NUMBER

QUESTION 1 4 marks Criterion A

Medium

A stadium has seats numbered consecutively from 1. Section A has seats 1 to 248, Section B continues from where A ends.

a. If Section B has 315 seats, what is the number of its last seat?

[2]

b. A seat is numbered 400. Which section is it in, and what position within that section (e.g. the 1st seat of the section, 2nd, etc.)?

[2]

Show complete worked solution

(a)

Section B: seats $249$ to $249+315-1=563$. Last seat number $=563$.

(b)

Since $249 \le 400 \le 563$, seat 400 is in Section B, at position $400-249+1=152$ (the 152nd seat of Section B).

QUESTION 2 4 marks Criterion A

Medium

Evaluate, showing full working: (a) the sum of all natural numbers from 1 to 50; (b) the sum of all even natural numbers from 2 to 50.

a. Find the sum of all natural numbers from 1 to 50, using the fact that pairing the first and last, second and second-last, etc. all give the same sum.

[2]

b. Find the sum of all even natural numbers from 2 to 50, using a similar pairing approach.

[2]

Show complete worked solution

(a)

Pairing gives 25 pairs, each summing to $1+50=51$. Total $=25\times51=1275$.

(b)

There are 25 even numbers (2 to 50). Pairing first+last: $2+50=52$, giving $12.5$ pairs... more directly: sum $= 2(1+2+\cdots+25) = 2\times\frac{25\times26}{2}=650$.

QUESTION 3 6 marks Criterion B

Medium

Investigate the pattern in the sum of the first $n$ natural numbers.

a. Calculate the sum of the first 3, first 4, and first 5 natural numbers.

[2]

b. Compare each sum to the formula $\frac{n(n+1)}{2}$ for the corresponding $n$. Does it match?

[2]

c. Use the formula to predict the sum of the first 100 natural numbers, then explain how confident you are in this prediction.

[2]

Show complete worked solution

(a)

$1+2+3=6$. $1+2+3+4=10$. $1+2+3+4+5=15$.

(b)

$n=3$: $\frac{3\times4}{2}=6$. $n=4$: $\frac{4\times5}{2}=10$. $n=5$: $\frac{5\times6}{2}=15$. All match.

(c)

$$\frac{100\times101}{2}=5050$$ Since the formula matched exactly for every tested case, and it is a well-known proven result (Gauss's method), this prediction can be trusted with high confidence.

QUESTION 4 5 marks Criterion C

Medium

Explain, using a diagrammatic or pairing argument (in words), why the sum of the first $n$ natural numbers equals $\frac{n(n+1)}{2}$.

a. Describe the pairing method: write the sum forwards and backwards, add corresponding terms, and explain what you notice.

[3]

b. Complete the explanation by solving for $S$.

[2]

Show complete worked solution

(a)

Writing $S = 1+2+\cdots+n$ and also $S=n+(n-1)+\cdots+1$, then adding term-by-term: each pair sums to $(n+1)$, and there are $n$ such pairs, giving $2S=n(n+1)$.

(b)

$$S = \frac{n(n+1)}{2}$$

QUESTION 5 5 marks Criterion D

Medium

A charity is stacking donated cans into a triangular display: 1 can on top, 2 in the next row, 3 in the row below that, and so on.

a. If the display has 12 rows, find the total number of cans, using the natural number sum formula.

[2]

b. The charity has 100 cans donated. What is the maximum number of complete rows they can build, and how many cans will be left over?

[3]

Show complete worked solution

(a)

$$\frac{12\times13}{2}=78 \text{ cans}$$

(b)

Testing: 13 rows needs $\frac{13\times14}{2}=91$ cans. 14 rows needs $\frac{14\times15}{2}=105$ cans (too many). So 13 complete rows can be built, using 91 cans, leaving $100-91=9$ cans left over.

QUESTION 6 6 marks Criterion A

Hard

A theatre has seats arranged in rows. Row 1 has 18 seats, and each subsequent row has 3 more seats than the previous row. The theatre has 22 rows.

a. Find the number of seats in Row 22.

[2]

b. The theatre manager wants to know the total seating capacity, but also needs to reserve the LAST 2 seats of every row for accessibility, leaving them empty. Find the total number of seats that will actually be occupied, across all 22 rows.

[4]

Show complete worked solution

(a)

$$u_{22} = 18+(22-1)(3) = 18+63=81 \text{ seats}$$

(b)

Total seats $=\frac{22}{2}(18+81)=11(99)=1089$. Reserved seats $=22\times2=44$. Occupied seats $=1089-44=1045$.

QUESTION 7 7 marks Criterion B

Hard

Prove, using algebra, that the sum of any 4 consecutive natural numbers is never divisible by 4.

a. Let the 4 consecutive natural numbers be $n, n+1, n+2, n+3$. Write and simplify an expression for their sum.

[2]

b. Explain, by considering the remainder when $4n+6$ is divided by 4, why this sum can NEVER be exactly divisible by 4, for ANY natural number $n$.

[3]

c. Verify this conclusion using the 4 consecutive numbers 7, 8, 9, 10.

[2]

Show complete worked solution

(a)

$$n+(n+1)+(n+2)+(n+3) = 4n+6$$

(b)

$4n+6 = 4(n+1)+2$ — this shows $4n+6$ is always 2 MORE than a multiple of 4 (i.e. it leaves remainder 2 when divided by 4), for every possible value of $n$. Since the remainder is never 0, the sum can never be exactly divisible by 4.

(c)

Sum $=7+8+9+10=34$. $34\div4=8.5$ (remainder 2, matching the proof) — confirms 34 is not divisible by 4.

QUESTION 8 5 marks Criterion C

Hard

A younger student asks why $0$ is considered a 'whole number' but not always a 'natural number', and why this distinction even matters.

a. Clearly define, using precise mathematical language, the difference between the sets of Natural Numbers and Whole Numbers, and where 0 fits.

[3]

b. Explain, with a concrete real-world example, WHY this distinction can matter practically (e.g. why counting something might exclude zero, but measuring something might need to include it).

[2]

Show complete worked solution

(a)

Let $\mathbb{N} = \{1,2,3,4,\ldots\}$ denote the natural numbers (some definitions include 0, but the traditional MYP convention excludes it), and let $W=\{0,1,2,3,\ldots\}$ denote the whole numbers. The key distinction is that $0 \in W$ but $0 \notin \mathbb{N}$ under this convention — whole numbers include everything natural numbers do, PLUS zero.

(b)

Under the stated convention, $\mathbb N=\{1,2,3,\ldots\}$ and the whole numbers are $\{0,1,2,3,\ldots\}$. For example, a class may have $0$ absences, so zero must be allowed in that count. By contrast, when positions in a queue are numbered from the first person, the position numbers are $1,2,3,\ldots$. The convention must be stated because some textbooks include zero in $\mathbb N$.

QUESTION 9 7 marks Criterion D

Hard

A city's public transport authority is planning a new bus route. Data shows the number of daily riders has followed a pattern: 240 riders in Week 1, increasing by exactly 35 riders each subsequent week, for the first 8 weeks of operation.

a. Model the number of riders in Week $n$ (for $1\le n\le8$) using an arithmetic formula, and predict the ridership in Week 8.

[3]

b. The authority wants to predict ridership in Week 20 using the SAME linear model. Calculate this prediction, then critically evaluate whether extending a linear (constant increase) model this far into the future is realistic for real-world ridership growth, identifying at least one limitation of the model.

[4]

Show complete worked solution

(a)

$u_n = 240+(n-1)(35)$. Week 8: $u_8=240+7(35)=240+245=485$ riders.

(b)

$u_{20}=240+19(35)=240+665=905$ riders. This linear extrapolation is likely UNREALISTIC over such a long timeframe — real ridership growth typically slows and plateaus as it approaches the practical capacity of the route/buses, is affected by seasonal variation, and cannot increase forever at a constant rate. The model is likely only valid for the initial growth phase, not for long-term prediction.

QUESTION 10 6 marks Criterion A

Hard

A theatre has seats arranged in rows. Row 1 has 18 seats, and each subsequent row has 3 more seats than the previous row. The theatre has 22 rows.

a. Find the number of seats in Row 22.

[2]

b. The theatre manager wants to know the total seating capacity, but also needs to reserve the LAST 2 seats of every row for accessibility, leaving them empty. Find the total number of seats that will actually be occupied, across all 22 rows.

[4]

Show complete worked solution

(a)

$$u_{22} = 18+(22-1)(3) = 18+63=81 \text{ seats}$$

(b)

Total seats $=\frac{22}{2}(18+81)=11(99)=1089$. Reserved seats $=22\times2=44$. Occupied seats $=1089-44=1045$.

QUESTION 11 7 marks Criterion B

Hard

Prove, using algebra, that the sum of any 4 consecutive natural numbers is never divisible by 4.

a. Let the 4 consecutive natural numbers be $n, n+1, n+2, n+3$. Write and simplify an expression for their sum.

[2]

b. Explain, by considering the remainder when $4n+6$ is divided by 4, why this sum can NEVER be exactly divisible by 4, for ANY natural number $n$.

[3]

c. Verify this conclusion using the 4 consecutive numbers 7, 8, 9, 10.

[2]

Show complete worked solution

(a)

$$n+(n+1)+(n+2)+(n+3) = 4n+6$$

(b)

$4n+6 = 4(n+1)+2$ — this shows $4n+6$ is always 2 MORE than a multiple of 4 (i.e. it leaves remainder 2 when divided by 4), for every possible value of $n$. Since the remainder is never 0, the sum can never be exactly divisible by 4.

(c)

Sum $=7+8+9+10=34$. $34\div4=8.5$ (remainder 2, matching the proof) — confirms 34 is not divisible by 4.

QUESTION 12 5 marks Criterion C

Hard

A younger student asks why $0$ is considered a 'whole number' but not always a 'natural number', and why this distinction even matters.

a. Clearly define, using precise mathematical language, the difference between the sets of Natural Numbers and Whole Numbers, and where 0 fits.

[3]

b. Explain, with a concrete real-world example, WHY this distinction can matter practically (e.g. why counting something might exclude zero, but measuring something might need to include it).

[2]

Show complete worked solution

(a)

Let $\mathbb{N} = \{1,2,3,4,\ldots\}$ denote the natural numbers (some definitions include 0, but the traditional MYP convention excludes it), and let $W=\{0,1,2,3,\ldots\}$ denote the whole numbers. The key distinction is that $0 \in W$ but $0 \notin \mathbb{N}$ under this convention — whole numbers include everything natural numbers do, PLUS zero.

(b)

Under the stated convention, $\mathbb N=\{1,2,3,\ldots\}$ and the whole numbers are $\{0,1,2,3,\ldots\}$. For example, a class may have $0$ absences, so zero must be allowed in that count. By contrast, when positions in a queue are numbered from the first person, the position numbers are $1,2,3,\ldots$. The convention must be stated because some textbooks include zero in $\mathbb N$.

QUESTION 13 7 marks Criterion D

Hard

A city's public transport authority is planning a new bus route. Data shows the number of daily riders has followed a pattern: 240 riders in Week 1, increasing by exactly 35 riders each subsequent week, for the first 8 weeks of operation.

a. Model the number of riders in Week $n$ (for $1\le n\le8$) using an arithmetic formula, and predict the ridership in Week 8.

[3]

b. The authority wants to predict ridership in Week 20 using the SAME linear model. Calculate this prediction, then critically evaluate whether extending a linear (constant increase) model this far into the future is realistic for real-world ridership growth, identifying at least one limitation of the model.

[4]

Show complete worked solution

(a)

$u_n = 240+(n-1)(35)$. Week 8: $u_8=240+7(35)=240+245=485$ riders.

(b)

$u_{20}=240+19(35)=240+665=905$ riders. This linear extrapolation is likely UNREALISTIC over such a long timeframe — real ridership growth typically slows and plateaus as it approaches the practical capacity of the route/buses, is affected by seasonal variation, and cannot increase forever at a constant rate. The model is likely only valid for the initial growth phase, not for long-term prediction.

QUESTION 1 3 marks Criterion A

Medium

The temperature at the summit of a mountain at 6 am was $-14^\circ$C. By 2 pm it had risen by $19^\circ$C, then fell by $8^\circ$C by 8 pm.

a. Find the temperature at 2 pm.

[1]

b. Find the temperature at 8 pm.

[1]

c. Find the overall change in temperature from 6 am to 8 pm.

[1]

Show complete worked solution

(a)

$$-14 + 19 = 5^\circ\text{C}$$

(b)

$$5 - 8 = -3^\circ\text{C}$$

(c)

$$-3 - (-14) = 11^\circ\text{C increase}$$

QUESTION 2 4 marks Criterion A

Medium

A submarine is at a depth of $-120$ m (120 m below sea level). It rises $45$ m, then descends $68$ m.

a. Find the submarine's new depth after both movements, showing each step.

[2]

b. A second submarine starts at $-95$ m and needs to reach the same final depth as the first submarine. Find how far it must descend.

[2]

Show complete worked solution

(a)

After rising: $-120+45=-75$ m. After descending: $-75-68=-143$ m.

(b)

$$-143 - (-95) = -48 \text{ m, i.e. it must descend a further 48 m}$$

QUESTION 3 3 marks Criterion A

Medium

Evaluate each of the following, showing full working:

a. $(-6) + (-9) - (-14)$

[1]

b. $(-3) \times (-4) \times (-2)$

[1]

c. $\dfrac{(-28)}{(-4)} + (-7)$

[1]

Show complete worked solution

(a)

$$-6-9+14 = -1$$

(b)

$$12 \times (-2) = -24$$

(c)

$$7 + (-7) = 0$$

QUESTION 4 5 marks Criterion B

Medium

Investigate the sign of the result when multiplying several negative integers together.

a. Calculate $(-2)\times(-3)$, $(-2)\times(-3)\times(-4)$, and $(-2)\times(-3)\times(-4)\times(-5)$.

[3]

b. State a rule connecting the number of negative factors to the sign of the product.

[2]

Show complete worked solution

(a)

$(-2)\times(-3)=6$ (positive). $6\times(-4)=-24$ (negative). $-24\times(-5)=120$ (positive).

(b)

If the number of negative factors is even, the product is positive. If the number of negative factors is odd, the product is negative.

QUESTION 5 5 marks Criterion C

Medium

A classmate says: "Subtracting a negative number always makes the answer bigger."

a. Test this claim with three different examples of your own, showing full working.

[3]

b. Explain, in your own words, why subtracting a negative number has this effect.

[2]

Show complete worked solution

(a)

E.g. $5-(-3)=8$ (bigger than 5). $-2-(-6)=4$ (bigger than $-2$). $-10-(-1)=-9$ (bigger than $-10$). In every case the result is bigger than the starting number.

(b)

Subtracting a negative is the same as adding its positive opposite (e.g. $-(-3)=+3$), and adding a positive number always increases the value — so the claim is correct.

QUESTION 6 4 marks Criterion D

Medium

A company's profit/loss (in thousands of dollars) over 4 quarters was: Q1: $-18$, Q2: $+32$, Q3: $-9$, Q4: $+41$.

a. Find the company's total profit or loss for the year.

[2]

b. The company needs a total annual profit of at least \$40{,}000 to avoid layoffs. Based on your answer, will layoffs be avoided? Justify your answer.

[2]

Show complete worked solution

(a)

$$-18+32-9+41 = 46 \text{ (i.e. a profit of \$46{,}000)}$$

(b)

Yes — the total profit of \$46{,}000 exceeds the required \$40{,}000, so layoffs will be avoided, with \$6{,}000 to spare.

QUESTION 7 6 marks Criterion A

Hard

A submarine's depth changes are recorded (negative = below sea level): starts at $-180$m, rises $65$m, descends $340$m, then rises $95$m.

a. Find the submarine's final depth.

[2]

b. The submarine's hull is only rated to withstand pressure safely down to $-500$m. During the ENTIRE sequence of movements (not just the final position), find the submarine's LOWEST point reached, and determine if the hull rating was ever exceeded.

[4]

Show complete worked solution

(a)

$$-180+65-340+95 = -360 \text{ m}$$

(b)

Tracking each stage: start $-180$, after rise: $-115$, after descent: $-455$, after final rise: $-360$. The lowest point reached was $-455$m (after the descent stage), which is within the $-500$m safety rating — so the hull rating was never exceeded, even though the descent came close.

QUESTION 8 8 marks Criterion B

Hard

Investigate whether the parity (odd/even) of the sum of $n$ consecutive integers depends on the starting number, or only on $n$ itself — for the case where $n$ is EVEN.

a. Find the sum of the 2 consecutive integers starting at 5 (i.e. 5,6), and separately the sum starting at 10 (i.e. 10,11). State the parity of each sum.

[2]

b. Now find the sum of 4 consecutive integers starting at 3 (3,4,5,6), and separately starting at 7 (7,8,9,10). State the parity of each.

[2]

c. Based on parts (a) and (b), conjecture a rule: for an EVEN count $n$ of consecutive integers, does the sum's parity depend on the starting number? Prove your conjecture algebraically, using $n$ consecutive integers starting at $a$: $a, a+1, \ldots, a+n-1$.

[4]

Show complete worked solution

(a)

$5+6=11$ (odd). $10+11=21$ (odd). Both sums are odd, despite starting from very different numbers.

(b)

$3+4+5+6=18$ (even). $7+8+9+10=34$ (even). Both sums are even, again regardless of the different starting points.

(c)

Conjecture: for even $n$, the sum's parity does NOT depend on the starting number $a$ — it depends only on $n$. Proof: the sum is $S=\frac{n}{2}(2a+n-1)$. Since $n$ is even, $n-1$ is odd, so $2a+n-1$ is always ODD (even plus odd is odd), regardless of $a$. Since an odd number times anything has the same parity as that 'anything', $S$ has the same parity as $\frac{n}{2}$ — which depends only on $n$, never on $a$. This proves the starting number never affects the sum's parity when $n$ is even.

QUESTION 9 5 marks Criterion C

Hard

A student says: 'A negative number times a negative number is positive, so negative numbers basically don't exist in the final answer — they always disappear.'

a. Define clearly: under what SPECIFIC condition does multiplying two negative numbers give a positive result, and give a counterexample showing negative numbers do NOT always 'disappear'.

[3]

b. State a precise, general rule for when a product of several negative numbers is positive versus negative.

[2]

Show complete worked solution

(a)

Two negative numbers multiplied together give a POSITIVE result specifically because $(-a)\times(-b) = ab$ for positive $a,b$ (the two sign flips cancel). However, this is only true for MULTIPLICATION of exactly two negatives — e.g. $(-3)+(-5)=-8$ (addition of two negatives stays negative), and $(-3)\times(-3)\times(-3)=-27$ (three negatives multiplied gives a negative result), clearly disproving the student's overgeneralization.

(b)

A product of several negative numbers is POSITIVE if there is an EVEN number of negative factors, and NEGATIVE if there is an ODD number of negative factors.

QUESTION 10 6 marks Criterion D

Hard

A company's quarterly profit/loss (in thousands of dollars) over 2 years (8 quarters) was: $-45, 62, -18, 71, -30, 55, -12, 68$.

a. Find the company's total profit or loss over the full 2 years.

[2]

b. The company's board wants to identify a TREND: are losses (negative quarters) getting smaller in magnitude over time, suggesting improving stability? List the magnitude of EACH loss quarter in order, and evaluate whether there's a clear improving trend, being explicit about the limitations of drawing conclusions from only 4 data points.

[4]

Show complete worked solution

(a)

$$-45+62-18+71-30+55-12+68=151 \text{ (a profit of \$151{,}000)}$$

(b)

Loss magnitudes in order: $45, 18, 30, 12$. There is a GENERAL downward trend in loss size (45, 18, 30, 12), though it isn't perfectly smooth (30 is a slight increase from 18). With only 4 loss-quarters of data, this is too small a sample to confidently confirm a genuine long-term improving trend — more quarters of data would be needed to rule out random fluctuation versus a real underlying pattern.

QUESTION 11 6 marks Criterion A

Hard

A submarine's depth changes are recorded (negative = below sea level): starts at $-180$m, rises $65$m, descends $340$m, then rises $95$m.

a. Find the submarine's final depth.

[2]

b. The submarine's hull is only rated to withstand pressure safely down to $-500$m. During the ENTIRE sequence of movements (not just the final position), find the submarine's LOWEST point reached, and determine if the hull rating was ever exceeded.

[4]

Show complete worked solution

(a)

$$-180+65-340+95 = -360 \text{ m}$$

(b)

Tracking each stage: start $-180$, after rise: $-115$, after descent: $-455$, after final rise: $-360$. The lowest point reached was $-455$m (after the descent stage), which is within the $-500$m safety rating — so the hull rating was never exceeded, even though the descent came close.

QUESTION 12 8 marks Criterion B

Hard

Investigate whether the parity (odd/even) of the sum of $n$ consecutive integers depends on the starting number, or only on $n$ itself — for the case where $n$ is EVEN.

a. Find the sum of the 2 consecutive integers starting at 5 (i.e. 5,6), and separately the sum starting at 10 (i.e. 10,11). State the parity of each sum.

[2]

b. Now find the sum of 4 consecutive integers starting at 3 (3,4,5,6), and separately starting at 7 (7,8,9,10). State the parity of each.

[2]

c. Based on parts (a) and (b), conjecture a rule: for an EVEN count $n$ of consecutive integers, does the sum's parity depend on the starting number? Prove your conjecture algebraically, using $n$ consecutive integers starting at $a$: $a, a+1, \ldots, a+n-1$.

[4]

Show complete worked solution

(a)

$5+6=11$ (odd). $10+11=21$ (odd). Both sums are odd, despite starting from very different numbers.

(b)

$3+4+5+6=18$ (even). $7+8+9+10=34$ (even). Both sums are even, again regardless of the different starting points.

(c)

Conjecture: for even $n$, the sum's parity does NOT depend on the starting number $a$ — it depends only on $n$. Proof: the sum is $S=\frac{n}{2}(2a+n-1)$. Since $n$ is even, $n-1$ is odd, so $2a+n-1$ is always ODD (even plus odd is odd), regardless of $a$. Since an odd number times anything has the same parity as that 'anything', $S$ has the same parity as $\frac{n}{2}$ — which depends only on $n$, never on $a$. This proves the starting number never affects the sum's parity when $n$ is even.

QUESTION 13 5 marks Criterion C

Hard

A student says: 'A negative number times a negative number is positive, so negative numbers basically don't exist in the final answer — they always disappear.'

a. Define clearly: under what SPECIFIC condition does multiplying two negative numbers give a positive result, and give a counterexample showing negative numbers do NOT always 'disappear'.

[3]

b. State a precise, general rule for when a product of several negative numbers is positive versus negative.

[2]

Show complete worked solution

(a)

Two negative numbers multiplied together give a POSITIVE result specifically because $(-a)\times(-b) = ab$ for positive $a,b$ (the two sign flips cancel). However, this is only true for MULTIPLICATION of exactly two negatives — e.g. $(-3)+(-5)=-8$ (addition of two negatives stays negative), and $(-3)\times(-3)\times(-3)=-27$ (three negatives multiplied gives a negative result), clearly disproving the student's overgeneralization.

(b)

A product of several negative numbers is POSITIVE if there is an EVEN number of negative factors, and NEGATIVE if there is an ODD number of negative factors.

QUESTION 14 6 marks Criterion D

Hard

A company's quarterly profit/loss (in thousands of dollars) over 2 years (8 quarters) was: $-45, 62, -18, 71, -30, 55, -12, 68$.

a. Find the company's total profit or loss over the full 2 years.

[2]

b. The company's board wants to identify a TREND: are losses (negative quarters) getting smaller in magnitude over time, suggesting improving stability? List the magnitude of EACH loss quarter in order, and evaluate whether there's a clear improving trend, being explicit about the limitations of drawing conclusions from only 4 data points.

[4]

Show complete worked solution

(a)

$$-45+62-18+71-30+55-12+68=151 \text{ (a profit of \$151{,}000)}$$

(b)

Loss magnitudes in order: $45, 18, 30, 12$. There is a GENERAL downward trend in loss size (45, 18, 30, 12), though it isn't perfectly smooth (30 is a slight increase from 18). With only 4 loss-quarters of data, this is too small a sample to confidently confirm a genuine long-term improving trend — more quarters of data would be needed to rule out random fluctuation versus a real underlying pattern.

QUESTION 1 3 marks Criterion A

Medium

Simplify each expression, showing full working:

a. $2^3 \times 2^4$

[1]

b. $\dfrac{5^6}{5^2}$

[1]

c. $(3^2)^3$

[1]

Show complete worked solution

(a)

$$2^3\times2^4 = 2^{3+4} = 2^7 = 128$$

(b)

$$5^{6-2} = 5^4 = 625$$

(c)

$$3^{2\times3} = 3^6 = 729$$

QUESTION 2 4 marks Criterion A

Medium

A single bacterium in a petri dish doubles in number every hour.

a. Write an expression, using index notation, for the number of bacteria after $n$ hours.

[1]

b. Find the number of bacteria after 8 hours.

[1]

c. Find how many more hours it takes to go from 256 bacteria to over 10000 bacteria.

[2]

Show complete worked solution

(a)

$$2^n$$

(b)

$$2^8 = 256$$

(c)

$2^{13}=8192$ and $2^{14}=16384$, so it takes until hour 14 to exceed 10000 — that's $14-8=6$ more hours.

QUESTION 3 5 marks Criterion B

Medium

Investigate the pattern in the last digit of increasing powers of 2.

a. List the last digit of $2^1, 2^2, 2^3, \ldots, 2^8$.

[2]

b. Describe the pattern you observe.

[1]

c. Using the pattern, predict the last digit of $2^{23}$ without calculating the full value.

[2]

Show complete worked solution

(a)

$2,4,8,6,2,4,8,6$ (for powers 1 through 8).

(b)

The last digits repeat in a cycle of 4: $2,4,8,6$, then repeat.

(c)

$23 \div 4 = 5$ remainder $3$, so $2^{23}$ has the same last digit as $2^3$, which is $8$.

QUESTION 4 5 marks Criterion C

Medium

A student simplifies $2^3 \times 3^2$ as $6^5$.

a. Evaluate both $2^3\times3^2$ and $6^5$ directly to show whether the student's simplification is correct.

[2]

b. Explain, using the index laws, why the rule $a^m \times b^n = (ab)^{m+n}$ that the student seems to be using is not valid, and state when index addition rules DO apply.

[3]

Show complete worked solution

(a)

$2^3\times3^2 = 8\times9=72$. $6^5=7776$. These are not equal, so the student's simplification is incorrect.

(b)

The index law $a^m\times a^n=a^{m+n}$ only applies when the BASE is the same on both terms. Here the bases (2 and 3) are different, so the exponents cannot simply be combined — each power must be evaluated separately before multiplying.

QUESTION 5 6 marks Criterion D

Medium

A savings amount doubles every 3 years due to compound growth. An initial deposit of \$500 is made.

a. Write an expression for the amount after $t$ periods of 3 years, and find the amount after 4 such periods (12 years).

[3]

b. The saver wants to reach \$50000. Find the smallest whole number of 3-year periods needed, and state the total number of years this takes.

[3]

Show complete worked solution

(a)

Amount $=500\times2^t$. After 4 periods: $500\times2^4=500\times16=8000$.

(b)

$500\times2^t \ge 50000 \Rightarrow 2^t\ge100$. Since $2^6=64$ and $2^7=128$, $t=7$ periods are needed, i.e. $7\times3=21$ years.

QUESTION 6 7 marks Criterion A

Hard

A single-cell organism divides into 2 every hour. A colony starts with 3 organisms.

a. Write an expression, using index notation, for the population after $h$ hours, and find the population after 10 hours.

[3]

b. A rival colony starts with just 1 organism, but divides into 3 every hour instead of 2. Find after how many WHOLE hours the rival colony's population first EXCEEDS the original colony's population (assuming both start at $h=0$).

[4]

Show complete worked solution

(a)

$P(h)=3\times2^h$. After 10 hours: $3\times2^{10}=3\times1024=3072$.

(b)

Original: $3\times2^h$. Rival: $1\times3^h=3^h$. Testing values: $h=3$: original$=24$, rival$=27$ (rival exceeds!). $h=2$: original$=12$, rival$=9$ (rival still behind). So the rival colony first exceeds the original after 3 whole hours.

QUESTION 7 8 marks Criterion B

Hard

Investigate the units digit of increasing powers of 7, i.e. $7^1, 7^2, 7^3, \ldots$

a. Find the units digit of $7^1$ through $7^8$.

[3]

b. State the repeating cycle length, and predict the units digit of $7^{35}$ using the cycle, showing your reasoning.

[3]

c. Prove that $7^{35}$ is NOT a multiple of 10, using your units-digit finding, without calculating the full value of $7^{35}$.

[2]

Show complete worked solution

(a)

$7^1=7$(7), $7^2=49$(9), $7^3=343$(3), $7^4=2401$(1), $7^5=16807$(7), $7^6=\ldots$(9), $7^7=\ldots$(3), $7^8=\ldots$(1). Units digits: $7,9,3,1,7,9,3,1$.

(b)

The cycle length is 4 ($7,9,3,1$ repeating). $35\div4=8$ remainder $3$, so $7^{35}$ has the same units digit as $7^3$, which is $\boxed{3}$.

(c)

A number is a multiple of 10 only if its units digit is 0. Since $7^{35}$ has units digit 3 (never 0, as the full 4-digit cycle $7,9,3,1$ never includes 0), $7^{35}$ can never be a multiple of 10.

QUESTION 8 5 marks Criterion C

Hard

A student writes $2^{10}$ as '$2\times10=20$', confusing index notation with multiplication notation.

a. Precisely define what $2^{10}$ means (using repeated multiplication notation), and calculate its correct value.

[3]

b. Explain, using the relative SIZES of $2^{10}=1024$ versus $2\times10=20$, why confusing these two notations could lead to serious errors in a real scientific or financial calculation.

[2]

Show complete worked solution

(a)

$2^{10}$ means $2$ multiplied by itself $10$ times: $\underbrace{2\times2\times2\times\cdots\times2}_{10\text{ factors}} = 1024$, which is very different from $2\times10=20$.

(b)

The two values differ by a factor of over 50 ($1024$ vs $20$) — in a real context like calculating compound growth, data storage (e.g. computer memory, where $2^{10}$ 'kilobytes' is a standard unit), or population growth, this confusion could lead to a drastically wrong final answer, potentially by orders of magnitude.

QUESTION 9 6 marks Criterion D

Hard

A hard drive's storage capacity doubles with each new model generation. The Generation 1 drive holds $2^8$ MB. Each new generation adds exactly 1 to the exponent.

a. Find the storage capacity, in MB, of the Generation 5 drive.

[2]

b. A consumer wants a drive with AT LEAST 1,000,000 MB (1 TB) of storage. Find the generation number required, and comment on whether 'doubling every generation forever' is a realistic assumption for real hard drive technology over many decades.

[4]

Show complete worked solution

(a)

Generation 5 exponent $=8+4=12$. Capacity $=2^{12}=4096$ MB.

(b)

Need $2^{8+n}\ge1000000$. Testing: $2^{20}=1048576\ge1000000$ (generation where exponent$=20$, i.e. Generation $20-8+1=13$). This 'doubling' pattern (similar to Moore's Law in computing) has historically held reasonably well over shorter timeframes, but is not realistic to extrapolate indefinitely — physical and economic limits eventually slow or halt this kind of exponential growth in real technology.

QUESTION 10 7 marks Criterion A

Hard

A single-cell organism divides into 2 every hour. A colony starts with 3 organisms.

a. Write an expression, using index notation, for the population after $h$ hours, and find the population after 10 hours.

[3]

b. A rival colony starts with just 1 organism, but divides into 3 every hour instead of 2. Find after how many WHOLE hours the rival colony's population first EXCEEDS the original colony's population (assuming both start at $h=0$).

[4]

Show complete worked solution

(a)

$P(h)=3\times2^h$. After 10 hours: $3\times2^{10}=3\times1024=3072$.

(b)

Original: $3\times2^h$. Rival: $1\times3^h=3^h$. Testing values: $h=3$: original$=24$, rival$=27$ (rival exceeds!). $h=2$: original$=12$, rival$=9$ (rival still behind). So the rival colony first exceeds the original after 3 whole hours.

QUESTION 11 8 marks Criterion B

Hard

Investigate the units digit of increasing powers of 7, i.e. $7^1, 7^2, 7^3, \ldots$

a. Find the units digit of $7^1$ through $7^8$.

[3]

b. State the repeating cycle length, and predict the units digit of $7^{35}$ using the cycle, showing your reasoning.

[3]

c. Prove that $7^{35}$ is NOT a multiple of 10, using your units-digit finding, without calculating the full value of $7^{35}$.

[2]

Show complete worked solution

(a)

$7^1=7$(7), $7^2=49$(9), $7^3=343$(3), $7^4=2401$(1), $7^5=16807$(7), $7^6=\ldots$(9), $7^7=\ldots$(3), $7^8=\ldots$(1). Units digits: $7,9,3,1,7,9,3,1$.

(b)

The cycle length is 4 ($7,9,3,1$ repeating). $35\div4=8$ remainder $3$, so $7^{35}$ has the same units digit as $7^3$, which is $\boxed{3}$.

(c)

A number is a multiple of 10 only if its units digit is 0. Since $7^{35}$ has units digit 3 (never 0, as the full 4-digit cycle $7,9,3,1$ never includes 0), $7^{35}$ can never be a multiple of 10.

QUESTION 12 5 marks Criterion C

Hard

A student writes $2^{10}$ as '$2\times10=20$', confusing index notation with multiplication notation.

a. Precisely define what $2^{10}$ means (using repeated multiplication notation), and calculate its correct value.

[3]

b. Explain, using the relative SIZES of $2^{10}=1024$ versus $2\times10=20$, why confusing these two notations could lead to serious errors in a real scientific or financial calculation.

[2]

Show complete worked solution

(a)

$2^{10}$ means $2$ multiplied by itself $10$ times: $\underbrace{2\times2\times2\times\cdots\times2}_{10\text{ factors}} = 1024$, which is very different from $2\times10=20$.

(b)

The two values differ by a factor of over 50 ($1024$ vs $20$) — in a real context like calculating compound growth, data storage (e.g. computer memory, where $2^{10}$ 'kilobytes' is a standard unit), or population growth, this confusion could lead to a drastically wrong final answer, potentially by orders of magnitude.

QUESTION 13 6 marks Criterion D

Hard

A hard drive's storage capacity doubles with each new model generation. The Generation 1 drive holds $2^8$ MB. Each new generation adds exactly 1 to the exponent.

a. Find the storage capacity, in MB, of the Generation 5 drive.

[2]

b. A consumer wants a drive with AT LEAST 1,000,000 MB (1 TB) of storage. Find the generation number required, and comment on whether 'doubling every generation forever' is a realistic assumption for real hard drive technology over many decades.

[4]

Show complete worked solution

(a)

Generation 5 exponent $=8+4=12$. Capacity $=2^{12}=4096$ MB.

(b)

Need $2^{8+n}\ge1000000$. Testing: $2^{20}=1048576\ge1000000$ (generation where exponent$=20$, i.e. Generation $20-8+1=13$). This 'doubling' pattern (similar to Moore's Law in computing) has historically held reasonably well over shorter timeframes, but is not realistic to extrapolate indefinitely — physical and economic limits eventually slow or halt this kind of exponential growth in real technology.

QUESTION 1 6 marks Criterion B

Medium

Consider the digits and operation symbols: $$6 \quad 2 \quad 4 \quad 1$$ used with $+$ and $\times$ in the expression $6 + 2 \times 4 - 1$, where brackets may be inserted anywhere.

a. Evaluate $6 + 2 \times 4 - 1$ without any brackets.

[1]

b. Insert brackets in at least 3 different positions and evaluate each resulting expression. Record your results in a table.

[3]

c. Describe any pattern you notice in how bracket position affects the result. Which position gives the largest value?

[2]

Show complete worked solution

(a)

$$6+2\times4-1 = 6+8-1 = 13$$

(b)

Example results: $(6+2)\times4-1 = 31$; $6+2\times(4-1)=12$; $(6+2)\times(4-1)=24$; $6+(2\times4-1)=13$. (Working shown for each.)

(c)

Bracketing the addition together with a following multiplication tends to increase the result (since addition happens before a larger multiplier is applied), while bracketing to delay multiplication tends to decrease it. Of the options tested, $(6+2)\times4-1=31$ gives the largest value.

QUESTION 2 5 marks Criterion C

Medium

Consider the expression $$15 - 3 \times 2^2 + 8 \div 4$$

a. Evaluate the expression, showing each step of your working clearly and stating which operation you perform first, second, etc.

[3]

b. Explain, in words, why the order of operations convention (BIDMAS/PEMDAS) is necessary — what would go wrong without an agreed order?

[2]

Show complete worked solution

(a)

$2^2=4$.

$3\times4=12$ and $8\div4=2$.

$15-12+2=\boxed{5}$.

(b)

Without an agreed order, the same expression could be evaluated in different sequences, giving different answers depending on who solves it. A universal convention ensures every mathematician gets the same, unambiguous result for the same expression.

QUESTION 3 5 marks Criterion D

Medium

A café bill is calculated as follows: 3 coffees at \$4.50 each, 2 pastries at \$3.20 each, then a 10\% service charge is added to the subtotal, and finally a \$2 discount voucher is subtracted.

a. Write a single expression (using correct order of operations) for the total bill, and evaluate it.

[3]

b. The café's till shows a total of \$21.89 (before the voucher was applied). Explain whether this matches your calculation, and state the correct final amount the customer should pay.

[2]

Show complete worked solution

(a)

$$\text{Total} = (3\times4.50 + 2\times3.20)\times1.10 - 2$$ Subtotal $= 13.50+6.40=19.90$. After service charge: $19.90\times1.10=21.89$. After discount: $21.89-2=19.89$.

(b)

Yes, \$21.89 matches the subtotal-plus-service-charge stage of the calculation, before the \$2 voucher is subtracted. The correct final amount the customer should pay is \$19.89.

QUESTION 4 7 marks Criterion A

Hard

A recipe scaling calculation requires evaluating: $$\frac{(24+36) \times 3 - 15}{5} + 2^3$$

a. Evaluate the expression fully, showing every step in the correct order.

[3]

b. A second version of the formula removes the outer brackets around $(24+36)$, becoming $24+36\times3-15 \div 5+2^3$. Evaluate THIS version, and explain precisely why removing the brackets changes the result so significantly.

[4]

Show complete worked solution

(a)

Brackets: $24+36=60$. Multiply: $60\times3=180$. Subtract: $180-15=165$. Divide: $165\div5=33$. Power: $2^3=8$. Add: $33+8=41$.

(b)

New order: powers first ($2^3=8$), then multiply/divide ($36\times3=108$, $15\div5=3$), then add/subtract left to right: $24+108-3+8=137$. This differs greatly from 41 because without brackets, $36$ is multiplied by 3 alone (not the combined $60$), and $15$ is divided by 5 alone (not subtracted from a running total first) — brackets fundamentally change WHICH numbers get combined by which operations.

QUESTION 5 6 marks Criterion B

Hard

Investigate how many DIFFERENT results are possible by inserting a single pair of brackets somewhere into the expression $2+3\times4-1$ (including the option of no brackets at all).

a. Evaluate the expression with NO brackets, then with brackets around $(2+3)$, then around $(4-1)$, then around $(3\times4-1)$.

[4]

b. How many GENUINELY DIFFERENT numerical results did you find from your 4 tests? Explain why one of your bracket placements didn't actually change the answer.

[2]

Show complete worked solution

(a)

No brackets: $2+12-1=13$. $(2+3)\times4-1=20-1=19$. $2+3\times(4-1)=2+9=11$. $2+3\times4-1$ with brackets around $(3\times4-1)$: this doesn't change anything since multiplication already happens first, giving $2+11=13$ (same as no brackets).

(b)

3 genuinely different results: 13, 19, 11. The brackets around $(3\times4-1)$ didn't change anything because multiplication ALREADY happens before subtraction by the standard order of operations rule — adding brackets around an operation that would happen first anyway has no effect.

QUESTION 6 6 marks Criterion C

Hard

A student evaluates $20-4\times3+2^2$ by working strictly left to right (ignoring order of operations entirely), getting a different answer than the correct method.

a. Calculate the student's (incorrect) left-to-right answer, and the mathematically CORRECT answer using proper order of operations.

[4]

b. Explain precisely WHERE the two methods first diverge, and why mathematicians universally agree to use the standard order of operations rather than simple left-to-right reading.

[2]

Show complete worked solution

(a)

Student's left-to-right: $20-4=16$, $16\times3=48$, $48+2=50$, $50^2=2500$. Correct method: $2^2=4$ first, then $4\times3=12$, then $20-12+4=12$.

(b)

The methods diverge immediately at the SECOND step — left-to-right treats $-4$ then $\times3$ sequentially without prioritizing multiplication, and (critically) applies squaring to the WRONG number entirely (50 instead of just 2). A universal, agreed-upon order of operations ensures every person (and every calculator) evaluates any given expression identically, avoiding ambiguity.

QUESTION 7 6 marks Criterion D

Hard

An electrician charges according to the formula: (call-out fee) + (hourly rate $\times$ hours) $-$ (loyalty discount), specifically $$45 + 38 \times 3.5 - 20$$ dollars for a particular job.

a. Calculate the total charge, following the correct order of operations.

[2]

b. The customer misreads the formula and calculates $(45+38)\times3.5-20$ instead (incorrectly bracketing), getting a different total. Find their (incorrect) total, the SIZE of the discrepancy in dollars, and explain the real-world consequence of this kind of order-of-operations error on an actual bill.

[4]

Show complete worked solution

(a)

$$45+38\times3.5-20 = 45+133-20=158$$

(b)

Customer's version: $(45+38)\times3.5-20 = 83\times3.5-20=290.5-20=270.5$. Discrepancy: $270.5-158=\$112.50$ (a significant overestimate). In a real billing context, this kind of misreading could cause a customer to be alarmed by an incorrectly high expected charge, or a business to genuinely OVERCHARGE a customer if the formula itself were programmed incorrectly — showing why precise, unambiguous mathematical formulas matter in real invoicing systems.

QUESTION 8 7 marks Criterion A

Hard

A recipe scaling calculation requires evaluating: $$\frac{(24+36) \times 3 - 15}{5} + 2^3$$

a. Evaluate the expression fully, showing every step in the correct order.

[3]

b. A second version of the formula removes the outer brackets around $(24+36)$, becoming $24+36\times3-15 \div 5+2^3$. Evaluate THIS version, and explain precisely why removing the brackets changes the result so significantly.

[4]

Show complete worked solution

(a)

Brackets: $24+36=60$. Multiply: $60\times3=180$. Subtract: $180-15=165$. Divide: $165\div5=33$. Power: $2^3=8$. Add: $33+8=41$.

(b)

New order: powers first ($2^3=8$), then multiply/divide ($36\times3=108$, $15\div5=3$), then add/subtract left to right: $24+108-3+8=137$. This differs greatly from 41 because without brackets, $36$ is multiplied by 3 alone (not the combined $60$), and $15$ is divided by 5 alone (not subtracted from a running total first) — brackets fundamentally change WHICH numbers get combined by which operations.

QUESTION 9 6 marks Criterion B

Hard

Investigate how many DIFFERENT results are possible by inserting a single pair of brackets somewhere into the expression $2+3\times4-1$ (including the option of no brackets at all).

a. Evaluate the expression with NO brackets, then with brackets around $(2+3)$, then around $(4-1)$, then around $(3\times4-1)$.

[4]

b. How many GENUINELY DIFFERENT numerical results did you find from your 4 tests? Explain why one of your bracket placements didn't actually change the answer.

[2]

Show complete worked solution

(a)

No brackets: $2+12-1=13$. $(2+3)\times4-1=20-1=19$. $2+3\times(4-1)=2+9=11$. $2+3\times4-1$ with brackets around $(3\times4-1)$: this doesn't change anything since multiplication already happens first, giving $2+11=13$ (same as no brackets).

(b)

3 genuinely different results: 13, 19, 11. The brackets around $(3\times4-1)$ didn't change anything because multiplication ALREADY happens before subtraction by the standard order of operations rule — adding brackets around an operation that would happen first anyway has no effect.

QUESTION 10 6 marks Criterion C

Hard

A student evaluates $20-4\times3+2^2$ by working strictly left to right (ignoring order of operations entirely), getting a different answer than the correct method.

a. Calculate the student's (incorrect) left-to-right answer, and the mathematically CORRECT answer using proper order of operations.

[4]

b. Explain precisely WHERE the two methods first diverge, and why mathematicians universally agree to use the standard order of operations rather than simple left-to-right reading.

[2]

Show complete worked solution

(a)

Student's left-to-right: $20-4=16$, $16\times3=48$, $48+2=50$, $50^2=2500$. Correct method: $2^2=4$ first, then $4\times3=12$, then $20-12+4=12$.

(b)

The methods diverge immediately at the SECOND step — left-to-right treats $-4$ then $\times3$ sequentially without prioritizing multiplication, and (critically) applies squaring to the WRONG number entirely (50 instead of just 2). A universal, agreed-upon order of operations ensures every person (and every calculator) evaluates any given expression identically, avoiding ambiguity.

QUESTION 11 6 marks Criterion D

Hard

An electrician charges according to the formula: (call-out fee) + (hourly rate $\times$ hours) $-$ (loyalty discount), specifically $$45 + 38 \times 3.5 - 20$$ dollars for a particular job.

a. Calculate the total charge, following the correct order of operations.

[2]

b. The customer misreads the formula and calculates $(45+38)\times3.5-20$ instead (incorrectly bracketing), getting a different total. Find their (incorrect) total, the SIZE of the discrepancy in dollars, and explain the real-world consequence of this kind of order-of-operations error on an actual bill.

[4]

Show complete worked solution

(a)

$$45+38\times3.5-20 = 45+133-20=158$$

(b)

Customer's version: $(45+38)\times3.5-20 = 83\times3.5-20=290.5-20=270.5$. Discrepancy: $270.5-158=\$112.50$ (a significant overestimate). In a real billing context, this kind of misreading could cause a customer to be alarmed by an incorrectly high expected charge, or a business to genuinely OVERCHARGE a customer if the formula itself were programmed incorrectly — showing why precise, unambiguous mathematical formulas matter in real invoicing systems.

QUESTION 1 3 marks Criterion A

Medium

Evaluate each expression, showing full working:

a. $|-15| + |8|$

[1]

b. $|-6| \times |-7| - |4|$

[1]

c. $|12 - 20| + |20 - 12|$

[1]

Show complete worked solution

(a)

$$15+8=23$$

(b)

$$6\times7-4 = 42-4=38$$

(c)

$$|-8|+|8| = 8+8=16$$

QUESTION 2 6 marks Criterion A

Medium

A factory's quality control accepts a component if its measured length is within $0.5$ mm of the target length of $120$ mm.

a. Write an absolute value inequality describing the acceptable range of measured lengths $x$.

[2]

b. State the minimum and maximum acceptable lengths.

[2]

c. A component measures $120.7$ mm. Is it accepted? Show your reasoning using the absolute value expression.

[2]

Show complete worked solution

(a)

$$|x - 120| \le 0.5$$

(b)

Minimum: $119.5$ mm. Maximum: $120.5$ mm.

(c)

$|120.7-120|=0.7$. Since $0.7 > 0.5$, the component is rejected.

QUESTION 3 6 marks Criterion B

Medium

Investigate whether $|a+b| = |a| + |b|$ is always true.

a. Test this with $a=5, b=3$ (both positive), then with $a=-5, b=-3$ (both negative).

[2]

b. Now test with $a=5, b=-3$ (different signs).

[2]

c. State when the equation $|a+b|=|a|+|b|$ holds true, and when it does not.

[2]

Show complete worked solution

(a)

$a=5,b=3$: $|5+3|=8$ and $|5|+|3|=8$ — equal. $a=-5,b=-3$: $|-8|=8$ and $|-5|+|-3|=8$ — equal.

(b)

$|5+(-3)|=|2|=2$. $|5|+|-3|=5+3=8$. These are NOT equal ($2 \ne 8$).

(c)

The equality holds when $a$ and $b$ have the same sign or when either number is zero. Equivalently, it holds when $ab\ge0$. If $a$ and $b$ are nonzero and have opposite signs, then $|a+b|<|a|+|b|$.

QUESTION 4 5 marks Criterion C

Medium

A classmate simplifies $|x - 7|$ as $x - 7$ for all values of $x$.

a. Test the classmate's simplification using $x=10$ and $x=3$. Is it correct in both cases?

[2]

b. Explain the error in the classmate's reasoning, and state the correct piecewise definition of $|x-7|$.

[3]

Show complete worked solution

(a)

$x=10$: actual $|10-7|=3$, classmate's version gives $10-7=3$ — matches. $x=3$: actual $|3-7|=|-4|=4$, classmate's version gives $3-7=-4$ — does NOT match.

(b)

The classmate forgot that absolute value always gives a non-negative result — when $x-7$ is negative (i.e. $x<7$), the absolute value flips its sign. Correct definition: $|x-7| = x-7$ when $x\ge7$, and $|x-7|=7-x$ when $x<7$.

QUESTION 5 6 marks Criterion D

Medium

A hiking trail follows a straight path. A checkpoint is located at position 0 km. A hiker's position (in km, where negative means behind the checkpoint) is recorded every hour: $-3, -1, 2, 4, 1$.

a. Find the hiker's distance from the checkpoint (using absolute value) at each recorded time.

[3]

b. Find the total distance the hiker travelled between consecutive recordings (i.e. the sum of $|$change in position$|$ each hour).

[3]

Show complete worked solution

(a)

$|-3|=3$, $|-1|=1$, $|2|=2$, $|4|=4$, $|1|=1$ km, respectively.

(b)

Changes: $-1-(-3)=2$, $2-(-1)=3$, $4-2=2$, $1-4=-3$. Total distance $=|2|+|3|+|2|+|-3|=2+3+2+3=10$ km.

QUESTION 6 6 marks Criterion A

Hard

An engineering tolerance specification requires a machined part's diameter to satisfy $|d - 50| \le 0.3$ mm, where $d$ is the measured diameter and 50mm is the target.

a. Find the minimum and maximum acceptable diameters.

[2]

b. A batch of 6 parts is measured: $49.8, 50.4, 49.6, 50.1, 50.35, 49.9$ mm. Determine which parts are accepted and which are rejected, showing the absolute-value check for the two BORDERLINE cases (closest to the tolerance limits).

[4]

Show complete worked solution

(a)

$$49.7 \text{ mm} \le d \le 50.3 \text{ mm}$$

(b)

Check each value using $|x-50|\le0.3$. The accepted values are $49.8$, $50.1$ and $49.9$, with deviations $0.2$, $0.1$ and $0.1$. The values $50.4$, $49.6$ and $50.35$ are rejected because their deviations $0.4$, $0.4$ and $0.35$ exceed $0.3$.

QUESTION 7 8 marks Criterion B

Hard

Investigate the equation $|x-3|+|x+2|=9$, which involves TWO absolute value expressions.

a. Test $x=6$ in the equation. Does it satisfy it?

[2]

b. Test $x=5$ and $x=-4$. Do either of these satisfy the equation?

[3]

c. Based on your findings, and thinking about $|x-3|$ as 'distance from 3' and $|x+2|$ as 'distance from $-2$', explain geometrically (in terms of distances on a number line) why BOTH $x=5$ and $x=-4$ work, and conjecture whether there might be a whole RANGE of solutions, not just these two isolated values.

[3]

Show complete worked solution

(a)

$|6-3|+|6+2|=3+8=11\ne9$. Does not satisfy.

(b)

$x=5$: $|5-3|+|5+2|=2+7=9$. $x=-4$: $|-4-3|+|-4+2|=7+2=9$. Both satisfy!

(c)

The points $-2$ and 3 are 5 units apart. For $-2\le x\le3$, the distance sum is 5, so no point in that interval works. Solving outside the interval gives $x=-4$ on the left and $x=5$ on the right. Therefore $$\boxed{x=-4\text{ or }x=5},$$ two isolated values rather than continuous ranges.

QUESTION 8 5 marks Criterion C

Hard

A student solving $|x|=-5$ writes '$x=5$ or $x=-5$' (treating it the same as $|x|=5$).

a. Explain, using the DEFINITION of absolute value (that it represents a distance, which cannot be negative), why $|x|=-5$ actually has NO solutions at all.

[3]

b. Contrast this with $|x|=5$, which DOES have solutions. State them, and explain the key difference between the two equations that makes one solvable and the other not.

[2]

Show complete worked solution

(a)

By definition, $|x|$ represents the distance of $x$ from zero on a number line — and a distance can NEVER be negative (the smallest possible value of $|x|$ is 0, when $x=0$ itself). Since $-5$ is negative, there is NO value of $x$ for which $|x|$ could ever equal $-5$; the equation has no solution.

(b)

$|x|=5$ has solutions $x=5$ and $x=-5$ (both are exactly 5 units from zero). The key difference: $5$ is a valid (non-negative) value for a distance, while $-5$ is not — this is precisely why $|x|=5$ is solvable but $|x|=-5$ is not.

QUESTION 9 5 marks Criterion D

Hard

A drone's horizontal position relative to a base station is tracked in metres (negative = west, positive = east). Over a flight, its recorded positions at 5 checkpoints were: $-40, 25, -15, 60, -10$.

a. Find the drone's DISTANCE from the base station (using absolute value) at each checkpoint.

[2]

b. Safety regulations require the drone to stay within 50m of the base station at ALL times during flight. Based on your distances, identify any checkpoint(s) where the regulation was violated, and calculate by how much.

[3]

Show complete worked solution

(a)

$40, 25, 15, 60, 10$ metres respectively.

(b)

Checkpoint 4 (position $60$) violates the regulation, since its distance (60m) exceeds the 50m limit, by $60-50=10$ metres. All other checkpoints (40, 25, 15, 10m) are within the safe range.

QUESTION 10 6 marks Criterion A

Hard

An engineering tolerance specification requires a machined part's diameter to satisfy $|d - 50| \le 0.3$ mm, where $d$ is the measured diameter and 50mm is the target.

a. Find the minimum and maximum acceptable diameters.

[2]

b. A batch of 6 parts is measured: $49.8, 50.4, 49.6, 50.1, 50.35, 49.9$ mm. Determine which parts are accepted and which are rejected, showing the absolute-value check for the two BORDERLINE cases (closest to the tolerance limits).

[4]

Show complete worked solution

(a)

$$49.7 \text{ mm} \le d \le 50.3 \text{ mm}$$

(b)

Check each value using $|x-50|\le0.3$. The accepted values are $49.8$, $50.1$ and $49.9$, with deviations $0.2$, $0.1$ and $0.1$. The values $50.4$, $49.6$ and $50.35$ are rejected because their deviations $0.4$, $0.4$ and $0.35$ exceed $0.3$.

QUESTION 11 8 marks Criterion B

Hard

Investigate the equation $|x-3|+|x+2|=9$, which involves TWO absolute value expressions.

a. Test $x=6$ in the equation. Does it satisfy it?

[2]

b. Test $x=5$ and $x=-4$. Do either of these satisfy the equation?

[3]

c. Based on your findings, and thinking about $|x-3|$ as 'distance from 3' and $|x+2|$ as 'distance from $-2$', explain geometrically (in terms of distances on a number line) why BOTH $x=5$ and $x=-4$ work, and conjecture whether there might be a whole RANGE of solutions, not just these two isolated values.

[3]

Show complete worked solution

(a)

$|6-3|+|6+2|=3+8=11\ne9$. Does not satisfy.

(b)

$x=5$: $|5-3|+|5+2|=2+7=9$. $x=-4$: $|-4-3|+|-4+2|=7+2=9$. Both satisfy!

(c)

The points $-2$ and 3 are 5 units apart. For $-2\le x\le3$, the distance sum is 5, so no point in that interval works. Solving outside the interval gives $x=-4$ on the left and $x=5$ on the right. Therefore $$\boxed{x=-4\text{ or }x=5},$$ two isolated values rather than continuous ranges.

QUESTION 12 5 marks Criterion C

Hard

A student solving $|x|=-5$ writes '$x=5$ or $x=-5$' (treating it the same as $|x|=5$).

a. Explain, using the DEFINITION of absolute value (that it represents a distance, which cannot be negative), why $|x|=-5$ actually has NO solutions at all.

[3]

b. Contrast this with $|x|=5$, which DOES have solutions. State them, and explain the key difference between the two equations that makes one solvable and the other not.

[2]

Show complete worked solution

(a)

By definition, $|x|$ represents the distance of $x$ from zero on a number line — and a distance can NEVER be negative (the smallest possible value of $|x|$ is 0, when $x=0$ itself). Since $-5$ is negative, there is NO value of $x$ for which $|x|$ could ever equal $-5$; the equation has no solution.

(b)

$|x|=5$ has solutions $x=5$ and $x=-5$ (both are exactly 5 units from zero). The key difference: $5$ is a valid (non-negative) value for a distance, while $-5$ is not — this is precisely why $|x|=5$ is solvable but $|x|=-5$ is not.

QUESTION 13 5 marks Criterion D

Hard

A drone's horizontal position relative to a base station is tracked in metres (negative = west, positive = east). Over a flight, its recorded positions at 5 checkpoints were: $-40, 25, -15, 60, -10$.

a. Find the drone's DISTANCE from the base station (using absolute value) at each checkpoint.

[2]

b. Safety regulations require the drone to stay within 50m of the base station at ALL times during flight. Based on your distances, identify any checkpoint(s) where the regulation was violated, and calculate by how much.

[3]

Show complete worked solution

(a)

$40, 25, 15, 60, 10$ metres respectively.

(b)

Checkpoint 4 (position $60$) violates the regulation, since its distance (60m) exceeds the 50m limit, by $60-50=10$ metres. All other checkpoints (40, 25, 15, 10m) are within the safe range.

QUESTION 1 3 marks Criterion A

Medium

Evaluate each expression, showing full working:

a. $\sqrt{144} + \sqrt{81}$

[1]

b. $\sqrt{225} - \sqrt{49}$

[1]

c. $(\sqrt{36})^2 + 5^2$

[1]

Show complete worked solution

(a)

$$12+9=21$$

(b)

$$15-7=8$$

(c)

$$36+25=61$$

QUESTION 2 5 marks Criterion A

Medium

A square garden plot has an area of $169$ m$^2$.

a. Find the side length of the plot.

[1]

b. Find the perimeter of the plot, and the cost of fencing it at \$18.50 per metre.

[2]

c. The gardener wants to double the AREA of the plot (keeping it square). Find the new side length, correct to 2 decimal places.

[2]

Show complete worked solution

(a)

$$\sqrt{169}=13\text{ m}$$

(b)

Perimeter $=4\times13=52$ m. Cost $=52\times18.50=\$962$.

(c)

New area $=169\times2=338$ m$^2$. New side $=\sqrt{338}\approx18.38$ m.

QUESTION 3 5 marks Criterion B

Medium

Investigate what happens to the difference between consecutive square numbers as the numbers get larger.

a. Calculate $2^2-1^2$, $3^2-2^2$, $4^2-3^2$, and $5^2-4^2$.

[2]

b. Describe the pattern in these differences (3, 5, 7, 9, ...).

[1]

c. Using the pattern, predict the difference $11^2-10^2$ without calculating both squares, then verify your prediction.

[2]

Show complete worked solution

(a)

$4-1=3$. $9-4=5$. $16-9=7$. $25-16=9$.

(b)

The differences form the sequence of consecutive odd numbers, increasing by 2 each time.

(c)

The pattern of differences (3,5,7,9,...) for $n^2-(n-1)^2$ equals $2n-1$. For $n=11$: $2(11)-1=21$. Verifying: $121-100=21$ — matches.

QUESTION 4 5 marks Criterion C

Medium

A student claims that $\sqrt{a^2+b^2}$ is the same as $a+b$.

a. Test the claim using $a=3, b=4$, showing full working for both sides.

[2]

b. Explain why $\sqrt{a^2+b^2} \ne a+b$ in general, referring to what happens when you square both expressions.

[3]

Show complete worked solution

(a)

$\sqrt{3^2+4^2}=\sqrt{9+16}=\sqrt{25}=5$. $a+b=3+4=7$. Since $5\ne7$, the claim is false.

(b)

If $\sqrt{a^2+b^2}=a+b$, then squaring gives $a^2+b^2=a^2+2ab+b^2$, so $ab=0$. We must also have $a+b\ge0$ because a square root is nonnegative. Therefore equality occurs exactly when one number is zero and the other is nonnegative. In all other real cases the two expressions are different.

QUESTION 5 4 marks Criterion D

Medium

A television screen is advertised by its diagonal length. A screen has width $80$ cm and height $45$ cm.

a. Using $\text{diagonal}^2 = \text{width}^2 + \text{height}^2$, find the diagonal length of the screen, correct to 1 decimal place.

[2]

b. TV sizes are usually rounded to the nearest inch (1 inch $\approx$ 2.54 cm). What size (in inches) would this TV be advertised as?

[2]

Show complete worked solution

(a)

$$\text{diagonal} = \sqrt{80^2+45^2} = \sqrt{6400+2025} = \sqrt{8425} \approx 91.8 \text{ cm}$$

(b)

$91.8 \div 2.54 \approx 36.1$ inches, so it would be advertised as a 36-inch TV.

QUESTION 6 7 marks Criterion A

Hard

A square-shaped park has an area of 5,625 m$^2$. A square garden bed inside it has area exactly $\frac{1}{25}$ of the park's area.

a. Find the side length of the park.

[2]

b. Find the area, then the side length, of the garden bed.

[3]

c. Verify your garden side length is consistent with the park's side length, using the fact that if garden AREA is $\frac{1}{25}$ of park area, garden SIDE should be $\frac{1}{5}$ of park side (since area scales with the SQUARE of side length).

[2]

Show complete worked solution

(a)

$$\sqrt{5625}=75 \text{ m}$$

(b)

Garden area $=5625\div25=225$ m$^2$. Garden side length $=\sqrt{225}=15$ m.

(c)

Park side $\div5 = 75\div5=15$ m, matching the garden side found in part (b) exactly — confirming the relationship (since $(\frac{1}{5})^2=\frac{1}{25}$, consistent with the area ratio given).

QUESTION 7 7 marks Criterion B

Hard

Investigate the claim that the difference between consecutive perfect squares equals the corresponding sequence of odd numbers, and use this to find a large square WITHOUT a calculator.

a. Verify that $n^2-(n-1)^2=2n-1$ algebraically.

[2]

b. Given $30^2=900$, use the identity from part (a) to find $31^2$ WITHOUT directly multiplying $31\times31$.

[2]

c. Continue this technique to find $32^2$ and $33^2$, building each from the previous answer.

[3]

Show complete worked solution

(a)

$$n^2-(n-1)^2 = n^2-(n^2-2n+1) = 2n-1$$

(b)

$31^2 = 30^2+(2\times31-1) = 900+61=961$.

(c)

$32^2=31^2+(2\times32-1)=961+63=1024$. $33^2=32^2+(2\times33-1)=1024+65=1089$.

QUESTION 8 5 marks Criterion C

Hard

A student claims that $\sqrt{a^2}=a$ is always true for any number $a$.

a. Test the claim with $a=6$ (positive) and $a=-6$ (negative), calculating $\sqrt{a^2}$ in each case.

[3]

b. State the CORRECT general rule connecting $\sqrt{a^2}$ to $a$, valid for both positive and negative $a$, using absolute value notation.

[2]

Show complete worked solution

(a)

$a=6$: $\sqrt{6^2}=\sqrt{36}=6=a$ (works). $a=-6$: $\sqrt{(-6)^2}=\sqrt{36}=6$, but $a=-6$, so $\sqrt{a^2}\ne a$ here (fails).

(b)

$$\sqrt{a^2} = |a|$$ (the square root of a squared number always gives the ABSOLUTE VALUE of the original number, since a square root itself is always defined to be non-negative).

QUESTION 9 6 marks Criterion D

Hard

A construction crew needs to brace a rectangular wall frame diagonally to prevent it from leaning. The wall frame is 4.8 m wide and 3.6 m tall.

a. Using the Pythagorean relationship (diagonal)$^2$=(width)$^2$+(height)$^2$, find the exact length of diagonal bracing needed.

[3]

b. The crew has bracing material sold only in whole-metre lengths, and needs 10cm extra at each end for secure fastening. Determine the minimum length of bracing material (in whole metres) they must purchase, and explain your rounding decision.

[3]

Show complete worked solution

(a)

$$\text{diagonal}=\sqrt{4.8^2+3.6^2}=\sqrt{23.04+12.96}=\sqrt{36}=6 \text{ m}$$

(b)

Required length including fastening allowance: $6+0.1+0.1=6.2$ m. Since material is sold in whole metres, they must round UP (not to the nearest whole metre) to ensure enough material — purchasing 7 m, since 6 m would be insufficient for the 6.2 m actually needed.

QUESTION 10 7 marks Criterion A

Hard

A square-shaped park has an area of 5,625 m$^2$. A square garden bed inside it has area exactly $\frac{1}{25}$ of the park's area.

a. Find the side length of the park.

[2]

b. Find the area, then the side length, of the garden bed.

[3]

c. Verify your garden side length is consistent with the park's side length, using the fact that if garden AREA is $\frac{1}{25}$ of park area, garden SIDE should be $\frac{1}{5}$ of park side (since area scales with the SQUARE of side length).

[2]

Show complete worked solution

(a)

$$\sqrt{5625}=75 \text{ m}$$

(b)

Garden area $=5625\div25=225$ m$^2$. Garden side length $=\sqrt{225}=15$ m.

(c)

Park side $\div5 = 75\div5=15$ m, matching the garden side found in part (b) exactly — confirming the relationship (since $(\frac{1}{5})^2=\frac{1}{25}$, consistent with the area ratio given).

QUESTION 11 7 marks Criterion B

Hard

Investigate the claim that the difference between consecutive perfect squares equals the corresponding sequence of odd numbers, and use this to find a large square WITHOUT a calculator.

a. Verify that $n^2-(n-1)^2=2n-1$ algebraically.

[2]

b. Given $30^2=900$, use the identity from part (a) to find $31^2$ WITHOUT directly multiplying $31\times31$.

[2]

c. Continue this technique to find $32^2$ and $33^2$, building each from the previous answer.

[3]

Show complete worked solution

(a)

$$n^2-(n-1)^2 = n^2-(n^2-2n+1) = 2n-1$$

(b)

$31^2 = 30^2+(2\times31-1) = 900+61=961$.

(c)

$32^2=31^2+(2\times32-1)=961+63=1024$. $33^2=32^2+(2\times33-1)=1024+65=1089$.

QUESTION 12 5 marks Criterion C

Hard

A student claims that $\sqrt{a^2}=a$ is always true for any number $a$.

a. Test the claim with $a=6$ (positive) and $a=-6$ (negative), calculating $\sqrt{a^2}$ in each case.

[3]

b. State the CORRECT general rule connecting $\sqrt{a^2}$ to $a$, valid for both positive and negative $a$, using absolute value notation.

[2]

Show complete worked solution

(a)

$a=6$: $\sqrt{6^2}=\sqrt{36}=6=a$ (works). $a=-6$: $\sqrt{(-6)^2}=\sqrt{36}=6$, but $a=-6$, so $\sqrt{a^2}\ne a$ here (fails).

(b)

$$\sqrt{a^2} = |a|$$ (the square root of a squared number always gives the ABSOLUTE VALUE of the original number, since a square root itself is always defined to be non-negative).

QUESTION 13 6 marks Criterion D

Hard

A construction crew needs to brace a rectangular wall frame diagonally to prevent it from leaning. The wall frame is 4.8 m wide and 3.6 m tall.

a. Using the Pythagorean relationship (diagonal)$^2$=(width)$^2$+(height)$^2$, find the exact length of diagonal bracing needed.

[3]

b. The crew has bracing material sold only in whole-metre lengths, and needs 10cm extra at each end for secure fastening. Determine the minimum length of bracing material (in whole metres) they must purchase, and explain your rounding decision.

[3]

Show complete worked solution

(a)

$$\text{diagonal}=\sqrt{4.8^2+3.6^2}=\sqrt{23.04+12.96}=\sqrt{36}=6 \text{ m}$$

(b)

Required length including fastening allowance: $6+0.1+0.1=6.2$ m. Since material is sold in whole metres, they must round UP (not to the nearest whole metre) to ensure enough material — purchasing 7 m, since 6 m would be insufficient for the 6.2 m actually needed.

QUESTION 1 7 marks Criterion B

Medium

Goldbach's conjecture (still unproven in general) suggests that every even number greater than 2 can be written as the sum of two primes.

a. Verify Goldbach's conjecture for the even numbers 10, 16, and 24, by writing each as a sum of two primes. (There may be more than one way — find at least one for each.)

[3]

b. Write 9 as a sum of two primes. Explain why this example works although Goldbach's conjecture concerns even numbers, and why it does not establish a rule for every odd number.

[2]

c. Based on your findings, do you think Goldbach's conjecture is likely true for all even numbers greater than 2? Justify your answer.

[2]

Show complete worked solution

(a)

$10 = 3+7 = 5+5$. $16 = 3+13 = 5+11$. $24 = 5+19 = 11+13$. (Any one valid pair per number is sufficient.)

(b)

$9=2+7$, so 9 can be written as a sum of two primes. This works because 2 is the only even prime. The sum of two odd primes is even, so an odd number can be formed from two primes only when one addend is 2 and the other, $n-2$, is prime. For example, 11 cannot be written as a sum of two primes because $11-2=9$ is not prime. Goldbach's conjecture therefore makes no claim about odd numbers.

(c)

Based on the pattern holding for every even number tested (and it has been verified by computer for extremely large numbers), the conjecture appears very likely to be true, even though it has never been formally proven for all even numbers.

QUESTION 2 5 marks Criterion C

Medium

A student claims that 91 is a prime number because it is not divisible by 2, 3, or 5.

a. Explain, using the definition of a prime number, why checking divisibility by only 2, 3, and 5 is not sufficient to prove 91 is prime.

[2]

b. Using correct mathematical working, determine whether 91 is actually prime or composite.

[3]

Show complete worked solution

(a)

A prime number has no factors other than 1 and itself. To prove a number is prime, every possible factor up to its square root must be checked — not just the first few primes. Since $\sqrt{91}\approx9.54$, factors up to 9 (i.e. 2, 3, 5, 7) all need checking.

(b)

Testing 7: $91 \div 7 = 13$ exactly. So $91 = 7 \times 13$, meaning 91 is composite, not prime — the student's claim is incorrect.

QUESTION 3 7 marks Criterion D

Medium

A warehouse packs identical boxes of stock into shipping crates. There are 273 boxes to pack.

a. Express 273 as a product of its prime factors.

[2]

b. The warehouse wants to pack the boxes into crates so that each crate holds the same number of boxes, with more than 1 box per crate, and more than 1 crate used. Using your answer to (a), list all the possible numbers of boxes per crate.

[3]

c. The crates being used can hold a maximum of 25 boxes each. Which packing option(s) from part (b) are actually usable, and how many crates would be needed?

[2]

Show complete worked solution

(a)

$$273 = 3 \times 7 \times 13$$

(b)

From the prime factors $3, 7, 13$, the possible factors of 273 (other than 1 and 273) are: $3, 7, 13, 21 (3\times7), 39 (3\times13), 91 (7\times13)$. Any of these could be the number of boxes per crate.

(c)

The usable numbers of boxes per crate are $3$, $7$, $13$ and $21$. The corresponding numbers of crates are $$273\div3=91,\qquad273\div7=39,\qquad273\div13=21,\qquad273\div21=13.$$ Therefore the possible arrangements are $\boxed{3\text{ boxes in }91\text{ crates}}$, $\boxed{7\text{ boxes in }39\text{ crates}}$, $\boxed{13\text{ boxes in }21\text{ crates}}$ or $\boxed{21\text{ boxes in }13\text{ crates}}$.

QUESTION 4 6 marks Criterion A

Hard

Twin primes are pairs of primes that differ by exactly 2 (e.g. 11 and 13).

a. Find all twin prime pairs where both primes are less than 30.

[3]

b. For each twin prime pair you found (both under 30), find the number exactly BETWEEN the two primes (their average), and state one property this 'middle number' always seems to share.

[3]

Show complete worked solution

(a)

$(3,5), (5,7), (11,13), (17,19), (29,31)$ — noting 31 is just over 30, so if strictly BOTH must be under 30: $(3,5),(5,7),(11,13),(17,19)$.

(b)

$(3,5)\to4$. $(5,7)\to6$. $(11,13)\to12$. $(17,19)\to18$. Every middle number (4,6,12,18) is EVEN.

QUESTION 5 6 marks Criterion B

Hard

Investigate why (except for the pair 2 and 3) every twin prime pair must have an EVEN number exactly between them.

a. Let the twin prime pair be $p$ and $p+2$. Write an expression for the number exactly between them.

[1]

b. For a twin prime pair where BOTH $p$ and $p+2$ are odd (true for all twin primes except the pair 2,3), explain why $p+1$ must always be even.

[3]

c. Explain why this middle number ($p+1$) can NEVER itself be prime (except in the trivial case where it equals 2), connecting to your finding.

[2]

Show complete worked solution

(a)

$$\frac{p+(p+2)}{2}=p+1$$

(b)

If $p$ is odd, then $p+1$ is the number immediately after an odd number, which is always EVEN (odd numbers and even numbers strictly alternate on the number line) — so $p+1$ is guaranteed to be even whenever $p$ itself is odd.

(c)

Since $p+1$ is always even (for $p>2$), and the only even prime number is 2 itself, any even number greater than 2 must be composite (divisible by 2 and something else). So the middle number of a twin prime pair (other than the 2,3 pair) can never be prime.

QUESTION 6 5 marks Criterion C

Hard

A student claims: 'I checked and 1 is not divisible by anything except itself, just like a prime — so 1 should count as a prime number.'

a. State the precise mathematical definition of a prime number (in terms of exactly how many factors it has), and explain specifically why 1 does not meet this definition.

[3]

b. Explain one important mathematical reason why excluding 1 from being 'prime' matters — specifically, how it affects the uniqueness of prime factorization (e.g. would $12=2^2\times3$ still be a UNIQUE way to write 12 as a product of primes, if 1 were allowed to be inserted anywhere)?

[2]

Show complete worked solution

(a)

A prime number is defined as a number with EXACTLY two distinct factors: 1 and itself. The number 1 has only ONE factor (itself, since $1\div1=1$ and there's no separate '1 and itself' distinction) — it fails the 'exactly two factors' requirement, so by definition 1 is neither prime nor composite; it's a special case of its own.

(b)

If 1 were considered prime, prime factorizations would no longer be unique — e.g. $12=2^2\times3$ could also be written as $12=1\times2^2\times3$, or $1\times1\times2^2\times3$, and so on infinitely, since multiplying by 1 never changes the value. Excluding 1 preserves the important property that every whole number greater than 1 has exactly ONE prime factorization.

QUESTION 7 6 marks Criterion D

Hard

A cybersecurity system generates encryption keys using the product of two large prime numbers, since factoring the product back into its two original primes is extremely difficult for large numbers.

a. A simplified (much smaller, for practice) version of this system uses the product $91$. Find its two prime factors.

[2]

b. A slightly larger product, $221$, is given. Find its two prime factors, and briefly explain (in general terms) why factoring becomes dramatically HARDER as the numbers get larger, even though the basic method (testing primes) stays conceptually the same.

[4]

Show complete worked solution

(a)

$$91 = 7 \times 13$$

(b)

$221 = 13\times17$. As numbers get larger, the NUMBER of possible prime factors to test grows enormously (up to $\sqrt{n}$ for a number $n$), and there's no known fast shortcut — for genuinely huge numbers (hundreds of digits, as used in real encryption), even powerful computers would take an impractically long time to find the factors by testing, which is exactly the security principle real encryption systems rely on.

QUESTION 8 6 marks Criterion A

Hard

Twin primes are pairs of primes that differ by exactly 2 (e.g. 11 and 13).

a. Find all twin prime pairs where both primes are less than 30.

[3]

b. For each twin prime pair you found (both under 30), find the number exactly BETWEEN the two primes (their average), and state one property this 'middle number' always seems to share.

[3]

Show complete worked solution

(a)

$(3,5), (5,7), (11,13), (17,19), (29,31)$ — noting 31 is just over 30, so if strictly BOTH must be under 30: $(3,5),(5,7),(11,13),(17,19)$.

(b)

$(3,5)\to4$. $(5,7)\to6$. $(11,13)\to12$. $(17,19)\to18$. Every middle number (4,6,12,18) is EVEN.

QUESTION 9 6 marks Criterion B

Hard

Investigate why (except for the pair 2 and 3) every twin prime pair must have an EVEN number exactly between them.

a. Let the twin prime pair be $p$ and $p+2$. Write an expression for the number exactly between them.

[1]

b. For a twin prime pair where BOTH $p$ and $p+2$ are odd (true for all twin primes except the pair 2,3), explain why $p+1$ must always be even.

[3]

c. Explain why this middle number ($p+1$) can NEVER itself be prime (except in the trivial case where it equals 2), connecting to your finding.

[2]

Show complete worked solution

(a)

$$\frac{p+(p+2)}{2}=p+1$$

(b)

If $p$ is odd, then $p+1$ is the number immediately after an odd number, which is always EVEN (odd numbers and even numbers strictly alternate on the number line) — so $p+1$ is guaranteed to be even whenever $p$ itself is odd.

(c)

Since $p+1$ is always even (for $p>2$), and the only even prime number is 2 itself, any even number greater than 2 must be composite (divisible by 2 and something else). So the middle number of a twin prime pair (other than the 2,3 pair) can never be prime.

QUESTION 10 5 marks Criterion C

Hard

A student claims: 'I checked and 1 is not divisible by anything except itself, just like a prime — so 1 should count as a prime number.'

a. State the precise mathematical definition of a prime number (in terms of exactly how many factors it has), and explain specifically why 1 does not meet this definition.

[3]

b. Explain one important mathematical reason why excluding 1 from being 'prime' matters — specifically, how it affects the uniqueness of prime factorization (e.g. would $12=2^2\times3$ still be a UNIQUE way to write 12 as a product of primes, if 1 were allowed to be inserted anywhere)?

[2]

Show complete worked solution

(a)

A prime number is defined as a number with EXACTLY two distinct factors: 1 and itself. The number 1 has only ONE factor (itself, since $1\div1=1$ and there's no separate '1 and itself' distinction) — it fails the 'exactly two factors' requirement, so by definition 1 is neither prime nor composite; it's a special case of its own.

(b)

If 1 were considered prime, prime factorizations would no longer be unique — e.g. $12=2^2\times3$ could also be written as $12=1\times2^2\times3$, or $1\times1\times2^2\times3$, and so on infinitely, since multiplying by 1 never changes the value. Excluding 1 preserves the important property that every whole number greater than 1 has exactly ONE prime factorization.

QUESTION 11 6 marks Criterion D

Hard

A cybersecurity system generates encryption keys using the product of two large prime numbers, since factoring the product back into its two original primes is extremely difficult for large numbers.

a. A simplified (much smaller, for practice) version of this system uses the product $91$. Find its two prime factors.

[2]

b. A slightly larger product, $221$, is given. Find its two prime factors, and briefly explain (in general terms) why factoring becomes dramatically HARDER as the numbers get larger, even though the basic method (testing primes) stays conceptually the same.

[4]

Show complete worked solution

(a)

$$91 = 7 \times 13$$

(b)

$221 = 13\times17$. As numbers get larger, the NUMBER of possible prime factors to test grows enormously (up to $\sqrt{n}$ for a number $n$), and there's no known fast shortcut — for genuinely huge numbers (hundreds of digits, as used in real encryption), even powerful computers would take an impractically long time to find the factors by testing, which is exactly the security principle real encryption systems rely on.

QUESTION 1 6 marks Criterion B

Medium

Investigate the relationship between the HCF of two numbers and the HCF of their sum with one of the numbers.

a. Find HCF$(12, 18)$, then find HCF$(12, 12+18)$ = HCF$(12, 30)$. What do you notice?

[2]

b. Test this again with the pair 15 and 25: find HCF$(15,25)$ and HCF$(15, 15+25)$.

[2]

c. Write a general statement (conjecture) describing this pattern, using $a$ and $b$ to represent the two numbers.

[2]

Show complete worked solution

(a)

HCF$(12,18)=6$. HCF$(12,30)=6$. Both give the same HCF.

(b)

HCF$(15,25)=5$. HCF$(15,40)=5$. Again the same HCF.

(c)

Conjecture: HCF$(a,b)$ = HCF$(a, a+b)$ for any two positive integers $a$ and $b$. This works because any common factor of $a$ and $b$ must also divide $a+b$, and any common factor of $a$ and $a+b$ must also divide $(a+b)-a=b$.

QUESTION 2 5 marks Criterion C

Medium

Two methods can be used to find the HCF of 84 and 126: listing factors, or prime factorization.

a. Using prime factorization, find HCF$(84, 126)$, showing full working.

[3]

b. Explain, in your own words, why the prime factorization method works — what does taking the 'lowest power of each common prime' actually represent?

[2]

Show complete worked solution

(a)

$84 = 2^2 \times 3 \times 7$. $126 = 2 \times 3^2 \times 7$. Common factors (lowest power of each shared prime): $2^1 \times 3^1 \times 7^1 = 42$. So HCF $= 42$.

(b)

Each prime factor with its lowest shared power represents the largest amount of that prime that BOTH numbers can be divided by. Multiplying these together gives the largest number that divides both — by definition, the HCF.

QUESTION 3 5 marks Criterion D

Medium

A school is arranging chairs for an assembly. There are 96 boys and 72 girls, and the principal wants to arrange them into identical rows, with boys and girls not mixed within a row, using the maximum possible number of chairs per row.

a. Find the maximum number of chairs per row that allows both 96 boys and 72 girls to be arranged in complete rows of that size.

[3]

b. How many rows of boys and how many rows of girls will there be in total?

[2]

Show complete worked solution

(a)

This requires HCF$(96,72)$. $96=2^5\times3$, $72=2^3\times3^2$. HCF $=2^3\times3=24$. Maximum chairs per row $=24$.

(b)

Boy rows $=96\div24=4$. Girl rows $=72\div24=3$. Total rows $=7$.

QUESTION 4 6 marks Criterion A

Hard

A school is preparing identical care packages for a charity drive, using 168 notebooks and 252 pencils, with NOTHING left over.

a. Find the maximum number of identical packages that can be made, using prime factorization.

[3]

b. Find the number of notebooks and pencils in EACH package, and verify your answer by checking $84\times(\text{notebooks per package})=168$.

[3]

Show complete worked solution

(a)

$168=2^3\times3\times7$. $252=2^2\times3^2\times7$. HCF (lowest powers of shared primes): $2^2\times3\times7=84$. Maximum packages $=84$.

(b)

Notebooks per package $=168\div84=2$. Pencils per package $=252\div84=3$. Check: $84\times2=168$.

QUESTION 5 7 marks Criterion B

Hard

Investigate the relationship between HCF$(a,b)$ and HCF$(a,b,c)$ when a third number $c$ is introduced, using $a=24, b=36, c=60$.

a. Find HCF$(24,36)$ first (ignoring $c$ for now).

[2]

b. Now find HCF$(12, 60)$ — i.e. the HCF of your previous answer and the third number $c=60$. Compare this to what you'd get by finding HCF$(24,36,60)$ directly (all three prime-factorized together).

[3]

c. Based on this, state a general rule for finding the HCF of THREE numbers using the HCF of only two numbers at a time.

[2]

Show complete worked solution

(a)

$24=2^3\times3$, $36=2^2\times3^2$. HCF$=2^2\times3=12$.

(b)

$60=2^2\times3\times5$. HCF$(12,60)=12$ (since $12=2^2\times3$ divides evenly into $60=2^2\times3\times5$). Direct 3-way HCF: comparing $2^3\times3$, $2^2\times3^2$, $2^2\times3\times5$ — lowest shared powers give $2^2\times3=12$. Both methods agree: 12.

(c)

HCF$(a,b,c) = $ HCF$($HCF$(a,b), c)$ — you can find the HCF of any two of the numbers first, then find the HCF of that result with the third number, and you'll always get the same answer as computing all three together directly.

QUESTION 6 5 marks Criterion C

Hard

A student says: 'The HCF of two numbers can never be bigger than the SMALLER of the two numbers, but it also can't be bigger than half of either number, since factors are always small.'

a. Test the first part of the claim (HCF can't exceed the smaller number) using HCF$(15,15)$ — two IDENTICAL numbers.

[2]

b. Test the second part of the claim ('can't be bigger than half of either number') using HCF$(20,20)$. Does this disprove the second part of the claim?

[3]

Show complete worked solution

(a)

HCF$(15,15)=15$ — this equals (not exceeds) the smaller number, confirming the first part isn't violated, though it shows the HCF CAN equal the smaller number exactly.

(b)

HCF$(20,20)=20$, which is NOT less than or equal to half of 20 (which is 10) — in fact it equals the FULL number, not half. This disproves the second part of the student's claim: the HCF of two equal numbers is the number itself, which is clearly larger than half of it (for any number greater than 0).

QUESTION 7 6 marks Criterion D

Hard

A tile installer has two types of square tiles to cover sections of a floor with NO cutting: one section requires tiles that fit evenly into a $180$cm length, and another section requires tiles fitting evenly into a $252$cm length. The installer wants to use the SAME tile size for both sections, as large as possible.

a. Find the largest tile side length (in cm) that fits evenly into both 180cm and 252cm.

[3]

b. Find how many tiles fit along each section's length, and calculate the TOTAL number of tiles needed if both sections are also 36cm wide (i.e. each section is a single row of square tiles).

[3]

Show complete worked solution

(a)

$180=2^2\times3^2\times5$. $252=2^2\times3^2\times7$. HCF$=2^2\times3^2=36$. Largest tile size: 36 cm.

(b)

Section 1: $180\div36=5$ tiles. Section 2: $252\div36=7$ tiles. Total tiles needed $=5+7=12$.

QUESTION 8 6 marks Criterion A

Hard

A school is preparing identical care packages for a charity drive, using 168 notebooks and 252 pencils, with NOTHING left over.

a. Find the maximum number of identical packages that can be made, using prime factorization.

[3]

b. Find the number of notebooks and pencils in EACH package, and verify your answer by checking $84\times(\text{notebooks per package})=168$.

[3]

Show complete worked solution

(a)

$168=2^3\times3\times7$. $252=2^2\times3^2\times7$. HCF (lowest powers of shared primes): $2^2\times3\times7=84$. Maximum packages $=84$.

(b)

Notebooks per package $=168\div84=2$. Pencils per package $=252\div84=3$. Check: $84\times2=168$.

QUESTION 9 7 marks Criterion B

Hard

Investigate the relationship between HCF$(a,b)$ and HCF$(a,b,c)$ when a third number $c$ is introduced, using $a=24, b=36, c=60$.

a. Find HCF$(24,36)$ first (ignoring $c$ for now).

[2]

b. Now find HCF$(12, 60)$ — i.e. the HCF of your previous answer and the third number $c=60$. Compare this to what you'd get by finding HCF$(24,36,60)$ directly (all three prime-factorized together).

[3]

c. Based on this, state a general rule for finding the HCF of THREE numbers using the HCF of only two numbers at a time.

[2]

Show complete worked solution

(a)

$24=2^3\times3$, $36=2^2\times3^2$. HCF$=2^2\times3=12$.

(b)

$60=2^2\times3\times5$. HCF$(12,60)=12$ (since $12=2^2\times3$ divides evenly into $60=2^2\times3\times5$). Direct 3-way HCF: comparing $2^3\times3$, $2^2\times3^2$, $2^2\times3\times5$ — lowest shared powers give $2^2\times3=12$. Both methods agree: 12.

(c)

HCF$(a,b,c) = $ HCF$($HCF$(a,b), c)$ — you can find the HCF of any two of the numbers first, then find the HCF of that result with the third number, and you'll always get the same answer as computing all three together directly.

QUESTION 10 5 marks Criterion C

Hard

A student says: 'The HCF of two numbers can never be bigger than the SMALLER of the two numbers, but it also can't be bigger than half of either number, since factors are always small.'

a. Test the first part of the claim (HCF can't exceed the smaller number) using HCF$(15,15)$ — two IDENTICAL numbers.

[2]

b. Test the second part of the claim ('can't be bigger than half of either number') using HCF$(20,20)$. Does this disprove the second part of the claim?

[3]

Show complete worked solution

(a)

HCF$(15,15)=15$ — this equals (not exceeds) the smaller number, confirming the first part isn't violated, though it shows the HCF CAN equal the smaller number exactly.

(b)

HCF$(20,20)=20$, which is NOT less than or equal to half of 20 (which is 10) — in fact it equals the FULL number, not half. This disproves the second part of the student's claim: the HCF of two equal numbers is the number itself, which is clearly larger than half of it (for any number greater than 0).

QUESTION 11 6 marks Criterion D

Hard

A tile installer has two types of square tiles to cover sections of a floor with NO cutting: one section requires tiles that fit evenly into a $180$cm length, and another section requires tiles fitting evenly into a $252$cm length. The installer wants to use the SAME tile size for both sections, as large as possible.

a. Find the largest tile side length (in cm) that fits evenly into both 180cm and 252cm.

[3]

b. Find how many tiles fit along each section's length, and calculate the TOTAL number of tiles needed if both sections are also 36cm wide (i.e. each section is a single row of square tiles).

[3]

Show complete worked solution

(a)

$180=2^2\times3^2\times5$. $252=2^2\times3^2\times7$. HCF$=2^2\times3^2=36$. Largest tile size: 36 cm.

(b)

Section 1: $180\div36=5$ tiles. Section 2: $252\div36=7$ tiles. Total tiles needed $=5+7=12$.

QUESTION 1 7 marks Criterion B

Medium

Investigate the relationship between the HCF and LCM of two numbers, and their product.

a. For the pairs (12,18), (20,30), and (15,25), calculate HCF, LCM, and the product HCF$\times$LCM for each pair.

[4]

b. Compare each product HCF$\times$LCM to the product of the original two numbers ($a \times b$). What do you notice?

[2]

c. State a general rule connecting HCF, LCM, and the product of two numbers $a$ and $b$.

[1]

Show complete worked solution

(a)

(12,18): HCF$=6$, LCM$=36$, product$=216$. (20,30): HCF$=10$, LCM$=60$, product$=600$. (15,25): HCF$=5$, LCM$=75$, product$=375$.

(b)

$12\times18=216$, $20\times30=600$, $15\times25=375$ — each matches the HCF$\times$LCM product exactly.

(c)

$$\text{HCF}(a,b) \times \text{LCM}(a,b) = a \times b$$

QUESTION 2 5 marks Criterion C

Medium

Two buses leave a depot at the same time. Bus A returns to the depot every 18 minutes, and Bus B returns every 24 minutes.

a. Explain, in your own words, why the LCM of 18 and 24 gives the time until both buses are next at the depot together.

[2]

b. Calculate the LCM of 18 and 24, showing your full working using prime factorization.

[3]

Show complete worked solution

(a)

The LCM is the smallest number that both 18 and 24 divide into exactly — meaning it's the first time that is simultaneously a whole multiple of Bus A's cycle AND Bus B's cycle, so both buses complete a whole number of trips and arrive together.

(b)

$18=2\times3^2$, $24=2^3\times3$. LCM (highest power of each prime): $2^3\times3^2=72$. So LCM $=72$ minutes.

QUESTION 3 7 marks Criterion D

Medium

Three warning lights at a construction site flash at regular intervals: the red light every 8 seconds, the yellow light every 12 seconds, and the blue light every 18 seconds. All three flash together at exactly 9:00:00 am.

a. Find the LCM of 8, 12, and 18 to determine how many seconds until all three lights flash together again.

[3]

b. At what time will the lights next flash together?

[1]

c. A site supervisor claims the lights will flash together again exactly 5 times within the next 6 minutes. Is this claim correct? Justify your answer.

[3]

Show complete worked solution

(a)

$8=2^3$, $12=2^2\times3$, $18=2\times3^2$. LCM $=2^3\times3^2=72$ seconds.

(b)

$9{:}00{:}00$ am $+$ 72 seconds $= 9{:}01{:}12$ am.

(c)

6 minutes $=360$ seconds. Number of 72-second intervals in 360 seconds: $360\div72=5$. So the lights DO flash together exactly 5 times in the next 6 minutes (at 72, 144, 216, 288, and 360 seconds) — the supervisor's claim is correct.

QUESTION 4 6 marks Criterion A

Hard

Three lighthouses flash at regular intervals: Lighthouse A every 18 seconds, Lighthouse B every 24 seconds, Lighthouse C every 30 seconds. All three flash together at exactly midnight.

a. Find the LCM of 18, 24, and 30 using prime factorization.

[3]

b. Find how many times all three lighthouses flash together SIMULTANEOUSLY within the first hour (3600 seconds) after midnight, not counting the initial flash at midnight itself.

[3]

Show complete worked solution

(a)

$18=2\times3^2$, $24=2^3\times3$, $30=2\times3\times5$. LCM (highest power of each prime): $2^3\times3^2\times5=360$.

(b)

Number of 360-second intervals in 3600 seconds: $3600\div360=10$. So they flash together 10 times in the first hour (after midnight, not counting midnight itself).

QUESTION 5 7 marks Criterion B

Hard

Investigate the relationship between LCM$(a,b)$ when $a$ and $b$ share NO common factors (other than 1) versus when they do.

a. Find LCM$(8,9)$ (which share no common factor besides 1) and compare it to the simple product $8\times9$.

[2]

b. Now find LCM$(8,12)$ (which DO share a common factor) and compare it to the product $8\times12$.

[2]

c. State a general rule: when does LCM$(a,b)$ equal the simple product $a\times b$, and explain why, using the formula LCM$(a,b)=\frac{a\times b}{\text{HCF}(a,b)}$.

[3]

Show complete worked solution

(a)

HCF$(8,9)=1$. LCM$=\frac{8\times9}{1}=72$, which equals the product $8\times9=72$ exactly.

(b)

HCF$(8,12)=4$. LCM$=\frac{8\times12}{4}=24$, which is LESS than the product $8\times12=96$.

(c)

LCM$(a,b)$ equals $a\times b$ exactly when HCF$(a,b)=1$ (i.e. $a$ and $b$ are 'coprime', sharing no common factors). This follows directly from the formula: dividing by HCF$=1$ leaves the product unchanged. Whenever the numbers share a common factor greater than 1, the LCM will always be LESS than the simple product.

QUESTION 6 5 marks Criterion C

Hard

A student calculates LCM$(6,10)$ by simply multiplying them: '$6\times10=60$, so LCM=60' — without checking for a common factor.

a. Verify whether 60 is INDEED the smallest common multiple of 6 and 10, by listing the first several multiples of each and finding where they first coincide.

[3]

b. Explain the student's error, and show the correct method using HCF, to confirm LCM$(6,10)=30$.

[2]

Show complete worked solution

(a)

Multiples of 6: $6,12,18,24,30,36,42,48,54,60$. Multiples of 10: $10,20,30,40,50,60$. The multiples first coincide at 30, not 60 — so the TRUE LCM is 30, not 60.

(b)

The student assumed LCM always equals the simple product, forgetting to check for a shared common factor. HCF$(6,10)=2$, so LCM$=\frac{6\times10}{2}=30$ — matching the true smallest common multiple found by listing.

QUESTION 7 6 marks Criterion D

Hard

A factory produces Product X every 15 minutes and Product Y every 40 minutes, starting simultaneously at the beginning of a shift. A quality inspector visits the factory floor every 25 minutes, ALSO starting at the beginning of the shift.

a. Find the LCM of 15, 40, and 25, to determine when all three events (Product X, Product Y, and an inspection) will next coincide exactly.

[3]

b. Convert 600 minutes into hours and minutes, and comment on whether this coincidence is likely to be OBSERVED in practice, given that a typical factory shift is 8 hours long.

[3]

Show complete worked solution

(a)

$15=3\times5$, $40=2^3\times5$, $25=5^2$. LCM$=2^3\times3\times5^2=600$ minutes.

(b)

$600$ minutes $=10$ hours $0$ minutes. Since a typical 8-hour shift is SHORTER than 10 hours, this exact triple-coincidence would NOT actually occur within a single shift — it would only happen if the same schedule continued into a second shift, or if shifts were unusually long.

QUESTION 8 6 marks Criterion A

Hard

Three lighthouses flash at regular intervals: Lighthouse A every 18 seconds, Lighthouse B every 24 seconds, Lighthouse C every 30 seconds. All three flash together at exactly midnight.

a. Find the LCM of 18, 24, and 30 using prime factorization.

[3]

b. Find how many times all three lighthouses flash together SIMULTANEOUSLY within the first hour (3600 seconds) after midnight, not counting the initial flash at midnight itself.

[3]

Show complete worked solution

(a)

$18=2\times3^2$, $24=2^3\times3$, $30=2\times3\times5$. LCM (highest power of each prime): $2^3\times3^2\times5=360$.

(b)

Number of 360-second intervals in 3600 seconds: $3600\div360=10$. So they flash together 10 times in the first hour (after midnight, not counting midnight itself).

QUESTION 9 7 marks Criterion B

Hard

Investigate the relationship between LCM$(a,b)$ when $a$ and $b$ share NO common factors (other than 1) versus when they do.

a. Find LCM$(8,9)$ (which share no common factor besides 1) and compare it to the simple product $8\times9$.

[2]

b. Now find LCM$(8,12)$ (which DO share a common factor) and compare it to the product $8\times12$.

[2]

c. State a general rule: when does LCM$(a,b)$ equal the simple product $a\times b$, and explain why, using the formula LCM$(a,b)=\frac{a\times b}{\text{HCF}(a,b)}$.

[3]

Show complete worked solution

(a)

HCF$(8,9)=1$. LCM$=\frac{8\times9}{1}=72$, which equals the product $8\times9=72$ exactly.

(b)

HCF$(8,12)=4$. LCM$=\frac{8\times12}{4}=24$, which is LESS than the product $8\times12=96$.

(c)

LCM$(a,b)$ equals $a\times b$ exactly when HCF$(a,b)=1$ (i.e. $a$ and $b$ are 'coprime', sharing no common factors). This follows directly from the formula: dividing by HCF$=1$ leaves the product unchanged. Whenever the numbers share a common factor greater than 1, the LCM will always be LESS than the simple product.

QUESTION 10 5 marks Criterion C

Hard

A student calculates LCM$(6,10)$ by simply multiplying them: '$6\times10=60$, so LCM=60' — without checking for a common factor.

a. Verify whether 60 is INDEED the smallest common multiple of 6 and 10, by listing the first several multiples of each and finding where they first coincide.

[3]

b. Explain the student's error, and show the correct method using HCF, to confirm LCM$(6,10)=30$.

[2]

Show complete worked solution

(a)

Multiples of 6: $6,12,18,24,30,36,42,48,54,60$. Multiples of 10: $10,20,30,40,50,60$. The multiples first coincide at 30, not 60 — so the TRUE LCM is 30, not 60.

(b)

The student assumed LCM always equals the simple product, forgetting to check for a shared common factor. HCF$(6,10)=2$, so LCM$=\frac{6\times10}{2}=30$ — matching the true smallest common multiple found by listing.

QUESTION 11 6 marks Criterion D

Hard

A factory produces Product X every 15 minutes and Product Y every 40 minutes, starting simultaneously at the beginning of a shift. A quality inspector visits the factory floor every 25 minutes, ALSO starting at the beginning of the shift.

a. Find the LCM of 15, 40, and 25, to determine when all three events (Product X, Product Y, and an inspection) will next coincide exactly.

[3]

b. Convert 600 minutes into hours and minutes, and comment on whether this coincidence is likely to be OBSERVED in practice, given that a typical factory shift is 8 hours long.

[3]

Show complete worked solution

(a)

$15=3\times5$, $40=2^3\times5$, $25=5^2$. LCM$=2^3\times3\times5^2=600$ minutes.

(b)

$600$ minutes $=10$ hours $0$ minutes. Since a typical 8-hour shift is SHORTER than 10 hours, this exact triple-coincidence would NOT actually occur within a single shift — it would only happen if the same schedule continued into a second shift, or if shifts were unusually long.

QUESTION 1 9 marks Criterion A

Hard

Work without a calculator. Show each simplification clearly and justify every classification as rational or irrational.

a. Simplify $\sqrt{50}-\sqrt{8}$. State whether the result is rational or irrational and justify your answer.

[3]

b. Evaluate $\dfrac{\sqrt{18}\times\sqrt{8}}{6}$. State whether the result is rational or irrational.

[3]

c. A student claims, 'The sum of two irrational numbers is always irrational.' Give a counterexample and explain why it disproves the claim.

[3]

Show complete worked solution

(a)

Rewrite each surd using its largest square factor: $\sqrt{50}=\sqrt{25\times2}=5\sqrt2$ and $\sqrt8=\sqrt{4\times2}=2\sqrt2$.

Subtract the like surds: $5\sqrt2-2\sqrt2=3\sqrt2$.

$\sqrt2$ is irrational. Multiplying an irrational number by the non-zero rational number 3 does not make it rational.

Final answer: $\boxed{3\sqrt2}$, which is irrational.

(b)

Combine the radicals: $\sqrt{18}\times\sqrt8=\sqrt{18\times8}=\sqrt{144}$.

Evaluate the square root and divide: $\dfrac{\sqrt{144}}6=\dfrac{12}6=2$.

An integer is rational because it can be written as a fraction of integers: $2=\dfrac21$.

Final answer: $\boxed{2}$, which is rational.

(c)

Choose the two numbers $a=\sqrt5$ and $b=3-\sqrt5$. Both are irrational. In particular, if $b$ were rational, then $3-b=\sqrt5$ would be rational, which is impossible.

Add them: $a+b=\sqrt5+(3-\sqrt5)=3$.

The result 3 is rational because $3=\dfrac31$.

Conclusion: Two irrational numbers can have a rational sum, so the claim containing the word "always" is false.

QUESTION 2 10 marks Criterion B

Hard

For positive integers $n$, define $$u_n=(\sqrt{5}+n)(\sqrt{5}-n).$$ Investigate the values of this expression.

a. Calculate $u_1$, $u_2$ and $u_3$.

[2]

b. Calculate the next two values. Describe the pattern in the first differences.

[3]

c. Expand the expression to obtain a rule for $u_n$. Use the rule to explain why every value of $u_n$ is rational.

[3]

d. Each factor in $u_n$ is irrational. What conclusion can you draw about the product of two irrational numbers?

[2]

Show complete worked solution

(a)

Use $(a+b)(a-b)=a^2-b^2$ with $a=\sqrt5$ and $b=n$: $u_n=(\sqrt5)^2-n^2=5-n^2$.

$u_1=5-1^2=4$.

$u_2=5-2^2=1$ and $u_3=5-3^2=-4$.

Final answer: $\boxed{u_1=4,\ u_2=1,\ u_3=-4}$.

(b)

$u_4=5-4^2=5-16=-11$ and $u_5=5-5^2=5-25=-20$.

The sequence is $4,1,-4,-11,-20$.

Subtract consecutive terms: $1-4=-3$, $-4-1=-5$, $-11-(-4)=-7$, and $-20-(-11)=-9$.

Pattern: The first differences are consecutive negative odd numbers: $\boxed{-3,-5,-7,-9}$.

(c)

Expand: $(\sqrt5+n)(\sqrt5-n)=5-n\sqrt5+n\sqrt5-n^2$.

The middle terms cancel, giving $\boxed{u_n=5-n^2}$.

For integer $n$, $n^2$ is an integer, so $5-n^2$ is also an integer.

Conclusion: Every integer can be written over 1; therefore every $u_n$ is rational.

(d)

For integer $n$, both $\sqrt5+n$ and $\sqrt5-n$ are irrational.

However, their product is $5-n^2$, which is an integer and hence rational.

Conclusion: The product of two irrational numbers $\boxed{\text{can be rational}}$; it is not necessarily irrational.

QUESTION 3 8 marks Criterion A

Hard

Simplify the following surd expressions without a calculator. State whether each final answer is rational or irrational.

a. Simplify $\dfrac{\sqrt{75}+\sqrt{12}}{\sqrt{3}}$.

[4]

b. Rationalise and simplify $\dfrac{3}{\sqrt5-2}$.

[4]

Show complete worked solution

(a)

$\sqrt{75}=5\sqrt3$ and $\sqrt{12}=2\sqrt3$. Therefore $$\frac{\sqrt{75}+\sqrt{12}}{\sqrt3}=\frac{5\sqrt3+2\sqrt3}{\sqrt3}=\frac{7\sqrt3}{\sqrt3}=\boxed{7}.$$ The answer is rational because $7=\frac71$.

(b)

Multiply the numerator and denominator by the conjugate $\sqrt5+2$: $$\frac3{\sqrt5-2}\times\frac{\sqrt5+2}{\sqrt5+2}=\frac{3(\sqrt5+2)}{5-4}=\boxed{3\sqrt5+6}.$$ This is irrational because it contains the non-zero irrational term $3\sqrt5$.

QUESTION 4 9 marks Criterion B

Hard

Investigate two operations involving irrational numbers and use the results to evaluate a general claim.

a. Simplify $\sqrt3+\sqrt{12}$ and classify the answer.

[3]

b. Simplify $\sqrt2\times\sqrt8$ and classify the answer.

[3]

c. Use your answers to explain why an operation on two irrational numbers does not always have the same type of result.

[3]

Show complete worked solution

(a)

Since $\sqrt{12}=2\sqrt3$, $$\sqrt3+\sqrt{12}=\sqrt3+2\sqrt3=\boxed{3\sqrt3}.$$ The result is irrational.

(b)

$$\sqrt2\times\sqrt8=\sqrt{16}=\boxed4.$$ The result is rational.

(c)

The sum in part (a) is irrational, while the product in part (b) is rational. Therefore, knowing only that both inputs are irrational is not enough to decide whether their sum or product will be rational.

QUESTION 5 8 marks Criterion A

Hard

Locate $\sqrt{45}$ accurately on the number line using square-number bounds.

a. Show that $\sqrt{45}$ lies between 6 and 7.

[2]

b. Find two consecutive hundredths between which $\sqrt{45}$ lies.

[4]

c. Give $\sqrt{45}$ correct to two decimal places and justify the rounding.

[2]

Show complete worked solution

(a)

$6^2=36$ and $7^2=49$. Since $36<45<49$, taking positive square roots gives $$\boxed{6<\sqrt{45}<7}.$$

(b)

Test nearby hundredths: $$6.70^2=44.89,\qquad 6.71^2=45.0241.$$ Since $44.89<45<45.0241$, $$\boxed{6.70<\sqrt{45}<6.71}.$$

(c)

$6.705^2=44.957025<45$, so $\sqrt{45}>6.705$. Together with the bounds above, it rounds upward to $$\boxed{6.71}$$ correct to two decimal places.

QUESTION 6 8 marks Criterion A

Medium

Expand and simplify each expression exactly. Explain why the answers are rational even though the expressions contain surds.

a. $(\sqrt{18}+\sqrt8)^2$

[4]

b. $(\sqrt{18}-\sqrt8)^2$

[4]

Show complete worked solution

(a)

$\sqrt{18}=3\sqrt2$ and $\sqrt8=2\sqrt2$. Hence $$(\sqrt{18}+\sqrt8)^2=(5\sqrt2)^2=25\times2=\boxed{50}.$$

(b)

Using the same simplifications, $$(\sqrt{18}-\sqrt8)^2=(3\sqrt2-2\sqrt2)^2=(\sqrt2)^2=\boxed2.$$ Both answers are integers, so both are rational.

QUESTION 7 8 marks Criterion C

Hard

A student is asked to prove that $\sqrt3$ is irrational. Complete a proof by contradiction.

a. Assume $\sqrt3=\dfrac ab$, where $a$ and $b$ are integers with no common factor. Show that both $a$ and $b$ must be divisible by 3.

[6]

b. Explain the contradiction and state the conclusion.

[2]

Show complete worked solution

(a)

Assume $\sqrt3=\frac ab$ in lowest terms. Squaring gives $3=\frac{a^2}{b^2}$, so $$a^2=3b^2.$$ Thus $a^2$ is divisible by 3, so $a$ is divisible by 3. Write $a=3k$. Substitution gives $$9k^2=3b^2\implies b^2=3k^2.$$ Therefore $b$ is also divisible by 3.

(b)

If both $a$ and $b$ are divisible by 3, then they have a common factor, contradicting the statement that $\frac ab$ is in lowest terms. Therefore the assumption is false and $\boxed{\sqrt3\text{ is irrational}}$.

QUESTION 8 8 marks Criterion B

Hard

Construct examples that meet the stated conditions and justify each construction.

a. Find two irrational numbers whose sum is 10.

[4]

b. Find two irrational numbers whose product is 12.

[4]

Show complete worked solution

(a)

Choose $\sqrt2$ and $10-\sqrt2$. Their sum is $$\sqrt2+(10-\sqrt2)=\boxed{10}.$$ Both numbers are irrational: if $10-\sqrt2$ were rational, subtracting it from 10 would make $\sqrt2$ rational, which is impossible.

(b)

Choose $\sqrt6$ and $\sqrt{24}=2\sqrt6$. Both are irrational, and $$\sqrt6\times\sqrt{24}=\sqrt{144}=\boxed{12}.$$

QUESTION 9 6 marks Criterion A

Medium

Simplify each expression exactly and identify the algebraic structure that removes the irrational terms.

a. $\dfrac{\sqrt{27}-\sqrt3}{\sqrt3}$

[3]

b. $(\sqrt5+1)(\sqrt5-1)$

[3]

Show complete worked solution

(a)

$\sqrt{27}=3\sqrt3$. Thus $$\frac{\sqrt{27}-\sqrt3}{\sqrt3}=\frac{3\sqrt3-\sqrt3}{\sqrt3}=\frac{2\sqrt3}{\sqrt3}=\boxed2.$$ Like surds combine before division.

(b)

This is a product of conjugates: $$(\sqrt5+1)(\sqrt5-1)=(\sqrt5)^2-1^2=5-1=\boxed4.$$ The middle irrational terms cancel.

QUESTION 10 8 marks Criterion D

Hard

A square courtyard has side length 7 m. Its diagonal is to be marked with a straight strip.

a. Find the exact length of the diagonal and classify it as rational or irrational.

[4]

b. Using $\sqrt2\approx1.414$, determine whether a 9.85 m strip is long enough. State the shortfall or excess.

[4]

Show complete worked solution

(a)

By Pythagoras, $$d^2=7^2+7^2=98.$$ Therefore $$d=\sqrt{98}=\boxed{7\sqrt2\text{ m}}.$$ Since $\sqrt2$ is irrational, $7\sqrt2$ is irrational.

(b)

$$d\approx7(1.414)=9.898\text{ m}.$$ Since $9.898>9.85$, the strip is not long enough. The shortfall is $$9.898-9.85=0.048\text{ m}\approx\boxed{0.05\text{ m}}.$$

QUESTION 1 8 marks Criterion A

Hard

Convert each recurring decimal by forming and subtracting suitable equations. Give every fraction in simplest form.

a. Convert $0.\overline{27}$ to a fraction.

[3]

b. Convert $0.1\overline{6}$, meaning $0.16666\ldots$, to a fraction.

[3]

c. Find $0.\overline{27}-0.1\overline{6}$ as a fraction in simplest form.

[2]

Show complete worked solution

(a)

Let $x=0.272727\ldots$.

Two digits repeat, so multiply by 100: $100x=27.272727\ldots$.

Subtract the first equation from the second: $100x-x=27.272727\ldots-0.272727\ldots$, so $99x=27$.

$x=\dfrac{27}{99}=\dfrac{3}{11}$ after dividing numerator and denominator by 9.

Final answer: $\boxed{0.\overline{27}=\dfrac{3}{11}}$.

(b)

Let $x=0.16666\ldots$.

Move past the non-recurring digit: $10x=1.6666\ldots$.

Move one more place, because one digit repeats: $100x=16.6666\ldots$.

Subtract: $100x-10x=16.6666\ldots-1.6666\ldots$, so $90x=15$.

Final answer: $x=\dfrac{15}{90}=\boxed{\dfrac16}$.

(c)

Use the fractions found above: $0.\overline{27}-0.1\overline6=\dfrac3{11}-\dfrac16$.

The lowest common denominator of 11 and 6 is 66.

$\dfrac3{11}=\dfrac{18}{66}$ and $\dfrac16=\dfrac{11}{66}$.

Final answer: $\dfrac{18}{66}-\dfrac{11}{66}=\boxed{\dfrac7{66}}$.

QUESTION 2 10 marks Criterion B

Hard

For a digit $n$ from 1 to 9, let $d_n=0.\overline{n}$, so the digit $n$ repeats forever. Investigate this family of recurring decimals.

a. Write $d_1$, $d_4$ and $d_7$ as fractions. Record the results in a clear table.

[3]

b. State a general rule for $d_n$.

[2]

c. Prove your rule algebraically.

[3]

d. Use the rule when $n=9$. Explain the conclusion.

[2]

Show complete worked solution

(a)

For $d_1=0.111\ldots$, $10d_1-d_1=1$, so $9d_1=1$ and $d_1=\dfrac19$.

For $d_4=0.444\ldots$, $10d_4-d_4=4$, so $d_4=\dfrac49$.

For $d_7=0.777\ldots$, $10d_7-d_7=7$, so $d_7=\dfrac79$.

Table: $\begin{array}{c|c}n&d_n\\\hline1&1/9\\4&4/9\\7&7/9\end{array}$.

(b)

Compare each digit with the numerator: $1\mapsto\dfrac19$, $4\mapsto\dfrac49$, and $7\mapsto\dfrac79$.

The denominator stays 9 while the repeated digit becomes the numerator.

Conjecture: $\boxed{d_n=\dfrac n9}$ for $n=1,2,\ldots,9$.

(c)

Let $x=0.nnn\ldots$.

Multiply by 10: $10x=n.nnn\ldots$.

Subtract the original equation: $10x-x=n.nnn\ldots-0.nnn\ldots=n$.

Therefore $9x=n$, so $\boxed{x=\dfrac n9}$. This proves the rule for every allowed digit $n$.

(d)

Substitute $n=9$ into the proved rule: $d_9=\dfrac99$.

$\dfrac99=1$.

Conclusion: $\boxed{0.999\ldots=1}$. They are different decimal representations of the same number.

QUESTION 3 7 marks Criterion A

Hard

Convert each recurring decimal to a fraction by showing the equations used to remove the recurring digits.

a. Convert $0.\overline{36}$ to a fraction in simplest form.

[3]

b. Convert $2.1\overline6$, meaning $2.16666\ldots$, to a fraction.

[4]

Show complete worked solution

(a)

Let $x=0.363636\ldots$. Then $100x=36.363636\ldots$. Subtract: $$100x-x=36\implies99x=36.$$ Hence $$x=\frac{36}{99}=\boxed{\frac4{11}}.$$

(b)

Let $y=2.16666\ldots$. Then $10y=21.6666\ldots$ and $100y=216.6666\ldots$. Subtract: $$100y-10y=195\implies90y=195.$$ Thus $$y=\frac{195}{90}=\boxed{\frac{13}{6}}.$$

QUESTION 4 8 marks Criterion A

Hard

Work with the decimal $0.2\overline{45}=0.2454545\ldots$.

a. Convert the decimal to a fraction in simplest form.

[5]

b. Verify the result by separating the fraction into a terminating part and a recurring part.

[3]

Show complete worked solution

(a)

Let $x=0.2454545\ldots$. Then $10x=2.454545\ldots$ and $1000x=245.454545\ldots$. Subtract: $$1000x-10x=243\implies990x=243.$$ Therefore $$x=\frac{243}{990}=\boxed{\frac{27}{110}}.$$

(b)

$$\frac{27}{110}=\frac{22}{110}+\frac5{110}=0.2+\frac1{22}.$$ Since $\frac1{22}=0.0454545\ldots$, the sum is $0.2454545\ldots$, which verifies the answer.

QUESTION 5 5 marks Criterion A

Medium

Let $x=1.\overline{27}=1.272727\ldots$.

a. Express $x$ as a fraction in simplest form.

[3]

b. Find $x^2$ as a fraction in simplest form.

[2]

Show complete worked solution

(a)

$100x=127.272727\ldots$. Subtracting $x=1.272727\ldots$ gives $$99x=126,$$ so $$x=\frac{126}{99}=\boxed{\frac{14}{11}}.$$

(b)

$$x^2=\left(\frac{14}{11}\right)^2=\boxed{\frac{196}{121}}.$$ The numerator and denominator have no common factor.

QUESTION 6 5 marks Criterion A

Medium

Use fractions to calculate with the recurring decimals. Do not round.

a. Convert $0.\overline7$ and $0.\overline2$ to fractions.

[3]

b. Find $0.\overline7-0.\overline2$ exactly.

[2]

Show complete worked solution

(a)

For a one-digit recurring decimal, $0.\overline n=\frac n9$. Therefore $$0.\overline7=\frac79,\qquad0.\overline2=\frac29.$$

(b)

$$\frac79-\frac29=\boxed{\frac59}.$$ This equals $0.\overline5$.

QUESTION 7 8 marks Criterion A

Hard

Consider $x=0.12\overline3=0.123333\ldots$.

a. Convert $x$ to a fraction in simplest form.

[5]

b. Calculate $x+0.0\overline6$ exactly.

[3]

Show complete worked solution

(a)

$100x=12.3333\ldots$ and $1000x=123.3333\ldots$. Subtract: $$1000x-100x=111\implies900x=111.$$ Hence $$x=\frac{111}{900}=\boxed{\frac{37}{300}}.$$

(b)

$0.0\overline6=0.06666\ldots=\frac1{15}$. Therefore $$\frac{37}{300}+\frac1{15}=\frac{37}{300}+\frac{20}{300}=\frac{57}{300}=\boxed{\frac{19}{100}}.$$

QUESTION 8 7 marks Criterion B

Hard

The two-digit block $ab$ repeats in $0.\overline{ab}$.

a. Given that $0.\overline{ab}=\dfrac7{11}$, determine the two-digit block $ab$.

[3]

b. Prove that for any two-digit integer $n$, $0.\overline n=\dfrac n{99}$.

[4]

Show complete worked solution

(a)

Write the denominator as 99: $$\frac7{11}=\frac{7\times9}{11\times9}=\frac{63}{99}=0.636363\ldots.$$ Hence the repeating block is $\boxed{63}$.

(b)

Let $x=0.nnn\ldots$, where the block has two digits. Then $100x=n.nnn\ldots$. Subtracting gives $$100x-x=n\implies99x=n.$$ Therefore $$\boxed{x=\frac n{99}}.$$

QUESTION 9 7 marks Criterion C

Hard

Compare the two decimal representations $0.24999\ldots$ and $0.25$.

a. Convert $0.24999\ldots$ to a fraction.

[5]

b. Explain what this proves about the two decimal representations.

[2]

Show complete worked solution

(a)

Let $x=0.24999\ldots$. Then $10x=2.4999\ldots$ and $100x=24.9999\ldots$. Subtract: $$100x-10x=22.5\implies90x=22.5.$$ Thus $$x=\frac{22.5}{90}=\frac{225}{900}=\boxed{\frac14}.$$

(b)

Since $\frac14=0.25$, it follows that $$\boxed{0.24999\ldots=0.25}.$$ They are two decimal representations of the same number; there is no gap between them.

QUESTION 10 9 marks Criterion D

Hard

Calculate exactly with two recurring decimals, showing every conversion.

a. Convert $0.\overline{18}$ and $0.1\overline8$ to fractions.

[5]

b. Find the sum and the positive difference of the two numbers.

[4]

Show complete worked solution

(a)

$0.\overline{18}=\frac{18}{99}=\frac2{11}$. For $y=0.18888\ldots$, $10y=1.8888\ldots$ and $100y=18.8888\ldots$. Hence $$90y=17\implies y=\frac{17}{90}.$$

(b)

Sum: $$\frac2{11}+\frac{17}{90}=\frac{180+187}{990}=\boxed{\frac{367}{990}}.$$ Difference: $$\frac{17}{90}-\frac2{11}=\frac{187-180}{990}=\boxed{\frac7{990}}.$$

QUESTION 1 9 marks Criterion A

Hard

A rectangular display panel has measured length 18.47 m and width 6.936 m. Retain unrounded values in your calculations unless instructed otherwise.

a. Round each measurement to 3 significant figures.

[2]

b. Using the original measurements, calculate the area and perimeter. Give both answers to 3 significant figures.

[4]

c. A student first rounds both measurements to 3 significant figures and then calculates the area. Find the student's unrounded area, compare it with the area from part (b), and comment on premature rounding.

[3]

Show complete worked solution

(a)

Length: For 18.47, keep the first three significant digits 1, 8 and 4. The next digit is 7, so round 4 up: $18.47\approx\boxed{18.5\text{ m}}$.

Width: For 6.936, keep 6, 9 and 3. The next digit is 6, so round 3 up: $6.936\approx\boxed{6.94\text{ m}}$.

Check: Each answer contains exactly three significant figures.

(b)

Use $A=lw$: $A=18.47\times6.936=128.10792\text{ m}^2$.

To 3 significant figures, $A=\boxed{128\text{ m}^2}$.

Use $P=2(l+w)$: $P=2(18.47+6.936)=2(25.406)=50.812\text{ m}$.

Final answer: To 3 significant figures, $P=\boxed{50.8\text{ m}}$.

(c)

Use the rounded measurements from part (a): $A_{\rm rounded}=18.5\times6.94=128.39\text{ m}^2$.

Compare with the unrounded calculation: $128.39-128.10792=0.28208\text{ m}^2$.

Both values become $128\text{ m}^2$ when rounded to 3 significant figures.

Comment: Premature rounding changed the working value. Original measurements should be retained until the final line so that rounding errors do not accumulate.

QUESTION 2 9 marks Criterion C

Hard

A science report records a sample mass of 0.004976 kg, an energy reading of 690.4 J, and a sample volume of $2.8\times10^{-6}$ m$^3$. Present numerical information clearly and use appropriate significant figures.

a. Write the mass to 3 significant figures in scientific notation.

[2]

b. Write the energy reading to 2 significant figures. Explain why scientific notation communicates the precision more clearly than writing 690 J.

[3]

c. A student reports the density as 1777.142857 kg/m$^3$. Evaluate this report and give a better final answer.

[4]

Show complete worked solution

(a)

Write the number as $4.976\times10^{-3}$ kg.

The first three significant digits are 4, 9 and 7. The next digit is 6, so round 7 up to 8.

Final answer: $\boxed{4.98\times10^{-3}\text{ kg}}$ to 3 significant figures.

(b)

The first two significant digits in 690.4 are 6 and 9. The next digit is 0, so 9 is unchanged.

Write the result as $\boxed{6.9\times10^2\text{ J}}$.

Communication: This notation displays exactly two significant figures. Writing 690 J alone can be ambiguous because the reader cannot tell whether the final zero is significant.

(c)

Use $\rho=\dfrac{m}{V}$.

$\rho=\dfrac{0.004976}{2.8\times10^{-6}}=1777.142857\ldots\text{ kg/m}^3$.

The volume $2.8\times10^{-6}$ has only 2 significant figures, so the final density should be reported to 2 significant figures.

Evaluation: The student has copied too many calculator digits and implied false precision.

Final answer: $\boxed{\rho=1.8\times10^3\text{ kg/m}^3}$ to 2 significant figures.

QUESTION 3 6 marks Criterion A

Medium

Round each number to the stated number of significant figures and show which digit determines the rounding.

a. Round $0.0060847$ to 3 significant figures.

[3]

b. Round $508600$ to 3 significant figures and write the answer in scientific notation.

[3]

Show complete worked solution

(a)

The first three significant digits are 6, 0 and 8. The next digit is 4, so the 8 stays unchanged: $$0.0060847\approx\boxed{0.00608}\quad(3\text{ s.f.}).$$

(b)

The first three significant digits are 5, 0 and 8. The next digit is 6, so 508 rounds to 509: $$508600\approx509000=\boxed{5.09\times10^5}.$$

QUESTION 4 7 marks Criterion C

Hard

A square tile has side length recorded as 7.4 cm, correct to 2 significant figures.

a. Write the lower and upper bounds for the actual side length.

[3]

b. Find the lower and upper bounds for the area of the tile.

[4]

Show complete worked solution

(a)

Rounding to 2 significant figures here means rounding to the nearest $0.1$ cm. Half a unit is $0.05$ cm, so $$\boxed{7.35\le l<7.45\text{ cm}}.$$

(b)

Area increases as the positive side length increases. Therefore $$7.35^2\le A<7.45^2.$$ Calculating gives $$\boxed{54.0225\le A<55.5025\text{ cm}^2}.$$

QUESTION 5 7 marks Criterion A

Hard

A sample has mass 58.4 g and volume 7.26 cubic centimetres.

a. Calculate its density and give the answer to an appropriate number of significant figures.

[4]

b. The volume is later reported as 7.3 cubic centimetres. Recalculate the density and explain the precision used.

[3]

Show complete worked solution

(a)

$$\text{density}=\frac{58.4}{7.26}=8.044077\ldots$$ Both measurements have 3 significant figures, so $$\boxed{\text{density}=8.04\text{ g/cm}^3}.$$

(b)

$$\frac{58.4}{7.3}=8.000\ldots$$ The least precise measurement now has 2 significant figures, so the density should be reported as $$\boxed{8.0\text{ g/cm}^3}.$$

QUESTION 6 6 marks Criterion C

Hard

A measured value of 12.46 is reported as 12.5. Analyse the rounding error.

a. Find the absolute error.

[2]

b. Find the percentage error, correct to 2 significant figures.

[4]

Show complete worked solution

(a)

$$\text{absolute error}=|12.5-12.46|=\boxed{0.04}.$$

(b)

$$\text{percentage error}=\frac{0.04}{12.46}\times100=0.3210\ldots\%.$$ To 2 significant figures, this is $$\boxed{0.32\%}.$$

QUESTION 7 8 marks Criterion D

Hard

A cyclist travels 18.75 km in 0.426 hours.

a. Calculate the average speed, giving an appropriate number of significant figures.

[4]

b. A student first rounds the time to 0.43 h. Find the resulting speed and explain why intermediate rounding should be avoided.

[4]

Show complete worked solution

(a)

$$v=\frac{18.75}{0.426}=44.0140\ldots\text{ km/h}.$$ The time has 3 significant figures, so $$\boxed{v=44.0\text{ km/h}}.$$

(b)

Using $0.43$ gives $$\frac{18.75}{0.43}=43.6046\ldots\approx43.6\text{ km/h}.$$ This differs from $44.0$ km/h because information was discarded too early. Keep the original values until the final answer.

QUESTION 8 8 marks Criterion D

Hard

A solid cylinder has radius 2.36 cm and height 8.4 cm.

a. Calculate its volume, giving an appropriate number of significant figures.

[5]

b. If both radius and height are doubled, determine the scale factor for volume and estimate the new volume.

[3]

Show complete worked solution

(a)

$$V=\pi r^2h=\pi(2.36)^2(8.4)=146.978\ldots\text{ cm}^3.$$ The height has 2 significant figures, so $$\boxed{V=1.5\times10^2\text{ cm}^3}.$$

(b)

Volume is proportional to $r^2h$. Doubling all linear dimensions gives a factor $2^2\times2=8$. Thus $$V_{new}\approx8(1.5\times10^2)=\boxed{1.2\times10^3\text{ cm}^3}.$$

QUESTION 9 6 marks Criterion C

Hard

Explain and apply the rules for zeros in significant figures.

a. How many significant figures are in $0.04050$? Explain.

[3]

b. Round $0.040496$ to 4 significant figures and explain the final zero.

[3]

Show complete worked solution

(a)

The leading zeros only locate the decimal point and are not significant. The zero between 4 and 5 and the final zero after the decimal are significant. Therefore $0.04050$ has $\boxed{4}$ significant figures.

(b)

The first four significant digits are 4, 0, 4 and 9. The next digit is 6, so 4049 rounds to 4050: $$\boxed{0.04050}.$$ The final zero must be written because it communicates the stated precision.

QUESTION 10 5 marks Criterion A

Medium

Perform the calculation using full precision and report the result correctly.

a. Calculate $3.846\times2.71$ and give the answer to the correct number of significant figures.

[3]

b. State why the answer should not be reported as 10.42266.

[2]

Show complete worked solution

(a)

Using the unrounded values, $$3.846\times2.71=10.42266.$$ The factor 2.71 has 3 significant figures, so the result is $$\boxed{10.4}.$$

(b)

The displayed digits suggest more precision than the least precise input supports. The multiplication result must be rounded to 3 significant figures, matching the limiting measurement.

QUESTION 1 10 marks Criterion D

Hard

A school plans to cover part of a rectangular roof, 38.7 m by 16.4 m, with solar panels. Each panel measures 1.72 m by 1.13 m. The school reserves 18% of the roof area for walkways and equipment.

a. Estimate the number of panels by rounding each roof and panel dimension to 1 significant figure and using 80% as the usable proportion. Show the structure of your estimate.

[4]

b. Use the original measurements and 82% usable area to calculate an area-based maximum number of complete panels.

[3]

c. Explain why 267 panels is not guaranteed to fit and state what additional information is needed before ordering.

[3]

Show complete worked solution

(a)

Round to 1 significant figure: $38.7\approx40$, $16.4\approx20$, $1.72\approx2$, and $1.13\approx1$.

Use 80% as a decimal: $80\%=0.8$.

Estimated usable roof area $=40\times20\times0.8=640\text{ m}^2$.

Estimated panel area $=2\times1=2\text{ m}^2$.

Final estimate: $\dfrac{640}{2}=\boxed{320\text{ panels}}$.

(b)

Exact roof area $=38.7\times16.4=634.68\text{ m}^2$.

Usable area $=0.82\times634.68=520.4376\text{ m}^2$.

One panel has area $1.72\times1.13=1.9436\text{ m}^2$.

$\dfrac{520.4376}{1.9436}=267.7699\ldots$.

Final answer: Only whole panels can be used, so the area-based maximum is $\boxed{267\text{ panels}}$.

(c)

Reasoning: Dividing areas assumes that every part of the usable roof can be tiled perfectly by panel rectangles.

Missing constraints: Panel orientation, exact roof edges, walkway shape, gaps, access rules and roof obstructions can reduce the number that fits.

Conclusion: A scaled layout with the exact usable dimensions and required spacing is needed before ordering. Therefore 267 is an area-based upper estimate, not a guaranteed fit.

QUESTION 2 10 marks Criterion D

Hard

A tank needs 11,000 L of water. A pump is rated at 47.8 L per minute and is planned to run for 3 hours 52 minutes. The manager must decide whether this plan has a safe margin.

a. Estimate the volume pumped by using 50 L per minute and 4 hours. Based on this estimate, state whether the target appears to be reached.

[3]

b. Calculate the volume using the stated rate and exact planned time. Find the margin above or below 11,000 L.

[3]

c. During operation, the average rate may be 2% lower than the rated value. Calculate the resulting volume and evaluate whether the original estimate is reliable for this decision.

[4]

Show complete worked solution

(a)

Convert 4 hours to minutes: $4\times60=240$ minutes.

Use the rounded rate: $50\times240=12,000$ L.

Decision from estimate: $12,000>11,000$, so the estimate suggests that the target will be reached.

(b)

Convert the planned time: $3\times60+52=232$ minutes.

Use the stated rate: $47.8\times232=11,089.6$ L.

Find the margin: $11,089.6-11,000=89.6$ L.

Final answer: The calculated volume is $\boxed{11,089.6\text{ L}}$, which is $\boxed{89.6\text{ L above}}$ the target.

(c)

A 2% reduction means the pump works at 98% of the rated rate, so use the factor 0.98.

Reduced volume $=47.8\times0.98\times232=10,867.808$ L.

Shortfall $=11,000-10,867.808=132.192$ L.

Evaluation: The rounded estimate of 12,000 L was too optimistic for a safety decision. The manager should increase the running time or provide a larger margin.

Final conclusion: With a 2% lower rate, the plan $\boxed{\text{does not reach the target}}$.

QUESTION 3 6 marks Criterion A

Hard

Estimate and then evaluate $\dfrac{48.7\times0.196}{\sqrt{15.8}}$.

a. Use one-significant-figure values to estimate the result.

[3]

b. Calculate the value and use the estimate to comment on its reasonableness.

[3]

Show complete worked solution

(a)

Use $48.7\approx50$, $0.196\approx0.2$ and $\sqrt{15.8}\approx\sqrt{16}=4$: $$\frac{50\times0.2}{4}=\frac{10}{4}=\boxed{2.5}.$$

(b)

$$\frac{48.7\times0.196}{\sqrt{15.8}}=2.4013\ldots\approx\boxed{2.40}.$$ This is close to 2.5 and has the same order of magnitude, so it is reasonable.

QUESTION 4 6 marks Criterion D

Hard

A student buys 3 notebooks at $2.79 each, 2 folders at $4.35 each, pens for $1.89, and a calculator for $6.25.

a. Estimate the total by rounding each price to the nearest dollar.

[3]

b. Find the exact total and state whether the estimate is high or low.

[3]

Show complete worked solution

(a)

$$3(3)+2(4)+2+6=9+8+2+6=\boxed{\$25}.$$

(b)

$$3(2.79)+2(4.35)+1.89+6.25=8.37+8.70+1.89+6.25=\boxed{\$25.21}.$$ The estimate is low by $\boxed{\$0.21}$.

QUESTION 5 6 marks Criterion D

Hard

A school has 720 students. Each student uses about 1.8 drink bottles per week during a 38-week school year.

a. Use rounded values to estimate the annual number of bottles.

[3]

b. Calculate using the given values and compare with the estimate.

[3]

Show complete worked solution

(a)

Use $720\approx700$, $1.8\approx2$ and $38\approx40$: $$700\times2\times40=\boxed{56,000\text{ bottles}}.$$

(b)

$$720\times1.8\times38=\boxed{49,248\text{ bottles}}.$$ The estimate is higher by $56,000-49,248=6,752$, but it gives the correct scale: about fifty thousand bottles.

QUESTION 6 7 marks Criterion C

Hard

A rectangle is measured as 12.6 cm by 8.3 cm, with each measurement correct to the nearest 0.1 cm.

a. Write bounds for both dimensions.

[3]

b. Find lower and upper bounds for the area.

[4]

Show complete worked solution

(a)

$$12.55\le l<12.65,\qquad8.25\le w<8.35.$$ Each boundary is $0.05$ cm from its recorded value.

(b)

All lengths are positive, so multiply corresponding bounds: $$A_{min}=12.55(8.25)=103.5375,$$ $$A_{max}=12.65(8.35)=105.6275.$$ Hence $$\boxed{103.5375\le A<105.6275\text{ cm}^2}.$$

QUESTION 7 6 marks Criterion A

Hard

Estimate $\sqrt{78}$ without using a square-root key.

a. Find consecutive tenths between which the value lies.

[3]

b. Test 8.83 and use the result to give a three-decimal-place estimate.

[3]

Show complete worked solution

(a)

$8.8^2=77.44$ and $8.9^2=79.21$. Since $77.44<78<79.21$, $$\boxed{8.8<\sqrt{78}<8.9}.$$

(b)

$8.83^2=77.9689$, so $\sqrt{78}>8.83$. Test the rounding midpoint: $8.8315^2=77.99539225<78$, while $8.832^2=78.004224>78$. Hence $8.8315<\sqrt{78}<8.832$, so $$\sqrt{78}\approx\boxed{8.832}$$ to three decimal places.

QUESTION 8 6 marks Criterion D

Hard

An item costs \$84.75 before 7.5% sales tax.

a. Estimate the final price using convenient rounded numbers.

[3]

b. Calculate the final price to the nearest cent and compare it with the estimate.

[3]

Show complete worked solution

(a)

Use $84.75\approx85$. Also, $7.5\%$ of 85 is about $6.4$. Therefore the final price is about $$85+6.4=\boxed{\$91.4}.$$

(b)

$$84.75(1.075)=91.10625.$$ To the nearest cent the price is $\boxed{\$91.11}$. The estimate is higher by about $\$0.29$, so it is close.

QUESTION 9 6 marks Criterion C

Hard

A calculator display for $\dfrac{19.8\times5.1}{0.49}$ is recorded as 20.6.

a. Use estimation to show that the recorded display cannot be correct.

[3]

b. Calculate the correct value to 3 significant figures and suggest a likely error.

[3]

Show complete worked solution

(a)

$$\frac{19.8\times5.1}{0.49}\approx\frac{20\times5}{0.5}=\frac{100}{0.5}=\boxed{200}.$$ A value near 20.6 is about ten times too small.

(b)

$$\frac{19.8\times5.1}{0.49}=206.0816\ldots\approx\boxed{206}.$$ A misplaced decimal point or incorrect entry of the denominator could explain 20.6.

QUESTION 10 7 marks Criterion D

Hard

A stadium event has $4.8\times10^4$ visitors. Each visitor uses about 0.65 L of water.

a. Estimate the total water use to one significant figure.

[3]

b. Calculate the amount using the given values and find the number of 1000 L tanks required.

[4]

Show complete worked solution

(a)

Use $4.8\times10^4\approx5\times10^4$ and $0.65\approx0.7$: $$(5\times10^4)(0.7)=3.5\times10^4\text{ L}\approx\boxed{4\times10^4\text{ L}}$$ to one significant figure.

(b)

$$(4.8\times10^4)(0.65)=3.12\times10^4=31,200\text{ L}.$$ This is 31.2 tanks, so the event must provide the next whole number: $\boxed{32\text{ tanks}}$.

QUESTION 1 10 marks Criterion A

Hard

Work with numbers in scientific notation. Give each final numerical answer in normalized scientific notation.

a. Calculate $(6\times10^7)(4\times10^{-3})$.

[2]

b. Calculate $\dfrac{9.6\times10^8}{3.2\times10^{-4}}$.

[3]

c. Solve $(1.5\times10^{-6})x=4.5\times10^2$.

[3]

d. Arrange $7.2\times10^{-5}$, $6.8\times10^{-4}$ and $9.1\times10^{-6}$ in ascending order.

[2]

Show complete worked solution

(a)

Multiply the coefficients: $6\times4=24$.

Add the powers of 10: $10^7\times10^{-3}=10^{7+(-3)}=10^4$.

$24\times10^4$ is not normalized because 24 is greater than 10. Move the decimal one place left and increase the exponent by 1.

Final answer: $24\times10^4=\boxed{2.4\times10^5}$.

(b)

Divide the coefficients: $9.6\div3.2=3$.

Subtract the exponents: $10^8\div10^{-4}=10^{8-(-4)}=10^{12}$.

Final answer: $\boxed{3\times10^{12}}$.

(c)

Divide both sides by $1.5\times10^{-6}$: $x=\dfrac{4.5\times10^2}{1.5\times10^{-6}}$.

Divide coefficients: $4.5\div1.5=3$.

Subtract exponents: $2-(-6)=8$.

Final answer: $\boxed{x=3\times10^8}$.

(d)

For positive numbers in scientific notation, a more negative exponent gives a smaller number.

The exponents are $-6$, $-5$ and $-4$, so compare them in that order.

Final answer: $\boxed{9.1\times10^{-6}<7.2\times10^{-5}<6.8\times10^{-4}}$.

QUESTION 2 10 marks Criterion D

Hard

A telescope produces $2.4\times10^9$ bytes of data per minute during an observation lasting $3.5\times10^2$ minutes. Compression reduces the storage required by a factor of $1.6\times10^1$. Use $1$ GB $=10^9$ bytes.

a. Calculate the number of bytes produced before compression. Give the answer in scientific notation.

[3]

b. Calculate the compressed size in bytes and then in GB.

[3]

c. A 64 GB drive already contains $8.7\times10^9$ bytes. Determine whether the compressed observation fits. State the remaining margin and justify the decision.

[4]

Show complete worked solution

(a)

Multiply the coefficients: $2.4\times3.5=8.4$.

Add the exponents: $10^9\times10^2=10^{11}$.

Final answer: $(2.4\times10^9)(3.5\times10^2)=\boxed{8.4\times10^{11}\text{ bytes}}$.

(b)

Divide by the compression factor: $\dfrac{8.4\times10^{11}}{1.6\times10^1}=\dfrac{8.4}{1.6}\times10^{11-1}$.

$8.4\div1.6=5.25$, so the compressed size is $\boxed{5.25\times10^{10}\text{ bytes}}$.

Convert bytes to GB: $\dfrac{5.25\times10^{10}}{10^9}=5.25\times10^1=\boxed{52.5\text{ GB}}$.

(c)

Convert the existing data: $8.7\times10^9$ bytes $=8.7$ GB.

Free space $=64-8.7=55.3$ GB.

Compare: $52.5<55.3$, so the compressed observation fits.

Remaining margin $=55.3-52.5=\boxed{2.8\text{ GB}}$.

Conclusion: The file fits under the stated decimal definition of GB, provided no extra temporary working space is required.

QUESTION 3 6 marks Criterion A

Medium

Calculate with numbers in scientific notation. Give each answer in standard scientific notation.

a. $(3.6\times10^5)(2.5\times10^{-3})$

[3]

b. $(8.4\times10^7)\div(2.1\times10^3)$

[3]

Show complete worked solution

(a)

Multiply the coefficients and add the exponents: $$(3.6\times2.5)\times10^{5+(-3)}=9.0\times10^2.$$ Thus the answer is $\boxed{9.0\times10^2}$.

(b)

Divide the coefficients and subtract the exponents: $$\frac{8.4}{2.1}\times10^{7-3}=\boxed{4.0\times10^4}.$$

QUESTION 4 5 marks Criterion A

Hard

Evaluate $\dfrac{7.2\times10^{-4}}{1.8\times10^2}$.

a. Calculate the result in scientific notation.

[3]

b. Write the result as an ordinary decimal.

[2]

Show complete worked solution

(a)

$$\frac{7.2}{1.8}\times10^{-4-2}=4\times10^{-6}=\boxed{4.0\times10^{-6}}.$$

(b)

Move the decimal point six places left: $$4.0\times10^{-6}=\boxed{0.0000040}.$$

QUESTION 5 6 marks Criterion C

Hard

Add and subtract quantities written with different powers of ten.

a. $(4.75\times10^6)+(8.3\times10^5)$

[3]

b. $(6.2\times10^5)-(7.5\times10^4)$

[3]

Show complete worked solution

(a)

Rewrite $8.3\times10^5$ as $0.83\times10^6$: $$(4.75+0.83)\times10^6=\boxed{5.58\times10^6}.$$

(b)

Rewrite $7.5\times10^4$ as $0.75\times10^5$: $$(6.2-0.75)\times10^5=\boxed{5.45\times10^5}.$$

QUESTION 6 7 marks Criterion D

Hard

A spacecraft is 384,000 km from Earth and has already travelled 216,000 km toward Earth.

a. Write both distances in scientific notation and find the remaining distance.

[4]

b. At a constant speed of $8.0\times10^3$ km/h, find the remaining travel time.

[3]

Show complete worked solution

(a)

$384,000=3.84\times10^5$ and $216,000=2.16\times10^5$. Remaining distance: $$(3.84-2.16)\times10^5=\boxed{1.68\times10^5\text{ km}}.$$

(b)

$$t=\frac{1.68\times10^5}{8.0\times10^3}=\frac{1.68}{8.0}\times10^2=0.21\times10^2=\boxed{21\text{ h}}.$$

QUESTION 7 6 marks Criterion A

Hard

Solve $(2.4\times10^{-5})x=7.2\times10^3$.

a. Find $x$ in scientific notation.

[4]

b. Find $x\div(6.0\times10^2)$.

[2]

Show complete worked solution

(a)

$$x=\frac{7.2\times10^3}{2.4\times10^{-5}}=\frac{7.2}{2.4}\times10^{3-(-5)}=\boxed{3.0\times10^8}.$$

(b)

$$\frac{3.0\times10^8}{6.0\times10^2}=0.5\times10^6=\boxed{5.0\times10^5}.$$

QUESTION 8 7 marks Criterion D

Hard

A microscopic cube has side length $3.0\times10^{-4}$ m.

a. Find the volume of one cube in scientific notation.

[4]

b. Find the total volume of $5.0\times10^6$ identical cubes.

[3]

Show complete worked solution

(a)

$$V=(3.0\times10^{-4})^3=3.0^3\times10^{-12}=27\times10^{-12}=\boxed{2.7\times10^{-11}\text{ m}^3}.$$

(b)

$$(2.7\times10^{-11})(5.0\times10^6)=13.5\times10^{-5}=1.35\times10^{-4}\text{ m}^3.$$ The inputs have two significant figures, so $$\boxed{1.4\times10^{-4}\text{ m}^3}.$$

QUESTION 9 6 marks Criterion B

Hard

Consider $6.2\times10^{-5}$, $8.1\times10^{-6}$, $4.9\times10^{-4}$ and $1.2\times10^{-5}$.

a. Arrange the numbers in ascending order.

[3]

b. How many times larger is the greatest number than the least? Give the answer to 3 significant figures.

[3]

Show complete worked solution

(a)

Compare their exponents first, then their coefficients: $$\boxed{8.1\times10^{-6}<1.2\times10^{-5}<6.2\times10^{-5}<4.9\times10^{-4}}.$$

(b)

$$\frac{4.9\times10^{-4}}{8.1\times10^{-6}}=\frac{4.9}{8.1}\times10^2=60.4938\ldots\approx\boxed{60.5}.$$

QUESTION 10 6 marks Criterion D

Hard

A telescope records data at $2.4\times10^8$ bytes per minute for $3.5\times10^2$ minutes.

a. Find the total number of bytes recorded.

[3]

b. One storage unit holds $1.5\times10^9$ bytes. Find the number of complete units required.

[3]

Show complete worked solution

(a)

$$(2.4\times10^8)(3.5\times10^2)=8.4\times10^{10}.$$ The telescope records $\boxed{8.4\times10^{10}\text{ bytes}}$.

(b)

$$\frac{8.4\times10^{10}}{1.5\times10^9}=\frac{8.4}{1.5}\times10=5.6\times10=\boxed{56}.$$ Therefore 56 storage units are required.

Natural numbers 13 questions

Integers 14 questions

Index notation 13 questions

Order of operations 11 questions

Absolute value 13 questions

Square numbers and square roots 13 questions

Primes and composites 11 questions

Highest common factor 11 questions

Lowest common multiple 11 questions

Rational and Irrational Numbers 10 questions

Recurring Decimals Conversion to Fractions 10 questions

Significant Figures 10 questions

Estimation 10 questions

Scientific Notation 10 questions