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Rates, Ratios and Proportions: Paper 1 Challenge Questions

This is the Challenge set. The questions state a condition and leave the method to you, and most end with a claim to agree or disagree with: a ratio that changes every year, a round trip whose average speed has a ceiling, an alloy that cannot be made. Work the standard Paper 1 set first. Detailed solutions can be requested by email.

10 questions · 55 marks · about 70 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 — run backwards: the ratio of the sums is given and a claim about the third number has to be judged. New to this tier: no part tells you to introduce k

Let $a$ , $b$ and $c$ be positive numbers such that $a : b = 2 : 3$ and $(a + b) : (b + c) = 5 : 9$ . Someone claims that $c$ is three times $a$ . Do you agree? Explain your answer. (4 marks)

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Agree. Put $a = 2 k$ , $b = 3 k$ . Then $a + b = 5 k$ , so $b + c = 9 k$ and $c = 6 k = 3 a$ .

Question 2 · Modelled on 2014 P2 Q12 — the ratio is stated for the reciprocals. New to this tier: a tempting wrong answer that must be explained away, not just avoided

Let $x$ , $y$ and $z$ be non-zero numbers such that $\frac{1}{x} : \frac{1}{y} : \frac{1}{z} = 2 : 3 : 4$ .

(a) A student writes $x : y : z = 4 : 3 : 2$ . Explain why the student is wrong, and find $x : y : z$ . (2)

(b) If $x - z = 15$ , find $y$ . (2)

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(a) $x : y : z = 6 : 4 : 3$ . Reversing the order does not invert a three-term ratio: $4 : 3 : 2$ gives reciprocals in the ratio $\frac{1}{4} : \frac{1}{3} : \frac{1}{2} = 3 : 4 : 6$ , not $2 : 3 : 4$ .
(b) $y = 20$ . $6 k - 3 k = 15$ gives $k = 5$ .

Question 3 · Modelled on 2019 P1 Q7 — both quantities grow by the same amount, so the difference is what stays fixed. New to this tier: proving how a ratio behaves for every future year

The ratio of Ann's age to her father's age is now $1 : 4$ . In $10$ years' time, the ratio will be $2 : 5$ .

(a) Find Ann's present age. (2)

(b) Ann claims that the ratio of her age to her father's age becomes larger every year. Do you agree? Explain your answer. (3)

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(a) $10$ . $\frac{k + 10}{4 k + 10} = \frac{2}{5}$ gives $k = 10$ .
(b) Agree. After $n$ years the ratio is $\frac{10 + n}{40 + n} = 1 - \frac{30}{40 + n}$ . The gap of $30$ years never changes while $40 + n$ grows, so the subtracted fraction shrinks and the ratio rises.

Question 4 · Modelled on 2013 P2 Q12 and 2023 P2 Q12 — two maps of the same place. New to this tier: a trap where the length ratio is applied to an area

The scales of map $A$ and map $B$ are $1 : 20000$ and $1 : 50000$ respectively. The area of a reservoir on map $A$ is $45 cm^{2}$ .

(a) Find the actual area of the reservoir in $km^{2}$ . (2)

(b) A student claims that the area of the reservoir on map $B$ is $18 cm^{2}$ . Is the claim correct? Explain your answer. (3)

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(a) $1.8 km^{2}$ . On map $A$ , $1 cm^{2}$ stands for $0. 2^{2} = 0.04 km^{2}$ .
(b) Not correct. On map $B$ , $1 cm^{2}$ stands for $0. 5^{2} = 0.25 km^{2}$ , so the area is $1.8 \div 0.25 = 7.2 cm^{2}$ . The $18$ comes from multiplying by $\frac{2}{5}$ ; areas need $(\frac{2}{5})^{2}$ .

Question 5 · Modelled on 2018 P1 Q9 and 2024 P2 Q12 — a round trip with a target average. New to this tier: an average speed that no return speed can reach

A driver travels from town $P$ to town $Q$ at an average speed of $45 km/h$ and then returns to town $P$ along the same road.

(a) Find the average speed of the return journey if the average speed for the whole round trip is $60 km/h$ . (3)

(b) The driver claims that, by returning fast enough, the average speed for the whole round trip can be $90 km/h$ . Do you agree? Explain your answer. (2)

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(a) $90 km/h$ . $\frac{d}{45} + \frac{d}{v} = \frac{2 d}{60}$ gives $\frac{1}{v} = \frac{1}{90}$ .
(b) Disagree. Averaging $90 km/h$ over $2 d$ allows $\frac{2 d}{90} = \frac{d}{45}$ hours in total, and the outward journey has already used exactly that. No time is left for the return.

Question 6 · Original, in the style of the 2018 P1 Q9 journey question — New to this tier: two rates that combine (boat and current), and a claim that sounds symmetric but is not

A boat travels $48 km$ downstream in $2$ hours and then travels the same $48 km$ upstream in $3$ hours. The speed of the boat in still water and the speed of the current are both constant.

(a) Find the speed of the current. (3)

(b) On another day the current flows twice as fast, while the speed of the boat in still water is unchanged. The skipper claims that the same round trip still takes $5$ hours, because the time gained downstream cancels the time lost upstream. Do you agree? Explain your answer. (3)

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(a) $4 km/h$ . Downstream $24 km/h$ , upstream $16 km/h$ ; the boat does $20$ and the current $4$ .
(b) Disagree. With an $8 km/h$ current the trip takes $\frac{48}{28} + \frac{48}{12} = 5 \frac{5}{7}$ hours. Downstream saves $\frac{2}{7}$ h, upstream costs a whole extra hour.

Question 7 · Original, in the style of the 2012 P2 Q11 rate question — New to this tier: the unknown is how long one worker stayed, and a claim that confuses time worked with work done

Alan alone can finish a job in $10$ days. Ben alone can finish the same job in $15$ days. The two of them start the job together. After some days Alan leaves, and Ben finishes the rest of the job alone. The whole job takes $9$ days.

(a) For how many days does Alan work on the job? (3)

(b) A payment of $$6000$ is shared between Alan and Ben in proportion to the amount of work each of them does. Someone claims that Alan receives $\frac{4}{9}$ of the payment. Do you agree? Explain your answer. (3)

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(a) $4$ days. Ben works all $9$ days and does $\frac{9}{15} = \frac{3}{5}$ of the job, leaving $\frac{2}{5}$ for Alan at $\frac{1}{10}$ per day.
(b) Disagree. Alan does $\frac{2}{5}$ of the job, so he receives $\frac{2}{5}$ of the payment ( $$2400$ ), not $\frac{4}{9}$ . Days worked are not work done, because the two work at different rates.

Question 8 · Modelled on 2019 P1 Q7 — three quantities, with one unchanged. New to this tier: deciding whether a target ratio can be reached at all

A bag contains red balls, blue balls and green balls. The numbers of red balls, blue balls and green balls are in the ratio $3 : 4 : 5$ . After $8$ red balls are put into the bag and $8$ green balls are taken out, the ratio of the number of red balls to the number of blue balls is $5 : 6$ .

(a) Find the number of green balls now in the bag. (3)

(b) Is it possible to take out only blue balls from the bag so that the numbers of red balls, blue balls and green balls are in the ratio $5 : 4 : 7$ ? Explain your answer. (3)

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(a) $112$ . $\frac{3 k + 8}{4 k} = \frac{5}{6}$ gives $k = 24$ , so green is $5 (24) - 8$ .
(b) Yes. Red and green are now $80$ and $112$ , and $80 : 112 = 5 : 7$ already. One part is $16$ , so blue must be $64$ : take out $96 - 64 = 32$ blue balls.

Question 9 · Modelled on 2014 P1 Q10 (when do two movers meet) — New to this tier: a closed track with no graph given, so the meeting condition has to be worked out from relative speed

Ken and Leo run on a $400 m$ circular track at constant speeds in the ratio $5 : 4$ . Ken runs at $5 m/s$ . They start at the same point at the same time.

(a) Suppose that they run in the same direction. How long does it take for Ken to overtake Leo for the first time? (2)

(b) How many laps has Ken completed when he overtakes Leo for the first time in (a)? (2)

(c) Suppose instead that they run in opposite directions. Someone claims that they first meet at the point on the track opposite to the starting point. Do you agree? Explain your answer. (3)

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(a) $400$ seconds. Ken gains $5 - 4 = 1 m$ every second and must gain one full lap.
(b) $5$ laps. $5 \times 400 = 2000 m$ .
(c) Disagree. They close the $400 m$ at $9 m/s$ , meeting after $\frac{400}{9}$ s, when Ken has run $\frac{2000}{9} \approx 222 m$ , not $200 m$ .

Question 10 · Modelled on 2016 P2 Q13 and 2019 P2 Q12 (mixtures) — New to this tier: the things being mixed are themselves ratios, and one target mixture cannot be made

Alloy $X$ contains copper and zinc in the ratio $3 : 2$ by weight. Alloy $Y$ contains copper and zinc in the ratio $1 : 4$ by weight.

(a) $10 kg$ of alloy $X$ and $6 kg$ of alloy $Y$ are melted together. Find the ratio of the weight of copper to the weight of zinc in the new alloy. (2)

(b) In what ratio by weight should alloy $X$ and alloy $Y$ be melted together so that the new alloy contains equal weights of copper and zinc? (2)

(c) A craftsman claims that alloy $X$ and alloy $Y$ can be melted together to form a new alloy which contains copper and zinc in the ratio $2 : 1$ . Do you agree? Explain your answer. (3)

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(a) $9 : 11$ . Copper $= 6 + 1.2 = 7.2 kg$ out of $16 kg$ .
(b) $3 : 1$ . $\frac{3}{5} x + \frac{1}{5} y = \frac{1}{2} (x + y)$ gives $x = 3 y$ .
(c) Disagree. $2 : 1$ means copper is $\frac{2}{3}$ of the alloy, but any mixture has a copper fraction between $\frac{1}{5}$ and $\frac{3}{5}$ , and $\frac{2}{3} > \frac{3}{5}$ .

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

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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