Chapter 9 — Two Full Practice Papers — ACT

right	scale	right	scale	right	scale
45	36	30	28	15	17
43	35	28	27	13	16
41	34	26	25	11	15
39	33	24	23	9	14
37	32	22	21	7	13
35	31	20	20	5	12
33	30	18	18	3	9
31	29	16	17	1	5

Q1Basic

Topic: Reading a pie chart

A regional council surveyed 240 households about what they do with their glass waste. The pie chart shows the results.

How many of the households separate their glass from other waste?

A) $210$

B) $168$

C) $80$

D) $72$

Show the worked solution

Answer: D

Explanation

The chart shows that 30% of the households separate their glass:

$240 \times 0.30 = 72$

Why each wrong option is wrong:

A, $210$ — subtracted 30 from 240, treating the 30% as 30 households.
B, $168$ — the households that do not separate glass, the other 70%.
C, $80$ — took a third of 240. Thirty percent is a little less than a third.

Takeaway: "of" means multiply. Check whether the question wants the group described or the rest.

Q2Basic

Topic: Solving a two-step linear equation

A taxi charges a flat fee of $5 plus $3 for each kilometre travelled. A passenger pays $20 for one journey. How many kilometres long was the journey?

F) $15$

G) $\frac{25}{3}$

H) $\frac{20}{3}$

J) $5$

Show the worked solution

Answer: J

Explanation

With $x$ kilometres the fare is $3x + 5$ dollars, so $3x + 5 = 20\mathclose{.}$ Undo the operations in the reverse of the order they were applied: the $+5$ first, then the multiplication by 3.

Subtract 5, then divide by 3:

$3x = 15 \qquad x = 5$

Why each wrong option is wrong:

F, $15$ — subtracted the 5 and stopped. 15 is $3x\mathclose{,}$ not $x\mathclose{.}$
G, $\frac{25}{3}$ — added the 5 instead of subtracting it, reaching $3x = 25\mathclose{.}$
H, $\frac{20}{3}$ — divided 20 by 3 without removing the 5 first.

Takeaway: undo in reverse order — the addition first, then the multiplication.

Q3Basic

Topic: A unit rate

A machine in a bottling plant fills 45 bottles every 3 minutes, working at a steady rate. At that rate, how many bottles does the machine fill in 8 minutes?

A) $360$

B) $120$

C) $50$

D) $15$

Show the worked solution

Answer: B

Explanation

Find what happens in one unit of time first, then scale up. Going straight from 3 minutes to 8 tempts you into multiplying by 8 directly.

$45 \div 3 = 15$ bottles per minute, so $15 \times 8 = 120\mathclose{.}$

Why each wrong option is wrong:

A, $360$ — multiplied the 45 by 8, as though 45 bottles were filled every minute.
C, $50$ — added: 5 more minutes, so 5 more bottles. A rate multiplies; it does not add.
D, $15$ — the rate for one minute. The question asks about eight.

Takeaway: find the rate for one unit first, then multiply.

Q4Basic

Topic: Angles in a triangle

A surveyor measures two interior angles of a triangular plot of land. The plot is shown, with the two measured angles marked.

What is the value of $x\mathclose{?}$

F) $130$

G) $85$

H) $60$

J) $50$

Show the worked solution

Answer: J

Explanation

The three angles of any triangle total $180°\mathclose{,}$ so the third is whatever is left after the other two.

$180 - 35 - 95 = 50°$

Check: $35 + 95 + 50 = 180\mathclose{.}$

Why each wrong option is wrong:

F, $130$ — added the two given angles and stopped.
G, $85$ — subtracted only the $95°$ from 180 and forgot the $35°\mathclose{.}$
H, $60$ — the difference between the two given angles.

Takeaway: subtract both known angles from $180°\mathclose{,}$ then add all three back.

Q5Basic

Topic: Expanding a bracket

Which expression is equivalent to $4(2x - 5)\mathclose{?}$

A) $8 x - 5$

B) $8 x - 20$

C) $6 x - 20$

D) $2 x - 20$

Show the worked solution

Answer: B

Explanation

The number outside the bracket multiplies both terms inside it, signs included.

$4 \times 2x = 8x \qquad 4 \times (-5) = -20$

Why each wrong option is wrong:

A, $8 x - 5$ — multiplied the $2x$ by 4 and left the $-5$ alone.
C, $6 x - 20$ — added the 4 to the 2 instead of multiplying.
D, $2 x - 20$ — multiplied the $-5$ by 4 and left the $2x$ alone.

Takeaway: the number outside multiplies both terms inside.

Q6Basic

Topic: An exponent rule

Which expression is equivalent to $x^5 \cdot x^3\mathclose{?}$

F) $x^{15}$

G) $x^{8}$

H) $x^{2}$

J) $2 x^{8}$

Show the worked solution

Answer: G

Explanation

Multiplying powers of the same base adds the exponents — you are just counting how many copies of $x$ are being multiplied together.

$x^5 \cdot x^3 = x^{5 + 3} = x^8$

Why each wrong option is wrong:

F, $x^{15}$ — multiplied the exponents. That is the rule for a power of a power, $(x^5)^3\mathclose{.}$
H, $x^{2}$ — subtracted the exponents, which is the rule for dividing.
J, $2 x^{8}$ — counted the two $x$ 's as a 2 in front. That is what adding $x^8 + x^8$ gives.

Takeaway: multiply → add exponents; divide → subtract; power of a power → multiply.

Q7Basic

Topic: Evaluating a function

The function $f$ is defined by $f(x) = x^2 - 3x\mathclose{.}$ What is the value of $f(5)\mathclose{?}$

A) $-10$

B) $10$

C) $25$

D) $40$

Show the worked solution

Answer: B

Explanation

Substitute 5 for every $x\mathclose{,}$ taking the power first:

$f(5) = 5^2 - 3(5) = 25 - 15 = 10$

Why each wrong option is wrong:

A, $-10$ — subtracted the wrong way round, $15 - 25\mathclose{.}$
C, $25$ — worked out $5^2$ and forgot the $-3x$ term.
D, $40$ — added the 15 instead of subtracting it.

Takeaway: apply the power before the subtraction.

Q8Basic

Topic: A part-to-whole ratio

Two friends, Ana and Ben, share a bag of 45 sweets in the ratio $4 : 5\mathclose{,}$ with Ben receiving the larger share. How many sweets does Ben receive?

F) $5$

G) $9$

H) $20$

J) $25$

Show the worked solution

Answer: J

Explanation

$4 + 5 = 9$ parts, so one part is $45 \div 9 = 5\mathclose{.}$ The larger share is $5 \times 5 = 25\mathclose{.}$

Why each wrong option is wrong:

F, $5$ — the value of one part.
G, $9$ — the number of parts, $4 + 5\mathclose{.}$
H, $20$ — the smaller share, 4 parts.

Takeaway: add the parts first, then multiply by the part asked for.

Q9Basic

Topic: Reading a linear function

A delivery service charges each customer a fixed handling fee and, in addition, a fixed amount for each kilometre that a parcel travels. The total charge, in dollars, for a parcel that travels $x$ kilometres is given by $f(x) = 4x + 7\mathclose{.}$ What is the value of $f(0)\mathclose{,}$ the handling fee in dollars?

A) $11$

B) $7$

C) $4$

D) $0$

Show the worked solution

Answer: B

Explanation

$f(0) = 4(0) + 7 = 7$

It is the handling fee — what you pay before travelling anywhere.

Why each wrong option is wrong:

A, $11$ — added the 4 and the 7. That is $f(1)\mathclose{.}$
C, $4$ — the rate per kilometre, the number that multiplies $x\mathclose{.}$
D, $0$ — assumed an input of 0 always gives an output of 0.

Takeaway: the constant term is the value at zero: the fixed part of a linear model.

Q10Basic

Topic: The mean of a small data set

A hiker recorded the distance walked, in kilometres, on each of the five days of a walking holiday: $8, \; 11, \; 6, \; 14$ and $11\mathclose{.}$ What is the mean distance walked per day, in kilometres?

F) $50$

G) $12.5$

H) $11$

J) $10$

Show the worked solution

Answer: J

Explanation

The mean is the total shared equally, so add everything and divide by how many there are.

$\frac{8 + 11 + 6 + 14 + 11}{5} = \frac{50}{5} = 10$

Why each wrong option is wrong:

F, $50$ — the total, never divided by 5.
G, $12.5$ — divided the total by 4. There are five days: the 11 appears twice and counts twice.
H, $11$ — the mode, the value that appears twice.

Takeaway: mean is total ÷ count. The mode (11) and the range (8) are different things.

Q11Medium

Topic: A system solved by substitution

A carpenter cuts short planks and long planks. Each long plank is 1 metre shorter than twice the length of a short plank. Laid end to end, 3 short planks and 1 long plank measure 14 metres altogether. What is the length, in metres, of one short plank?

A) $5$

B) $3$

C) $\frac{14}{5}$

D) $\frac{13}{5}$

Show the worked solution

Answer: B

Explanation

Let $x$ be the length of a short plank and $y$ the length of a long plank, in metres. Each sentence gives one equation:

$y = 2x - 1 \qquad\qquad 3x + y = 14$

The first already gives $y$ alone, so substitute it into the second:

$3x + (2x - 1) = 14 \quad\Longrightarrow\quad 5x - 1 = 14 \quad\Longrightarrow\quad 5x = 15 \quad\Longrightarrow\quad x = 3$

Check: a long plank is $2(3) - 1 = 5$ metres, and $3(3) + 5 = 14$ ✓

Why each wrong option is wrong:

A, $5$ — the length of a long plank. The question asks for a short one.
C, $\frac{14}{5}$ — left out the "1 metre shorter" when substituting, solving $5x = 14\mathclose{.}$
D, $\frac{13}{5}$ — moved the $-1$ across without changing its sign, solving $5x = 13\mathclose{.}$

Takeaway: when one equation already gives a variable alone, substitute. Then check which of the two things was asked for.

Q12Medium

Topic: A percentage increase

A cycle shop sells a touring bicycle for $250. At the start of the summer the shop raises the price of every bicycle it sells by 12%. A customer buys the touring bicycle after the increase. How much does the customer pay for it, in dollars?

F) $30$

G) $220$

H) $262$

J) $280$

Show the worked solution

Answer: J

Explanation

A 12% rise multiplies by $1.12$ — the original whole, plus another 12% of it:

$250 \times 1.12 = 280$

Why each wrong option is wrong:

F, $30$ — the increase on its own. The question asks for the new price.
G, $220$ — took 12% off instead of adding it.
H, $262$ — added 12 dollars. A percentage is a proportion of the price, not a fixed amount.

Takeaway: increase by $p\%$ in one step, multiplying by $1 + \frac{p}{100}\mathclose{.}$

Q13Medium

Topic: A composite function

The functions $f$ and $g$ are defined by $f(x) = x + 3$ and $g(x) = 2x\mathclose{.}$ What is the value of $g(f(4))\mathclose{?}$

A) $7$

B) $8$

C) $11$

D) $14$

Show the worked solution

Answer: D

Explanation

Work outwards from the inner bracket. $f(4) = 4 + 3 = 7\mathclose{,}$ and that 7 is then the input to $g\mathclose{,}$ giving $g(7) = 14\mathclose{.}$

Why each wrong option is wrong:

A, $7$ — found $f(4)$ and stopped.
B, $8$ — found $g(4)$ and ignored $f\mathclose{.}$
C, $11$ — applied $g$ first: $g(4) = 8\mathclose{,}$ then $f(8) = 11\mathclose{.}$ That is $f(g(4))\mathclose{.}$

Takeaway: work from the inside bracket outwards.

Q14Medium

Topic: The area of a circle

A landscaper is building a circular pond with a radius of 7 metres and needs a liner that covers the whole of its flat base. What is the area of the base, in square metres?

F) $7 \pi$

G) $14 \pi$

H) $49 \pi$

J) $196 \pi$

Show the worked solution

Answer: H

Explanation

Area uses the square of the radius; circumference does not. Check which the question wants before choosing the formula.

$A = \pi r^2 = 49\pi$

Why each wrong option is wrong:

F, $7 \pi$ — forgot to square the radius.
G, $14 \pi$ — the circumference, $2\pi r\mathclose{.}$
J, $196 \pi$ — used 14, the diameter, as the radius.

Takeaway: area squares the radius; circumference does not.

Q15Medium

Topic: Scientific notation

Which of the following is $0.00036$ written in scientific notation?

A) $3.6 \times 10^{-3}$

B) $36 \times 10^{-5}$

C) $3.6 \times 10^{-4}$

D) $3.6 \times 10^{4}$

Show the worked solution

Answer: C

Explanation

Move the decimal point until exactly one non-zero digit stands in front of it, then count how many places it moved. A number below 1 gives a negative power.

The point moves 4 places right to give 3.6, so the power is $-4\mathclose{.}$

Why each wrong option is wrong:

A, $3.6 \times 10^{-3}$ — one place too few. Expanded, this is 0.0036.
B, $36 \times 10^{-5}$ — the right size, but not scientific notation: the leading number must be below 10.
D, $3.6 \times 10^{4}$ — a positive power, which gives 36 000. A number below 1 needs a negative power.

Takeaway: a number below 1 gets a negative power. Expand your answer to check.

Q16Medium

Topic: Probability from a set of outcomes

For a classroom game, a teacher writes each of the integers from 1 to 30 on a separate slip of paper and puts all 30 slips into a bag. A student draws one slip from the bag at random. What is the probability that the number on the slip is a multiple of 6?

F) $\frac{1}{3}$

G) $\frac{1}{5}$

H) $\frac{1}{6}$

J) $\frac{1}{30}$

Show the worked solution

Answer: H

Explanation

The multiples of 6 up to 30 are $6, 12, 18, 24, 30$ — five of them.

$P = \frac{5}{30} = \frac{1}{6}$

Why each wrong option is wrong:

F, $\frac{1}{3}$ — counted the multiples of 3, of which there are ten.
G, $\frac{1}{5}$ — counted six multiples, perhaps including 0 or 36. There are five.
J, $\frac{1}{30}$ — the chance of drawing one particular slip.

Takeaway: count the outcomes that work, then divide by how many there are altogether.

Q17Medium

Topic: A linear inequality

What is the least integer value of $x$ that satisfies $9 - 2x < 4x - 15\mathclose{?}$

A) $6$

B) $5$

C) $4$

D) $-4$

Show the worked solution

Answer: B

Explanation

Add $2x$ to both sides and add 15:

$24 < 6x \quad\Longrightarrow\quad 4 < x$

Strictly greater than 4, so the least integer is 5.

Why each wrong option is wrong:

A, $6$ — one integer too far. 5 already satisfies it.
C, $4$ — the boundary itself. The inequality is strict, so 4 does not satisfy it.
D, $-4$ — gathered the terms the wrong way round and came out with the sign reversed.

Takeaway: a strict inequality excludes its boundary. Gather $x$ on the side that keeps it positive and no sign flip is needed.

Q18Medium

Topic: Converting a rate between units

A cyclist in a road race rides along one section of the course at a constant speed of 15 miles per hour. Given that 1 mile is 5280 feet and that 1 hour is 3600 seconds, what is the cyclist's speed along that section, in feet per second?

F) $79200$

G) $1320$

H) $22$

J) $0.25$

Show the worked solution

Answer: H

Explanation

Write both conversions as fractions and let the units cancel:

$15 \; \frac{\text{miles}}{\text{hour}} \times \frac{5280 \text{ feet}}{1 \text{ mile}} \times \frac{1 \text{ hour}}{3600 \text{ seconds}} = \frac{15 \times 5280}{3600} = 22$

Why each wrong option is wrong:

F, $79200$ — changed the miles into feet and left the hours alone. That is feet per hour.
G, $1320$ — converted correctly to feet per minute and stopped.
J, $0.25$ — divided by 60 and nothing else. That is miles per minute.

Takeaway: write both conversions as fractions and let the units cancel.

Q19Medium

Topic: Pythagoras

A rectangular gate is braced by a diagonal strut, so that the strut and two edges of the gate form the right triangle shown, where the lengths are in metres.

What is the value of $x\mathclose{?}$

A) $\sqrt{21}$

B) $3 \sqrt{7}$

C) $15$

D) $21$

Show the worked solution

Answer: C

Explanation

The strut runs corner to corner, so it is the hypotenuse of a right triangle whose legs are the two sides of the gate.

$\sqrt{81 + 144} = \sqrt{225} = 15$

A 9-12-15 triangle — three times 3-4-5.

Why each wrong option is wrong:

A, $\sqrt{21}$ — added the sides and then took a square root, without squaring them first.
B, $3 \sqrt{7}$ — subtracted the squares, $\sqrt{144 - 81} = \sqrt{63}\mathclose{.}$ That is for finding a leg when the hypotenuse is known.
D, $21$ — added the two sides. That is the distance along two edges, not across.

Takeaway: square, add, then root. Recognising the triple saves the arithmetic.

Q20Medium

Topic: A function transformation

The graph of $y = f(x)$ passes through the point $(3, 8)\mathclose{.}$ Through which point must the graph of $y = f(x) - 5$ pass?

F) $\left( -2, \ 8\right)$

G) $\left( 8, \ 3\right)$

H) $\left( 3, \ 3\right)$

J) $\left( 3, \ 13\right)$

Show the worked solution

Answer: H

Explanation

The $-5$ is outside the brackets, so it lowers every output by 5. The $x$ -coordinate is unchanged.

Why each wrong option is wrong:

F, $\left( -2, \ 8\right)$ — moved the point sideways. A change outside the brackets never does that.
G, $\left( 8, \ 3\right)$ — swapped the two coordinates.
J, $\left( 3, \ 13\right)$ — moved the point up 5 instead of down.

Takeaway: outside the brackets moves the graph vertically, in the direction the sign says.

Q21Medium

Topic: An arithmetic sequence

A shop builds a display of cans in rows. The top row holds 4 cans, and each row below it holds 9 more cans than the row above. How many cans are in the 10th row from the top?

A) $40$

B) $81$

C) $85$

D) $94$

Show the worked solution

Answer: C

Explanation

The rows hold $4, 13, 22, 31, \dots$ cans: an arithmetic sequence with first term 4 and common difference 9. The 10th row is nine steps after the first:

$4 + 9 \times 9 = 85$

Why each wrong option is wrong:

A, $40$ — multiplied the first row's 4 cans by 10.
B, $81$ — worked out the nine steps, 81, and forgot to add the cans in the top row.
D, $94$ — used $a + nd\mathclose{:}$ ten steps where there are only nine.

Takeaway: $a + (n-1)d$ — nine steps to reach the tenth term.

Q22Medium

Topic: A missing value from a mean

Dana's mean score on the four quizzes of a term is 21. Her scores on the first three quizzes were 18, 24 and 19. What was her score on the fourth quiz?

F) $21$

G) $23$

H) $61$

J) $84$

Show the worked solution

Answer: G

Explanation

The four total $4 \times 21 = 84\mathclose{;}$ the three given total 61; so the fourth is 23.

Why each wrong option is wrong:

F, $21$ — assumed the missing number equals the mean.
H, $61$ — the total of the three given numbers.
J, $84$ — the total of all four.

Takeaway: mean × count = total. Get the total first.

Q23Medium

Topic: Solving a quadratic by factoring

A rectangular rug is 4 feet longer than it is wide, and its area is 21 square feet. A border strip is to be sewn all the way round the edge of the rug. What length of border strip, in feet, is needed?

A) $10$

B) $20$

C) $21$

D) $28$

Show the worked solution

Answer: B

Explanation

Let the rug be $x$ feet wide, so that it is $x + 4$ feet long. The area is the product of the two:

$x(x + 4) = 21 \quad\Longrightarrow\quad x^2 + 4x - 21 = 0$

Two numbers multiplying to $-21$ and adding to $+4\mathclose{:}$ $+7$ and $-3\mathclose{.}$

$(x + 7)(x - 3) = 0 \quad\Longrightarrow\quad x = -7 \text{ or } x = 3$

A width cannot be negative, so the rug is 3 feet wide and $3 + 4 = 7$ feet long. The border goes all the way round:

$2(3 + 7) = 20$

Why each wrong option is wrong:

A, $10$ — one width plus one length, $3 + 7\mathclose{.}$ That is only half-way round.
C, $21$ — the area. It measures the floor the rug covers, in square feet, not its edge.
D, $28$ — four times the longer side, as if the rug were a square 7 feet each way.

Takeaway: build the equation from the area, factor it, and keep the root that can be a length. Then give what the question asks for — here the distance round, not a side.

Q24Medium

Topic: The volume of a cylinder

A cylindrical water tank on a farm has the inside radius and the inside height shown, in metres. A farmer fills the tank to the top with water.

What is the volume, in cubic metres, of the water in the tank when it is full?

F) $16 \pi$

G) $20 \pi$

H) $40 \pi$

J) $80 \pi$

Show the worked solution

Answer: J

Explanation

A cylinder's volume is the area of its circular base multiplied by its height, so only the radius gets squared.

$V = \pi r^2 h = \pi \times 16 \times 5 = 80\pi$

Why each wrong option is wrong:

F, $16 \pi$ — the area of the base, $\pi r^2\mathclose{,}$ never multiplied by the height.
G, $20 \pi$ — forgot to square the radius.
H, $40 \pi$ — used $2\pi r h\mathclose{,}$ the area of the curved side, not a volume.

Takeaway: only the radius is squared.

Q25Medium

Topic: Conditional probability from a two-way table

A survey of 200 people recorded whether they cycle and whether they own a car.

	Owns a car	No car	Total
Cycles	30	50	80
Does not cycle	90	30	120
Total	120	80	200

One person is chosen at random from those who cycle. What is the probability that they own a car?

A) $\frac{3}{20}$

B) $\frac{1}{4}$

C) $\frac{3}{8}$

D) $\frac{2}{5}$

Show the worked solution

Answer: C

Explanation

Restricted to cyclists, so the denominator is 80:

$\frac{30}{80} = \frac{3}{8}$

Why each wrong option is wrong:

A, $\frac{3}{20}$ — used the grand total, 200. That is the chance that a random person both cycles and owns a car.
B, $\frac{1}{4}$ — used 120, the car owners. That answers the reverse question.
D, $\frac{2}{5}$ — the chance of being a cyclist at all.

Takeaway: the group named after "from those who" is the denominator.

Q26Medium

Topic: A proportion

A map is drawn so that 3 centimetres on the map represent 8 kilometres on the ground. Two towns are 24 kilometres apart. How many centimetres apart are the two towns on the map?

F) $72$

G) $64$

H) $19$

J) $9$

Show the worked solution

Answer: J

Explanation

Set the two ratios equal with the same quantity on top of both, then scale.

$\frac{3 \text{ cm}}{8 \text{ km}} = \frac{c}{24 \text{ km}}$

24 kilometres is three times 8, so the map length is three times 3: 9 centimetres. More distance must mean more map, which rules out any smaller answer.

Why each wrong option is wrong:

F, $72$ — multiplied 3 by 24 and never divided by 8.
G, $64$ — set the proportion up upside down, $\frac{8}{3} = \frac{c}{24}\mathclose{.}$
H, $19$ — added: 16 more kilometres, so 16 more centimetres. A scale multiplies.

Takeaway: keep the same quantity on top of both fractions, and check the direction — more distance means more map.

Q27Medium

Topic: The domain of a function

For which value of $x$ is $f(x) = \dfrac{x + 2}{x - 7}$ undefined?

A) $-2$

B) $0$

C) $2$

D) $7$

Show the worked solution

Answer: D

Explanation

A fraction has no value only when its denominator is zero:

$x - 7 = 0 \quad\Longrightarrow\quad x = 7$

Setting the numerator to zero answers a different question — where the graph crosses the $x$ -axis.

Why each wrong option is wrong:

A, $-2$ — set the numerator to zero. There the function is 0, a perfectly good value.
B, $0$ — assumed zero is always excluded. $f(0) = -\frac{2}{7}\mathclose{,}$ which is fine.
C, $2$ — took the 2 from the numerator.

Takeaway: the bottom decides the domain; the top decides the zeros.

Q28Medium

Topic: The slope of a line through two points

A line is graphed in the standard $(x, y)$ coordinate plane, as shown, and passes through the two points marked on it.

What is the slope of the line?

F) $-2$

G) $- \frac{1}{2}$

H) $\frac{4}{5}$

J) $2$

Show the worked solution

Answer: F

Explanation

Read the two points off the graph: $(-2, 7)$ and $(3, -3)\mathclose{.}$

$m = \frac{-3 - 7}{3 - (-2)} = \frac{-10}{5} = -2$

The line falls, so a negative slope is right.

Why each wrong option is wrong:

G, $- \frac{1}{2}$ — the fraction upside down: run over rise.
H, $\frac{4}{5}$ — added the coordinates where they should be subtracted.
J, $2$ — subtracted the $y$ -values in one order and the $x$ -values in the other, which flips the sign.

Takeaway: subtract in the same order top and bottom, and watch the double negative.

Q29Medium

Topic: Rearranging a formula

The formula $A = \dfrac{1}{2}h(a + b)$ gives the area $A$ of a trapezoid whose parallel sides have lengths $a$ and $b$ and whose height is $h\mathclose{.}$ Which of the following expresses $b$ in terms of $A\mathclose{,}$ $h$ and $a\mathclose{?}$

A) $\frac{2 A}{h} - a$

B) $\frac{2 A - a}{h}$

C) $\frac{A}{2 h} - a$

D) $2 A h - a$

Show the worked solution

Answer: A

Explanation

Undo what was done to $b\mathclose{,}$ in reverse order. Multiply both sides by 2, divide by $h\mathclose{,}$ then subtract $a\mathclose{:}$

$2A = h(a + b) \quad\Longrightarrow\quad \frac{2A}{h} = a + b \quad\Longrightarrow\quad b = \frac{2A}{h} - a$

Test it with numbers: $a = 3\mathclose{,}$ $b = 5\mathclose{,}$ $h = 4$ gives $A = \frac12(4)(8) = 16\mathclose{,}$ and $\frac{2(16)}{4} - 3 = 5\mathclose{.}$ Correct.

Why each wrong option is wrong:

B, $\frac{2 A - a}{h}$ — subtracted the $a$ before dividing by $h\mathclose{.}$ The $a$ is inside the bracket, so the division has to come first.
C, $\frac{A}{2 h} - a$ — divided by 2 where the formula's half has to be undone by multiplying by 2.
D, $2 A h - a$ — multiplied by $h$ instead of dividing.

Takeaway: isolate the term, then divide by everything multiplying it. Test with small numbers.

Q30Medium

Topic: The median from a bar chart

The bar chart shows the number of parcels that a courier delivered on each of the six working days of one week.

What is the median number of parcels delivered per day?

F) $8.5$

G) $10$

H) $14$

J) $17$

Show the worked solution

Answer: F

Explanation

Sort the values first — this is the step people skip. With an even count there is no single middle value, so average the middle two.

Read the six heights off the chart: 3, 14, 8, 20, 6 and 9.

Sorted: $3, 6, 8, 9, 14, 20\mathclose{.}$ The middle two are 8 and 9, so the median is 8.5.

Why each wrong option is wrong:

G, $10$ — the mean, $60 \div 6\mathclose{.}$
H, $14$ — averaged the middle two of the list as written, 8 and 20, without sorting.
J, $17$ — the range, $20 - 3\mathclose{.}$

Takeaway: sort first. With an even count, average the middle two.

Q31Hard

Topic: An equation with no solution

In the equation $6x + 5 = kx - 4\mathclose{,}$ $k$ is a constant. The equation has no solution. What is the value of $k\mathclose{?}$

A) $-6$

B) $-4$

C) $5$

D) $6$

Show the worked solution

Answer: D

Explanation

No solution means the $x$ terms cancel and the constants disagree, so $k = 6$ and the equation becomes $5 = -4\mathclose{,}$ which is false.

Why each wrong option is wrong:

A, $-6$ — the right size with the wrong sign. With $k = -6$ the equation has one solution.
B, $-4$ — copied the other constant.
C, $5$ — matched a constant instead of the coefficient of $x\mathclose{.}$

Takeaway: no solution needs matching coefficients and differing constants; infinitely many needs both to match.

Q32Hard

Topic: Similar triangles

At the same moment on a sunny afternoon, a vertical post that is 3 metres tall casts a shadow 2 metres long on level ground, and a nearby tree casts a shadow 12 metres long on the same ground. How tall is the tree, in metres?

F) $6$

G) $8$

H) $13$

J) $18$

Show the worked solution

Answer: J

Explanation

The two triangles are similar, so height over shadow is the same for both:

$\frac{h}{12} = \frac{3}{2} \quad\Longrightarrow\quad h = 12 \times \frac{3}{2} = 18$

The post is taller than its shadow, so the tree must be taller than its 12-metre shadow. 18 is.

Why each wrong option is wrong:

F, $6$ — the scale factor: the tree's shadow is 6 times the post's. That 6 still has to multiply the post's 3 metres.
G, $8$ — set the ratio up upside down, which makes the tree shorter than its own shadow.
H, $13$ — added: the post is 1 metre taller than its shadow, so the tree was made 1 metre taller than its own. Similar triangles scale by a ratio.

Takeaway: put the same measurement on top of both fractions, then sanity-check the size.

Q33Hard

Topic: An exponential model

A biologist starts a culture with 60 cells and observes that the number of cells in the culture doubles every 3 hours. If the culture goes on growing in the same way, how many cells will it contain 12 hours after the biologist started it?

A) $240$

B) $480$

C) $720$

D) $960$

Show the worked solution

Answer: D

Explanation

Count the doubling periods, not the hours: 12 hours at one doubling every 3 hours is four doublings, not twelve.

$12 \div 3 = 4$ doublings, so $60 \times 2^4 = 960\mathclose{.}$

Why each wrong option is wrong:

A, $240$ — multiplied by 4, the number of doublings. Four doublings multiply by $2^4 = 16\mathclose{.}$
B, $480$ — counted three doublings. There are four: at 3, 6, 9 and 12 hours.
C, $720$ — multiplied the 60 by the 12 hours, as though the growth were steady.

Takeaway: divide the time by the period to count the doublings.

Q34Hard

Topic: Complex numbers

Given that $i^2 = -1\mathclose{,}$ what is the value of $\sqrt{-25} + \sqrt{-144}\mathclose{?}$

F) $13 i$

G) $-17$

H) $17$

J) $17 i$

Show the worked solution

Answer: J

Explanation

Rewrite each square root of a negative as a multiple of $i$ before combining them. The rule $\sqrt{a}\sqrt{b} = \sqrt{ab}$ does not survive a negative under the sign.

$\sqrt{-25} = 5i \qquad \sqrt{-144} = 12i \qquad 5i + 12i = 17i$

Why each wrong option is wrong:

F, $13 i$ — combined the two under one root, $\sqrt{-169}\mathclose{.}$ Roots add like terms; what is under them does not.
G, $-17$ — dropped the $i$ and kept a minus sign.
H, $17$ — ignored the negatives and added $5 + 12\mathclose{.}$

Takeaway: convert each root to a multiple of $i$ before combining. The rule $\sqrt{a}\sqrt{b} = \sqrt{ab}$ fails for negatives.

Q35Hard

Topic: Interpreting standard deviation

Two classes of 25 students each sat the same test, and both averaged 70. Their results are summarised below.

Class	Mean	Lowest	Highest
A	70	68	72
B	70	45	95

Which statement is true?

A) Class B has the greater standard deviation

B) Class A has the greater standard deviation

C) The standard deviations are equal, because the means are equal

D) Standard deviation cannot be judged from a range

Show the worked solution

Answer: A

Explanation

No score in class A is more than 2 marks from the mean, so its standard deviation cannot be more than 2.

In class B the lowest and the highest scores are each 25 marks from the mean. In a class of 25, those two scores alone make the standard deviation more than 7. B is far more spread out.

Why each wrong option is wrong:

B — has the classes the wrong way round.
C — equal means say nothing about how far the scores sit from the mean.
D — too cautious. A range does not give the standard deviation exactly, but here it settles the comparison: at most 2 for class A, more than 7 for class B.

Takeaway: equal means say nothing about spread. That is what standard deviation measures.

Q36Hard

Topic: Two successive percentage changes

The price of a share rises by 50% during one year and then falls by 40% during the next year. Over the two years together, the price of the share has:

F) fallen by 10%

G) risen by 10%

H) fallen by 90%

J) risen by 90%

Show the worked solution

Answer: F

Explanation

Percentage changes multiply rather than add, so take a convenient starting value of 100 and apply each multiplier in turn.

$100 \times 1.50 = 150 \qquad 150 \times 0.60 = 90$

The price ends at 90% of where it began: a fall of 10%. The 40% was taken from the larger number, 150, which is why it outweighs the 50% rise.

Why each wrong option is wrong:

G — subtracted the percentages, $50 - 40\mathclose{.}$ Successive changes multiply.
H — read the overall multiplier, $0.9\mathclose{,}$ as the amount lost. It is the amount kept.
J — read the $0.9$ as a 90% rise.

Takeaway: successive percentage changes multiply. Start from 100.

Q37Hard

Topic: Trigonometry in context

A kite string 40 metres long is pulled tight and makes an angle of $55°$ with the level ground, as shown.

Which expression gives the height $h$ of the kite above the ground, in metres?

A) $40\cos 55°$

B) $\dfrac{40}{\sin 55°}$

C) $40\sin 55°$

D) $40\tan 55°$

Show the worked solution

Answer: C

Explanation

The string is the hypotenuse and the height is opposite the angle, so use sine:

$\sin 55° = \frac{\text{height}}{40} \quad\Longrightarrow\quad \text{height} = 40\sin 55°$

Why each wrong option is wrong:

A — cosine gives the distance along the ground, the side next to the angle.
B — dividing by the sine gives more than 40 metres. The kite cannot be higher than its string is long.
D — tangent links the height to the ground distance, and the ground distance is not what is given.

Takeaway: pick the ratio linking the side you have to the side you want. A leg is always shorter than the hypotenuse.

Q38Hard

Topic: A radical equation

What is the solution of the equation $\sqrt{x + 12} = x\mathclose{?}$

F) $-3$

G) $1$

H) $3$

J) $4$

Show the worked solution

Answer: J

Explanation

Squaring gives $x + 12 = x^2\mathclose{,}$ so

$x^2 - x - 12 = 0 \quad\Longrightarrow\quad (x - 4)(x + 3) = 0 \quad\Longrightarrow\quad x = 4 \text{ or } x = -3$

Test both in the original. At $x = 4\mathclose{:}$ $\sqrt{16} = 4\mathclose{.}$ ✓ At $x = -3\mathclose{:}$ $\sqrt{9} = 3\mathclose{,}$ and $3 \ne -3\mathclose{,}$ so $-3$ is extraneous.

Why each wrong option is wrong:

F, $-3$ — the extraneous root, kept without testing it.
G, $1$ — added the two roots of the squared equation.
H, $3$ — factored $x^2 - x - 12$ as $(x - 3)(x + 4)\mathclose{,}$ with the two signs the wrong way round.

Takeaway: after squaring, every root is only a candidate. Substitute back.

Q39Hard

Topic: A logarithm

If $\log_2 x = 5\mathclose{,}$ what is the value of $x\mathclose{?}$

A) $2.5$

B) $10$

C) $25$

D) $32$

Show the worked solution

Answer: D

Explanation

A logarithm states which power the base must be raised to, so rewrite it as an ordinary power immediately. $\log_2 x = 5$ means $2^5 = x\mathclose{,}$ so $x = 32\mathclose{.}$

Why each wrong option is wrong:

A, $2.5$ — divided the 5 by the 2.
B, $10$ — multiplied the base by the result.
C, $25$ — computed $5^2\mathclose{,}$ with the base and the exponent swapped.

Takeaway: a logarithm is an exponent. Rewrite it as a power immediately.

Q40Hard

Topic: The law of cosines

In the triangle shown, two of the sides measure 7 and 15, and the angle between those two sides measures $60°\mathclose{.}$

What is the value of $x\mathclose{,}$ the length of the third side?

F) $8$

G) $13$

H) $\sqrt{274}$

J) $22$

Show the worked solution

Answer: G

Explanation

Two sides with the known angle between them is the signal for the law of cosines. Pythagoras would assume a right angle, and this one is $60°\mathclose{.}$

$c^2 = 49 + 225 - 2(7)(15)(0.5) = 274 - 105 = 169 \quad\Longrightarrow\quad c = 13$

Why each wrong option is wrong:

F, $8$ — subtracted the two sides, the other limit.
H, $\sqrt{274}$ — used Pythagoras, which needs a right angle between the two sides.
J, $22$ — added the two sides. That is a limit the third side can approach and never reach.

Takeaway: two sides and the angle between them means the law of cosines.

Q41Hard

Topic: Scalar multiplication of a matrix

What is $3\begin{bmatrix} 2 & -1 \\ 0 & 4 \end{bmatrix}\mathclose{?}$

A) $\left[\begin{matrix}6 & -3\\0 & 12\end{matrix}\right]$

B) $\left[\begin{matrix}6 & -1\\0 & 4\end{matrix}\right]$

C) $\left[\begin{matrix}5 & 2\\3 & 7\end{matrix}\right]$

D) $\left[\begin{matrix}6 & -3\\0 & 4\end{matrix}\right]$

Show the worked solution

Answer: A

Explanation

A scalar multiplies every entry, not just the first:

$3\begin{bmatrix} 2 & -1 \\ 0 & 4 \end{bmatrix} = \begin{bmatrix} 6 & -3 \\ 0 & 12 \end{bmatrix}$

Why each wrong option is wrong:

B, $\left[\begin{matrix}6 & -1\\0 & 4\end{matrix}\right]$ — multiplied only the top-left entry.
C, $\left[\begin{matrix}5 & 2\\3 & 7\end{matrix}\right]$ — added 3 to every entry instead of multiplying.
D, $\left[\begin{matrix}6 & -3\\0 & 4\end{matrix}\right]$ — multiplied the first row and left the second alone.

Takeaway: a scalar reaches all four entries. Count them.

Q42Hard

Topic: Two rates working together

An office has two printers. Working alone, the first printer finishes a large print job in 10 minutes, and the second printer, also working alone, finishes the same job in 15 minutes. If the two printers work on the job together at those rates, how many minutes do they take to finish it?

F) $6$

G) $12.5$

H) $25$

J) $150$

Show the worked solution

Answer: F

Explanation

Times do not add — two printers together must be faster than either alone, so the answer has to be under 10 minutes. What adds is the rates:

$\frac{1}{10} + \frac{1}{15} = \frac{3}{30} + \frac{2}{30} = \frac{1}{6} \text{ of the job per minute}$

so the whole job takes 6 minutes.

Why each wrong option is wrong:

G, $12.5$ — averaged the two times, which is still slower than the faster printer alone.
H, $25$ — added the two times. Two printers cannot be slower than one.
J, $150$ — multiplied the two times.

Takeaway: add the rates, then invert. The answer must be less than the faster worker alone.

Q43Hard

Topic: Counting combinations

A company that makes phone accessories sells a phone case in 5 colours, 3 patterns and 2 sizes. A customer chooses exactly one colour, one pattern and one size, and every combination of the three is available. How many different cases can the customer choose from?

A) $10$

B) $15$

C) $17$

D) $30$

Show the worked solution

Answer: D

Explanation

Each colour can go with each pattern, and each of those pairs with each size, so the three counts multiply:

$5 \times 3 \times 2 = 30$

Why each wrong option is wrong:

A, $10$ — added the three counts.
B, $15$ — multiplied two of the three and forgot the sizes.
C, $17$ — multiplied two and then added the third.

Takeaway: independent choices multiply.

Q44Hard

Topic: The area of a sector

The circle shown has centre $O$ and a radius of 10. The radii $OA$ and $OB$ form an angle of $72°\mathclose{.}$

What is the area of the smaller sector between the two radii?

F) $4 \pi$

G) $20 \pi$

H) $40 \pi$

J) $100 \pi$

Show the worked solution

Answer: G

Explanation

A sector is the same fraction of the circle's area as its angle is of a full turn. Here $\frac{72}{360} = \frac{1}{5}\mathclose{,}$ so

$\frac{1}{5} \times \pi (10)^2 = \frac{1}{5} \times 100\pi = 20\pi$

Why each wrong option is wrong:

F, $4 \pi$ — the arc length, $\frac{1}{5} \times 2\pi(10)\mathclose{.}$ An arc is a length; a sector is an area.
H, $40 \pi$ — used $2\pi r^2$ in place of $\pi r^2\mathclose{.}$
J, $100 \pi$ — the whole circle, with the fraction never applied.

Takeaway: a sector is $\frac{\theta}{360}$ of the whole. Area uses $r^2\mathclose{;}$ arc length does not.

Q45Hard

Topic: The discriminant

In the equation $x^2 + 10x + c = 0\mathclose{,}$ $c$ is a constant, and the equation has exactly one real solution. What is the value of $c\mathclose{?}$

A) $5$

B) $10$

C) $25$

D) $100$

Show the worked solution

Answer: C

Explanation

One solution means $b^2 - 4ac = 0\mathclose{:}$

$100 - 4c = 0 \quad\Longrightarrow\quad c = 25$

Then $x^2 + 10x + 25 = (x+5)^2\mathclose{,}$ zero only at $x = -5\mathclose{.}$

Why each wrong option is wrong:

A, $5$ — halved the 10 and stopped. $c$ is the square of that half.
B, $10$ — copied the coefficient of $x\mathclose{.}$
D, $100$ — worked out $b^2 = 100$ and never divided by 4.

Takeaway: exactly one solution means the discriminant is zero.

Paper 2

45 questions · 50 minutes · calculator permitted · no formula sheet

Do this one only after you have worked through what Paper 1 showed you.

Q46Basic

Topic: Finding a percentage

A school choir has 40 members. This year 8 of the members are in their final year at the school, and the rest will return next year. What percent of the choir's members are in their final year?

F) $8$

G) $20$

H) $80$

J) $500$

Show the worked solution

Answer: G

Explanation

"What percent of the choir's members" makes the whole choir the whole, so its 40 goes underneath:

$\frac{8}{40} \times 100 = 20\%$

Check the other way: 20% of 40 is 8.

Why each wrong option is wrong:

F, $8$ — repeated the 8. It is 8 members out of 40, not 8 out of 100.
H, $80$ — the percent who will return next year, $\frac{32}{40}\mathclose{.}$ The question asks about the ones who are leaving.
J, $500$ — divided upside down, $\frac{40}{8}\mathclose{.}$ A part cannot be 500% of its whole.

Takeaway: whatever follows "percent of" is the whole, and it goes underneath. Over 100% means you divided upside down.

Q47Basic

Topic: Solving a linear equation with brackets

Two friends each hire a kayak for an afternoon. Each of them pays the same hire charge and also pays $6 to borrow a life jacket. Together the two friends pay $26. What is the hire charge, in dollars, for one kayak?

A) $7$

B) $10$

C) $13$

D) $19$

Show the worked solution

Answer: A

Explanation

Each friend pays the hire charge, $x$ dollars, plus 6 dollars for the life jacket, and there are two friends:

$2(x + 6) = 26$

Either divide by the 2 first or expand the bracket fully. Doing half of each is what produces the wrong answers here.

Divide by 2 first: $x + 6 = 13\mathclose{,}$ so $x = 7\mathclose{.}$

Check: each friend pays $7 + 6 = 13$ dollars, and two of them pay 26 ✓

Why each wrong option is wrong:

B, $10$ — counted only one life jacket, solving $2x + 6 = 26\mathclose{.}$ Both friends borrow one.
C, $13$ — halved the 26 and forgot the life jacket. 13 dollars is everything one friend pays.
D, $19$ — reached $x + 6 = 13$ and added the 6 instead of subtracting it.

Takeaway: a charge that each person pays belongs inside the bracket. Then either divide by the bracket's multiplier first, or expand fully. Do not do half of each.

Q48Basic

Topic: The area of a rectangle in context

A decorator is tiling one rectangular wall of a kitchen. The wall measures 5 metres by 3 metres, and the tiles that the customer has chosen cost $24 per square metre of wall covered. What is the total cost, in dollars, of the tiles needed to cover the whole wall?

F) $15$

G) $39$

H) $120$

J) $360$

Show the worked solution

Answer: J

Explanation

Tiles cover a surface, so this is an area, not a perimeter. Then finish the question — the area is not the answer.

Area is $5 \times 3 = 15$ m², so the cost is $15 \times 24 = 360\mathclose{.}$

Why each wrong option is wrong:

F, $15$ — the area, in square metres. The question asks for the cost.
G, $39$ — added the price to the area instead of multiplying.
H, $120$ — multiplied the 5 metres by the price and left out the 3.

Takeaway: covering a surface is an area. Finish the question — the area is rarely the answer.

Q49Basic

Topic: The mode of a data set

A shoe shop recorded the sizes of the eight pairs of shoes it sold in one morning: $4,\ 7,\ 4,\ 9,\ 4,\ 7,\ 2$ and $5\mathclose{.}$ What is the mode of the sizes sold?

A) $4$

B) $4.5$

C) $5.25$

D) $7$

Show the worked solution

Answer: A

Explanation

The mode is the value that appears most often. Counting: 4 appears three times, 7 twice, and 9, 2 and 5 once each.

$\text{mode} = 4$

Why each wrong option is wrong:

B, $4.5$ — the median, halfway between the two middle values, 4 and 5.
C, $5.25$ — the mean, $42 \div 8\mathclose{.}$
D, $7$ — the second most common value. 7 appears twice, 4 three times.

Takeaway: mode is the most frequent value, not the middle one and not the average. Tally before answering.

Q50Basic

Topic: Simplifying an expression

Which expression is equivalent to $5(2x - 3) - 2(x + 4)\mathclose{?}$

F) $8 x - 7$

G) $12 x - 7$

H) $8 x - 23$

J) $12 x - 23$

Show the worked solution

Answer: H

Explanation

$10x - 15 - 2x - 8 = 8x - 23$

The $-2$ multiplies both terms of the second bracket.

Why each wrong option is wrong:

F, $8 x - 7$ — took the minus to the $x$ only, writing $-2x + 8\mathclose{.}$
G, $12 x - 7$ — added the second bracket instead of subtracting it.
J, $12 x - 23$ — took the minus to the 8 only, writing $+2x - 8\mathclose{.}$

Takeaway: a minus in front of a bracket changes every sign inside it.

Q51Basic

Topic: Probability from equally likely outcomes

In a board game, a player rolls one fair six-sided die on each turn and moves forward only when the number rolled is greater than 4. What is the probability that the player moves forward on a given turn?

A) $\frac{1}{6}$

B) $\frac{1}{3}$

C) $\frac{1}{2}$

D) $\frac{2}{3}$

Show the worked solution

Answer: B

Explanation

"Greater than 4" means 5 or 6 — two outcomes out of six.

$P = \frac{2}{6} = \frac{1}{3}$

Note that 4 itself is not included; "greater than" excludes the number named.

Why each wrong option is wrong:

A, $\frac{1}{6}$ — counted only the 6.
C, $\frac{1}{2}$ — counted 4, 5 and 6. "Greater than 4" does not include 4.
D, $\frac{2}{3}$ — counted 1, 2, 3 and 4: the numbers that are not greater than 4.

Takeaway: count the outcomes that qualify, then divide by the total. Read whether the boundary value counts.

Q52Basic

Topic: Function notation

If $g(x) = \dfrac{x + 6}{2}\mathclose{,}$ what is $g(-4)\mathclose{?}$

F) $-5$

G) $1$

H) $2$

J) $4$

Show the worked solution

Answer: G

Explanation

The whole numerator is divided by 2, not just part of it, so work out the top before dividing.

$g(-4) = \frac{-4 + 6}{2} = \frac{2}{2} = 1$

Why each wrong option is wrong:

F, $-5$ — subtracted the 6 instead of adding it, $\frac{-10}{2}\mathclose{.}$
H, $2$ — worked out the top, 2, and forgot to divide.
J, $4$ — divided only the $-4$ by 2 and then added the 6.

Takeaway: the whole numerator is divided, not just part of it.

Q53Basic

Topic: Converting a rate

A garden tap delivers water at a steady rate of 3 litres per minute. A gardener turns the tap on and leaves it running at that rate for 2 hours in order to fill a pond. How many litres of water does the tap deliver in the 2 hours?

A) $6$

B) $120$

C) $180$

D) $360$

Show the worked solution

Answer: D

Explanation

Convert the time into the unit the rate uses before multiplying. The rate is per minute, so the 2 hours must become minutes.

2 hours is 120 minutes, so $3 \times 120 = 360$ litres.

Why each wrong option is wrong:

A, $6$ — multiplied by the 2 hours without changing them to minutes.
B, $120$ — the number of minutes in 2 hours. The rate was never used.
C, $180$ — one hour's worth.

Takeaway: convert the time to the rate's unit before multiplying.

Q54Basic

Topic: The mode and the range

A weather station recorded the temperature at noon, in degrees Celsius, on each of six days: $5, \; 9, \; 5, \; 12, \; 7$ and $4\mathclose{.}$ What is the range of the six temperatures?

F) $5$

G) $7$

H) $8$

J) $12$

Show the worked solution

Answer: H

Explanation

The range is the largest value minus the smallest, and nothing else:

$12 - 4 = 8$

Here the mode is 5 and the mean is 7 — three different numbers from the same six values.

Why each wrong option is wrong:

F, $5$ — the mode, the value that appears twice.
G, $7$ — the mean, $42 \div 6\mathclose{.}$
J, $12$ — the largest value, with the smallest never subtracted.

Takeaway: range is largest minus smallest. The mode here is 5 and the mean is 7 — three different numbers.

Q55Basic

Topic: A ratio in two parts

A builder makes mortar by mixing cement and sand in the ratio $2 : 7$ by mass. For one job on a garden wall the builder uses 63 kilograms of sand. How many kilograms of cement must the builder mix with that sand to keep the ratio the same?

A) $9$

B) $14$

C) $18$

D) $126$

Show the worked solution

Answer: C

Explanation

Here 63 kg is one part of the ratio, not the whole. Find what a single part is worth, then multiply by the number of parts wanted.

One part is $63 \div 7 = 9$ kg, so cement is $2 \times 9 = 18$ kg.

Why each wrong option is wrong:

A, $9$ — the value of one part. Cement is two parts.
B, $14$ — treated the 63 kilograms as the whole mix and took $\frac{2}{9}$ of it. The 63 is the sand alone: seven of the nine parts.
D, $126$ — multiplied the 63 by 2 and never divided by 7.

Takeaway: here the 63 is one part, not the whole. Read which quantity the number belongs to.

Q56Medium

Topic: A system by elimination

A canteen serves juice in large jugs and small jugs. Together, 2 large jugs and 3 small jugs hold 16 litres of juice. The 2 large jugs hold 4 litres more than one small jug does. How many litres of juice does one small jug hold?

F) $3$

G) $3.5$

H) $6$

J) $20$

Show the worked solution

Answer: F

Explanation

Let a large jug hold $x$ litres and a small jug $y$ litres. Each sentence gives one equation:

$2x + 3y = 16 \qquad\qquad 2x = y + 4$

Write the second as $2x - y = 4\mathclose{.}$ Subtracting it from the first removes $x\mathclose{.}$ Take care with the $y$ terms: $3y - (-y) = 4y\mathclose{.}$

$4y = 12 \quad\Longrightarrow\quad y = 3$

Check: $2x = 3 + 4 = 7\mathclose{,}$ so the two large jugs hold 7 litres, and $7 + 3(3) = 16$ ✓

Why each wrong option is wrong:

G, $3.5$ — what one large jug holds, $7 \div 2\mathclose{.}$
H, $6$ — subtracted the equations but took $3y - y = 2y\mathclose{.}$ Subtracting $-y$ adds it.
J, $20$ — added the two amounts given, 16 and 4, and stopped.

Takeaway: one sentence, one equation. Matching terms mean subtract. Then check which of the two things was asked for.

Q57Medium

Topic: Working backwards from a discount

A camping shop reduces the price of every tent it sells by 25% in an end-of-season sale. During the sale, a customer pays $150 for a tent. What was the price of that tent, in dollars, before the sale began?

A) $112.5$

B) $175$

C) $187.5$

D) $200$

Show the worked solution

Answer: D

Explanation

After a 25% reduction the customer pays 75% of the original, so undo it by dividing. Adding 25% back on does not return you to the start.

150 is 75% of the original, so the original is $150 \div 0.75 = 200\mathclose{.}$

Why each wrong option is wrong:

A, $112.5$ — took 25% off a second time. The original price must be more than 150.
B, $175$ — added 25 dollars. The discount was a percentage of the original price.
C, $187.5$ — added 25% of 150 back on. The 25% was taken from the original price, which is larger.

Takeaway: to undo a percentage change, divide by the multiplier. Adding the same percentage back does not return you to the start.

Q58Medium

Topic: A transformation of a parabola

The graph of $y = x^2$ is shifted 3 units right. Which equation describes the new graph?

F) $y = (x - 3)^2$

G) $y = (x + 3)^2$

H) $y = x^2 - 3$

J) $y = x^2 + 3$

Show the worked solution

Answer: F

Explanation

Horizontal shifts go inside the brackets and act opposite to their sign. Right by 3 means $(x - 3)^2\mathclose{,}$ whose lowest point is at $x = 3\mathclose{.}$

Why each wrong option is wrong:

G, $y = (x + 3)^2$ — shifted left. $(x + 3)^2$ is lowest at $x = -3\mathclose{.}$
H, $y = x^2 - 3$ — moved the graph down 3. A number outside the brackets moves it vertically.
J, $y = x^2 + 3$ — moved the graph up 3.

Takeaway: inside and opposite for horizontal; outside and obvious for vertical.

Q59Medium

Topic: The equation of a circle

A circle has centre $(3, -2)$ and radius 5. Which equation describes it?

A) $(x+3)^2 + (y-2)^2 = 25$

B) $(x-3)^2 + (y+2)^2 = 25$

C) $(x-3)^2 + (y+2)^2 = 5$

D) $(x+3)^2 + (y+2)^2 = 25$

Show the worked solution

Answer: B

Explanation

$(x - h)^2 + (y - k)^2 = r^2$ with $h = 3\mathclose{,}$ $k = -2\mathclose{,}$ $r = 5\mathclose{.}$ The $-2$ becomes $(y + 2)$ and the right side is $25\mathclose{.}$

Why each wrong option is wrong:

A, $(x+3)^2 + (y-2)^2 = 25$ — kept both signs of the centre. This circle is centred at $(-3, 2)\mathclose{.}$
C, $(x-3)^2 + (y+2)^2 = 5$ — has the radius, 5, where its square belongs.
D, $(x+3)^2 + (y+2)^2 = 25$ — changed the sign for the $-2$ but not for the 3. Its centre is $(-3, -2)\mathclose{.}$

Takeaway: the signs flip and the right-hand side is $r^2\mathclose{.}$

Q60Medium

Topic: A weighted average

A swimming coach times two groups of swimmers over one length of the pool. The first group has 12 swimmers, and their mean time is 50 seconds. The second group has 8 swimmers, and their mean time is 60 seconds. What is the mean time, in seconds, of all the swimmers in the two groups?

F) $54$

G) $55$

H) $56$

J) $110$

Show the worked solution

Answer: F

Explanation

Averages of unequal groups cannot themselves be averaged. Convert each group to a total, add the totals, and divide by the combined count.

$\frac{12(50) + 8(60)}{20} = \frac{1080}{20} = 54$

The larger group is the faster one, so the answer sits nearer 50 than 60.

Why each wrong option is wrong:

G, $55$ — averaged the two averages, as though the groups were the same size.
H, $56$ — gave the 8 swimmers the 50 seconds and the 12 swimmers the 60: the group sizes the wrong way round.
J, $110$ — added the two averages.

Takeaway: averages of unequal groups cannot be averaged. Go back to totals.

Q61Medium

Topic: A geometric sequence

On the first day of a promotion a shop gives out 2 free samples, and on each day after that it gives out 3 times as many samples as on the day before. How many samples does the shop give out on the 7th day?

A) $26$

B) $486$

C) $1458$

D) $4374$

Show the worked solution

Answer: C

Explanation

The daily numbers are $2, 6, 18, 54, \dots\mathclose{:}$ each is 3 times the one before, so they form a geometric sequence with first term 2 and ratio 3. The 7th day is six multiplications after the first:

$2 \times 3^6 = 2 \times 729 = 1458$

Why each wrong option is wrong:

A, $26$ — treated the numbers as arithmetic, adding the first gap of 4 six times.
B, $486$ — used $3^5\mathclose{.}$ That is the 6th day.
D, $4374$ — used $3^7\mathclose{:}$ seven multiplications where there are only six.

Takeaway: $a \cdot r^{\,n-1}$ — the exponent is one less than the term number.

Q62Medium

Topic: The mean from a frequency table

A shop recorded how many drinks each customer bought.

Drinks bought	1	2	3
Number of customers	4	10	6

What is the mean number of drinks per customer?

F) $2$

G) $2.1$

H) $10$

J) $20$

Show the worked solution

Answer: G

Explanation

Total drinks:

$1(4) + 2(10) + 3(6) = 4 + 20 + 18 = 42$

Customers: $4 + 10 + 6 = 20\mathclose{.}$

$\text{mean} = \frac{42}{20} = 2.1$

Why each wrong option is wrong:

F, $2$ — averaged the top row, 1, 2 and 3, ignoring how many customers bought each number.
H, $10$ — the largest number in the bottom row.
J, $20$ — the number of customers.

Takeaway: multiply each value by its frequency, add, then divide by the total frequency. Averaging the top row alone ignores how many people gave each answer.

Q63Medium

Topic: Reading a decay factor

A factory buys a machine, and the value of the machine falls every year after that. The value $V\mathclose{,}$ in dollars, of the machine $t$ years after it was bought is modelled by $V = 18000(0.85)^t\mathclose{.}$ Which of the following statements is best supported by the model?

A) The machine loses 15% of its value each year

B) The machine loses 85% of its value each year

C) The machine loses $85 each year

D) The machine gains 85% of its value each year

Show the worked solution

Answer: A

Explanation

The multiplier is what the machine keeps, so what it loses is the rest. Subtract from 1 rather than reading $0.85$ as the loss.

A multiplier of $0.85$ keeps 85%, so 15% is lost each year.

Why each wrong option is wrong:

B — reads the multiplier as the loss. $0.85$ is what is kept.
C — treats a percentage as a fixed number of dollars. In its first year alone the machine loses $2700.
D — a multiplier below 1 makes the value fall, not rise.

Takeaway: the multiplier is what is kept. Subtract from 1 for what is lost.

Q64Medium

Topic: Probability without replacement

A box holds 4 red pens and 6 blue pens, which are identical apart from their colour. A student takes one pen out of the box at random and, without putting it back, takes out a second pen at random. What is the probability that both pens are red?

F) $\frac{2}{15}$

G) $\frac{4}{25}$

H) $\frac{2}{5}$

J) $\frac{4}{5}$

Show the worked solution

Answer: F

Explanation

Without replacement, both the count of reds and the total drop by one before the second draw.

$\frac{4}{10} \times \frac{3}{9} = \frac{12}{90} = \frac{2}{15}$

Why each wrong option is wrong:

G, $\frac{4}{25}$ — put the first pen back: $\frac{4}{10} \times \frac{4}{10}\mathclose{.}$
H, $\frac{2}{5}$ — the chance for the first pen alone.
J, $\frac{4}{5}$ — added the two chances. "Both" multiplies.

Takeaway: without replacement, both the top and the bottom drop by one on the second draw.

Q65Medium

Topic: An inequality needing a sign flip, on a number line

The number line shows every value of $x$ that satisfies one of the inequalities below. The open circle marks a value that is left out, and the arrow shows that the values carry on without end.

Which inequality is it?

A) $-3x + 4 > 19$

B) $-3x + 4 < 19$

C) $3x + 4 > 19$

D) $3x + 4 < 19$

Show the worked solution

Answer: A

Explanation

Read the picture first: an open circle at $-5$ and a heavy line running off to the left. Every number less than $-5$ is a solution, and $-5$ itself is not:

$x < -5$

Now solve the options. For A, subtract 4 from both sides:

$-3x > 15$

Divide by $-3\mathclose{.}$ Dividing by a negative number reverses the inequality:

$x < -5$ ✓

Check with a value on the line, $x = -6\mathclose{:}$ $-3(-6) + 4 = 22\mathclose{,}$ and $22 > 19$ ✓

Why each wrong option is wrong:

B, $-3x + 4 < 19$ — solves to $x > -5\mathclose{:}$ the same open circle, with the line running off to the right. It is what A looks like if the sign is not reversed.
C, $3x + 4 > 19$ — has no minus sign on the $x\mathclose{,}$ so nothing reverses: $3x > 15$ gives $x > 5\mathclose{,}$ a different circle altogether.
D, $3x + 4 < 19$ — chosen because its sign "points left" like the arrow. It solves to $x < 5\mathclose{,}$ which does run left — but from 5, not from $-5\mathclose{.}$

Takeaway: turn the picture into an inequality for $x$ first, then solve each option. The direction of the arrow belongs to $x\mathclose{,}$ and it is the opposite of the option's own sign whenever you divide by a negative.

Q66Medium

Topic: Probability from a bar chart

A bag holds green, yellow and red counters, which are identical apart from their colour. The bar chart shows the number of counters of each colour.

One counter is drawn from the bag at random. What is the probability that the counter drawn is not red?

F) $\frac{3}{20}$

G) $\frac{2}{5}$

H) $\frac{3}{5}$

J) $\frac{2}{3}$

Show the worked solution

Answer: G

Explanation

The chart shows 3 green, 5 yellow and 12 red counters. Not red means green or yellow: $3 + 5 = 8$ counters out of 20.

$\frac{8}{20} = \frac{2}{5}$

Or use the complement: red is $\frac{12}{20} = \frac{3}{5}\mathclose{,}$ so not red is $1 - \frac{3}{5} = \frac{2}{5}\mathclose{.}$

Why each wrong option is wrong:

F, $\frac{3}{20}$ — counted the green counters and forgot that yellow is not red either.
H, $\frac{3}{5}$ — the probability that it is red.
J, $\frac{2}{3}$ — two of the three colours are not red. But the colours are not equally likely: more than half the counters are red.

Takeaway: the denominator is everything possible. For a "not" question, $1 - P(A)$ is usually the quicker route.

Q67Medium

Topic: The vertex of a parabola

The graph of $y = x^2 - 6x + 5$ is shown in the standard $(x, y)$ coordinate plane.

At what value of $x$ does the graph reach its minimum?

A) $-6$

B) $-3$

C) $3$

D) $6$

Show the worked solution

Answer: C

Explanation

The vertex of $y = ax^2 + bx + c$ sits at $x = -\frac{b}{2a}\mathclose{,}$ and $b$ carries its own sign — here two minus signs cancel.

$x = -\frac{b}{2a} = -\frac{-6}{2} = 3$

Why each wrong option is wrong:

A, $-6$ — read $b$ straight off the equation.
B, $-3$ — halved $b$ and lost the minus the formula carries. With $b = -6\mathclose{,}$ $-\frac{b}{2a}$ is $+3\mathclose{.}$
D, $6$ — changed the sign of $b$ and never divided by $2a\mathclose{.}$

Takeaway: $b$ carries its own sign, and the formula has a minus of its own. Two minuses make a plus.

Q68Medium

Topic: Degrees and radians

What is $135°$ expressed in radians?

F) $135 \pi$

G) $\frac{4 \pi}{3}$

H) $\frac{3 \pi}{4}$

J) $\frac{3}{4}$

Show the worked solution

Answer: H

Explanation

Multiply by $\frac{\pi}{180}\mathclose{.}$ The $\pi$ stays in the answer; if it has vanished, the conversion went the wrong way.

$135 \times \frac{\pi}{180} = \frac{3\pi}{4}$

Why each wrong option is wrong:

F, $135 \pi$ — multiplied by $\pi$ and never divided by 180.
G, $\frac{4 \pi}{3}$ — the fraction upside down, 180 over 135.
J, $\frac{3}{4}$ — the right fraction with the $\pi$ dropped.

Takeaway: multiply by $\frac{\pi}{180}\mathclose{.}$ The $\pi$ stays in the answer.

Q69Medium

Topic: The determinant of a matrix

What is the determinant of $\begin{bmatrix} 4 & 3 \\ 2 & 5 \end{bmatrix}\mathclose{?}$

A) $-14$

B) $14$

C) $20$

D) $26$

Show the worked solution

Answer: B

Explanation

The determinant is the product along the main diagonal minus the product along the other one:

$ad - bc = (4)(5) - (3)(2) = 20 - 6 = 14$

Why each wrong option is wrong:

A, $-14$ — subtracted in the wrong order, $bc - ad\mathclose{.}$
C, $20$ — the main diagonal only.
D, $26$ — added the two products.

Takeaway: main diagonal first, then subtract the other.

Q70Medium

Topic: A Venn diagram

A club meets every week to play board games. Of its members, 45 play chess, 30 play draughts, and 18 play both chess and draughts. Every member of the club plays at least one of the two games. How many members does the club have?

F) $39$

G) $57$

H) $75$

J) $93$

Show the worked solution

Answer: G

Explanation

The 18 who play both have been counted twice, once in each total, so subtract them once:

$45 + 30 - 18 = 57$

Check by parts: 27 chess only, 12 draughts only, 18 both.

Why each wrong option is wrong:

F, $39$ — subtracted the 18 twice. That counts only the members who play exactly one game.
H, $75$ — added the two totals and left the 18 counted twice.
J, $93$ — added the 18 a third time.

Takeaway: subtract the overlap once — it was counted twice.

Q71Medium

Topic: Slope from a table

A linear function's values are recorded below.

$x$	1	2	3	4
$f(x)$	9	13	17	21

What is the slope of the function?

A) $\frac{1}{4}$

B) $3$

C) $4$

D) $12$

Show the worked solution

Answer: C

Explanation

Slope is the change in output divided by the change in input, and on a linear table it is the same between any two rows.

$\frac{21 - 9}{4 - 1} = \frac{12}{3} = 4$

Why each wrong option is wrong:

A, $\frac{1}{4}$ — run over rise, the fraction upside down.
B, $3$ — the change in $x$ alone.
D, $12$ — the change in $f(x)$ from the first column to the last, not yet divided by the change in $x\mathclose{.}$

Takeaway: slope is change in output over change in input, and on a linear table it is the same between any two rows.

Q72Medium

Topic: A perfect-square trinomial

If $x^2 + kx + 36$ is a perfect square for some positive constant $k\mathclose{,}$ what is the value of $k\mathclose{?}$

F) $6$

G) $12$

H) $18$

J) $36$

Show the worked solution

Answer: G

Explanation

A perfect square is $(x + a)^2 = x^2 + 2ax + a^2\mathclose{.}$ Here $a^2 = 36$ so $a = 6\mathclose{,}$ and the middle term is $2a = 12\mathclose{.}$

Why each wrong option is wrong:

F, $6$ — the square root of 36. That is $a\mathclose{;}$ the middle coefficient is $2a\mathclose{.}$
H, $18$ — halved the constant instead of taking its square root and doubling.
J, $36$ — copied the constant.

Takeaway: in a perfect square the middle coefficient is twice the square root of the constant.

Q73Medium

Topic: A three-part inequality

If $-7 < 2x + 1 \le 9\mathclose{,}$ which of the following gives all possible values of $x\mathclose{?}$

A) $-4 < x \le 4$

B) $-4 \le x \le 4$

C) $-4 \le x < 4$

D) $-4 < x < 4$

Show the worked solution

Answer: A

Explanation

Do the same thing to all three parts at once, and copy each inequality sign down exactly as it was — the ends decide between the options here.

Subtract 1 from all three parts and divide by 2:

$-4 < x \le 4$

Why each wrong option is wrong:

B, $-4 \le x \le 4$ — made both ends inclusive. $x = -4$ gives $2x + 1 = -7\mathclose{,}$ and $-7 < -7$ is false.
C, $-4 \le x < 4$ — swapped the two signs, including $-4$ and excluding 4.
D, $-4 < x < 4$ — made both ends strict. $x = 4$ gives 9, and $9 \le 9$ is true, so 4 belongs.

Takeaway: operate on all three parts, and carry each inequality sign down unchanged.

Q74Medium

Topic: A word problem with two unknowns

A cinema sells adult tickets at $14 each and child tickets at $9 each. For one evening showing of a film it sold 60 tickets altogether, and the money taken for those tickets came to $710. How many child tickets did the cinema sell for that showing?

F) $26$

G) $30$

H) $34$

J) $60$

Show the worked solution

Answer: F

Explanation

With $a + c = 60$ and $14a + 9c = 710\mathclose{,}$ substituting $a = 60 - c$ gives $840 - 5c = 710\mathclose{,}$ so $c = 26$ and $a = 34\mathclose{.}$

Why each wrong option is wrong:

G, $30$ — assumed half of each. That would take $690, not $710.
H, $34$ — the number of adult tickets.
J, $60$ — the total number of tickets.

Takeaway: name both unknowns, write one equation for the count and one for the money.

Q75Medium

Topic: Comparing two linear models

A town has two gyms. The first charges a membership fee of $25 a month plus $4 for each class a member attends. The second charges a flat fee of $65 a month, and its classes are free. For how many classes attended in a month do the two gyms cost a member the same amount?

A) $10$

B) $16.25$

C) $40$

D) $90$

Show the worked solution

Answer: A

Explanation

$25 + 4n = 65 \quad\Longrightarrow\quad 4n = 40 \quad\Longrightarrow\quad n = 10$

Why each wrong option is wrong:

B, $16.25$ — divided the 65 by 4 without first removing the 25 monthly fee.
C, $40$ — the difference between the two fees, not yet divided by 4.
D, $90$ — added the two fees.

Takeaway: set the two totals equal. Remove the fixed part before dividing.

Q76Hard

Topic: A system with no solution

The system $\begin{cases} 4x - 2y = 8 \\ 6x + ky = 5 \end{cases}$ has no solution. What is the value of $k\mathclose{?}$

F) $-3$

G) $-2$

H) $- \frac{3}{2}$

J) $3$

Show the worked solution

Answer: F

Explanation

The first line has slope 2. The second has slope $-\frac{6}{k}\mathclose{,}$ so $-\frac{6}{k} = 2$ gives $k = -3\mathclose{.}$ The intercepts then differ, so the lines are parallel and distinct.

Why each wrong option is wrong:

G, $-2$ — copied the coefficient of $y$ from the first equation. The slopes are then 2 and 3, and the lines cross.
H, $- \frac{3}{2}$ — the ratio of the $x$ -coefficients, $\frac{6}{4}\mathclose{,}$ with a minus sign. That ratio has to multiply the $-2\mathclose{:}$ $-2 \times \frac{3}{2} = -3\mathclose{.}$
J, $3$ — the right size with the sign lost. With $k = 3$ the slopes are 2 and $-2\mathclose{,}$ and the lines cross.

Takeaway: no solution means equal slopes and different intercepts. Check the intercept too.

Q77Hard

Topic: The distance between two points

Two points are plotted in the standard $(x, y)$ coordinate plane, as shown.

What is the distance between the two points?

A) $\sqrt{7}$

B) $5$

C) $7$

D) $25$

Show the worked solution

Answer: B

Explanation

The points are $(-3, 2)$ and $(1, -1)\mathclose{.}$ The distance formula is Pythagoras on the horizontal and vertical gaps. Watch the double negative in $1 - (-3)\mathclose{.}$

Gaps of 4 and 3, so $\sqrt{16 + 9} = 5\mathclose{.}$ A 3-4-5 triangle.

Why each wrong option is wrong:

A, $\sqrt{7}$ — added the gaps and then took a root, without squaring.
C, $7$ — added the two gaps, 4 and 3: the distance along the grid, not straight across.
D, $25$ — squared and added, then forgot the square root.

Takeaway: square, add, then root. Watch the double negatives in the subtraction.

Q78Hard

Topic: An exponential model in function notation

The number of fish in a lake is modelled by $f(x) = 3200(0.75)^x\mathclose{,}$ where $x$ is the number of years since counting began. According to the model, by what percent does the number of fish fall over the first two years?

F) $25$

G) $43.75$

H) $50$

J) $56.25$

Show the worked solution

Answer: G

Explanation

The multiplier $0.75$ is applied once for each year, so after two years

$3200 \times 0.75 \times 0.75 = 3200 \times 0.5625 = 1800$

fish are left. That is $56.25\%$ of the starting number, so the fall is $100 - 56.25 = 43.75\%\mathclose{.}$

The 3200 does not matter: any model with a multiplier of $0.75$ keeps $0.75^2 = 0.5625$ of its value after two steps.

Why each wrong option is wrong:

F, $25$ — the fall in one year.
H, $50$ — added two falls of 25%. The second year's 25% is taken from a smaller number, so the two together are less than 50%.
J, $56.25$ — the percent that is left, not the percent lost.

Takeaway: below 1 is decay. Over two periods, square the multiplier; what is lost is 1 minus the result.

Q79Hard

Topic: Powers of the imaginary unit

Given $i^2 = -1\mathclose{,}$ what is $i^{43}\mathclose{?}$

A) $1$

B) $i$

C) $- i$

D) $-1$

Show the worked solution

Answer: C

Explanation

The powers of $i$ repeat every four, so divide the exponent by 4 and use the remainder.

$43 \div 4$ is 10 remainder 3, and $i^3 = i^2 \cdot i = -i\mathclose{.}$

Why each wrong option is wrong:

A, $1$ — what a remainder of 0 gives. $i^{40} = 1\mathclose{,}$ but three more factors of $i$ follow.
B, $i$ — what a remainder of 1 gives.
D, $-1$ — what a remainder of 2 gives: that is $i^2\mathclose{.}$

Takeaway: the powers of $i$ cycle with period 4. Divide the exponent by 4 and use the remainder.

Q80Hard

Topic: The law of sines

In triangle $PQR\mathclose{,}$ shown below, angle $P$ measures $60°\mathclose{,}$ angle $Q$ measures $45°\mathclose{,}$ and the side opposite $Q$ has a length of 10.

What is the value of $x\mathclose{,}$ the length of the side opposite $P\mathclose{?}$

F) $\frac{10 \sqrt{6}}{3}$

G) $5 \sqrt{3}$

H) $5 \sqrt{6}$

J) $\frac{40}{3}$

Show the worked solution

Answer: H

Explanation

$\frac{p}{\sin 60°} = \frac{10}{\sin 45°} \quad\Longrightarrow\quad p = \frac{10 \times \frac{\sqrt 3}{2}}{\frac{\sqrt 2}{2}} = \frac{10\sqrt 3}{\sqrt 2} = 5\sqrt{6} \approx 12.2$

Larger angle, longer side — as it must be.

Why each wrong option is wrong:

F, $\frac{10 \sqrt{6}}{3}$ — the ratio upside down. It makes the side facing the larger angle the shorter one.
G, $5 \sqrt{3}$ — multiplied by $\sin 60°$ and never divided by $\sin 45°\mathclose{.}$
J, $\frac{40}{3}$ — used the angles themselves, $10 \times \frac{60}{45}\mathclose{,}$ in place of their sines.

Takeaway: the law of sines pairs a side with its opposite angle. The bigger angle always faces the longer side.

Q81Hard

Topic: Scaling an area

A rectangle's length and width are both tripled. By what factor does its area increase?

A) $3$

B) $6$

C) $9$

D) $27$

Show the worked solution

Answer: C

Explanation

Both dimensions triple, and area is a product of two lengths, so the factor applies twice:

$3 \times 3 = 9$

Why each wrong option is wrong:

A, $3$ — treated area like a length.
B, $6$ — added the two factors of 3 where they multiply.
D, $27$ — $3^3\mathclose{,}$ the factor for a volume.

Takeaway: lengths by $k\mathclose{,}$ areas by $k^2\mathclose{,}$ volumes by $k^3\mathclose{.}$

Q82Hard

Topic: A rational equation

What is the solution to $\dfrac{15}{x} - 2 = \dfrac{9}{x}\mathclose{?}$

F) $3$

G) $6$

H) $12$

J) $24$

Show the worked solution

Answer: F

Explanation

$\frac{15}{x} - \frac{9}{x} = 2 \quad\Longrightarrow\quad \frac{6}{x} = 2 \quad\Longrightarrow\quad x = 3$

Why each wrong option is wrong:

G, $6$ — subtracted the numerators and stopped at $\frac{6}{x} = 2\mathclose{,}$ reading off the 6.
H, $12$ — reached $\frac{6}{x} = 2$ and multiplied the 6 by 2 instead of dividing.
J, $24$ — added the numerators.

Takeaway: combine fractions over the shared denominator first. Then check the answer does not make it zero.

Q83Hard

Topic: Combining logarithms

What is the value of $\log_3(9) + \log_3(27)\mathclose{?}$

A) $5$

B) $6$

C) $36$

D) $243$

Show the worked solution

Answer: A

Explanation

$\log_3 9 = 2$ and $\log_3 27 = 3\mathclose{,}$ so the sum is 5. Equivalently $\log_3(9 \times 27) = \log_3(243) = 5\mathclose{.}$

Why each wrong option is wrong:

B, $6$ — multiplied the two logs, $2 \times 3\mathclose{.}$ They add.
C, $36$ — added what is inside the logs.
D, $243$ — multiplied what is inside, correctly, and stopped. The answer is the power of 3 that gives 243.

Takeaway: adding logs multiplies what is inside. With small powers, evaluate each separately and use the rule as a check.

Q84Hard

Topic: The magnitude of a vector

What is the magnitude of the vector $\langle -8, 15 \rangle\mathclose{?}$

F) $\sqrt{23}$

G) $7$

H) $17$

J) $23$

Show the worked solution

Answer: H

Explanation

Magnitude is Pythagoras on the two components, and squaring removes the minus sign:

$\sqrt{(-8)^2 + 15^2} = \sqrt{64 + 225} = \sqrt{289} = 17$

Why each wrong option is wrong:

F, $\sqrt{23}$ — added the sizes and then took a root, without squaring.
G, $7$ — added the two components.
J, $23$ — added their sizes, 8 and 15.

Takeaway: magnitude is Pythagoras on the components, and is never negative.

Q85Hard

Topic: Perpendicular lines

A line is perpendicular to $y = -2x + 1$ and passes through $(4, 3)\mathclose{.}$ What is its $y$ -intercept?

A) $-5$

B) $-2$

C) $\frac{1}{2}$

D) $1$

Show the worked solution

Answer: D

Explanation

The perpendicular slope is $\frac{1}{2}\mathclose{.}$ Then $3 = \frac{1}{2}(4) + b$ gives $b = 1\mathclose{.}$

Why each wrong option is wrong:

A, $-5$ — changed the sign of the slope but did not flip it, using 2: $3 = 2(4) + b\mathclose{.}$
B, $-2$ — the slope of the original line.
C, $\frac{1}{2}$ — the new slope, not the intercept.

Takeaway: flip and negate for perpendicular, then substitute the point.

Q86Hard

Topic: Interpreting a linear model

A taxi company charges a fixed amount for every journey and a further amount for each mile travelled. The fare $C\mathclose{,}$ in dollars, for a journey of $m$ miles is given by $C = 6 + 1.5m\mathclose{.}$ Which of the following is the best interpretation of the 6 in this equation?

F) A fixed charge applied before any distance is travelled

G) The cost of each mile travelled

H) The number of miles included in the base fare

J) The total cost of a six-mile journey

Show the worked solution

Answer: F

Explanation

Setting $m = 0$ gives $C = 6\mathclose{,}$ so the 6 is what you pay before moving. The 1.5 is the cost per mile.

Why each wrong option is wrong:

G — that is the $1.5\mathclose{,}$ the number multiplying $m\mathclose{.}$
H — the 6 is in dollars, not miles.
J — a six-mile journey costs $6 + 1.5(6) = 15$ dollars.

Takeaway: the constant is the value at zero; the coefficient is the rate.

Q87Hard

Topic: A quadratic from its roots

Which of the following quadratic expressions has zeros at $x = -4$ and $x = 6\mathclose{?}$

A) $x^{2} + 2 x - 24$

B) $x^{2} - 10 x - 24$

C) $x^{2} - 2 x - 24$

D) $x^{2} - 2 x + 24$

Show the worked solution

Answer: C

Explanation

Zeros of $-4$ and $6$ give factors $(x + 4)$ and $(x - 6)\mathclose{:}$

$(x+4)(x-6) = x^2 - 2x - 24$

Why each wrong option is wrong:

A, $x^{2} + 2 x - 24$ — is $(x - 4)(x + 6)\mathclose{,}$ whose zeros are 4 and $-6\mathclose{:}$ both signs reversed.
B, $x^{2} - 10 x - 24$ — added the sizes of the zeros, 4 and 6. Their sum is $-4 + 6 = 2\mathclose{,}$ so the middle term is $-2x\mathclose{.}$
D, $x^{2} - 2 x + 24$ — has the right middle term, but the zeros multiply to $-24\mathclose{,}$ not $+24\mathclose{.}$

Takeaway: a root of $r$ gives a factor of $(x - r)\mathclose{,}$ so the signs flip.

Q88Hard

Topic: Comparing spread from a summary table

Two machines were each tested 20 times, and their output is summarised below.

Machine	Mean	Lowest	Highest
A	50	48	52
B	50	20	80

Which statement is true?

F) Machine B's output has the greater standard deviation

G) Machine A's output has the greater standard deviation

H) The two standard deviations are equal, because the means are equal

J) Standard deviation cannot be compared from a range

Show the worked solution

Answer: F

Explanation

Both means are 50, which says nothing about spread.

No result from machine A is more than 2 from the mean, so its standard deviation cannot be more than 2. Machine B's lowest and highest results are each 30 from the mean, and in 20 tests those two results alone make the standard deviation more than 9.

Why each wrong option is wrong:

G — has the machines the wrong way round.
H — equal means say nothing about how far the results sit from the mean.
J — too cautious. A range does not give the standard deviation exactly, but here it settles the comparison: at most 2 for machine A, more than 9 for machine B.

Takeaway: equal means do not imply equal spread. That is precisely what standard deviation measures, and a range is enough to compare two sets roughly.

Q89Hard

Topic: A percentage of a percentage

A survey of one town found that 40% of its households have a garden and that 30% of the households that have a garden also have a pond in it. No household without a garden has a pond. What percent of all the households in the town have a pond?

A) $10$

B) $12$

C) $30$

D) $70$

Show the worked solution

Answer: B

Explanation

"Of those gardens" means the 30% applies to the 40%, not to the town, and "of" means multiply:

$0.40 \times 0.30 = 0.12 = 12\%$

Why each wrong option is wrong:

A, $10$ — subtracted them.
C, $30$ — applied the 30% to every household, not just to those with a garden.
D, $70$ — added the two percentages.

Takeaway: "of" means multiply. A percentage of a percentage is smaller than either.

Q90Hard

Topic: Several ideas together

The rectangular garden shown is twice as long as it is wide, and a fence that runs all the way round its edge is 54 metres long.

What is the area of the garden, in square metres?

F) $9$

G) $27$

H) $81$

J) $162$

Show the worked solution

Answer: J

Explanation

With width $w$ and length $2w\mathclose{,}$ the perimeter is $6w = 54\mathclose{,}$ so $w = 9$ and the length is 18. The area is $9 \times 18 = 162\mathclose{.}$

Why each wrong option is wrong:

F, $9$ — the width, where the question asks for the area.
G, $27$ — half the perimeter. That is the length plus the width, not an area.
H, $81$ — squared the width, as though the garden were 9 by 9.

Takeaway: name the smaller quantity, write the other in terms of it, then answer the question actually asked — here area, not a side.

right	scale	right	scale	right	scale
45	36	30	28	15	17
43	35	28	27	13	16
41	34	26	25	11	15
39	33	24	23	9	14
37	32	22	21	7	13
35	31	20	20	5	12
33	30	18	18	3	9
31	29	16	17	1	5

right	scale	right	scale	right	scale
45	36	30	28	15	17
43	35	28	27	13	16
41	34	26	25	11	15
39	33	24	23	9	14
37	32	22	21	7	13
35	31	20	20	5	12
33	30	18	18	3	9
31	29	16	17	1	5

Chapter 9 — Two Full Practice Papers

Scoring

Paper 1

Paper 2

right	scale	right	scale	right	scale
45	36	30	28	15	17
43	35	28	27	13	16
41	34	26	25	11	15
39	33	24	23	9	14
37	32	22	21	7	13
35	31	20	20	5	12
33	30	18	18	3	9
31	29	16	17	1	5