quadratic) solver, or factorising as $(x-1)(x-4)=0$:
$$\boxed{x=1 \text{ or } x=4}$$
(b) When $x=1$: $y=1-3=-2$, giving the point $(1,-2)$.
When $x=4$: $y=4-3=1$, giving the point $(4,1)$.
$$\boxed{\text{The line and curve intersect at } (1,-2) \text{ and } (4,1)}$$
QUESTION 17
6 marks
Easy
Find the coordinates of the point(s) of intersection of the line $y=2x+3$ and the curve $y=x^2$.
(a) Form a single equation in $x$, and hence find the possible values of $x$. [4]
(b) State the coordinates of the point(s) of intersection. [2]
Show complete worked solution
(a) Setting the two expressions for $y$ equal:
$$x^2=2x+3$$
$$x^2-2x-3=0$$
Using the GDC's polynomial solver, or factorising as $(x-3)(x+1)=0$:
$$\boxed{x=-1 \text{ or } x=3}$$
(b) When $x=-1$: $y=(-1)^2=1$, giving the point $(-1,1)$.
When $x=3$: $y=(3)^2=9$, giving the point $(3,9)$.
$$\boxed{\text{The line and curve intersect at } (-1,1) \text{ and } (3,9)}$$
QUESTION 18
7 marks
Easy
A laboratory technician wants to make 12 litres of a 30% acid solution by mixing a 20% acid solution with a 50% acid solution.
Let $x$ be the number of litres of the 20% solution used, and $y$ be the number of litres of the 50% solution used.
(a) Write down a system of two linear equations in $x$ and $y$: one for the total volume, and one for the total amount of pure acid. [3]
(b) Solve the system to find how many litres of each solution should be used. [4]
Show complete worked solution
(a) The total volume of the mixture must be 12 litres:
$$x+y=12$$
The total amount of pure acid must equal 30% of 12 litres, i.e. 3.6 litres. The 20% solution contributes $0.2x$ litres of pure acid and the 50% solution contributes $0.5y$ litres of pure acid:
$$0.2x+0.5y=3.6$$
$$\boxed{\begin{aligned}x+y&=12\\0.2x+0.5y&=3.6\end{aligned}}$$
(b) From the first equation, $x=12-y$. Substituting into the second equation:
$$0.2(12-y)+0.5y=3.6$$
$$2.4-0.2y+0.5y=3.6$$
$$0.3y=1.2$$
$$y=4$$
Then $x=12-4=8$.
$$\boxed{\text{8 litres of the 20\% solution should be mixed with 4 litres of the 50\% solution}}$$
QUESTION 19
6 marks
Easy
A mobile phone company offers two plans. Plan A costs \$20 per month plus \$0.05 per minute of calls. Plan B costs \$35 per month plus \$0.02 per minute of calls.
Let $x$ be the number of minutes used in a month, and $y$ be the total monthly cost in dollars.
(a) Write down a system of two linear equations in $x$ and $y$, one for each plan. [2]
(b) Use technology to solve the system, and hence state the number of minutes for which the two plans cost the same amount, and what that cost is. [4]
Show complete worked solution
(a) For Plan A:
$$y=20+0.05x$$
For Plan B:
$$y=35+0.02x$$
$$\boxed{\begin{aligned}y&=20+0.05x\\y&=35+0.02x\end{aligned}}$$
(b) Setting the two expressions for $y$ equal:
$$20+0.05x=35+0.02x$$
$$0.03x=15$$
$$x=500$$
Then $y=20+0.05(500)=20+25=45$.
$$\boxed{\text{The two plans cost the same, \$45, at 500 minutes of calls}}$$
QUESTION 20
5 marks
Easy
Use technology to solve the following system of linear equations for $x$ and $y$:
$$3.5x+2y=17$$
$$x-1.5y=-5.5$$
Show complete worked solution
Entering the coefficients into the GDC's simultaneous equation solver:
$$\begin{pmatrix}3.5&2\\1&-1.5\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}17\\-5.5\end{pmatrix}$$
gives:
$$x=2, \quad y=5$$
Check: $3.5(2)+2(5)=7+10=17$ ✓, and $2-1.5(5)=2-7.5=-5.5$ ✓
$$\boxed{x=2, \quad y=5}$$
QUESTION 21
7 marks
Easy
A total of \$8000 is invested across two accounts: one paying 4% annual simple interest, and another paying 6% annual simple interest. After one year, the total interest earned from both accounts is \$380.
Let $x$ be the amount in dollars invested at 4%, and $y$ be the amount in dollars invested at 6%.
(a) Write down a system of two linear equations in $x$ and $y$. [3]
(b) Solve the system to find the amount invested in each account. [4]
Show complete worked solution
(a) The total amount invested:
$$x+y=8000$$
The total interest earned after one year:
$$0.04x+0.06y=380$$
$$\boxed{\begin{aligned}x+y&=8000\\0.04x+0.06y&=380\end{aligned}}$$
(b) From the first equation, $x=8000-y$. Substituting into the second equation:
$$0.04(8000-y)+0.06y=380$$
$$320-0.04y+0.06y=380$$
$$0.02y=60$$
$$y=3000$$
Then $x=8000-3000=5000$.
$$\boxed{\text{\$5000 is invested at 4\% and \$3000 is invested at 6\%}}$$
QUESTION 22
6 marks
Easy
Two candles are lit at the same instant. Candle A has an initial height of 20 cm and burns at a rate of 2 cm per hour. Candle B has an initial height of 15 cm and also burns at a rate of 2 cm per hour.
Let $h$ be the height of a candle, in cm, after $t$ hours.
(a) Write down a system of two linear equations in $h$ and $t$, one for each candle's height. [2]
(b) Attempt to solve the system to find a time at which the two candles have equal height. Explain, with reference to the algebra, what happens, and interpret your finding physically. [4]
Show complete worked solution
(a) For candle A:
$$h=20-2t \quad \Leftrightarrow \quad h+2t=20$$
For candle B:
$$h=15-2t \quad \Leftrightarrow \quad h+2t=15$$
$$\boxed{\begin{aligned}h+2t&=20\\h+2t&=15\end{aligned}}$$
(b) Subtracting the second equation from the first:
$$(h+2t)-(h+2t)=20-15$$
$$0=5$$
This is a false statement, independent of $t$, so the system has no solution.
$$\boxed{\text{No solution exists; since both candles burn at the same rate, the 5 cm gap between their heights never changes, so they are never equal in height}}$$
QUESTION 23
5 marks
Easy
Use technology to solve the following system of linear equations for $x$ and $y$:
$$7x-3y=25$$
$$4x+5y=21$$
Show complete worked solution
Entering the coefficients into the GDC's simultaneous equation solver:
$$\begin{pmatrix}7&-3\\4&5\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}25\\21\end{pmatrix}$$
gives:
$$x=4, \quad y=1$$
Check: $7(4)-3(1)=28-3=25$ ✓, and $4(4)+5(1)=16+5=21$ ✓
$$\boxed{x=4, \quad y=1}$$
QUESTION 24
8 marks
Medium
A fruit stand sells apples, bananas and oranges by weight. The table below shows three separate purchases made by different customers, and the total price paid for each.
Let $a$, $b$ and $o$ be the price per kg, in dollars, of apples, bananas and oranges respectively.
(a) Write down a system of three linear equations in $a$, $b$ and $o$. [3]
(b) Use technology to solve the system for $a$, $b$ and $o$. [3]
(c) Hence find the total cost of buying 2 kg of each fruit. [2]
| Purchase | Apples (kg) | Bananas (kg) | Oranges (kg) | Total price |
|---|---|---|---|---|
| 1 | 2 | 3 | 1 | \$7.70 |
| 2 | 1 | 2 | 3 | \$9.30 |
| 3 | 3 | 1 | 2 | \$9.40 |
Show complete worked solution
(a) From the three purchases:
$$2a+3b+o=7.70$$
$$a+2b+3o=9.30$$
$$3a+b+2o=9.40$$
$$\boxed{\begin{aligned}2a+3b+o&=7.70\\a+2b+3o&=9.30\\3a+b+2o&=9.40\end{aligned}}$$
(b) Writing the system in matrix form:
$$\begin{pmatrix}2&3&1\\1&2&3\\3&1&2\end{pmatrix}\begin{pmatrix}a\\b\\o\end{pmatrix}=\begin{pmatrix}7.70\\9.30\\9.40\end{pmatrix}$$
Solving using the GDC's matrix (or simultaneous equation) solver:
$$\boxed{a=1.50, \quad b=0.90, \quad o=2.00}$$
(c) The cost of 2 kg of each fruit is:
$$2a+2b+2o=2(1.50)+2(0.90)+2(2.00)=3.00+1.80+4.00=8.80$$
$$\boxed{\text{The total cost is \$8.80}}$$
QUESTION 25
7 marks
Medium
A theatre sells adult, child and senior tickets. The table below shows the number of each type of ticket sold, and the total revenue, for three different performances.
Let $a$, $c$ and $s$ be the price in dollars of an adult, child and senior ticket respectively.
(a) Write down a system of three linear equations in $a$, $c$ and $s$. [2]
(b) Use technology to solve the system for $a$, $c$ and $s$. [5]
| Performance | Adult | Child | Senior | Revenue |
|---|---|---|---|---|
| 1 | 2 | 1 | 1 | \$46 |
| 2 | 1 | 3 | 2 | \$58 |
| 3 | 3 | 2 | 1 | \$68 |
Show complete worked solution
(a) From the three performances:
$$2a+c+s=46$$
$$a+3c+2s=58$$
$$3a+2c+s=68$$
$$\boxed{\begin{aligned}2a+c+s&=46\\a+3c+2s&=58\\3a+2c+s&=68\end{aligned}}$$
(b) Writing the system in matrix form:
$$\begin{pmatrix}2&1&1\\1&3&2\\3&2&1\end{pmatrix}\begin{pmatrix}a\\c\\s\end{pmatrix}=\begin{pmatrix}46\\58\\68\end{pmatrix}$$
Solving using the GDC's matrix (or simultaneous equation) solver:
$$\boxed{a=14, \quad c=8, \quad s=10}$$
Check (Performance 1): $2(14)+8+10=28+8+10=46$ ✓
QUESTION 26
7 marks
Medium
Use technology to solve the following system of linear equations for $x$, $y$ and $z$:
$$x+2y-z=-3$$
$$2x-y+3z=14$$
$$-x+3y+2z=1$$
Show complete worked solution
Writing the system in matrix form:
$$\begin{pmatrix}1&2&-1\\2&-1&3\\-1&3&2\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}-3\\14\\1\end{pmatrix}$$
Solving using the GDC's matrix equation solver (or simultaneous equation solver):
$$x=2, \quad y=-1, \quad z=3$$
Check: $2+2(-1)-3=2-2-3=-3$ ✓; $2(2)-(-1)+3(3)=4+1+9=14$ ✓; $-2+3(-1)+2(3)=-2-3+6=1$ ✓
$$\boxed{x=2, \quad y=-1, \quad z=3}$$
QUESTION 27
7 marks
Medium
Use technology to solve the following system of linear equations for $x$, $y$ and $z$:
$$3x-y+2z=11$$
$$x+4y-z=7$$
$$-2x+y+3z=-16$$
Show complete worked solution
Writing the system in matrix form:
$$\begin{pmatrix}3&-1&2\\1&4&-1\\-2&1&3\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}11\\7\\-16\end{pmatrix}$$
Solving using the GDC's matrix equation solver (or simultaneous equation solver):
$$x=5, \quad y=0, \quad z=-2$$
Check: $3(5)-0+2(-2)=15-4=11$ ✓; $5+4(0)-(-2)=5+2=7$ ✓; $-2(5)+0+3(-2)=-10-6=-16$ ✓
$$\boxed{x=5, \quad y=0, \quad z=-2}$$
QUESTION 28
7 marks
Medium
Two payment plans for a piece of gym equipment are advertised. Plan 1 has total cost $y$ (in dollars) after $x$ months given by $y-3x=15$. Plan 2 has total cost $y$ (in dollars) after $x$ months given by $2y-6x=30$.
(a) Show algebraically that this system of equations does not have a unique solution, and state what type of solution set it has. [3]
(b) Explain what this means in the context of the two payment plans. [4]
Show complete worked solution
(a) Multiplying the first equation by 2:
$$2(y-3x)=2(15)$$
$$2y-6x=30$$
This is identical to the second equation. Since one equation is simply a scalar multiple of the other, the system does not have a unique solution.
$$\boxed{\text{The system has infinitely many solutions, since the two equations are equivalent}}$$
(b) Since the two equations represent the same relationship between cost and time, Plan 1 and Plan 2 describe exactly the same cost structure for every value of $x$ (they were simply written in different but equivalent forms).
$$\boxed{\text{Plan 1 and Plan 2 cost exactly the same amount after any given number of months } x \text{, since the two equations are algebraically equivalent}}$$
QUESTION 29
7 marks
Medium
A chemist wants a mixture of two solutions, X and Y, such that the total volume is 10 litres, and such that the total volume, when scaled by 0.6, gives 4.8 litres of a certain reagent-equivalent.
Let $x$ and $y$ be the volumes in litres of solution X and solution Y respectively used in the mixture.
(a) Write down a system of two linear equations in $x$ and $y$ representing these two requirements. [2]
(b) Determine whether this system has a unique solution, no solution, or infinitely many solutions, showing your reasoning clearly. [5]
Show complete worked solution
(a) The total volume requirement:
$$x+y=10$$
The second requirement:
$$0.6x+0.6y=4.8$$
$$\boxed{\begin{aligned}x+y&=10\\0.6x+0.6y&=4.8\end{aligned}}$$
(b) Dividing the second equation by 0.6:
$$\frac{0.6x+0.6y}{0.6}=\frac{4.8}{0.6}$$
$$x+y=8$$
This states that $x+y=8$, which directly contradicts the first equation, $x+y=10$ (the same expression, $x+y$, cannot equal both 10 and 8).
$$\boxed{\text{The system has no solution; the two requirements are inconsistent (the coefficient matrix is singular but the equations represent parallel, non-coincident lines)}}$$
QUESTION 30
9 marks
Medium
Consider the system of equations
$$x+y+z=6$$
$$x-y+2z=3$$
$$2x+3z=9$$
(a) Show that the third equation can be obtained by adding the first two equations, and hence explain why this system does not have a unique solution. [3]
(b) Letting $z=t$, express $x$ and $y$ in terms of the parameter $t$ to give the general solution of the system. [6]
Show complete worked solution
(a) Adding the first two equations:
$$(x+y+z)+(x-y+2z)=6+3$$
$$2x+3z=9$$
This is exactly the third equation. So the third equation gives no new information beyond the first two; effectively there are only 2 independent equations in 3 unknowns.
$$\boxed{\text{The system does not have a unique solution; since one equation is redundant, there are infinitely many solutions forming a one-parameter family}}$$
(b) Let $z=t$. Substituting into the first two equations:
$$x+y=6-t \quad \text{(i)}$$
$$x-y=3-2t \quad \text{(ii)}$$
Adding (i) and (ii):
$$2x=9-3t$$
$$x=4.5-1.5t$$
Subtracting (ii) from (i):
$$2y=(6-t)-(3-2t)=3+t$$
$$y=1.5+0.5t$$
$$\boxed{x=4.5-1.5t, \quad y=1.5+0.5t, \quad z=t, \text{ for any value of } t}$$
QUESTION 31
8 marks
Medium
Consider the system of equations
$$x+y+z=5$$
$$x-y+z=2$$
$$2x+2y+2z=8$$
Determine whether this system has a unique solution, no solution, or infinitely many solutions. Show full working to justify your conclusion.
Show complete worked solution
Multiplying the first equation by 2:
$$2(x+y+z)=2(5)$$
$$2x+2y+2z=10$$
Comparing this with the third equation, $2x+2y+2z=8$: the left-hand sides are identical, but the right-hand sides are different ($10\neq8$).
This is a contradiction: the same expression $2x+2y+2z$ cannot equal both 10 and 8 at the same time. (Equivalently, entering the coefficient matrix into the GDC shows it is singular, since row 3 is a scalar multiple of row 1, but the augmented system is inconsistent.)
$$\boxed{\text{The system has no solution, since the first and third equations are inconsistent}}$$
QUESTION 32
8 marks
Medium
A line has equation $y=2x+6$ and a curve has equation $y=x^2-x-4$.
(a) Use technology, or algebraic substitution, to find the coordinates of the points of intersection of the line and the curve. [5]
(b) Find the distance between the two points of intersection, giving your answer in exact form. [3]
Show complete worked solution
(a) Setting the two expressions for $y$ equal:
$$x^2-x-4=2x+6$$
$$x^2-3x-10=0$$
Using the GDC's polynomial solver, or factorising as $(x-5)(x+2)=0$:
$$x=-2 \text{ or } x=5$$
When $x=-2$: $y=2(-2)+6=2$, giving the point $(-2,2)$.
When $x=5$: $y=2(5)+6=16$, giving the point $(5,16)$.
$$\boxed{\text{The points of intersection are } (-2,2) \text{ and } (5,16)}$$
(b) Using the distance formula:
$$d=\sqrt{(5-(-2))^2+(16-2)^2}=\sqrt{7^2+14^2}=\sqrt{49+196}=\sqrt{245}$$
$$\sqrt{245}=\sqrt{49\times5}=7\sqrt5$$
$$\boxed{d=7\sqrt5 \approx15.7 \text{ units}}$$
QUESTION 33
8 marks
Medium
A company's weekly revenue, in dollars, from producing and selling $x$ units of a product is modelled by $R(x)=-x^2+50x$. Its weekly cost, in dollars, is modelled by $C(x)=200+10x$.
(a) By setting $R(x)=C(x)$, form a quadratic equation in $x$ and write it in the form $x^2+bx+c=0$. [3]
(b) Use technology to solve this equation, giving your answers to 3 significant figures. [3]
(c) Interpret your answers to part (b) in the context of the company's weekly production. [2]
Show complete worked solution
(a) Setting revenue equal to cost:
$$-x^2+50x=200+10x$$
$$-x^2+40x-200=0$$
Multiplying by $-1$:
$$x^2-40x+200=0$$
$$\boxed{x^2-40x+200=0}$$
(b) Using the GDC's polynomial (or quadratic) solver, or the quadratic formula $x=\dfrac{40\pm\sqrt{40^2-4(1)(200)}}{2}=\dfrac{40\pm\sqrt{800}}{2}=20\pm10\sqrt2$:
$$x=5.86 \text{ or } x=34.1 \text{ (each to 3 s.f.)}$$
$$\boxed{x\approx5.86, \; 34.1}$$
(c) These are the break-even production levels, where weekly revenue equals weekly cost.
$$\boxed{\text{The company breaks even when it produces and sells approximately 5.86 units or approximately 34.1 units per week}}$$
QUESTION 34
9 marks
Medium
A vending machine contains only \$1, \$2 and \$5 coins. It contains 50 coins in total, worth \$158 altogether. The number of \$5 coins is 6 more than twice the number of \$1 coins.
Let $o$, $t$ and $f$ be the number of \$1, \$2 and \$5 coins respectively in the machine.
(a) Write down a system of three linear equations in $o$, $t$ and $f$. [3]
(b) Use technology to solve the system for $o$, $t$ and $f$. [4]
(c) State, in a sentence, the number of each type of coin in the machine. [2]
Show complete worked solution
(a) The total number of coins:
$$o+t+f=50$$
The total value of the coins:
$$o+2t+5f=158$$
The relationship between $f$ and $o$:
$$f=2o+6 \quad \Leftrightarrow \quad -2o+f=6$$
$$\boxed{\begin{aligned}o+t+f&=50\\o+2t+5f&=158\\-2o+f&=6\end{aligned}}$$
(b) Writing the system in matrix form:
$$\begin{pmatrix}1&1&1\\1&2&5\\-2&0&1\end{pmatrix}\begin{pmatrix}o\\t\\f\end{pmatrix}=\begin{pmatrix}50\\158\\6\end{pmatrix}$$
Solving using the GDC's matrix (or simultaneous equation) solver:
$$\boxed{o=8, \quad t=20, \quad f=22}$$
(c)
$$\boxed{\text{The machine contains 8 \$1 coins, 20 \$2 coins and 22 \$5 coins}}$$
QUESTION 35
8 marks
Medium
A print shop charges a fixed rate per black-and-white copy, per colour copy, and per binding. Three customer orders and their total prices are shown below.
Let $x$, $y$ and $z$ be the price in dollars of one B&W copy, one colour copy, and one binding respectively.
(a) Write down a system of three linear equations in $x$, $y$ and $z$. [2]
(b) Use technology to solve the system for $x$, $y$ and $z$. [6]
| Order | B&W copies | Colour copies | Bindings | Total price |
|---|---|---|---|---|
| 1 | 50 | 10 | 5 | \$12.50 |
| 2 | 30 | 20 | 2 | \$9.50 |
| 3 | 100 | 5 | 8 | \$18.25 |
Show complete worked solution
(a) From the three orders:
$$50x+10y+5z=12.50$$
$$30x+20y+2z=9.50$$
$$100x+5y+8z=18.25$$
$$\boxed{\begin{aligned}50x+10y+5z&=12.50\\30x+20y+2z&=9.50\\100x+5y+8z&=18.25\end{aligned}}$$
(b) Writing the system in matrix form:
$$\begin{pmatrix}50&10&5\\30&20&2\\100&5&8\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}12.50\\9.50\\18.25\end{pmatrix}$$
Solving using the GDC's matrix equation solver:
$$\boxed{x=0.05, \quad y=0.25, \quad z=1.50}$$
That is, a B&W copy costs 5 cents, a colour copy costs 25 cents, and a binding costs \$1.50.
QUESTION 36
8 marks
Medium
Consider the system of equations
$$x+y+z=6$$
$$x-y+z=1$$
$$2x+2y+2z=12$$
(a) Show that this system does not have a unique solution, and state the type of solution set it has. [3]
(b) Letting $z=t$, find the general solution of the system in terms of $t$. [5]
Show complete worked solution
(a) Multiplying the first equation by 2:
$$2(x+y+z)=2(6)$$
$$2x+2y+2z=12$$
This is identical to the third equation, so the third equation is redundant (it gives no new information). There are effectively only 2 independent equations in 3 unknowns.
$$\boxed{\text{The system does not have a unique solution; it has infinitely many solutions, forming a one-parameter family}}$$
(b) Let $z=t$. Substituting into the first and second equations:
$$x+y=6-t \quad \text{(i)}$$
$$x-y=1-t \quad \text{(ii)}$$
Adding (i) and (ii):
$$2x=7-2t$$
$$x=3.5-t$$
Subtracting (ii) from (i):
$$2y=(6-t)-(1-t)=5$$
$$y=2.5$$
$$\boxed{x=3.5-t, \quad y=2.5, \quad z=t, \text{ for any value of } t}$$
QUESTION 37
7 marks
Medium
A line has equation $y=3x+2$ and a curve has equation $y=x^2-3x+2$.
(a) Use technology, or algebraic substitution, to find the coordinates of the points of intersection of the line and the curve. [5]
(b) Verify your answer to part (a) by substituting one of the points back into both original equations. [2]
Show complete worked solution
(a) Setting the two expressions for $y$ equal:
$$x^2-3x+2=3x+2$$
$$x^2-6x=0$$
$$x(x-6)=0$$
$$x=0 \text{ or } x=6$$
When $x=0$: $y=3(0)+2=2$, giving $(0,2)$.
When $x=6$: $y=3(6)+2=20$, giving $(6,20)$.
$$\boxed{\text{The points of intersection are } (0,2) \text{ and } (6,20)}$$
(b) Checking $(6,20)$ in the line equation: $y=3(6)+2=18+2=20$ ✓
Checking $(6,20)$ in the curve equation: $y=6^2-3(6)+2=36-18+2=20$ ✓
$$\boxed{\text{The point } (6,20) \text{ satisfies both original equations, confirming the intersection is correct}}$$
QUESTION 38
7 marks
Medium
The three interior angles of a triangle are $A$, $B$ and $C$ degrees. Angle $A$ is twice angle $B$. Angle $C$ is $30\degree$ more than angle $B$.
(a) Write down a system of three linear equations in $A$, $B$ and $C$. [3]
(b) Use technology to solve the system for $A$, $B$ and $C$. [4]
Show complete worked solution
(a) The angles of a triangle sum to $180\degree$:
$$A+B+C=180$$
Angle $A$ is twice angle $B$:
$$A=2B \quad \Leftrightarrow \quad A-2B=0$$
Angle $C$ is $30$ more than angle $B$:
$$C=B+30 \quad \Leftrightarrow \quad -B+C=30$$
$$\boxed{\begin{aligned}A+B+C&=180\\A-2B&=0\\-B+C&=30\end{aligned}}$$
(b) Writing the system in matrix form:
$$\begin{pmatrix}1&1&1\\1&-2&0\\0&-1&1\end{pmatrix}\begin{pmatrix}A\\B\\C\end{pmatrix}=\begin{pmatrix}180\\0\\30\end{pmatrix}$$
Solving using the GDC's matrix (or simultaneous equation) solver:
$$\boxed{A=75\degree, \quad B=37.5\degree, \quad C=67.5\degree}$$
Check: $75+37.5+67.5=180$ ✓
QUESTION 39
7 marks
Medium
Consider the system of equations
$$2x-y+3z=5$$
$$4x-2y+6z=11$$
$$x+y-z=4$$
(a) By comparing the first and second equations, determine whether this system has a unique solution, no solution, or infinitely many solutions. [3]
(b) Justify your conclusion fully. [4]
Show complete worked solution
(a) Multiplying the first equation by 2:
$$2(2x-y+3z)=2(5)$$
$$4x-2y+6z=10$$
$$\boxed{\text{The system has no solution}}$$
(b) The left-hand side of this equation, $4x-2y+6z$, is identical to the left-hand side of the second equation. But the second equation states $4x-2y+6z=11$, while doubling the first equation gives $4x-2y+6z=10$. Since $10\neq11$, these two statements are contradictory: the same expression cannot equal both 10 and 11.
Attempting to solve the full 3-variable system using the GDC's matrix solver confirms the coefficient matrix is singular (row 2 equals $2\times$row 1), and the system is inconsistent, regardless of the third equation.
$$\boxed{\text{The system has no solution, since the first and second equations are inconsistent}}$$
QUESTION 40
9 marks
Medium
A farm has only chickens, goats and cows, 30 animals in total. Chickens have 2 legs, while goats and cows each have 4 legs; the animals have 100 legs in total. Chickens are worth \$5 each, goats \$40 each, and cows \$80 each, with a total value of \$1170.
Let $x$, $y$ and $z$ be the number of chickens, goats and cows respectively.
(a) Write down a system of three linear equations in $x$, $y$ and $z$. [3]
(b) Use technology to solve the system for $x$, $y$ and $z$. [4]
(c) State, in a sentence, the number of each type of animal on the farm. [2]
Show complete worked solution
(a) The total number of animals:
$$x+y+z=30$$
The total number of legs:
$$2x+4y+4z=100$$
The total value:
$$5x+40y+80z=1170$$
$$\boxed{\begin{aligned}x+y+z&=30\\2x+4y+4z&=100\\5x+40y+80z&=1170\end{aligned}}$$
(b) Writing the system in matrix form:
$$\begin{pmatrix}1&1&1\\2&4&4\\5&40&80\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}30\\100\\1170\end{pmatrix}$$
Solving using the GDC's matrix equation solver:
$$\boxed{x=10, \quad y=12, \quad z=8}$$
(c)
$$\boxed{\text{The farm has 10 chickens, 12 goats and 8 cows}}$$
QUESTION 41
8 marks
Medium
A ball is launched vertically so that its height above the ground, in metres, after $t$ seconds is modelled by $h(t)=-5t^2+20t$. At the same instant, a drone is flying so that its height above the ground, in metres, after $t$ seconds is modelled by $d(t)=5t+3$.
(a) By setting $h(t)=d(t)$, form a quadratic equation in $t$, writing it in the form $at^2+bt+c=0$ with integer coefficients. [3]
(b) Use technology to solve this equation, giving your answers to 3 significant figures. [3]
(c) Given that $t\geq0$, state the times at which the ball and the drone are at the same height. [2]
Show complete worked solution
(a) Setting the two height expressions equal:
$$-5t^2+20t=5t+3$$
$$-5t^2+15t-3=0$$
Multiplying by $-1$:
$$5t^2-15t+3=0$$
$$\boxed{5t^2-15t+3=0}$$
(b) Using the GDC's polynomial (or quadratic) solver, or the quadratic formula $t=\dfrac{15\pm\sqrt{15^2-4(5)(3)}}{2(5)}=\dfrac{15\pm\sqrt{165}}{10}$:
$$t=0.215 \text{ or } t=2.78 \text{ (each to 3 s.f.)}$$
$$\boxed{t\approx0.215, \; 2.78}$$
(c) Both solutions satisfy $t\geq0$, so both are valid.
$$\boxed{\text{The ball and the drone are at the same height at approximately } t=0.215 \text{ s and } t=2.78 \text{ s after launch}}$$
QUESTION 42
12 marks
Hard
A factory manufactures $x$, $y$ and $z$ units per week of three products, subject to the following resource constraints (in hours):
$$x+2y+z=100 \quad \text{(machine hours)}$$
$$2x+y+3z=140 \quad \text{(labour hours)}$$
$$3x+3y+4z=240 \quad \text{(total combined hours)}$$
(a) Show that the third equation provides no new information beyond the first two, and hence explain why the system does not have a unique solution. [3]
(b) Letting $z=t$, find the general solution of the system, expressing $x$ and $y$ in terms of $t$. [5]
(c) Given that $x$, $y$ and $z$ must all be non-negative (the factory cannot produce a negative number of units), find the range of values that $t$ can take. [4]
Show complete worked solution
(a) Adding the first two equations:
$$(x+2y+z)+(2x+y+3z)=100+140$$
$$3x+3y+4z=240$$
This is exactly the third equation, so it is a linear combination of the first two and provides no independent constraint.
$$\boxed{\text{The system does not have a unique solution; with only 2 independent equations in 3 unknowns, there are infinitely many solutions}}$$
(b) Let $z=t$. Substituting into the first two equations:
$$x+2y=100-t \quad \text{(i)}$$
$$2x+y=140-3t \quad \text{(ii)}$$
From (i): $x=100-t-2y$. Substituting into (ii):
$$2(100-t-2y)+y=140-3t$$
$$200-2t-4y+y=140-3t$$
$$200-2t-3y=140-3t$$
$$-3y=140-3t-200+2t=-60-t$$
$$y=\frac{60+t}{3}=20+\frac{t}{3}$$
Then:
$$x=100-t-2\left(20+\frac{t}{3}\right)=100-t-40-\frac{2t}{3}=60-\frac{5t}{3}$$
$$\boxed{x=60-\frac{5}{3}t, \quad y=20+\frac{1}{3}t, \quad z=t}$$
(c) Requiring $z=t\geq0$.
Requiring $y=20+\frac{1}{3}t\geq0$: since $t\geq0$, this is automatically satisfied (in fact $y\geq20$ always).
Requiring $x=60-\frac{5}{3}t\geq0$:
$$60\geq\frac{5}{3}t$$
$$t\leq60\times\frac{3}{5}=36$$
Combining all three conditions:
$$\boxed{0\leq t\leq36}$$
QUESTION 43
12 marks
Hard
A metal recycler blends $x$, $y$ and $z$ tonnes of three types of scrap for three separate targets:
$$x+y+z=500 \quad \text{(total mass target, tonnes)}$$
$$0.4x+0.4y+0.4z=180 \quad \text{(a purity-weighted target)}$$
$$x-y+2z=50 \quad \text{(a composition balance target)}$$
(a) By simplifying the second equation, show that this system of three equations has no solution. [4]
(b) Explain, in the context of the recycler's targets, why this occurs. [3]
(c) Suppose the purity-weighted target is corrected to $0.4x+0.4y+0.4z=200$. Show that the corrected system (using this new second equation, together with the first and third equations) has infinitely many solutions, and find the general solution in terms of a parameter $t=z$. [5]
Show complete worked solution
(a) Dividing the second equation by 0.4:
$$\frac{0.4x+0.4y+0.4z}{0.4}=\frac{180}{0.4}$$
$$x+y+z=450$$
This directly contradicts the first equation, $x+y+z=500$: the same expression $x+y+z$ cannot equal both 500 and 450.
$$\boxed{\text{The system has no solution, since the first and second equations are inconsistent}}$$
(b) The total mass target requires the three types of scrap to sum to 500 tonnes, but the purity-weighted target (once simplified) effectively requires them to sum to only 450 tonnes. These two targets cannot both be met simultaneously, so no blend $(x,y,z)$ satisfies every requirement.
$$\boxed{\text{The two targets are set inconsistently with each other; no combination of scrap masses can satisfy both the total mass requirement and the purity-weighted requirement at once}}$$
(c) With the corrected second equation, dividing by 0.4 gives $x+y+z=500$, which is now identical to the first equation. So the corrected system has only 2 independent equations (the first/second, and the third) in 3 unknowns, meaning it has infinitely many solutions.
Let $z=t$. From $x+y+z=500$: $x+y=500-t$ (i). From $x-y+2z=50$: $x-y=50-2t$ (ii).
Adding (i) and (ii):
$$2x=550-3t$$
$$x=275-1.5t$$
Subtracting (ii) from (i):
$$2y=(500-t)-(50-2t)=450+t$$
$$y=225+0.5t$$
$$\boxed{x=275-1.5t, \quad y=225+0.5t, \quad z=t, \text{ for any value of } t}$$
QUESTION 44
12 marks
Hard
A radio antenna is supported by three guy wires with tensions $T_1$, $T_2$, $T_3$ (in newtons), which satisfy the following system of equations, obtained from a static equilibrium analysis:
$$T_1+T_2+T_3=850$$
$$2T_1-T_3=200$$
$$T_2+2T_3=900$$
(a) Use technology to solve this system for $T_1$, $T_2$ and $T_3$. [4]
The wind-induced sway of the antenna's tip from vertical, $s(x)$ cm, at a height $x$ metres above the base (where $0\leq x\leq20$, the height of the antenna), is modelled by
$$s(x)=0.01x^3-0.3x^2+2x$$
(b) Use technology to find all real solutions of $s(x)=0$ in the given domain. [4]
(c) Hence determine, with a reason, all heights along the antenna at which the sway is zero. [4]
Show complete worked solution
(a) Writing the system in matrix form:
$$\begin{pmatrix}1&1&1\\2&0&-1\\0&1&2\end{pmatrix}\begin{pmatrix}T_1\\T_2\\T_3\end{pmatrix}=\begin{pmatrix}850\\200\\900\end{pmatrix}$$
Solving using the GDC's matrix (or simultaneous equation) solver:
$$\boxed{T_1=250 \text{ N}, \quad T_2=300 \text{ N}, \quad T_3=300 \text{ N}}$$
Check: $250+300+300=850$ ✓; $2(250)-300=500-300=200$ ✓; $300+2(300)=300+600=900$ ✓
(b) Using the GDC's polynomial solver (or by graphing $s(x)$ and finding the zeros) to solve
$$0.01x^3-0.3x^2+2x=0$$
Factoring out $x$:
$$x(0.01x^2-0.3x+2)=0$$
So $x=0$, or $0.01x^2-0.3x+2=0$. Multiplying the quadratic factor by 100: $x^2-30x+200=0$, which factorises as $(x-10)(x-20)=0$, giving $x=10$ or $x=20$.
$$\boxed{x=0, \quad x=10, \quad x=20}$$
(c) All three solutions lie within the given domain $0\leq x\leq20$.
$$\boxed{\text{The sway is zero at the base } (x=0\text{ m}), \text{ at } x=10\text{ m}, \text{ and at the very top of the antenna } (x=20\text{ m})}$$
QUESTION 45
10 marks
Hard
Use technology to solve the following system of linear equations for $x$, $y$ and $z$:
$$2x-3y+z=16$$
$$4x+y-2z=2$$
$$-x+2y+3z=5$$
Give full working, and verify your solution by substituting back into all three original equations.
Show complete worked solution
Writing the system in matrix form:
$$\begin{pmatrix}2&-3&1\\4&1&-2\\-1&2&3\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}16\\2\\5\end{pmatrix}$$
Solving using the GDC's matrix equation solver:
$$x=3, \quad y=-2, \quad z=4$$
Verification by direct substitution into each original equation:
Equation 1: $2(3)-3(-2)+4=6+6+4=16$ ✓
Equation 2: $4(3)+(-2)-2(4)=12-2-8=2$ ✓
Equation 3: $-(3)+2(-2)+3(4)=-3-4+12=5$ ✓
All three equations are satisfied, confirming the solution is correct.
$$\boxed{x=3, \quad y=-2, \quad z=4}$$
QUESTION 46
11 marks
Hard
A line has equation $y=2x+16$ and a curve has equation $y=x^2-2x-5$.
(a) Use technology, or algebraic substitution, to find the coordinates of the points of intersection of the line and the curve. [5]
(b) Find the exact length of the line segment joining the two points of intersection. [3]
(c) Find the coordinates of the midpoint of this line segment. [3]
Show complete worked solution
(a) Setting the two expressions for $y$ equal:
$$x^2-2x-5=2x+16$$
$$x^2-4x-21=0$$
Using the GDC's polynomial solver, or factorising as $(x-7)(x+3)=0$:
$$x=-3 \text{ or } x=7$$
When $x=-3$: $y=2(-3)+16=10$, giving $(-3,10)$.
When $x=7$: $y=2(7)+16=30$, giving $(7,30)$.
$$\boxed{\text{The points of intersection are } (-3,10) \text{ and } (7,30)}$$
(b) Using the distance formula:
$$d=\sqrt{(7-(-3))^2+(30-10)^2}=\sqrt{10^2+20^2}=\sqrt{100+400}=\sqrt{500}$$
$$\sqrt{500}=\sqrt{100\times5}=10\sqrt5$$
$$\boxed{d=10\sqrt5 \text{ units } (\approx22.4 \text{ units})}$$
(c) Using the midpoint formula:
$$M=\left(\frac{-3+7}{2}, \frac{10+30}{2}\right)=\left(\frac{4}{2}, \frac{40}{2}\right)=(2,20)$$
$$\boxed{M=(2,20)}$$
QUESTION 47
10 marks
Hard
The supply price for a product, in dollars, when $q$ units are produced is modelled by $p=0.5q^2-2q+10$. The demand price, in dollars, for $q$ units is modelled by $p=-3q+40$.
(a) By setting the two expressions for $p$ equal, form a quadratic equation in $q$, writing it in the form $q^2+bq+c=0$. [3]
(b) Use technology to solve this equation, giving your answers to 3 significant figures. [3]
(c) Given that $q\geq0$, find the equilibrium quantity and the equilibrium price, each to 3 significant figures. [4]
Show complete worked solution
(a) Setting the two expressions for $p$ equal:
$$0.5q^2-2q+10=-3q+40$$
$$0.5q^2+q-30=0$$
Multiplying by 2:
$$q^2+2q-60=0$$
$$\boxed{q^2+2q-60=0}$$
(b) Using the GDC's polynomial (or quadratic) solver, or the quadratic formula $q=\dfrac{-2\pm\sqrt{2^2-4(1)(-60)}}{2}=\dfrac{-2\pm\sqrt{244}}{2}$:
$$q=6.81 \text{ or } q=-8.81 \text{ (each to 3 s.f.)}$$
$$\boxed{q\approx6.81, \; -8.81}$$
(c) Since $q\geq0$ (a quantity cannot be negative), the solution $q\approx-8.81$ is rejected.
$$q\approx6.81$$
Substituting into the demand equation:
$$p=-3(6.81)+40=-20.43+40=19.57\approx19.6$$
$$\boxed{\text{The equilibrium quantity is } q\approx6.81 \text{ units, at an equilibrium price of } p\approx\$19.6}$$
QUESTION 48
11 marks
Hard
Three pipes, A, B and C, can each fill an empty tank on their own at constant (but different) rates. Let $a$, $b$ and $c$ be the fraction of the tank that pipes A, B and C respectively can fill in one hour.
Working together, pipes A and B can fill the tank in 2 hours. Working together, pipes B and C can fill the tank in 3 hours. Working together, pipes A and C can fill the tank in 2.4 hours.
(a) Explain why $a+b=\dfrac{1}{2}$, and write down the corresponding equations for the other two pairs of pipes, to form a system of three linear equations in $a$, $b$ and $c$. [3]
(b) Use technology to solve the system for $a$, $b$ and $c$. [4]
(c) Hence find how long it would take for all three pipes, working together, to fill the tank. [4]
Show complete worked solution
(a) If pipes A and B together fill the tank in 2 hours, then together they fill $\frac{1}{2}$ of the tank per hour, i.e. $a+b=\frac{1}{2}$.
Similarly, pipes B and C together fill the tank in 3 hours, so $b+c=\frac{1}{3}$.
And pipes A and C together fill the tank in 2.4 hours, so $a+c=\frac{1}{2.4}=\frac{5}{12}$.
$$\boxed{\begin{aligned}a+b&=\frac{1}{2}\\b+c&=\frac{1}{3}\\a+c&=\frac{5}{12}\end{aligned}}$$
(b) Adding all three equations:
$$2(a+b+c)=\frac{1}{2}+\frac{1}{3}+\frac{5}{12}=\frac{6}{12}+\frac{4}{12}+\frac{5}{12}=\frac{15}{12}=\frac{5}{4}$$
$$a+b+c=\frac{5}{8}$$
Then, using each pairwise equation:
$$c=(a+b+c)-(a+b)=\frac{5}{8}-\frac{1}{2}=\frac{5}{8}-\frac{4}{8}=\frac{1}{8}$$
$$a=(a+b+c)-(b+c)=\frac{5}{8}-\frac{1}{3}=\frac{15}{24}-\frac{8}{24}=\frac{7}{24}$$
$$b=(a+b+c)-(a+c)=\frac{5}{8}-\frac{5}{12}=\frac{15}{24}-\frac{10}{24}=\frac{5}{24}$$
(This can equivalently be confirmed using the GDC's matrix/simultaneous equation solver.)
$$\boxed{a=\frac{7}{24}, \quad b=\frac{5}{24}, \quad c=\frac{1}{8}}$$
(c) Working together, all three pipes fill $a+b+c=\frac{5}{8}$ of the tank per hour. The time to fill the whole tank is the reciprocal of this rate:
$$\text{time}=\frac{1}{a+b+c}=\frac{1}{5/8}=\frac{8}{5}=1.6 \text{ hours}$$
$$\boxed{\text{Working together, all three pipes fill the tank in 1.6 hours}}$$
QUESTION 49
12 marks
Hard
Consider the following system of equations in $x$, $y$ and $z$, where $k$ is a constant:
$$x+y+z=6$$
$$x+2y+3z=10$$
$$2x+3y+kz=16$$
(a) By finding the determinant of the coefficient matrix in terms of $k$, find the value of $k$ for which this system does not have a unique solution. [4]
(b) For this value of $k$, determine whether the system has no solution or infinitely many solutions. Justify your answer fully. [4]
(c) Hence, or otherwise, write down the general solution of the system for this value of $k$, letting $z=t$. [4]
Show complete worked solution
(a) The coefficient matrix is
$$\begin{pmatrix}1&1&1\\1&2&3\\2&3&k\end{pmatrix}$$
Expanding the determinant along the first row:
$$\det=1(2k-9)-1(k-6)+1(3-4)$$
$$=2k-9-k+6-1$$
$$=k-4$$
The system does not have a unique solution when the determinant is zero:
$$k-4=0$$
$$\boxed{k=4}$$
(b) With $k=4$, consider adding the first two equations:
$$(x+y+z)+(x+2y+3z)=6+10$$
$$2x+3y+4z=16$$
This is exactly the third equation (with $k=4$). So the third equation gives no new information; it is a linear combination of the first two, and the system is consistent but has only 2 independent equations in 3 unknowns.
$$\boxed{\text{When } k=4 \text{, the system has infinitely many solutions (not no solution), since the third equation is consistent with, and dependent on, the first two}}$$
(c) Let $z=t$. Substituting into the first two equations:
$$x+y=6-t \quad \text{(i)}$$
$$x+2y=10-3t \quad \text{(ii)}$$
Subtracting (i) from (ii):
$$y=(10-3t)-(6-t)=4-2t$$
Then from (i):
$$x=6-t-y=6-t-(4-2t)=2+t$$
$$\boxed{x=2+t, \quad y=4-2t, \quad z=t, \text{ for any value of } t}$$
QUESTION 50
11 marks
Hard
A pension fund manager invests a total of \$500,000 across three asset classes: bonds, stocks, and real estate. The table below summarises the assumed annual return rate for each asset class, and two planning constraints for the total portfolio.
The total expected annual return across the whole portfolio must be \$24,000, and the portfolio must satisfy the risk-balance condition $2x-y+z=500{,}000$.
(a) Write down a system of three linear equations in $x$, $y$ and $z$. [3]
(b) Use technology to solve the system for $x$, $y$ and $z$. [4]
(c) The fund manager requires that no more than \$150,000 be invested in stocks. Determine whether this solution satisfies that requirement. [4]
| Asset class | Amount invested (\$) | Annual return rate |
|---|---|---|
| Bonds | $x$ | 3% |
| Stocks | $y$ | 8% |
| Real estate | $z$ | 5% |
Show complete worked solution
(a) The total amount invested:
$$x+y+z=500{,}000$$
The total expected annual return:
$$0.03x+0.08y+0.05z=24{,}000$$
The risk-balance condition:
$$2x-y+z=500{,}000$$
$$\boxed{\begin{aligned}x+y+z&=500{,}000\\0.03x+0.08y+0.05z&=24{,}000\\2x-y+z&=500{,}000\end{aligned}}$$
(b) Writing the system in matrix form:
$$\begin{pmatrix}1&1&1\\0.03&0.08&0.05\\2&-1&1\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}500{,}000\\24{,}000\\500{,}000\end{pmatrix}$$
Solving using the GDC's matrix (or simultaneous equation) solver:
$$\boxed{x=\$200{,}000, \quad y=\$100{,}000, \quad z=\$200{,}000}$$
Check: $200{,}000+100{,}000+200{,}000=500{,}000$ ✓; $0.03(200{,}000)+0.08(100{,}000)+0.05(200{,}000)=6000+8000+10{,}000=24{,}000$ ✓
(c) The amount invested in stocks is $y=\$100{,}000$, which is less than the maximum of \$150,000.
$$\boxed{\text{Yes, the solution satisfies the requirement, since } \$100{,}000 \leq \$150{,}000}$$