IGCSE Maths 0580 Ratio and Proportion Past Paper Questions and Worked Solutions

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IGCSE Maths (0580) · Number · 1.11

Every ratio and proportion question from 14 Cambridge IGCSE Mathematics (0580) past papers (May/June 2025 to May/June 2026), syllabus section 1.11. 14 questions covering Core and Extended tiers, ordered from Core to Extended. Each question is labelled with its question paper code (for example 0580/11/M/J/26) and every final answer matches the official Cambridge mark scheme. Try each question first, then click “Show solution” to check your working. This page is part of our IGCSE Maths 0580 Number past paper questions.

0580 Ratio and Proportion past paper questions: Core (Papers 1 and 3)

14 marksCore

May/June 2026 · Paper 1 · 0580/11/M/J/26 · Q14

A bag contains 45 balls. The balls are red or green in the ratio red : green = 2 : 3. Carlos puts 4 more red balls and x more green balls into the bag. The balls are now in the ratio red : green = 1 : 2. Find the value of x.

Show solution
  1. The ratio 2 : 3 has 2 + 3 = 5 parts. One part = 45 ÷ 5 = 9 balls.
  2. Red = 2 × 9 = 18 balls and green = 3 × 9 = 27 balls.
  3. After adding 4 red: red = 18 + 4 = 22.
  4. The new ratio is 1 : 2, so there are twice as many green as red: green = 2 × 22 = 44.
  5. Extra green balls x = 44 − 27 = 17
Answer 17
22 marksCore

May/June 2026 · Paper 1 · 0580/12/M/J/26 · Q14

The total number of apples and oranges in a bowl is 20. The ratio apples : oranges = 7 : 3. Work out the number of apples.

Show solution
  1. The ratio 7 : 3 has 7 + 3 = 10 parts.
  2. One part = 20 ÷ 10 = 2 pieces of fruit.
  3. Apples = 7 parts = 7 × 2 = 14
Answer 14
32 marksCore

May/June 2026 · Paper 1 · 0580/13/M/J/26 · Q5(a)

Sam takes 75 minutes to make a cake. The time is in the ratio preparing : cooking : decorating = 6 : 10 : 9. Work out the number of minutes Sam takes to decorate the cake.

Show solution
  1. Add the parts of the ratio: 6 + 10 + 9 = 25 parts.
  2. One part = 75 ÷ 25 = 3 minutes.
  3. Decorating = 9 parts = 9 × 3 = 27 minutes
Answer 27 minutes
44 marksCore

Feb/Mar 2026 · Paper 3 · 0580/32/F/M/26 · Q6

The diagram shows a circle with a sector marked.

0580/32 February/March 2026 Q6 – circle with a minor sector and a reflex sector

(a) Measure the reflex sector angle.

(b) On the diagram, draw a line that divides the reflex sector into two smaller sectors with angles in the ratio 2 : 13.

Show solution
Part (a)
  1. It is easier to measure the small angle and subtract it from 360°.
  2. The small angle measures 30°, so the reflex angle = 360 − 30 = 330°
Answer 330°
Part (b)
  1. The ratio 2 : 13 has 2 + 13 = 15 parts.
  2. One part = 330 ÷ 15 = 22°.
  3. The two sectors are 2 × 22 = 44° and 13 × 22 = 286°
  4. Measure 44° from one edge of the reflex sector and draw a radius to split it.
0580/32 February/March 2026 Q6(b) solution – reflex sector split into 44° and 286°44°286°
Answer 44° 286°
53 marksCore

Oct/Nov 2025 · Paper 1 · 0580/11/O/N/25 · Q22

A recipe for 12 chocolate cookies uses 100 g flour, 55 g butter, 55 g sugar, 20 g cocoa and 1 egg.

(a) Sunita makes 24 chocolate cookies. Work out how much flour she needs.

(b) Albie has 28 g of cocoa. Does he have enough cocoa to make 18 cookies? Explain how you decide.

Show solution
Part (a)
  1. The recipe makes 12 cookies. 24 ÷ 12 = 2, so double every amount.
  2. Flour = 2 × 100 = 200 g
Answer 200 g
Part (b)
  1. 18 ÷ 12 = 1.5, so multiply every amount by 1.5.
  2. Cocoa needed = 1.5 × 20 = 30 g.
  3. No, because 18 cookies need 30 g of cocoa and he only has 28 g.
Answer No, because 18 cookies need 30 g of cocoa and he only has 28 g.
63 marksCore

Oct/Nov 2025 · Paper 1 · 0580/12/O/N/25 · Q15

Three sisters share $n in the ratio 10 : 3 : 2. The smallest share is $120. Find the value of n.

Show solution
  1. The smallest share is 2 parts, so 2 parts = $120.
  2. One part = 120 ÷ 2 = $60.
  3. Total parts = 10 + 3 + 2 = 15.
  4. n = 15 × 60 = 900
Answer 900
73 marksCore

Oct/Nov 2025 · Paper 1 · 0580/13/O/N/25 · Q16

Kai, Jo and Liz share some money in the ratio Kai : Jo : Liz = 4 : 7 : 13. Jo receives $1400. Find the total amount of money they share.

Show solution
  1. Jo’s share is 7 parts, and Jo receives $1400. So 7 parts = $1400.
  2. One part = 1400 ÷ 7 = $200.
  3. Total parts = 4 + 7 + 13 = 24.
  4. Total money = 24 × 200 = $4800
Answer $4800
82 marksCore

Oct/Nov 2025 · Paper 3 · 0580/31/O/N/25 · Q10

Anna and Bruno share some money in the ratio Anna : Bruno = 3 : 7. Bruno receives $36 more than Anna. Calculate the total amount of money they share.

Show solution
  1. Anna has 3 parts and Bruno has 7 parts, so Bruno has 7 − 3 = 4 parts more.
  2. Those 4 parts are worth $36, so 1 part = 36 ÷ 4 = $9.
  3. Total = 3 + 7 = 10 parts = 10 × 9 = $90
Answer $90
92 marksCore

Oct/Nov 2025 · Paper 3 · 0580/33/O/N/25 · Q13

Divide $136 in the ratio 3 : 5.

Show solution
  1. The ratio 3 : 5 has 3 + 5 = 8 parts.
  2. One part = 136 ÷ 8 = $17.
  3. 3 parts = 3 × 17 = $51 and 5 parts = 5 × 17 = $85.
  4. $51, $85
  5. Check: 51 + 85 = 136 ✓
Answer $51, $85
103 marksCore

May/June 2025 · Paper 1 · 0580/13/M/J/25 · Q7

70 students study one of French, Spanish and German. The ratio number who study French : number who study Spanish = 3 : 7. 15 students study French. Find the number of students who study German.

Show solution
  1. French : Spanish = 3 : 7, and French is 15 students, so 3 parts = 15.
  2. One part = 15 ÷ 3 = 5 students.
  3. Spanish = 7 parts = 7 × 5 = 35 students.
  4. German = total − French − Spanish = 70 − 15 − 35 = 20
Answer 20

0580 Ratio and Proportion past paper questions: Extended (Papers 2 and 4)

113 marksExtended

May/June 2026 · Paper 4 · 0580/43/M/J/26 · Q5

Ruta and Zoli share some money in the ratio 7 : 3. Ruta receives $420 more than Zoli. Work out how much money Ruta and Zoli each receive.

Show solution
  1. Ruta has 7 parts and Zoli has 3 parts, so Ruta has 7 − 3 = 4 parts more.
  2. These 4 parts are worth $420, so 1 part = 420 ÷ 4 = $105.
  3. Ruta = 7 × 105 = $735
  4. Zoli = 3 × 105 = $315
  5. Check: 735 − 315 = 420 ✓
Answer $735 $315
125 marksExtended

Feb/Mar 2026 · Paper 2 · 0580/22/F/M/26 · Q5(a,c)

A theatre has 3 types of seating. A stalls ticket costs $30, a circle ticket costs $24 and a private box ticket costs $72.

(a) Find the ratio of the cost of the tickets stalls : circle : private box. Give your answer in its simplest form.

(c) There are 1200 seats in the stalls. One evening, 5⁄6 of the tickets for seats in the stalls are sold. On this evening, the number of tickets sold for seats is in the ratio stalls : circle = 10 : 7. Find the number of tickets sold for seats in the circle.

Show solution
Part (a)
  1. Write the ratio of the costs: 30 : 24 : 72.
  2. The highest common factor of 30, 24 and 72 is 6. Divide every part by 6.
  3. 5 : 4 : 12
Answer 5 : 4 : 12
Part (c)
  1. Stall tickets sold = 5⁄6 × 1200 = 1200 ÷ 6 × 5 = 1000.
  2. Stalls : circle = 10 : 7, so 10 parts = 1000 and 1 part = 100.
  3. Circle = 7 parts = 7 × 100 = 700
Answer 700
132 marksExtended

Oct/Nov 2025 · Paper 2 · 0580/21/O/N/25 · Q1

Divide $90 in the ratio 2 : 3.

Show solution
  1. Add the parts of the ratio: 2 + 3 = 5 parts.
  2. One part = 90 ÷ 5 = $18.
  3. Shares: 2 × 18 = $36 and 3 × 18 = $54.
  4. $36, $54
  5. Check: 36 + 54 = 90 ✓
Answer $36, $54
142 marksExtended

Oct/Nov 2025 · Paper 4 · 0580/42/O/N/25 · Q1

Write the ratio 60 grams : 3 kilograms in the form 1 : n.

Show solution
  1. Both parts of a ratio must be in the same units. 3 kg = 3 × 1000 = 3000 g.
  2. The ratio is 60 : 3000.
  3. To make the first part 1, divide both parts by 60: 60 ÷ 60 = 1 and 3000 ÷ 60 = 50.
  4. 1 : 50
Answer 1 : 50

Question papers used on this page

All ratio and proportion questions above, listed by 0580 question paper code.

QP codeFile namePaperQuestions
0580/11/M/J/260580_s26_qp_11May/June 2026 · Paper 1 (Core)Q14
0580/12/M/J/260580_s26_qp_12May/June 2026 · Paper 1 (Core)Q14
0580/13/M/J/260580_s26_qp_13May/June 2026 · Paper 1 (Core)Q5(a)
0580/43/M/J/260580_s26_qp_43May/June 2026 · Paper 4 (Extended)Q5
0580/22/F/M/260580_m26_qp_22Feb/Mar 2026 · Paper 2 (Extended)Q5(a,c)
0580/32/F/M/260580_m26_qp_32Feb/Mar 2026 · Paper 3 (Core)Q6
0580/11/O/N/250580_w25_qp_11Oct/Nov 2025 · Paper 1 (Core)Q22
0580/12/O/N/250580_w25_qp_12Oct/Nov 2025 · Paper 1 (Core)Q15
0580/13/O/N/250580_w25_qp_13Oct/Nov 2025 · Paper 1 (Core)Q16
0580/21/O/N/250580_w25_qp_21Oct/Nov 2025 · Paper 2 (Extended)Q1
0580/31/O/N/250580_w25_qp_31Oct/Nov 2025 · Paper 3 (Core)Q10
0580/33/O/N/250580_w25_qp_33Oct/Nov 2025 · Paper 3 (Core)Q13
0580/42/O/N/250580_w25_qp_42Oct/Nov 2025 · Paper 4 (Extended)Q1
0580/13/M/J/250580_s25_qp_13May/June 2025 · Paper 1 (Core)Q7

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