Ratio & proportion – IGCSE Maths exam questions with mark scheme

1Grade 1–3FoundationCalculator2 marks

In a school orchestra there are 36 string players and 84 other players.

(a) Write the ratio string players : other players in its simplest form. [1]

(b) What fraction of the orchestra are string players? Give your answer in its simplest form. [1]

(a) $3 : 7$ (must be fully simplified; do not accept $6:14$ or $12:28$)B1
(b) $\frac{3}{10}$ (accept $\frac{36}{120}$ simplified; do not accept $\frac37$; ft their (a) $a:b$ giving $\frac{a}{a+b}$ if fully simplified)B1
Answer: (a) $3:7$ (b) $\frac{3}{10}$
Examiner tip: In (b) the fraction uses the total of the ratio as denominator: $3 + 7 = 10$, not $7$.

2Grade 1–3FoundationCalculator3 marks

Ayo, Bella and Chen share 168 stickers in the ratio $2 : 5 : 7$

Work out the number of stickers each person gets. [3]

$2 + 5 + 7 = 14$ and $168 \div 14\ (=12)$M1
for multiplying one share of 12 by 2, 5 or 7 (e.g. $2 \times 12 = 24$)M1
Ayo 24, Bella 60, Chen 84 (all three correct; check they total 168)A1
Answer: Ayo 24, Bella 60, Chen 84
Examiner tip: Dividing 168 by 3 (the number of people) instead of 14 (the number of parts) scores no marks.

3Grade 4–5Calculator3 marks

Rice is sold in two sizes of bag.

A small bag holds 400 g and costs £2.36
A large bag holds 650 g and costs £3.71

Which bag is the better value for money?
You must show all your working. [3]

for a correct method to find a comparable value for one bag, e.g. $236 \div 400\ (=0.59)$ pence per gram, or $400 \div 236\ (=1.69\ldots)$ g per penny, or the cost of 2600 g of eachM1
for a correct comparable value for both bags, e.g. $0.59$ and $0.5707\ldots$ (p per g), or $1.69\ldots$ and $1.75\ldots$ (g per p), or £15.34 and £14.84 for 2600 gM1
Large bag, supported by correct comparable values (values must be accurate enough to show the difference)C1
Answer: The large bag (0.571 p/g compared with 0.59 p/g)
Examiner tip: A conclusion of “large bag” with no working, or with values rounded so far that both look the same, gets no marks.

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4Grade 4–5Calculator4 marks

A map has a scale of $1 : 25\,000$

(a) On the map, the distance between two villages is 7.4 cm.
Work out the real distance between the villages. Give your answer in kilometres. [2]

(b) A footpath is 3.2 km long in real life.
How long is the footpath on the map? Give your answer in centimetres. [2]

(a) $7.4 \times 25\,000\ (= 185\,000$ cm$)$M1
(a) $1.85$ (km)A1
(b) $3.2 \times 100\,000\ (= 320\,000$ cm$)$ or $25\,000$ cm $= 0.25$ km seenM1
(b) $12.8$ (cm)A1
Answer: (a) 1.85 km (b) 12.8 cm
Examiner tip: Most lost marks come from unit conversion: there are 100 000 cm in 1 km, not 1000.

5Grade 4–5Calculator3 marks

Here is a list of ingredients for making 8 flapjacks.

Oats	120 g
Butter	75 g
Syrup	45 g

Kemi has 400 g of oats, 200 g of butter and 150 g of syrup.

Work out the greatest number of flapjacks Kemi can make. [3]

for a correct method for one ingredient, e.g. $400 \div 120 \times 8\ (=26.6\ldots)$ or $75 \div 8\ (=9.375$ g per flapjack$)$M1
for a complete method for all three ingredients, e.g. $26.6\ldots,\ 21.3\ldots,\ 26.6\ldots$ flapjacks, or a correct comparison showing butter is the limiting ingredientM1
21A1
Answer: 21
Examiner tip: Rounding 21.33… up to 22 loses the final mark — Kemi does not have enough butter for a 22nd flapjack.

6Grade 4–5Calculator4 marks

In a chess club, $\frac{2}{7}$ of the members are boys. All the other members are girls.

(a) Write the ratio boys : girls in the form $1 : n$ [2]

(b) There are 24 more girls than boys in the club.
Work out the total number of members. [2]

(a) $2 : 5$ seen (or $\frac57$ for girls)M1
(a) $1 : 2.5$ (accept $1 : \frac52$)A1
(b) $24 \div 3\ (= 8)$ (difference of 3 parts) or $\frac37$ of total $= 24$M1
(b) 56A1
Answer: (a) $1 : 2.5$ (b) 56
Examiner tip: Writing $2 : 7$ mixes a part with the whole — the girls are $7 - 2 = 5$ parts.

7Grade 6–7HigherCalculator3 marks

The ratio $a : b = 3 : 4$
The ratio $b : c = 6 : 5$

(a) Find the ratio $a : b : c$ Give your answer in its simplest form. [2]

(b) $c$ is 7 more than $a$.
Work out the value of $a + b + c$ [1]

(a) for writing both ratios with the same value for $b$, e.g. $9 : 12$ and $12 : 10$M1
(a) $9 : 12 : 10$A1
(b) 217 (ft their (a) if $c - a$ is a whole number of parts)B1
Answer: (a) $9 : 12 : 10$ (b) 217
Examiner tip: You cannot just join the ratios as $3 : 4 : 5$ — make the shared quantity $b$ equal in both first (LCM of 4 and 6 is 12).

8Grade 6–7HigherCalculator3 marks

A bag contains only red counters and blue counters.
The ratio of red counters to blue counters is $5 : 3$

Mo puts 12 more blue counters into the bag. Now the ratio of red counters to blue counters is $5 : 4$

How many counters were in the bag at the start? [3]

for setting up a correct equation or argument, e.g. $\frac{5x}{3x+12} = \frac54$ or noticing red stays the same so blue goes from 3 parts to 4 parts (with red 5 parts in both)M1
for finding the value of one part, e.g. $x = 12$, or 12 blue counters $=$ 1 partM1
96A1
Answer: 96
Examiner tip: The number of red counters does not change — use that fixed amount to link the two ratios.

9Grade 4–5Calculator4 marks

The graph shows the cost, in pounds, of buying different lengths of a fabric.

(a) Explain how the graph shows that the cost is directly proportional to the length. [1]

(b) Use the graph to find the cost of 3 metres of the fabric. [1]

(a) the graph is a straight line through the origin (both needed: straight line and passes through $(0, 0)$)C1
(b) £13.50 (accept 13.5)B1
(c) for finding the cost per metre, e.g. $18 \div 4\ (= 4.5)$, or a correct scaling method, e.g. $18.6 \div 4 \times 18$M1
(c) £83.70 (accept 83.7)A1
Answer: (a) straight line through $(0,0)$ (b) £13.50 (c) £83.70
Examiner tip: “It goes up steadily” is not enough — you must say it is a straight line that passes through the origin.

10Grade 8–9HigherCalculator4 marks

Alloy A is made from gold and silver in the ratio $3 : 1$ by mass.
Alloy B is made from gold and silver in the ratio $2 : 3$ by mass.

Alloy A and alloy B are melted together in the ratio $2 : 5$ by mass to make alloy C.

Show that alloy C contains equal masses of gold and silver. [4]

for finding the gold fraction in either alloy, e.g. $\frac34$ for A or $\frac25$ for B (or choosing a convenient mass, e.g. 28 g of A gives 21 g gold)M1
for finding the mass (or fraction) of A and B in alloy C, e.g. $\frac27$ and $\frac57$, or 140 g of C contains 40 g of A and 100 g of BM1
for a complete method to find the gold in C, e.g. $\frac27\times\frac34 + \frac57\times\frac25$ or $30 + 40 = 70$ g gold out of 140 gM1
for a fully correct conclusion, e.g. gold $= \frac12$ so silver $= \frac12$, ratio $1 : 1$, or 70 g gold and 70 g silverA1
Answer: Gold $= \frac{3}{14}+\frac{4}{14}=\frac12$ of alloy C, so gold : silver $= 1 : 1$
Examiner tip: Adding the ratios to get $5 : 4$ ignores how much of each alloy is used — work with actual masses or fractions.

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