MYP 4 & 5 · Maths
ALGEBRA (NOTATION AND EQUATIONS)
108 questions across 9 sub-topics
Use the Sub-Topic filter above to focus on one.
Algebraic notation
Algebraic substitution
Linear equations
Rational equations
Linear inequations
Problem solving
Money and investment problems
Motion problems
Mixture problems
Algebraic notation 12 questions
QUESTION 1
7 marks
Criterion A
Easy
A stationery shop sells notebooks for $p$ dollars each and pens for $q$ dollars each.
a.
Write down an expression for the cost of $5$ notebooks.
[1] b.
Write down an expression for the cost of $3$ notebooks and $4$ pens.
[1] c.
A customer buys $n$ notebooks and receives a discount of \$2 off the total notebook cost. Write down an expression, in terms of $n$ and $p$, for the amount the customer pays for the notebooks.
[2] d.
Simplify the expression $3(2p + q) - (p - 5q)$ fully, showing each step.
[3] Show complete worked solution
(a)
$5p$
(b)
$3p + 4q$
(c)
$np - 2$
(d)
$$3(2p+q) - (p-5q) = 6p + 3q - p + 5q = 5p + 8q$$
QUESTION 2
6 marks
Criterion C
Medium
Three students simplify the expression $5x - 2(x - 3)$ and get three different answers: Student P gets $3x - 3$, Student Q gets $3x + 6$, Student R gets $7x - 6$.
a.
Determine the correct simplified expression, showing full working.
[3] b.
Identify which student, if any, obtained the correct answer.
[2] c.
Explain the most likely error made by Student R.
[1] Show complete worked solution
(a)
$$5x - 2(x-3) = 5x - 2x + 6 = 3x + 6$$
(b)
Student Q obtained the correct answer, $3x + 6$.
(c)
Student R most likely distributed the $-2$ inconsistently across the brackets — applying the wrong sign to one of the two terms instead of multiplying both $x$ and $-3$ by $-2$.
QUESTION 3
5 marks
Criterion B
Hard
The diagram below shows Shape 1 and Shape 2 of a pattern. Each shape is a row of $n$ white unit squares surrounded by a border of grey unit squares.
a.
By continuing the pattern, determine the number of grey border squares in Shape 3 (where $n = 3$).
[1] b.
Using your results for Shapes 1, 2 and 3, determine an expression, in terms of $n$, for the number of grey border squares in Shape $n$.
[2] c.
Hence, calculate the number of border squares needed for Shape $20$.
[2] Show complete worked solution
(a)
Shape 3 has a $5 \times 3$ grid of unit squares ($15$ in total) with $3$ white squares in the middle row, so the number of border squares is $15 - 3 = 12$.
(b)
Shape $n$ is a grid of $(n+2)$ columns by $3$ rows, so the total number of squares is $3(n+2)$, of which $n$ are white. $$\text{Border squares} = 3(n+2) - n = 3n + 6 - n = 2n + 6$$
(c)
$$2(20) + 6 = 46$$
QUESTION 4
3 marks
Criterion A
Easy
Write an algebraic expression for each of the following.
a.
Seven more than twice a number $n$.
[1] b.
The product of $x$ and $y$, decreased by $4$.
[1] c.
Half of the sum of $a$ and $b$.
[1] Show complete worked solution
(a)
$2n + 7$
(b)
$xy - 4$
(c)
$\dfrac{a+b}{2}$
QUESTION 5
3 marks
Criterion A
Easy
Simplify each expression.
a.
$4a \times 3b$
[1] b.
$20xy \div 4x$
[2] Show complete worked solution
(a)
$12ab$
(b)
$5y$ (assuming $x \ne 0$)
QUESTION 6
3 marks
Criterion A
Medium
Consider the expression $7m - 3n + 2m + 5n - m$.
a.
Simplify the expression fully.
[2] b.
State whether $8m$ and $2n$ are like terms, and explain why.
[1] Show complete worked solution
(a)
$$7m - 3n + 2m + 5n - m = 8m + 2n$$
(b)
No — $8m$ and $2n$ involve different variables ($m$ and $n$), so they are unlike terms and cannot be combined further.
QUESTION 7
4 marks
Criterion D
Medium
A mobile phone plan costs a fixed \$20 per month plus \$0.10 per text message sent.
a.
Write down an expression, in terms of $t$, for the monthly cost if $t$ text messages are sent.
[2] b.
Calculate the cost in a month where $85$ texts are sent.
[2] Show complete worked solution
(a)
$$\text{Cost} = 20 + 0.10t$$
(b)
$$20 + 0.10(85) = 20 + 8.5 = 28.5$$ The cost is \$28.50.
QUESTION 8
5 marks
Criterion A
Medium
Simplify the following, giving your answer in index form.
a.
$a^3 \times a^5$
[1] b.
$\dfrac{b^7}{b^2}$
[2] c.
$(c^2)^4$
[2] Show complete worked solution
(a)
$a^8$
(b)
$b^5$
(c)
$c^8$
QUESTION 9
4 marks
Criterion A
Hard
Evaluate the following without a calculator.
a.
$5^0 + 3^{-1}$
[2] b.
$\left(\dfrac{2}{3}\right)^{-2}$
[2] Show complete worked solution
(a)
$$1 + \dfrac{1}{3} = \dfrac{4}{3}$$
(b)
$$\left(\dfrac{3}{2}\right)^2 = \dfrac{9}{4}$$
QUESTION 10
4 marks
Criterion C
Medium
A student simplifies $3x^2 \times 2x^3$ and writes the answer as $6x^6$.
a.
Determine the correct simplified expression.
[2] b.
Explain the error made by the student.
[2] Show complete worked solution
(a)
$$3x^2 \times 2x^3 = 6x^5$$
(b)
When multiplying powers of the same base, the exponents should be added ($2+3=5$), not multiplied. The student incorrectly computed $2 \times 3 = 6$ as the new exponent instead of $2+3=5$.
QUESTION 11
5 marks
Criterion B
Hard
A sequence of numbers is generated using the rule $T_n = 3n - 1$, where $T_n$ is the $n$th term.
a.
Determine $T_1$, $T_2$ and $T_3$.
[1] b.
Determine which term number gives a value of $101$.
[2] c.
Explain how you know that $102$ can never be a term in this sequence.
[2] Show complete worked solution
(a)
$T_1=2$, $T_2=5$, $T_3=8$
(b)
$$3n - 1 = 101 \Rightarrow 3n = 102 \Rightarrow n = 34$$
(c)
Every term satisfies $T_n = 3n-1$, so every term leaves a remainder of $2$ when divided by $3$ (since $3n$ is always a multiple of $3$). But $102$ is a multiple of $3$ (remainder $0$), so it can never equal $3n-1$ for an integer $n$.
QUESTION 12
4 marks
Criterion D
Medium
A water tank is filled at a rate of $r$ litres per minute.
a.
Write down an expression for the volume of water, in litres, in the tank after $12$ minutes, if the tank started empty.
[2] b.
If the tank contains $460$ litres after $20$ minutes (starting from empty), determine the rate $r$.
[2] Show complete worked solution
(a)
$12r$ litres
(b)
$$20r = 460 \Rightarrow r = 23 \text{ litres per minute}$$
Algebraic substitution 12 questions
QUESTION 1
5 marks
Criterion A
Medium
The formula for the perimeter of a rectangle is $P = 2(l + w)$, where $l$ is the length and $w$ is the width.
a.
Calculate $P$ when $l = 12$ and $w = 7$.
[1] b.
The formula for converting a Celsius temperature $C$ to Fahrenheit $F$ is $F = \dfrac{9}{5}C + 32$. Calculate $F$ when $C = 25$.
[2] c.
Given that $F = 98.6$, determine the value of $C$.
[2] Show complete worked solution
(a)
$$P = 2(12 + 7) = 2(19) = 38$$
(b)
$$F = \dfrac{9}{5}(25) + 32 = 45 + 32 = 77$$
(c)
$$98.6 = \dfrac{9}{5}C + 32 \Rightarrow 66.6 = \dfrac{9}{5}C \Rightarrow C = 66.6 \times \dfrac{5}{9} = 37$$
QUESTION 2
6 marks
Criterion D
Hard
The kinetic energy of a moving object is given by $KE = \dfrac{1}{2}mv^2$, where $m$ is the mass in kg and $v$ is the speed in m/s, and $KE$ is measured in joules (J).
a.
Calculate the kinetic energy of a $60$ kg cyclist travelling at $8$ m/s.
[2] b.
A car of mass $1200$ kg has kinetic energy $150\,000$ J. Determine its speed, correct to one decimal place.
[2] c.
Explain what happens to the kinetic energy of an object if its speed is doubled, while its mass stays constant.
[2] Show complete worked solution
(a)
$$KE = \dfrac{1}{2}(60)(8)^2 = 0.5 \times 60 \times 64 = 1920 \text{ J}$$
(b)
$$150000 = \dfrac{1}{2}(1200)v^2 \Rightarrow v^2 = \dfrac{150000}{600} = 250 \Rightarrow v = \sqrt{250} \approx 15.8 \text{ m/s}$$
(c)
Since $KE$ is proportional to $v^2$, doubling the speed means the new kinetic energy is proportional to $(2v)^2 = 4v^2$ — so the kinetic energy becomes four times as large, not twice as large.
QUESTION 3
6 marks
Criterion C
Hard
Consider the expression $2x^2 - 3xy + y^2$ where $x = -3$ and $y = 2$.
a.
Calculate the value of the expression, showing full working.
[3] b.
A student believes that $x^2 = -9$ when $x = -3$. Explain why this is incorrect, and state the correct value of $x^2$.
[2] c.
Explain how brackets should be used when substituting a negative value into an expression, to avoid this type of error.
[1] Show complete worked solution
(a)
$$2(-3)^2 - 3(-3)(2) + (2)^2 = 2(9) + 18 + 4 = 18 + 18 + 4 = 40$$
(b)
Squaring a negative number always gives a positive result, since a negative multiplied by a negative is positive: $(-3)^2 = (-3) \times (-3) = 9$, not $-9$. The value $-9$ would only be correct for $-3^2$ (square first, then negate).
(c)
The negative value should always be enclosed in brackets when it replaces the variable, e.g. substitute $x = (-3)$ so that $x^2$ becomes $(-3)^2$ — this makes clear that the whole negative number is being squared.
QUESTION 4
3 marks
Criterion A
Easy
Given $a = 5$, $b = -2$, and $c = 3$, evaluate the following.
a.
$a + bc$
[1] b.
$2a^2 - b$
[2] Show complete worked solution
(a)
$$5 + (-2)(3) = 5 - 6 = -1$$
(b)
$$2(5)^2 - (-2) = 50 + 2 = 52$$
QUESTION 5
4 marks
Criterion A
Medium
The area of a trapezium is given by $A = \dfrac{1}{2}(a+b)h$, where $a$ and $b$ are the parallel side lengths and $h$ is the height.
a.
Calculate $A$ when $a=8$, $b=14$, $h=6$.
[2] b.
If $A=45$, $a=6$, $h=5$, determine $b$.
[2] Show complete worked solution
(a)
$$A = \dfrac{1}{2}(8+14)(6) = \dfrac{1}{2}(22)(6) = 66$$
(b)
$$45 = \dfrac{1}{2}(6+b)(5) \Rightarrow 45 = 2.5(6+b) \Rightarrow 18 = 6+b \Rightarrow b = 12$$
QUESTION 6
4 marks
Criterion D
Medium
The density of an object is given by $\rho = \dfrac{m}{V}$, where $m$ is mass in grams and $V$ is volume in cm$^3$.
a.
Calculate the density of an object with mass $250$ g and volume $50$ cm$^3$.
[2] b.
A different object has density $2.7$ g/cm$^3$ and mass $540$ g. Determine its volume.
[2] Show complete worked solution
(a)
$$\rho = \dfrac{250}{50} = 5 \text{ g/cm}^3$$
(b)
$$2.7 = \dfrac{540}{V} \Rightarrow V = \dfrac{540}{2.7} = 200 \text{ cm}^3$$
QUESTION 7
4 marks
Criterion A
Hard
Given $x = -4$, $y = \dfrac{1}{2}$, evaluate:
a.
$\dfrac{x^2}{2y}$
[2] b.
$3x - \dfrac{1}{y}$
[2] Show complete worked solution
(a)
$$\dfrac{(-4)^2}{2(0.5)} = \dfrac{16}{1} = 16$$
(b)
$$3(-4) - \dfrac{1}{0.5} = -12 - 2 = -14$$
QUESTION 8
4 marks
Criterion C
Medium
Consider the expression $-x^2 + 4$ evaluated at $x = -3$.
a.
Calculate the value of the expression.
[2] b.
Explain why the answer is negative even though $x^2$ always produces a non-negative result.
[2] Show complete worked solution
(a)
$$-(-3)^2 + 4 = -9 + 4 = -5$$
(b)
The negative sign in $-x^2$ is applied after squaring, not before: $x^2 = (-3)^2 = 9$ is indeed non-negative, but then it is negated to give $-9$, and $-9 + 4 = -5$ is still negative overall.
QUESTION 9
5 marks
Criterion A
Medium
For the quadratic equation $2x^2 - 5x - 3 = 0$, the discriminant is given by $\Delta = b^2 - 4ac$.
a.
State the values of $a$, $b$ and $c$.
[1] b.
Calculate the value of $\Delta$.
[2] c.
Explain what the value of $\Delta$ tells you about the number of real solutions to the equation.
[2] Show complete worked solution
(a)
$a=2$, $b=-5$, $c=-3$
(b)
$$\Delta = (-5)^2 - 4(2)(-3) = 25 + 24 = 49$$
(c)
Since $\Delta = 49 > 0$ and is a perfect square ($7^2$), the equation has two distinct, rational real solutions.
QUESTION 10
4 marks
Criterion D
Hard
The height, in metres, of a ball above the ground $t$ seconds after being thrown is given by $h(t) = -5t^2 + 20t + 1.5$.
a.
Calculate the height of the ball after $2$ seconds.
[2] b.
Calculate the initial height of the ball (at $t=0$), and explain what this value represents in context.
[2] Show complete worked solution
(a)
$$h(2) = -5(2)^2 + 20(2) + 1.5 = -20 + 40 + 1.5 = 21.5 \text{ m}$$
(b)
$$h(0) = 1.5 \text{ m}$$ This represents the height above the ground from which the ball was released (e.g. thrown from hand height).
QUESTION 11
3 marks
Criterion A
Medium
Determine whether $x = 3$ is a solution to the equation $2x^2 - 5x - 3 = 0$ by substitution.
a.
Substitute $x=3$ into the left-hand side and evaluate.
[2] b.
State whether $x=3$ is a solution, with a reason.
[1] Show complete worked solution
(a)
$$2(3)^2 - 5(3) - 3 = 18 - 15 - 3 = 0$$
(b)
Yes — since substituting $x=3$ gives $0$, which equals the right-hand side, $x=3$ is a solution.
QUESTION 12
4 marks
Criterion B
Hard
A student claims that $n^2 - n + 41$ is always a prime number for positive integer values of $n$.
a.
Verify the claim for $n=1$, $n=2$, and $n=3$ by substitution.
[2] b.
Substitute $n = 41$ to show that the claim is false for this value.
[2] Show complete worked solution
(a)
$n=1$: $41$ (prime). $n=2$: $4-2+41=43$ (prime). $n=3$: $9-3+41=47$ (prime).
(b)
$$41^2 - 41 + 41 = 41^2 = 1681 = 41 \times 41$$ Since $1681$ has $41$ as a factor (other than $1$ and itself), it is not prime — so the claim is false for $n=41$.
Linear equations 12 questions
QUESTION 1
7 marks
Criterion A
Medium
A rectangular vegetable garden has a width of $x$ metres. The length of the garden is $3$ m more than twice its width, as shown in the diagram. The garden is enclosed by $54$ m of fencing.
a.
Show that the perimeter of the garden satisfies the equation $6x + 6 = 54$.
[2] b.
Solve the equation to find the value of $x$.
[2] c.
Hence, determine the length of the garden and its area.
[3] Show complete worked solution
(a)
$$\text{Perimeter} = 2(\text{width} + \text{length}) = 2\big(x + (2x+3)\big) = 2(3x+3) = 6x+6$$ Since the garden is enclosed by $54$ m of fencing, the perimeter equals $54$ m, so $6x+6=54$.
(b)
$$6x + 6 = 54 \Rightarrow 6x = 48 \Rightarrow x = 8$$
(c)
Length $= 2(8) + 3 = 19$ m. Area $= \text{width} \times \text{length} = 8 \times 19 = 152$ m$^2$.
QUESTION 2
7 marks
Criterion D
Medium
A taxi company charges a callout fee of \$4, plus \$2.50 per kilometre travelled.
a.
Write down an equation for the total cost $C$, in dollars, of a journey of $d$ kilometres.
[2] b.
Kavya pays \$26.50 for a journey. Determine the distance she travelled.
[2] c.
On another day, Kavya's journey costs \$4 more than her \$26.50 journey. Form and solve an equation to determine the new distance travelled.
[3] Show complete worked solution
(a)
$$C = 4 + 2.5d$$
(b)
$$26.5 = 4 + 2.5d \Rightarrow 22.5 = 2.5d \Rightarrow d = 9 \text{ km}$$
(c)
New cost $= \$30.50$. $$30.5 = 4 + 2.5d \Rightarrow 26.5 = 2.5d \Rightarrow d = 10.6 \text{ km}$$
QUESTION 3
7 marks
Criterion A
Hard
Solve the following equations for $x$.
a.
$5x - 7 = 2x + 11$
[2] b.
$3(x - 4) = 2(x + 5)$
[2] c.
$\dfrac{2x+1}{3} - \dfrac{x-2}{4} = 2$
[3] Show complete worked solution
(a)
$$5x - 7 = 2x + 11 \Rightarrow 3x = 18 \Rightarrow x = 6$$
(b)
$$3x - 12 = 2x + 10 \Rightarrow x = 22$$
(c)
Multiplying every term by $12$: $$4(2x+1) - 3(x-2) = 24 \Rightarrow 8x+4-3x+6 = 24 \Rightarrow 5x+10=24 \Rightarrow 5x = 14 \Rightarrow x = 2.8$$
QUESTION 4
4 marks
Criterion A
Easy
Solve the following equations for $x$.
a.
$x + 9 = 23$
[1] b.
$5x = 35$
[1] c.
$4x + 3 = 19$
[2] Show complete worked solution
(a)
$x = 14$
(b)
$x = 7$
(c)
$$4x = 16 \Rightarrow x = 4$$
QUESTION 5
4 marks
Criterion A
Medium
Solve the following equations for $x$.
a.
$2(x+5) = 18$
[2] b.
$-3(2x-1) = 15$
[2] Show complete worked solution
(a)
$$2x + 10 = 18 \Rightarrow x = 4$$
(b)
$$-6x + 3 = 15 \Rightarrow -6x = 12 \Rightarrow x = -2$$
QUESTION 6
4 marks
Criterion D
Medium
A cinema charges \$8 per adult ticket and \$5 per child ticket. A family buys $2$ adult tickets and some child tickets, spending \$31 in total.
a.
Form an equation, in terms of the number of child tickets $c$, and solve it.
[2] b.
State how many people were in total in the family.
[2] Show complete worked solution
(a)
$$16 + 5c = 31 \Rightarrow 5c = 15 \Rightarrow c = 3$$
(b)
$2 + 3 = 5$ people.
QUESTION 7
4 marks
Criterion A
Hard
Solve the following equations for $x$.
a.
$\dfrac{x}{2} + 3 = \dfrac{x}{3} - 1$
[2] b.
$\dfrac{3(x-1)}{2} = x+4$
[2] Show complete worked solution
(a)
Multiplying by $6$: $$3x + 18 = 2x - 6 \Rightarrow x = -24$$
(b)
$$3x - 3 = 2x + 8 \Rightarrow x = 11$$
QUESTION 8
3 marks
Criterion C
Medium
A student solves $2(x - 3) = 10$ as follows: $2x - 3 = 10 \Rightarrow 2x = 13 \Rightarrow x = 6.5$.
a.
Determine the correct solution.
[2] b.
Identify the error made by the student.
[1] Show complete worked solution
(a)
$$2x - 6 = 10 \Rightarrow 2x = 16 \Rightarrow x = 8$$
(b)
The student did not distribute the $2$ across both terms in the brackets — they multiplied $2$ by $x$ but not by $-3$.
QUESTION 9
4 marks
Criterion A
Medium
A triangle has side lengths $x$, $x+2$, and $2x-3$ (in cm). The perimeter is $35$ cm.
a.
Form an equation and solve for $x$.
[2] b.
Determine the length of each side.
[2] Show complete worked solution
(a)
$$x + (x+2) + (2x-3) = 35 \Rightarrow 4x - 1 = 35 \Rightarrow x = 9$$
(b)
$9$ cm, $11$ cm, and $15$ cm.
QUESTION 10
4 marks
Criterion D
Hard
A jacket's price is reduced by $15\%$ in a sale, and the sale price is \$68.
a.
Form an equation, in terms of the original price $p$, and solve for $p$.
[2] b.
Determine the amount of the discount, in dollars.
[2] Show complete worked solution
(a)
$$0.85p = 68 \Rightarrow p = 80$$ The original price was \$80.
(b)
$$80 - 68 = 12$$ The discount was \$12.
QUESTION 11
3 marks
Criterion A
Medium
Solve the equation $3(x+2) = 3x + 6$.
a.
Attempt to solve the equation for $x$, showing your working.
[2] b.
Explain what this result means.
[1] Show complete worked solution
(a)
$$3x + 6 = 3x + 6 \Rightarrow 0 = 0$$
(b)
The equation is true for every value of $x$ (an identity) — it has infinitely many solutions, not just one.
QUESTION 12
4 marks
Criterion B
Hard
Solve the equation $2(x+3) = 2x + 5$.
a.
Attempt to solve the equation for $x$, showing your working.
[2] b.
Explain what this result means, and contrast it with the equation $2(x+3) = 2x + 6$.
[2] Show complete worked solution
(a)
$$2x + 6 = 2x + 5 \Rightarrow 6 = 5$$
(b)
$6 = 5$ is false for any value of $x$, so this equation has no solution. By contrast, $2(x+3) = 2x+6$ simplifies to $2x+6=2x+6$, which is true for every $x$ (infinitely many solutions) — the difference comes down to whether the constant terms match after expanding.
Rational equations 12 questions
QUESTION 1
6 marks
Criterion A
Medium
Solve the following equations for $x$, stating any restrictions on $x$.
a.
$\dfrac{3}{x} + \dfrac{2}{x} = 5$
[3] b.
$\dfrac{4}{x-1} = \dfrac{2}{x+3}$
[3] Show complete worked solution
(a)
$$\dfrac{5}{x} = 5 \Rightarrow x = 1 \qquad (x \ne 0)$$
(b)
Cross-multiplying: $$4(x+3) = 2(x-1) \Rightarrow 4x+12 = 2x-2 \Rightarrow 2x = -14 \Rightarrow x = -7 \qquad (x \ne 1,\ x \ne -3)$$
QUESTION 2
9 marks
Criterion D
Hard
Pipe A can fill a tank in $x$ hours, working alone. Pipe B, working alone, takes $4$ hours longer than Pipe A to fill the same tank. Working together, the two pipes fill the tank in $3$ hours.
a.
Write down an expression, in terms of $x$, for the fraction of the tank filled by Pipe A in one hour, and by Pipe B in one hour.
[2] b.
Show that the situation can be modelled by the equation $\dfrac{1}{x} + \dfrac{1}{x+4} = \dfrac{1}{3}$.
[3] c.
Hence, determine the time Pipe A takes to fill the tank alone, giving your answer correct to two decimal places.
[4] Show complete worked solution
(a)
Pipe A fills $\dfrac{1}{x}$ of the tank per hour. Pipe B fills $\dfrac{1}{x+4}$ of the tank per hour.
(b)
In one hour, working together, the pipes fill $\dfrac{1}{x} + \dfrac{1}{x+4}$ of the tank. Since they take $3$ hours together, they fill $\dfrac{1}{3}$ of the tank per hour, so $\dfrac{1}{x} + \dfrac{1}{x+4} = \dfrac{1}{3}$.
(c)
Multiplying every term by $3x(x+4)$: $$3(x+4) + 3x = x(x+4) \Rightarrow 6x + 12 = x^2 + 4x \Rightarrow x^2 - 2x - 12 = 0$$ Using the quadratic formula: $$x = \dfrac{2 \pm \sqrt{4 + 48}}{2} = \dfrac{2 \pm \sqrt{52}}{2} = 1 \pm \sqrt{13}$$ Since $x$ must be positive, $x = 1 + \sqrt{13} \approx 4.61$ hours.
QUESTION 3
4 marks
Criterion C
Medium
Consider the equation $\dfrac{5}{x-2} = 3$.
a.
Solve the equation for $x$.
[2] b.
Explain why $x = 2$ can never be a solution to this equation, regardless of the value on the right-hand side.
[2] Show complete worked solution
(a)
$$5 = 3(x-2) \Rightarrow 5 = 3x - 6 \Rightarrow 3x = 11 \Rightarrow x = \dfrac{11}{3}$$
(b)
Substituting $x = 2$ makes the denominator $x - 2 = 0$, and division by zero is undefined. So $x=2$ must always be excluded from the solution set of this equation.
QUESTION 4
4 marks
Criterion A
Easy
Solve the following equations for $x$.
a.
$\dfrac{x}{4} = 7$
[2] b.
$\dfrac{12}{x} = 3$
[2] Show complete worked solution
(a)
$x = 28$
(b)
$x = 4$
QUESTION 5
6 marks
Criterion A
Medium
Solve the following equations for $x$.
a.
$\dfrac{2}{x+1} = \dfrac{3}{x-2}$
[3] b.
$\dfrac{x}{3} - \dfrac{2}{x} = 0$
[3] Show complete worked solution
(a)
Cross-multiplying: $$2(x-2) = 3(x+1) \Rightarrow 2x-4 = 3x+3 \Rightarrow x = -7$$
(b)
Multiplying by $3x$: $$x^2 - 6 = 0 \Rightarrow x^2 = 6 \Rightarrow x = \pm\sqrt{6}$$
QUESTION 6
8 marks
Criterion D
Hard
A cyclist rides $60$ km at a certain speed, then returns the same $60$ km route at a speed $5$ km/h slower, taking a total time of $7$ hours for the round trip.
a.
Let the speed for the first leg be $x$ km/h. Write down expressions, in terms of $x$, for the time taken for each leg of the trip.
[2] b.
Show that the situation is modelled by the equation $\dfrac{60}{x} + \dfrac{60}{x-5} = 7$.
[2] c.
Hence, determine the speed $x$, correct to one decimal place.
[4] Show complete worked solution
(a)
$\dfrac{60}{x}$ hours and $\dfrac{60}{x-5}$ hours.
(b)
The total time for both legs is the sum of the two individual times, and this must equal the total trip time of $7$ hours: $\dfrac{60}{x} + \dfrac{60}{x-5} = 7$.
(c)
Multiplying by $x(x-5)$: $$60(x-5) + 60x = 7x(x-5) \Rightarrow 120x - 300 = 7x^2 - 35x \Rightarrow 7x^2 - 155x + 300 = 0$$ Using the quadratic formula: $$x = \dfrac{155 \pm \sqrt{155^2 - 4(7)(300)}}{14} = \dfrac{155 \pm \sqrt{15625}}{14} = \dfrac{155 \pm 125}{14}$$ giving $x = 20$ or $x \approx 2.1$. Since $x - 5$ must be positive, $x \approx 2.1$ is rejected, so $x = 20$ km/h.
QUESTION 7
5 marks
Criterion C
Medium
Consider the equation $\dfrac{2}{x} + \dfrac{3}{x-4} = 0$.
a.
State the values of $x$ that must be excluded from the domain, and explain why.
[2] b.
Solve the equation for $x$.
[3] Show complete worked solution
(a)
$x \ne 0$ and $x \ne 4$, since these values would make one of the denominators equal to zero, which is undefined.
(b)
Multiplying by $x(x-4)$: $$2(x-4) + 3x = 0 \Rightarrow 2x - 8 + 3x = 0 \Rightarrow 5x = 8 \Rightarrow x = 1.6$$
QUESTION 8
5 marks
Criterion A
Medium
Solve the following equations for $x$.
a.
$\dfrac{5}{2x} = \dfrac{1}{6}$
[2] b.
$\dfrac{x}{3} = \dfrac{12}{x}$
[3] Show complete worked solution
(a)
Cross-multiplying: $$30 = 2x \Rightarrow x = 15$$
(b)
$$x^2 = 36 \Rightarrow x = \pm 6$$
QUESTION 9
5 marks
Criterion A
Hard
Solve $\dfrac{1}{x-2} + \dfrac{1}{x+2} = \dfrac{4}{x^2-4}$.
a.
Combine the left-hand side into a single fraction, and hence form an equation without fractions.
[3] b.
Solve the resulting equation, and explain why your answer must be rejected.
[2] Show complete worked solution
(a)
Since $x^2-4=(x-2)(x+2)$: $$\dfrac{(x+2)+(x-2)}{(x-2)(x+2)} = \dfrac{2x}{x^2-4} = \dfrac{4}{x^2-4} \Rightarrow 2x = 4$$
(b)
$2x=4 \Rightarrow x=2$. However, $x=2$ makes the original denominators $x-2$ and $x^2-4$ equal to zero, so it is an extraneous solution — the equation actually has no solution.
QUESTION 10
5 marks
Criterion D
Medium
A runner's average pace (minutes per km) for a race is given by $\text{pace} = \dfrac{t}{d}$, where $t$ is total time in minutes and $d$ is distance in km.
a.
A runner completes a $10$ km race in $52$ minutes. Calculate the average pace.
[2] b.
Another runner wants an average pace of $4.5$ min/km, completing an unknown distance $d$ in $36$ minutes. Form and solve an equation for $d$.
[3] Show complete worked solution
(a)
$$\dfrac{52}{10} = 5.2 \text{ min/km}$$
(b)
$$4.5 = \dfrac{36}{d} \Rightarrow d = \dfrac{36}{4.5} = 8 \text{ km}$$
QUESTION 11
4 marks
Criterion C
Hard
A student solves $\dfrac{x+3}{x-1} = \dfrac{4}{x-1}$ by cancelling the denominators to get $x + 3 = 4$, giving $x = 1$.
a.
Explain why $x=1$ cannot actually be a solution to the original equation.
[2] b.
State the correct solution set for the original equation.
[2] Show complete worked solution
(a)
Substituting $x=1$ makes both denominators equal to zero, which is undefined — so $x=1$ cannot be a valid solution, even though it solves the simplified equation.
(b)
There is no solution — the only value that satisfies the simplified equation ($x=1$) is excluded by the domain restriction on the original equation.
QUESTION 12
4 marks
Criterion A
Medium
Solve $\dfrac{3}{x} + \dfrac{1}{2x} = \dfrac{7}{4}$.
a.
Combine the left-hand side into a single fraction.
[2] b.
Hence, solve for $x$.
[2] Show complete worked solution
(a)
$$\dfrac{3}{x} + \dfrac{1}{2x} = \dfrac{6}{2x} + \dfrac{1}{2x} = \dfrac{7}{2x}$$
(b)
$$\dfrac{7}{2x} = \dfrac{7}{4} \Rightarrow 2x = 4 \Rightarrow x = 2$$
Linear inequations 12 questions
QUESTION 1
6 marks
Criterion A
Medium
Solve the following inequalities for $x$.
a.
$3x - 5 \le 10$
[2] b.
$-2x + 4 > 12$, and explain why the inequality sign reverses at one step of your solution.
[2] c.
Write down the combined solution set that satisfies both inequalities from (a) and (b), using inequality notation.
[2] Show complete worked solution
(a)
$$3x \le 15 \Rightarrow x \le 5$$
(b)
$$-2x > 8 \Rightarrow x < -4$$ The inequality sign reverses when dividing both sides by a negative number ($-2$).
(c)
Since $x \le 5$ and $x < -4$, every value satisfying $x < -4$ automatically satisfies $x \le 5$ too, so the combined solution set is simply $x < -4$.
QUESTION 2
7 marks
Criterion D
Medium
Rohan has \$85 to spend at an amusement park. Entry costs \$15, and each ride costs \$5.
a.
Form an inequality, in terms of the number of rides $r$, to represent this situation, and solve it to determine the maximum number of rides Rohan can go on.
[3] b.
Rohan also wants to buy a \$12 souvenir. Form a new inequality including the cost of the souvenir, and determine the new maximum number of rides.
[2] c.
Explain why the answer to part (b) must be rounded down to $11$, rather than rounded to the nearest whole number ($12$).
[2] Show complete worked solution
(a)
$$15 + 5r \le 85 \Rightarrow 5r \le 70 \Rightarrow r \le 14$$ Rohan can go on a maximum of $14$ rides.
(b)
$$15 + 12 + 5r \le 85 \Rightarrow 5r \le 58 \Rightarrow r \le 11.6$$ Since $r$ must be a whole number, the maximum number of rides is $11$.
(c)
$r \le 11.6$ means $12$ rides would exceed Rohan's budget — he cannot go on a fraction of a ride, and rounding up would break the constraint $r \le 11.6$, so the answer must always be rounded down in this type of budget context.
QUESTION 3
7 marks
Criterion B
Hard
Consider the compound inequality $-7 < 2x + 3 \le 9$.
a.
Solve the compound inequality for $x$, showing full working.
[3] b.
Determine the number of integer values of $x$ that satisfy this inequality.
[2] c.
Determine the largest and smallest possible values of $2x^2$ for integer values of $x$ satisfying the inequality.
[2] Show complete worked solution
(a)
Subtracting $3$ from all three parts: $$-10 < 2x \le 6$$ Dividing by $2$: $$-5 < x \le 3$$
(b)
The integers satisfying $-5 < x \le 3$ are $-4, -3, -2, -1, 0, 1, 2, 3$ — a total of $8$ integers.
(c)
Checking $x^2$ across $-4$ to $3$: the largest $x^2$ occurs at $x=-4$ (since $|-4|>|3|$), giving $2(-4)^2 = 32$. The smallest $x^2$ occurs at $x=0$, giving $2(0)^2 = 0$.
QUESTION 4
4 marks
Criterion A
Easy
Solve the following inequalities for $x$.
a.
$x - 6 \ge 2$
[1] b.
$3x < 21$
[1] c.
$-x \le 5$
[2] Show complete worked solution
(a)
$x \ge 8$
(b)
$x < 7$
(c)
$$-x \le 5 \Rightarrow x \ge -5 \text{ (sign reverses on dividing by } -1\text{)}$$
QUESTION 5
4 marks
Criterion A
Medium
Solve the following inequalities for $x$.
a.
$7 - 2x \ge -3$
[2] b.
$\dfrac{x}{-3} > 4$
[2] Show complete worked solution
(a)
$$-2x \ge -10 \Rightarrow x \le 5$$
(b)
$$x < -12 \text{ (sign reverses on multiplying by } -3\text{)}$$
QUESTION 6
4 marks
Criterion D
Medium
Aisha earns \$14 per hour. She wants to save at least \$350 in a month, and already has \$60 saved.
a.
Form an inequality, in terms of hours worked $h$, and solve it.
[2] b.
Determine the minimum whole number of hours Aisha must work.
[2] Show complete worked solution
(a)
$$60 + 14h \ge 350 \Rightarrow 14h \ge 290 \Rightarrow h \ge 20.71...$$
(b)
$21$ hours (checking: $20$ hours gives $60+280=340 < 350$, but $21$ hours gives $60+294=354 \ge 350$).
QUESTION 7
6 marks
Criterion A
Hard
Solve the following inequalities for $x$.
a.
$\dfrac{2x-1}{3} \ge x - 2$
[3] b.
$\dfrac{x+4}{2} < \dfrac{x-1}{3}$
[3] Show complete worked solution
(a)
$$2x - 1 \ge 3x - 6 \Rightarrow -x \ge -5 \Rightarrow x \le 5$$
(b)
$$3(x+4) < 2(x-1) \Rightarrow 3x + 12 < 2x - 2 \Rightarrow x < -14$$
QUESTION 8
3 marks
Criterion C
Medium
A student solves $-4x + 8 < 20$ and writes $-4x < 12 \Rightarrow x < -3$.
a.
Determine the correct solution.
[2] b.
Identify the student's error.
[1] Show complete worked solution
(a)
$$-4x < 12 \Rightarrow x > -3 \text{ (sign reverses on dividing by } -4\text{)}$$
(b)
The student forgot to reverse the inequality sign when dividing both sides by a negative number.
QUESTION 9
4 marks
Criterion A
Medium
Solve $2 \le 3x - 4 < 17$.
a.
Solve for $x$, showing full working.
[3] b.
Determine the smallest integer value of $x$ satisfying the inequality.
[1] Show complete worked solution
(a)
Adding $4$ to all parts: $$6 \le 3x < 21$$ Dividing by $3$: $$2 \le x < 7$$
(b)
$x = 2$
QUESTION 10
4 marks
Criterion D
Hard
A company's profit, in dollars, from selling $n$ units of a product is given by $P = 8n - 450$.
a.
Form an inequality to represent making a profit (that is, $P>0$), and solve it.
[2] b.
Determine the minimum whole number of units that must be sold to make a profit.
[2] Show complete worked solution
(a)
$$8n - 450 > 0 \Rightarrow n > 56.25$$
(b)
$57$ units (since $56$ units gives $P = -2 < 0$, but $57$ units gives $P = 6 > 0$).
QUESTION 11
5 marks
Criterion B
Medium
Determine the set of integer values of $x$ that satisfy both $2x + 1 > -9$ and $x - 5 \le 0$.
a.
Solve each inequality separately.
[3] b.
List all integer values of $x$ satisfying both inequalities.
[2] Show complete worked solution
(a)
$$2x + 1 > -9 \Rightarrow 2x > -10 \Rightarrow x > -5$$ $$x - 5 \le 0 \Rightarrow x \le 5$$
(b)
$-4, -3, -2, -1, 0, 1, 2, 3, 4, 5$ — a total of $10$ integers.
QUESTION 12
4 marks
Criterion C
Hard
Consider the inequality $2(x+3) > 2x + 1$.
a.
Attempt to solve the inequality, showing your working.
[2] b.
Explain what this result means for the solution set.
[2] Show complete worked solution
(a)
$$2x + 6 > 2x + 1 \Rightarrow 6 > 1$$
(b)
Since $6 > 1$ is always true, regardless of the value of $x$, the inequality holds for every real number — the solution set is all real values of $x$.
Problem solving 12 questions
QUESTION 1
6 marks
Criterion A
Medium
The sum of three consecutive even integers is $132$.
a.
Let the smallest integer be $n$. Write down expressions for the other two consecutive even integers.
[2] b.
Form an equation and solve it to find the three integers.
[2] c.
Verify your answer by checking that the sum of the three integers is indeed $132$.
[2] Show complete worked solution
(a)
$n + 2$ and $n + 4$
(b)
$$n + (n+2) + (n+4) = 132 \Rightarrow 3n + 6 = 132 \Rightarrow n = 42$$ The integers are $42, 44, 46$.
(c)
$$42 + 44 + 46 = 132 \checkmark$$
QUESTION 2
7 marks
Criterion D
Hard
Maria is currently three times as old as her son Leo. In $12$ years, Maria will be twice as old as Leo.
a.
Let Leo's current age be $x$ years. Write down an expression for Maria's current age, and expressions for both of their ages in $12$ years' time.
[2] b.
Form an equation from the given information, and solve it to find Leo's current age.
[3] c.
Hence, determine Maria's current age, and her age in $12$ years' time.
[2] Show complete worked solution
(a)
Maria's current age: $3x$. In $12$ years: Leo is $x + 12$, Maria is $3x + 12$.
(b)
$$3x + 12 = 2(x + 12) \Rightarrow 3x + 12 = 2x + 24 \Rightarrow x = 12$$ Leo is currently $12$ years old.
(c)
Maria is currently $3(12) = 36$ years old, and will be $48$ in $12$ years. Check: Leo will be $24$, and $48 = 2 \times 24 \checkmark$
QUESTION 3
7 marks
Criterion C
Medium
A piggy bank contains only \$1 coins and \$2 coins. There are $18$ coins in total, worth \$29 altogether.
a.
Let $x$ be the number of \$1 coins. Form an equation in terms of $x$ and solve it to find the number of each type of coin.
[3] b.
Verify your solution satisfies both the total number of coins and the total value given in the problem.
[2] c.
Explain why a solution such as $x = 20$ would not be valid in this context, even if it satisfied the value equation.
[2] Show complete worked solution
(a)
Number of \$2 coins $= 18 - x$. $$x(1) + (18-x)(2) = 29 \Rightarrow x + 36 - 2x = 29 \Rightarrow -x = -7 \Rightarrow x = 7$$ There are $7$ \$1 coins and $11$ \$2 coins.
(b)
$7 + 11 = 18$ coins $\checkmark$. Value: $7(1) + 11(2) = 7 + 22 = 29 \checkmark$
(c)
There are only $18$ coins in total, so $x$ cannot exceed $18$. If $x = 20$, the number of \$2 coins would be $18 - 20 = -2$, which is impossible — a negative number of coins has no meaning.
QUESTION 4
3 marks
Criterion A
Easy
A number increased by $7$ is equal to $3$ times the number.
a.
Form an equation and solve to find the number.
[2] b.
Verify your solution.
[1] Show complete worked solution
(a)
$$x + 7 = 3x \Rightarrow 7 = 2x \Rightarrow x = 3.5$$
(b)
$3.5 + 7 = 10.5$ and $3 \times 3.5 = 10.5$ $\checkmark$
QUESTION 5
3 marks
Criterion A
Medium
The sum of two consecutive odd integers is $52$.
a.
Let the smaller integer be $n$. Form an equation and solve it.
[2] b.
State the two integers.
[1] Show complete worked solution
(a)
$$n + (n+2) = 52 \Rightarrow 2n + 2 = 52 \Rightarrow n = 25$$
(b)
$25$ and $27$
QUESTION 6
5 marks
Criterion D
Medium
A school play sold $210$ tickets in total. Adult tickets cost \$12 and student tickets cost \$7. Total revenue was \$2050.
a.
Let $a$ be the number of adult tickets sold. Write down an expression for the number of student tickets sold.
[2] b.
Form an equation and solve it to find the number of adult tickets sold.
[2] c.
State the number of student tickets sold.
[1] Show complete worked solution
(a)
$(210 - a)$
(b)
$$12a + 7(210-a) = 2050 \Rightarrow 12a + 1470 - 7a = 2050 \Rightarrow 5a = 580 \Rightarrow a = 116$$
(c)
$210 - 116 = 94$
QUESTION 7
5 marks
Criterion D
Hard
The sum of the ages of a father and his daughter is $60$ years. In $6$ years, the father will be three times as old as his daughter will be then.
a.
Let the daughter's current age be $x$. Write an expression for the father's current age.
[2] b.
Form an equation using the given information about their ages in $6$ years' time, and solve it.
[3] Show complete worked solution
(a)
$(60 - x)$ years
(b)
$$(60 - x + 6) = 3(x + 6) \Rightarrow 66 - x = 3x + 18 \Rightarrow 48 = 4x \Rightarrow x = 12$$ So the daughter is currently $12$ and the father is currently $48$. Check: in $6$ years, daughter is $18$, father is $54 = 3 \times 18$ $\checkmark$
QUESTION 8
4 marks
Criterion C
Medium
A rectangle's length is $5$ cm more than its width. A student sets up the equation $2w + 2(w+5) = 42$ to represent a perimeter of $42$ cm, and solves to find $w = 8$.
a.
Verify the student's solution is correct by substituting back into the original equation.
[2] b.
Determine the length and area of the rectangle, and explain why a width of $w=-2$ (the other algebraic root, if the equation were quadratic) would not be a valid solution in this context.
[2] Show complete worked solution
(a)
$$2(8) + 2(8+5) = 16 + 26 = 42 \checkmark$$
(b)
Length $=13$ cm, Area $= 8 \times 13 = 104$ cm$^2$. A width of $-2$ would not be valid because a physical length or width cannot be negative.
QUESTION 9
5 marks
Criterion A
Hard
A two-digit number has a tens digit that is $2$ more than its units digit. The sum of the number and its reversed-digit number is $132$.
a.
Let the units digit be $u$. Using the fact that a number with tens digit $t$ and units digit $u$ has value $10t+u$, form an equation and solve it.
[3] b.
Determine the original two-digit number.
[2] Show complete worked solution
(a)
Tens digit $= u+2$. Number $= 10(u+2)+u = 11u+20$. Reversed number $= 10u + (u+2) = 11u+2$. $$(11u+20) + (11u+2) = 132 \Rightarrow 22u + 22 = 132 \Rightarrow u = 5$$
(b)
Units digit $=5$, tens digit $=7$, so the number is $75$. Check: reversed is $57$, and $75+57=132$ $\checkmark$
QUESTION 10
4 marks
Criterion D
Medium
Four friends share the cost of a \$96 gift equally. Later, one friend drops out, and the remaining three friends must now pay an equal, higher share.
a.
Calculate the original cost per person, and the new cost per person after one friend drops out.
[2] b.
Determine how much extra each of the remaining three friends must pay compared to the original amount.
[2] Show complete worked solution
(a)
Original: $96 \div 4 = \$24$. New: $96 \div 3 = \$32$.
(b)
$$32 - 24 = 8$$ Each remaining friend pays \$8 more.
QUESTION 11
4 marks
Criterion B
Hard
Think of a number. Add $5$. Double the result. Subtract $10$. Divide by $2$.
a.
Let the original number be $n$. Show algebraically that the result of this process is always equal to $n$.
[2] b.
Explain why this \"number trick\" always returns the original number, regardless of what number is chosen.
[2] Show complete worked solution
(a)
$$\dfrac{2(n+5) - 10}{2} = \dfrac{2n + 10 - 10}{2} = \dfrac{2n}{2} = n$$
(b)
The operations algebraically cancel out: adding $5$ then doubling introduces a $+10$, which is exactly cancelled by the \"subtract $10$\" step, and the initial doubling is cancelled by the final halving — leaving just $n$, for any value of $n$.
QUESTION 12
3 marks
Criterion C
Medium
A charity run raises money through a \$25 entry fee per runner plus \$4 per km sponsored, for a $10$ km run. A student estimates that $200$ runners would raise about \$13000 in total.
a.
Calculate the exact amount raised per runner, and hence the exact total for $200$ runners.
[2] b.
Comment on whether the student's estimate was reasonable.
[1] Show complete worked solution
(a)
Per runner: $25 + 4(10) = \$65$. Total: $200 \times 65 = \$13000$.
(b)
The student's estimate of \$13000 was not just reasonable — it was exactly correct.
Money and investment problems 12 questions
QUESTION 1
9 marks
Criterion D
Hard
Two ride-hailing apps calculate the fare for a journey of $d$ kilometres using the charges shown in the table below.
| App | Base fee | Rate per km |
|---|---|---|
| RideX | \$2.50 | \$0.90 |
| CabGo | \$1.00 | \$1.20 |
a.
Write down, in terms of $d$, an expression for the cost $C_R$ of a RideX journey and the cost $C_G$ of a CabGo journey.
[2] b.
Determine algebraically the distance at which both apps charge the same fare. Show full working.
[3] c.
Hence, state which app is cheaper for a $10$ km journey, and by how much.
[2] d.
Priya has \$12 to spend on a single journey. Determine the maximum distance, to the nearest $0.1$ km, she can travel with each app.
[2] Show complete worked solution
(a)
$$C_R = 2.50 + 0.90d \qquad C_G = 1.00 + 1.20d$$
(b)
$$2.50 + 0.90d = 1.00 + 1.20d \Rightarrow 1.50 = 0.30d \Rightarrow d = 5$$ Both apps charge the same fare for a $5$ km journey.
(c)
At $d=10$: $C_R = 2.50 + 9.00 = 11.50$ and $C_G = 1.00 + 12.00 = 13.00$. RideX is cheaper, by \$1.50.
(d)
RideX: $$2.50 + 0.90d = 12 \Rightarrow d = \dfrac{9.50}{0.90} = 10.5\overline{5}$$ Rounding down (she cannot exceed her \$12 budget), the maximum distance is $10.5$ km. CabGo: $$1.00 + 1.20d = 12 \Rightarrow d = \dfrac{11.00}{1.20} = 9.1\overline{6}$$ so the maximum distance is $9.1$ km.
QUESTION 2
7 marks
Criterion A
Medium
Simple interest is calculated using the formula $I = \dfrac{PRT}{100}$, where $P$ is the principal (\$), $R$ is the annual interest rate (%), and $T$ is the time in years.
a.
Calculate the interest earned when \$2400 is invested at $5\%$ per annum for $3$ years.
[2] b.
Determine the total amount in the account after the $3$ years.
[2] c.
An investment of \$1500 earns \$270 in interest after a certain number of years at a rate of $6\%$ per annum. Determine the number of years.
[3] Show complete worked solution
(a)
$$I = \dfrac{2400 \times 5 \times 3}{100} = 360$$ Interest earned $= \$360$.
(b)
$$2400 + 360 = 2760$$ Total amount $= \$2760$.
(c)
$$270 = \dfrac{1500 \times 6 \times T}{100} = 90T \Rightarrow T = 3$$ The investment was held for $3$ years.
QUESTION 3
8 marks
Criterion D
Hard
Two savings accounts are available. Account X pays simple interest at $4.5\%$ per annum. Account Y pays a flat bonus of \$60 plus simple interest at $3\%$ per annum. Priya has \$2000 to invest for $2$ years.
a.
Calculate the total interest earned in Account X after $2$ years.
[2] b.
Calculate the total interest earned (including the bonus) in Account Y after $2$ years.
[2] c.
Compare the two accounts after $2$ years. Determine which account, if either, gives the better return.
[2] d.
Determine after how many years Account X would begin to earn strictly more total interest than Account Y.
[2] Show complete worked solution
(a)
$$\dfrac{2000 \times 4.5 \times 2}{100} = 180$$ Account X earns \$180 in interest.
(b)
$$\dfrac{2000 \times 3 \times 2}{100} + 60 = 120 + 60 = 180$$ Account Y earns \$180 in total (including the bonus).
(c)
Both accounts give exactly the same total return of \$180 after $2$ years — neither account is better at this point in time.
(d)
Account X: $90T$. Account Y: $60 + 60T$. $$90T > 60 + 60T \Rightarrow 30T > 60 \Rightarrow T > 2$$ Account X earns more total interest for any investment period longer than $2$ years.
QUESTION 4
4 marks
Criterion A
Easy
Calculate the following.
a.
The simple interest on \$800 at $4\%$ per annum for $2$ years.
[2] b.
The sale price of a \$150 jacket after a $30\%$ discount.
[2] Show complete worked solution
(a)
$$\dfrac{800 \times 4 \times 2}{100} = 64$$ Interest $= \$64$.
(b)
$$150 \times 0.7 = 105$$ Sale price $= \$105$.
QUESTION 5
4 marks
Criterion A
Medium
A shopkeeper buys a bicycle for \$240 and sells it for \$300.
a.
Calculate the profit made, and the profit as a percentage of the cost price.
[2] b.
The shopkeeper buys a second bicycle for \$180 and wants to make a $40\%$ profit. Determine the selling price.
[2] Show complete worked solution
(a)
Profit $= 300-240=\$60$. Percentage $= \dfrac{60}{240} \times 100 = 25\%$.
(b)
$$180 \times 1.4 = 252$$ Selling price $= \$252$.
QUESTION 6
4 marks
Criterion D
Medium
Diego earns \$18 per hour for a standard $8$-hour work day, and \$27 per hour for any overtime hours beyond that.
a.
Calculate Diego's total pay for a day in which he works $11$ hours.
[2] b.
On another day, Diego earns \$252. Determine how many hours he worked that day.
[2] Show complete worked solution
(a)
$$8(18) + 3(27) = 144 + 81 = 225$$ Total pay $= \$225$.
(b)
Standard pay $=8(18)=144$. Overtime pay $=252-144=108$. Overtime hours $=108 \div 27=4$. Total hours $=8+4=12$.
QUESTION 7
4 marks
Criterion A
Hard
A store offers a $20\%$ discount on a \$150 item, followed by an additional $10\%$ off the already-discounted price at checkout.
a.
Calculate the price after the first $20\%$ discount.
[2] b.
Calculate the final price after both discounts, and determine the overall percentage discount from the original \$150 price.
[2] Show complete worked solution
(a)
$$150 \times 0.8 = 120$$
(b)
$$120 \times 0.9 = 108$$ Overall discount $= 150-108=\$42$, which is $\dfrac{42}{150} \times 100 = 28\%$.
QUESTION 8
3 marks
Criterion C
Medium
A salesperson earns a $6\%$ commission on sales. A student calculates the commission on \$1450 of sales as $1450 \times 0.6 = 870$.
a.
Determine the correct commission earned.
[2] b.
Identify the student's error.
[1] Show complete worked solution
(a)
$$1450 \times 0.06 = 87$$ Commission $= \$87$.
(b)
The student used $0.6$ ($60\%$) instead of $0.06$ ($6\%$) — the decimal point was misplaced when converting the percentage.
QUESTION 9
7 marks
Criterion D
Hard
Elena borrows \$4000 at a simple interest rate of $7.5\%$ per annum, to be repaid in full after $3$ years.
a.
Calculate the total interest owed after $3$ years.
[2] b.
Calculate the total amount Elena must repay.
[2] c.
If Elena repays the loan in $24$ equal monthly instalments, determine the amount of each instalment, to the nearest cent.
[3] Show complete worked solution
(a)
$$\dfrac{4000 \times 7.5 \times 3}{100} = 900$$ Interest $= \$900$.
(b)
$$4000 + 900 = 4900$$
(c)
$$4900 \div 24 = 204.1\overline{6} \approx \$204.17$$
QUESTION 10
4 marks
Criterion A
Medium
An online store lists a price of \$45 (USD). The exchange rate is $1$ USD $= 83.20$ INR.
a.
Calculate the equivalent price in INR.
[2] b.
A customer has INR $3900$. Determine the maximum whole number of \$45 USD items they could buy.
[2] Show complete worked solution
(a)
$$45 \times 83.20 = 3744$$ Price $=$ INR $3744$.
(b)
$$3900 \div 3744 \approx 1.04$$ The customer can buy $1$ item.
QUESTION 11
4 marks
Criterion D
Hard
A gym offers two plans: Plan M costs \$45 per month with unlimited visits. Plan N costs \$8 per visit, with no monthly fee.
a.
Form an inequality to determine the number of visits per month, $v$, at which Plan M becomes cheaper than Plan N.
[2] b.
Determine the minimum whole number of visits per month at which Plan M is the cheaper option.
[2] Show complete worked solution
(a)
$$45 < 8v \Rightarrow v > 5.625$$
(b)
$6$ visits (since $5$ visits costs $\$40 < \$45$ on Plan N, but $6$ visits costs $\$48 > \$45$).
QUESTION 12
5 marks
Criterion B
Medium
\$1000 is invested at $10\%$ per annum, with interest added to the principal each year (compounding annually).
a.
Calculate the amount at the end of each of the first $3$ years.
[3] b.
Determine an expression for the amount after $n$ years, and use it to check your answer for year $3$.
[2] Show complete worked solution
(a)
Year 1: $1000 \times 1.1 = \$1100$. Year 2: $1100 \times 1.1 = \$1210$. Year 3: $1210 \times 1.1 = \$1331$.
(b)
$$\text{Amount} = 1000(1.1)^n$$ For $n=3$: $1000(1.1)^3 = 1331$, which matches part (a) $\checkmark$
Motion problems 12 questions
QUESTION 1
7 marks
Criterion A
Medium
A cyclist travels $45$ km in $2.5$ hours at a constant speed.
a.
Calculate the cyclist's average speed, in km/h.
[2] b.
Determine how long the cyclist would take to travel $63$ km at the same speed.
[2] c.
The cyclist then increases speed by $20\%$ for the return journey. Calculate the new speed, and determine the time taken for the $45$ km return journey, to the nearest minute.
[3] Show complete worked solution
(a)
$$\text{Speed} = \dfrac{45}{2.5} = 18 \text{ km/h}$$
(b)
$$\text{Time} = \dfrac{63}{18} = 3.5 \text{ hours} = 3 \text{ h } 30 \text{ min}$$
(c)
New speed $= 18 \times 1.2 = 21.6$ km/h. $$\text{Time} = \dfrac{45}{21.6} = 2.0833... \text{ hours} = 2 \text{ h } 5 \text{ min (to the nearest minute)}$$
QUESTION 2
7 marks
Criterion D
Hard
Two towns, A and B, are $210$ km apart. A car leaves Town A travelling towards Town B at $60$ km/h. At the same time, a truck leaves Town B travelling towards Town A at $45$ km/h.
a.
Write down an expression for the distance travelled by the car, and by the truck, after $t$ hours.
[2] b.
Form an equation and solve it to determine how long it takes for the car and truck to meet.
[3] c.
Hence, calculate how far from Town A the two vehicles meet.
[2] Show complete worked solution
(a)
Car: $60t$ km. Truck: $45t$ km.
(b)
Since the two vehicles are approaching each other, the sum of the distances they travel equals $210$ km: $$60t + 45t = 210 \Rightarrow 105t = 210 \Rightarrow t = 2 \text{ hours}$$
(c)
$$\text{Distance from A} = 60 \times 2 = 120 \text{ km}$$
QUESTION 3
5 marks
Criterion C
Medium
A hiker walks from a cabin to a lookout at a speed of $4$ km/h, along a $6$ km path, and returns along the same path at a speed of $2$ km/h.
a.
Calculate the total time taken for the round trip, and hence calculate the hiker's average speed for the whole journey, correct to two decimal places.
[3] b.
A student claims the average speed should be $\dfrac{4+2}{2}=3$ km/h. Explain why this method is incorrect for this situation.
[2] Show complete worked solution
(a)
Time out $= \dfrac{6}{4} = 1.5$ h. Time back $= \dfrac{6}{2} = 3$ h. Total time $= 4.5$ h. Total distance $= 12$ km. $$\text{Average speed} = \dfrac{12}{4.5} \approx 2.67 \text{ km/h}$$
(b)
Average speed is total distance divided by total time — it is not simply the average of the two individual speeds, because the hiker spends a different amount of time travelling at each speed (more time at the slower $2$ km/h pace than at the faster $4$ km/h pace), so the slower speed should count for more in the average.
QUESTION 4
4 marks
Criterion A
Easy
Consider a car travelling at constant speed.
a.
A car travels $180$ km in $3$ hours. Calculate its average speed.
[2] b.
At this speed, calculate the distance travelled in $45$ minutes.
[2] Show complete worked solution
(a)
$$\dfrac{180}{3} = 60 \text{ km/h}$$
(b)
$$60 \times 0.75 = 45 \text{ km}$$
QUESTION 5
4 marks
Criterion A
Medium
A sprinter runs at a speed of $8.5$ m/s.
a.
Convert this speed to km/h.
[2] b.
Determine the time, in seconds, for the sprinter to run $100$ m at this speed.
[2] Show complete worked solution
(a)
$$8.5 \times 3.6 = 30.6 \text{ km/h}$$
(b)
$$\dfrac{100}{8.5} \approx 11.8 \text{ s}$$
QUESTION 6
4 marks
Criterion D
Medium
A bus travels the $250$ km from City P to City Q at an average speed of $80$ km/h, departing at $09{:}15$.
a.
Calculate the travel time for the journey, to the nearest minute.
[2] b.
Determine the arrival time in City Q.
[2] Show complete worked solution
(a)
$$\dfrac{250}{80} = 3.125 \text{ hours} = 3 \text{ h } 8 \text{ min (to the nearest minute)}$$
(b)
$09{:}15 + 3\text{h }08\text{min} = 12{:}23$
QUESTION 7
5 marks
Criterion A
Hard
A car leaves a service station travelling at $70$ km/h. Half an hour later, a second car leaves the same service station along the same road, travelling at $90$ km/h in the same direction.
a.
Write expressions for the distance travelled by each car, $t$ hours after the second car departs.
[2] b.
Form an equation and solve it to determine how long after the second car departs it catches up to the first car.
[3] Show complete worked solution
(a)
Car 1: $70(t+0.5)$ km. Car 2: $90t$ km.
(b)
$$90t = 70(t+0.5) \Rightarrow 90t = 70t + 35 \Rightarrow 20t = 35 \Rightarrow t = 1.75 \text{ hours}$$
QUESTION 8
4 marks
Criterion C
Medium
A student says that if a car's speed doubles, the time taken to travel a fixed distance also doubles.
a.
Using the formula $\text{time} = \dfrac{\text{distance}}{\text{speed}}$, determine whether the student's claim is correct.
[2] b.
Illustrate your answer with a numerical example using a distance of $120$ km.
[2] Show complete worked solution
(a)
The claim is false — since time is inversely proportional to speed for a fixed distance, doubling the speed halves the time, rather than doubling it.
(b)
At $40$ km/h: time $= 120/40 = 3$ h. At $80$ km/h (double the speed): time $= 120/80 = 1.5$ h — the time was halved, not doubled.
QUESTION 9
4 marks
Criterion D
Hard
A delivery driver travels $84$ km at $56$ km/h, stops for a $20$-minute break, then continues for another $63$ km at $70$ km/h.
a.
Calculate the driving time for each of the two stages of the journey.
[2] b.
Calculate the total time for the entire trip, including the break, in hours and minutes.
[2] Show complete worked solution
(a)
Stage 1: $84/56 = 1.5$ h. Stage 2: $63/70 = 0.9$ h.
(b)
$$1.5 + 0.9 + \dfrac{20}{60} = 2.7\overline{3} \text{ hours} = 2 \text{ h } 44 \text{ min}$$
QUESTION 10
4 marks
Criterion A
Medium
The table shows the distance travelled by a train at a constant speed.
| Time (h) | Distance (km) |
|---|---|
| 1 | 95 |
| 2.5 | 237.5 |
| 4 | 380 |
a.
Verify that the train is travelling at a constant speed, and state this speed.
[2] b.
Determine how long the train would take to travel $475$ km at this speed.
[2] Show complete worked solution
(a)
$95/1=95$, $237.5/2.5=95$, $380/4=95$ — constant speed of $95$ km/h, confirmed.
(b)
$$475 \div 95 = 5 \text{ hours}$$
QUESTION 11
5 marks
Criterion B
Hard
A journey has a fixed distance of $240$ km.
a.
Calculate the time taken at each of the following speeds: $40$ km/h, $60$ km/h, $80$ km/h, $120$ km/h.
[2] b.
Describe, using your results, the relationship between speed and time for this fixed distance, and explain why doubling the speed always exactly halves the time.
[3] Show complete worked solution
(a)
$6$ h, $4$ h, $3$ h, $2$ h respectively.
(b)
Speed and time are inversely proportional here, since $\text{speed} \times \text{time} = 240$ (a constant). Doubling the speed must halve the time for their product to remain $240$ — this is confirmed in the table: $40 \to 80$ (doubled) gives $6\text{h} \to 3\text{h}$ (halved), and $60 \to 120$ (doubled) gives $4\text{h} \to 2\text{h}$ (halved).
QUESTION 12
4 marks
Criterion D
Medium
A car uses fuel at a rate of $6.5$ litres per $100$ km travelled.
a.
Calculate the amount of fuel used for a $340$ km trip.
[2] b.
The car's fuel tank holds $50$ litres. Determine the maximum distance the car can travel on a full tank before running out of fuel.
[2] Show complete worked solution
(a)
$$340 \times \dfrac{6.5}{100} = 22.1 \text{ litres}$$
(b)
$$50 \div 6.5 \times 100 \approx 769.2 \text{ km}$$
Mixture problems 12 questions
QUESTION 1
7 marks
Criterion A
Medium
A shopkeeper blends two types of rice: Basmati rice costing \$6 per kg, and regular rice costing \$3.50 per kg, to make a $20$ kg blend costing \$4.25 per kg.
a.
Let $x$ be the mass, in kg, of Basmati rice used. Write down an expression for the mass of regular rice used.
[2] b.
Form an equation in terms of $x$ and solve it to determine the mass of each type of rice used.
[3] c.
Verify your solution by checking the total cost of the blend.
[2] Show complete worked solution
(a)
$(20 - x)$ kg
(b)
$$6x + 3.5(20-x) = 4.25(20) \Rightarrow 6x + 70 - 3.5x = 85 \Rightarrow 2.5x = 15 \Rightarrow x = 6$$ Basmati rice: $6$ kg. Regular rice: $14$ kg.
(c)
$$6(6) + 3.5(14) = 36 + 49 = 85 = 4.25 \times 20 \checkmark$$
QUESTION 2
7 marks
Criterion D
Hard
A cleaning company has a $15\%$ concentration disinfectant solution and a $60\%$ concentration disinfectant solution. They need to prepare $10$ litres of a $24\%$ concentration solution by mixing the two.
a.
Let $x$ be the volume, in litres, of the $15\%$ solution used. Write down an expression for the volume of the $60\%$ solution used.
[2] b.
Form an equation and solve it to determine the volume of each solution required.
[3] c.
Determine the amount of pure disinfectant (that is, $100\%$ active ingredient) contained in the final $10$ L mixture.
[2] Show complete worked solution
(a)
$(10 - x)$ litres
(b)
$$0.15x + 0.60(10-x) = 0.24(10) \Rightarrow 0.15x + 6 - 0.6x = 2.4 \Rightarrow -0.45x = -3.6 \Rightarrow x = 8$$ The $15\%$ solution: $8$ L. The $60\%$ solution: $2$ L.
(c)
$$0.24 \times 10 = 2.4 \text{ L of pure disinfectant}$$
QUESTION 3
7 marks
Criterion C
Medium
A grocer mixes $x$ kg of cashews costing \$18 per kg with $(15-x)$ kg of peanuts costing \$6 per kg to make $15$ kg of a trail mix. A student writes the cost equation as $18x + 6x = C$, where $C$ is the total cost of the mix.
a.
Identify the error in the student's equation, and write down the correct equation for the total cost $C$.
[2] b.
If the total cost of the mix is \$156, use your corrected equation to determine the mass of cashews used.
[3] c.
Determine the cost per kg of the finished trail mix.
[2] Show complete worked solution
(a)
The second term should use $(15-x)$ kg of peanuts, not $x$ kg — the student mistakenly used $x$ for both quantities. The correct equation is $$C = 18x + 6(15-x)$$
(b)
$$18x + 6(15-x) = 156 \Rightarrow 18x + 90 - 6x = 156 \Rightarrow 12x = 66 \Rightarrow x = 5.5$$ The mass of cashews used is $5.5$ kg.
(c)
$$\dfrac{156}{15} = 10.40$$ The trail mix costs \$10.40 per kg.
QUESTION 4
4 marks
Criterion A
Easy
A fruit punch is made by mixing $x$ litres of orange juice costing \$3 per litre with $(5-x)$ litres of apple juice costing \$2 per litre, to make $5$ litres of punch costing \$2.60 per litre.
a.
Form an equation and solve for $x$.
[3] b.
State the volume of orange juice and apple juice used.
[1] Show complete worked solution
(a)
$$3x + 2(5-x) = 2.6(5) \Rightarrow 3x + 10 - 2x = 13 \Rightarrow x = 3$$
(b)
$3$ L orange juice, $2$ L apple juice.
QUESTION 5
4 marks
Criterion A
Medium
A jeweller mixes gold that is $75\%$ pure with gold that is $50\%$ pure to make $40$ g of a $60\%$ pure gold alloy.
a.
Let $x$ be the mass, in grams, of the $75\%$ pure gold used. Form an equation and solve for $x$.
[3] b.
State the mass of each type of gold used.
[1] Show complete worked solution
(a)
$$0.75x + 0.50(40-x) = 0.60(40) \Rightarrow 0.75x + 20 - 0.5x = 24 \Rightarrow 0.25x = 4 \Rightarrow x = 16$$
(b)
$16$ g of the $75\%$ pure gold, $24$ g of the $50\%$ pure gold.
QUESTION 6
5 marks
Criterion D
Medium
A cafe blends $x$ kg of premium beans costing \$22 per kg with $(10-x)$ kg of standard beans costing \$14 per kg, to make $10$ kg of a house blend costing \$17.20 per kg.
a.
Form an equation and solve for $x$.
[3] b.
The cafe sells the entire $10$ kg house blend at \$25 per kg. Calculate the total profit made on the blend.
[2] Show complete worked solution
(a)
$$22x + 14(10-x) = 17.2(10) \Rightarrow 22x + 140 - 14x = 172 \Rightarrow 8x = 32 \Rightarrow x = 4$$
(b)
Cost $=\$172$. Revenue $=10 \times 25=\$250$. Profit $=250-172=\$78$.
QUESTION 7
7 marks
Criterion D
Hard
A $500$ mL bottle contains a fruit cordial that is $40\%$ concentrated syrup. Water (which is $0\%$ syrup) is added to dilute it to a $25\%$ concentration.
a.
Let $w$ mL of water be added. Write down an expression for the total volume, and explain why the total amount of syrup does not change when water is added.
[2] b.
Form an equation and solve for $w$.
[3] c.
Determine the total volume of the diluted cordial.
[2] Show complete worked solution
(a)
Total volume $= (500+w)$ mL. The amount of syrup stays at $500 \times 0.40 = 200$ mL, since only water (no syrup) is being added.
(b)
$$200 = 0.25(500+w) \Rightarrow 200 = 125 + 0.25w \Rightarrow 75 = 0.25w \Rightarrow w = 300$$
(c)
$$500 + 300 = 800 \text{ mL}$$
QUESTION 8
4 marks
Criterion C
Medium
A solution that is $20\%$ acid is mixed with a solution that is $45\%$ acid, in some proportion, to make a new blended solution.
a.
Explain why the concentration of the resulting blend can never be higher than $45\%$ or lower than $20\%$, regardless of the proportions used.
[2] b.
A student claims a $50\%$ acid blend can be made by mixing these two solutions. Explain why this is impossible.
[2] Show complete worked solution
(a)
The blend's concentration is a weighted average of $20\%$ and $45\%$, and a weighted average of two values always lies between those two values (inclusive) — it can never exceed the larger value or fall below the smaller one.
(b)
$50\%$ lies outside the range $[20\%, 45\%]$, so no combination of the two given solutions can reach it — achieving $50\%$ would require a solution stronger than $45\%$ to begin with.
QUESTION 9
5 marks
Criterion A
Hard
Equal masses of two teas are mixed: Tea A costs \$9 per kg and Tea B costs \$13 per kg.
a.
Determine the cost per kg of the blend, using equal masses of each.
[2] b.
If instead the teas are mixed in the ratio $3:1$ (Tea A : Tea B) by mass, determine the new cost per kg of the blend.
[3] Show complete worked solution
(a)
Since the masses are equal, the blend cost is the simple average: $\dfrac{9+13}{2} = \$11$ per kg.
(b)
Weighted average: $$\dfrac{3(9) + 1(13)}{3+1} = \dfrac{27+13}{4} = \dfrac{40}{4} = \$10 \text{ per kg}$$
QUESTION 10
5 marks
Criterion A
Medium
A hardware store mixes $x$ litres of red paint costing \$11 per litre with $y$ litres of white paint costing \$5 per litre to make exactly $12$ litres of pink paint, at a total cost of \$96.
a.
Write down two equations in terms of $x$ and $y$ representing the total volume and the total cost.
[2] b.
Solve the equations simultaneously to determine the volume of red paint and white paint used.
[3] Show complete worked solution
(a)
$$x + y = 12 \qquad 11x + 5y = 96$$
(b)
From the first equation, $y = 12-x$. Substituting: $$11x + 5(12-x) = 96 \Rightarrow 11x + 60 - 5x = 96 \Rightarrow 6x = 36 \Rightarrow x = 6$$ So $x=6$ L red paint and $y=6$ L white paint.
QUESTION 11
5 marks
Criterion B
Hard
Two solutions, with concentrations $p\%$ and $q\%$ (where $p
a.
Show that the concentration, $C\%$, of the resulting blend is given by $C = \dfrac{mp + nq}{m+n}$.
[3] b.
Use this formula to determine the concentration when a $10\%$ solution and a $40\%$ solution are mixed in a ratio of $2:3$.
[2] Show complete worked solution
(a)
Taking $m$ litres of the first solution and $n$ litres of the second: the amount of pure substance is $\dfrac{mp}{100} + \dfrac{nq}{100}$ litres, and the total volume is $(m+n)$ litres. So $$C = \dfrac{\frac{mp+nq}{100}}{m+n} \times 100 = \dfrac{mp+nq}{m+n}$$
(b)
$$C = \dfrac{2(10) + 3(40)}{2+3} = \dfrac{20+120}{5} = \dfrac{140}{5} = 28\%$$
QUESTION 12
4 marks
Criterion D
Medium
A juice concentrate is mixed with water in a ratio of $1:4$ (concentrate : water) to make ready-to-drink juice.
a.
Determine the amount of concentrate and water needed to make $2.5$ litres of ready-to-drink juice.
[2] b.
A larger batch uses $1.8$ litres of concentrate. Determine the total volume of ready-to-drink juice produced.
[2] Show complete worked solution
(a)
Total parts $=5$. Concentrate ($1$ part) $=0.5$ L. Water ($4$ parts) $=2$ L.
(b)
$$1.8 \times 5 = 9 \text{ litres}$$