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Rates, Ratios and Proportions: Paper 2 Challenge Multiple-Choice

This is the Challenge set. Each question hides its ratio one step deeper than usual, in reciprocals, in equal products, in sums of pairs or in a head start, and one of the wrong options is always the right working applied to the wrong reading. Work the standard Paper 2 set first. Detailed solutions can be requested by email.

10 questions · about 20 minutes · answers below each question

All questions are original, written for Math and AI Academy and modelled on past-paper style. They are not reproductions of HKEAA examination questions.

Question 1 · Modelled on 2015 P2 Q11 and 2018 P2 Q10 — New to this tier: one of the given ratios is between reciprocals, and reading it the wrong way round leads straight to an option

Let

a

b

and

c

be non-zero numbers. If

\frac{1}{a} : \frac{1}{b} = 3 : 5

and

b : c = 2 : 3

, then

a : c =

$2 : 5$ .
$9 : 10$ .
$10 : 9$ .
$5 : 2$ .

Show answer

C. $\frac{1}{a} : \frac{1}{b} = 3 : 5$ means $a : b = 5 : 3$ . Then $a : b : c = 10 : 6 : 9$ , so $a : c = 10 : 9$ . Option A is what you get from reading the first ratio as $a : b = 3 : 5$ : the right method on the wrong ratio.

Question 2 · Modelled on 2014 P2 Q12 — New to this tier: equal products, not equal quotients, so the largest multiplier belongs to the smallest number

a

b

and

c

are positive numbers such that

2 a = 3 b = 5 c

, then

a : b : c =

$15 : 10 : 6$ .
$6 : 10 : 15$ .
$2 : 3 : 5$ .
$5 : 3 : 2$ .

Show answer

A. Let $2 a = 3 b = 5 c = 30 k$ . Then $a = 15 k$ , $b = 10 k$ , $c = 6 k$ . Option C copies the multipliers; option D reverses them, which is still not the reciprocal ratio.

Question 3 · Modelled on 2013 P2 Q12 and 2023 P2 Q12 — New to this tier: no actual area and no numerical scale, only how the two scales compare

On a map of scale

1 : 2 n

, the area of a park is

8 cm^{2}

. On another map of scale

1 : n

, the area of the same park is

$2 cm^{2}$ .
$4 cm^{2}$ .
$16 cm^{2}$ .
$32 cm^{2}$ .

Show answer

D. On the $1 : n$ map every length is twice as long as on the $1 : 2 n$ map, so every area is $2^{2} = 4$ times as large: $8 \times 4 = 32$ . Option C doubles the area; option A goes the wrong way.

Question 4 · Modelled on 2024 P2 Q12 — New to this tier: the two stages have equal distances, not given times, and no distance is stated

Sam cycles up a hill at an average speed of

12 km/h

and then cycles down along the same road at an average speed of

20 km/h

. His average speed for the whole journey is

$14 km/h$ .
$15 km/h$ .
$16 km/h$ .
$17 km/h$ .

Show answer

B. Let the road be $d$ km. Time $= \frac{d}{12} + \frac{d}{20} = \frac{2 d}{15}$ , so the average speed is $2 d \div \frac{2 d}{15} = 15$ . Option C is the average of the two speeds, which would need equal times, not equal distances.

Question 5 · Original, in the “which of the following must be true” format of Paper 2 — New to this tier: testing what may and may not be done to both terms of a ratio

Let

a

and

b

be positive numbers such that

a : b = 2 : 3

. Which of the following must be true?
I.

(a + 2) : (b + 3) = 2 : 3

II.

a^{2} : b^{2} = 4 : 9

III.

(a + 1) : (b + 1) = 3 : 4

I and II only
I and III only
II and III only
I, II and III

Show answer

A. With $a = 2 k$ , $b = 3 k$ : I is $\frac{2 k + 2}{3 k + 3} = \frac{2 ( k + 1 )}{3 ( k + 1 )} = \frac{2}{3}$ , true. II is $\frac{4 k ^{2}}{9 k ^{2}}$ , true. III holds only when $k = 1$ ; try $a = 4$ , $b = 6$ to get $5 : 7$ .

Question 6 · Modelled on 2018 P1 Q9 — New to this tier: no speed and no distance is given, only a ratio of speeds and a difference of times

Car

A

and car

B

travel along the same road from town

P

to town

Q

at constant speeds in the ratio

3 : 4

. If car

A

takes

15

minutes longer than car

B

, then the time taken by car

B

15 minutes.
20 minutes.
45 minutes.
60 minutes.

Show answer

C. For the same distance, times are in the inverse ratio of speeds: $A : B = 4 : 3$ . The difference of one part is $15$ minutes, so $B$ takes $3$ parts. Option D is the time taken by car $A$ .

Question 7 · Modelled on 2019 P1 Q7 — New to this tier: a swap, so one group gains exactly what the other loses and the total is unchanged

A bag contains red marbles and blue marbles in the ratio

3 : 5

. If

12

blue marbles are taken out and replaced by

12

red marbles, then the numbers of red marbles and blue marbles are equal. Find the original number of red marbles.

$24$
$36$
$48$
$60$

Show answer

B. $3 k + 12 = 5 k - 12$ gives $k = 12$ , so there were $36$ red marbles. Option C is the number of red marbles after the swap, and option D is the original number of blue marbles.

Question 8 · Modelled on 2014 P1 Q10 (when do two movers meet) — New to this tier: no graph; the head start and the closing speed have to be set up from the wording

Ben leaves home at 9:00 and walks along a straight road at

5 km/h

. Amy leaves the same home at 9:24 and cycles along the same road in the same direction at

15 km/h

. At what time does Amy catch up with Ben?

9:30
9:32
9:36
9:48

Show answer

C. By 9:24 Ben is $5 \times \frac{24}{60} = 2 km$ ahead. Amy closes the gap at $15 - 5 = 10 km/h$ , taking $\frac{2}{10}$ h $= 12$ minutes. Option B divides the $2 km$ by Amy’s full speed, as if Ben had stopped.

Question 9 · Modelled on 2025 P2 Q11 — New to this tier: the given ratio is between sums of pairs, and the single numbers must be untangled from it

a

b

and

c

are positive numbers such that

(a + b) : (b + c) : (c + a) = 4 : 5 : 6

, then

a : b : c =

$5 : 3 : 7$ .
$3 : 5 : 7$ .
$4 : 5 : 6$ .
$7 : 3 : 5$ .

Show answer

A. Let the sums be $4 k$ , $5 k$ , $6 k$ . Adding all three: $2 (a + b + c) = 15 k$ , so $a + b + c = 7.5 k$ . Then $a = 7.5 k - 5 k$ , $b = 7.5 k - 6 k$ , $c = 7.5 k - 4 k$ , giving $2.5 : 1.5 : 3.5 = 5 : 3 : 7$ .

Question 10 · Original — New to this tier: the size of one part is not given by a total or a difference, but by a condition after a transfer

A sum of money is shared among

A

B

and

C

in the ratio

2 : 3 : 5

. If

C

then gives

$30

A

A

and

B

have the same amount. Find the sum of money.

$$150$
$$200$
$$240$
$$300$

Show answer

D. $2 k + 30 = 3 k$ gives $k = 30$ , and the sum is $10 k = 300$ . The $$30$ leaving $C$ plays no part in the condition: only $A$ and $B$ are compared. Option A is the share of $C$ alone.

Continue practising this topic: Try the Paper 2 multiple-choice for the standard set, or move to Paper 1 long questions — also available as Paper 1 Challenge.

Where this sits in the syllabus

Rates, ratios and proportions is a junior secondary topic that the HKDSE keeps testing. Between 2012 and 2026 most Paper 2 papers carry a question on a : b : c, a map scale, an average speed or a mixture, and Paper 1 Section A(1) uses the topic for short word problems. The k-method practised here returns later in similar figures, variations and trigonometric ratios.

Full coverage of this topic, and the rest of the course, is on the Mathematics page. More sets are listed on the practice index.

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