IGCSE Maths 0580 Percentages Past Paper Questions and Worked Solutions

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IGCSE Maths (0580) · Number · 1.13

Every percentages question from 19 Cambridge IGCSE Mathematics (0580) past papers (May/June 2025 to May/June 2026), syllabus section 1.13. 26 questions covering Core and Extended tiers, ordered from Core to Extended. Each question is labelled with its question paper code (for example 0580/12/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 Percentages past paper questions: Core (Papers 1 and 3)

11 markCore

May/June 2026 · Paper 1 · 0580/12/M/J/26 · Q2(b)

Work out 50% of 526.

Show solution
  1. 50% means half.
  2. 526 ÷ 2 = 263
Answer 263
24 marksCore

May/June 2026 · Paper 1 · 0580/13/M/J/26 · Q15

(a) In a restaurant the cost of a pizza is $25. The restaurant reduces the cost of the pizza to $19. Work out the percentage decrease in the cost.

(b) In another restaurant, the cost of a pizza is $15. This restaurant adds a service charge of 10% to each pizza. Find the cost of a pizza, including the service charge.

Show solution
Part (a)
  1. Decrease = old price − new price = 25 − 19 = $6.
  2. Percentage decrease = decrease⁄original × 100 = 6⁄25 × 100.
  3. = 0.24 × 100 = 24%
Answer 24%
Part (b)
  1. 10% of $15 = 15 ÷ 10 = $1.50.
  2. Cost with service charge = 15 + 1.50 = $16.50
Answer $16.50
31 markCore

May/June 2026 · Paper 3 · 0580/31/M/J/26 · Q4

Find 18% of $65.

Show solution
  1. 18% as a decimal is 0.18.
  2. 18% of $65 = 0.18 × 65 = 11.7 = $11.70
  3. Tip: Money is written with 2 decimal places, so write $11.70, not $11.7.
Answer $11.70
43 marksCore

May/June 2026 · Paper 3 · 0580/31/M/J/26 · Q22

Joseph invests $36 000 at a rate of 6.2% per year compound interest. Calculate the total interest earned at the end of 4 years.

Show solution
  1. Compound interest: a 6.2% increase each year means multiply by 1.062 each year.
  2. Value after 4 years = 36 000 × 1.062⁴ = 36 000 × 1.27203… = 45 793.15…
  3. Interest = final value − amount invested = 45 793.15 − 36 000 = $9793.16
  4. Tip: To 3 significant figures this is $9790.
Answer $9793.16
52 marksCore

May/June 2026 · Paper 3 · 0580/32/M/J/26 · Q10

Dave invests $8500 at a rate of 4.5% per year simple interest. Calculate the total interest earned at the end of 3 years.

Show solution
  1. Simple interest is the same every year. 4.5% as a decimal is 0.045.
  2. Interest for 1 year = 8500 × 0.045 = $382.50.
  3. Interest for 3 years = 382.50 × 3 = $1147.50
Answer $1147.50
62 marksCore

May/June 2026 · Paper 3 · 0580/33/M/J/26 · Q16

Jo invests $34 500 for 5 years at a rate of 4.4% per year simple interest. Calculate the total interest earned during the 5 years.

Show solution
  1. Simple interest is the same every year. 4.4% as a decimal is 0.044.
  2. Interest for 1 year = 34 500 × 0.044 = $1518.
  3. Interest for 5 years = 1518 × 5 = $7590
Answer $7590
73 marksCore

Feb/Mar 2026 · Paper 3 · 0580/32/F/M/26 · Q22

Kim invests $14 700 at a rate of 4.8% per year compound interest. Calculate the value of her investment at the end of 5 years. Give your answer to the nearest dollar.

Show solution
  1. A 4.8% increase each year means multiply by 1.048 each year.
  2. Value after 5 years = 14 700 × 1.048⁵ = 14 700 × 1.26417… = 18 583.33…
  3. Round to the nearest dollar: $18 583
Answer $18 583
82 marksCore

Feb/Mar 2026 · Paper 3 · 0580/32/F/M/26 · Q24

A man buys a painting for $1200. He sells the painting for $1360. Calculate the percentage profit the man makes on the painting.

Show solution
  1. Profit = selling price − buying price = 1360 − 1200 = $160.
  2. Percentage profit = profit⁄original price × 100 = 160⁄1200 × 100 = 13.33…
  3. 13.3%
Answer 13.3%
93 marksCore

Oct/Nov 2025 · Paper 1 · 0580/11/O/N/25 · Q9

(a) A grid has 20 equal rectangles. Shade 3⁄4 of this grid. (b) Work out 15% of 80.

0580/11 October/November 2025 Q9(a) – grid of 20 rectangles
Show solution
Part (a)
  1. 3⁄4 of 20 = 20 ÷ 4 × 3 = 5 × 3 = 15.
  2. Shade 15 rectangles
Answer 15 rectangles
Part (b)
  1. Find 10% first: 10% of 80 = 80 ÷ 10 = 8.
  2. 5% is half of 10%: 8 ÷ 2 = 4.
  3. 15% = 10% + 5% = 8 + 4 = 12
Answer 12
103 marksCore

Oct/Nov 2025 · Paper 1 · 0580/13/O/N/25 · Q14

Victoria invests $2000 at a rate of 5% per year simple interest. Calculate the value of her investment after 4 years.

Show solution
  1. Simple interest is the same amount every year.
  2. Interest for one year = 5% of 2000 = 2000 ÷ 100 × 5 = $100.
  3. Interest for 4 years = 4 × 100 = $400.
  4. Value = 2000 + 400 = $2400
Answer $2400
113 marksCore

Oct/Nov 2025 · Paper 3 · 0580/31/O/N/25 · Q11

(a) Naz pays $76.25 per day to hire a car for 7 days. He also pays $146 for fuel. Calculate the total amount Naz pays.

(b) Li pays $1400 to hire a car. $364 of this is insurance. Calculate the insurance as a percentage of the $1400.

Show solution
Part (a)
  1. Car hire for 7 days: 76.25 × 7 = $533.75.
  2. Add the fuel: 533.75 + 146 = $679.75
Answer $679.75
Part (b)
  1. Write the insurance as a fraction of the total: 364⁄1400.
  2. Change to a percentage: 364⁄1400 × 100 = 26%
Answer 26%
123 marksCore

Oct/Nov 2025 · Paper 3 · 0580/31/O/N/25 · Q19

Jo invests $500 at a rate of 2% per year compound interest. Calculate the amount of interest earned at the end of 7 years.

Show solution
  1. A 2% increase each year means multiply by 1.02 each year.
  2. Value after 7 years = 500 × 1.02⁷ = 500 × 1.14868… = 574.34…
  3. Interest = 574.34 − 500 = $74.34
Answer $74.34
131 markCore

Oct/Nov 2025 · Paper 3 · 0580/32/O/N/25 · Q16

Find 57% of $128.

Show solution
  1. 57% as a decimal is 0.57.
  2. 0.57 × 128 = $72.96
Answer $72.96
145 marksCore

Oct/Nov 2025 · Paper 3 · 0580/32/O/N/25 · Q28

Ali invests $2950 in an account paying 12% compound interest. Bob invests $3000 in an account paying 12% simple interest. Show that after 3 years Ali has a larger investment than Bob.

Show solution
  1. Ali (compound interest): multiply by 1.12 for each of the 3 years.
  2. 2950 × 1.12³ = 2950 × 1.404928 = $4144.54
  3. Bob (simple interest): interest each year = 3000 × 0.12 = $360.
  4. Interest for 3 years = 3 × 360 = $1080, so his total = 3000 + 1080 = $4080
  5. Compare: 4144.54 > 4080, so Ali has the larger investment ✓
Answer $4144.54 $4080
152 marksCore

Oct/Nov 2025 · Paper 3 · 0580/33/O/N/25 · Q11

Calculate 19% of $46.25. Give your answer correct to the nearest cent.

Show solution
  1. 19% as a decimal is 0.19.
  2. 0.19 × 46.25 = 8.7875.
  3. A cent is 0.01 of a dollar, so round to 2 decimal places: $8.79
Answer $8.79
162 marksCore

Oct/Nov 2025 · Paper 3 · 0580/33/O/N/25 · Q23(a)

Jo buys a candle for $3.20. She sells the candle for $4.64. Find the percentage profit on the candle.

Show solution
  1. Profit = selling price − buying price = 4.64 − 3.20 = $1.44.
  2. Percentage profit = profit⁄original price × 100 = 1.44⁄3.20 × 100 = 45%
Answer 45%
173 marksCore

May/June 2025 · Paper 1 · 0580/13/M/J/25 · Q16

On Monday the cost of a concert ticket is $x. On Tuesday the cost of a ticket for the same concert is 20% more than the cost on Monday. Jack buys 4 tickets on Monday and 5 tickets on Tuesday. Jack pays $270 in total. Work out the value of x.

Show solution
  1. A 20% increase means multiply by 1.2. Tuesday’s price = 1.2x.
  2. Total cost: 4 tickets at x + 5 tickets at 1.2x = 270.
  3. 4x + 6x = 270, so 10x = 270.
  4. x = 27
  5. Check: 4 × 27 + 5 × 32.40 = 108 + 162 = 270 ✓
Answer 27

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

182 marksExtended

May/June 2026 · Paper 2 · 0580/23/M/J/26 · Q1

Write $30 as a percentage of $120.

Show solution
  1. Write the part as a fraction of the whole: 30⁄120.
  2. Simplify: 30⁄120 = 1⁄4.
  3. Change to a percentage by multiplying by 100: 1⁄4 × 100 = 25%
Answer 25%
193 marksExtended

May/June 2026 · Paper 4 · 0580/42/M/J/26 · Q4

A television costs $1260. Liesl pays a deposit of 35% of the cost of the television. She then pays monthly payments of $45.50 for the remaining cost of the television. Find the number of monthly payments that Liesl pays.

Show solution
  1. 35% as a decimal is 0.35. Deposit = 0.35 × 1260 = $441.
  2. Remaining cost = 1260 − 441 = $819.
  3. Number of payments = 819 ÷ 45.50 = 18
  4. Check: 18 × 45.50 = 819 ✓
Answer 18
204 marksExtended

May/June 2026 · Paper 4 · 0580/42/M/J/26 · Q10

Yasmin collects coins. 10% of her coins are gold. 25% of the remaining coins are silver. The other 54 coins are copper.

(a) Calculate the number of silver coins. (b) Calculate the total number of coins.

Show solution
Part (a)
  1. 10% are gold, so 90% are not gold.
  2. 25% of the non-gold coins are silver, so the other 75% of them are copper.
  3. Copper as a fraction of all coins = 0.9 × 0.75 = 0.675.
  4. Total coins = 54 ÷ 0.675 = 80, so the non-gold coins = 0.9 × 80 = 72.
  5. Silver = 25% of 72 = 0.25 × 72 = 18 coins
Answer 18 coins
Part (b)
  1. From part (a): total = 54 ÷ 0.675 = 80 coins
  2. Check: gold 8 + silver 18 + copper 54 = 80 ✓
Answer 80 coins
211 markExtended

May/June 2026 · Paper 4 · 0580/43/M/J/26 · Q2

Work out 55% of 12.2.

Show solution
  1. 55% as a decimal is 0.55.
  2. 0.55 × 12.2 = 6.71
Answer 6.71
223 marksExtended

May/June 2026 · Paper 4 · 0580/43/M/J/26 · Q19(a)

Lee invests $2000 in an account that pays simple interest at a rate of 3.5% per year. Work out the total value of the investment at the end of 4 years.

Show solution
  1. Simple interest is the same every year. 3.5% as a decimal is 0.035.
  2. Interest for 1 year = 2000 × 0.035 = $70. For 4 years: 4 × 70 = $280.
  3. Total value = 2000 + 280 = $2280
Answer $2280
233 marksExtended

Feb/Mar 2026 · Paper 2 · 0580/22/F/M/26 · Q5(b)

A stalls ticket costs $30 and a circle ticket costs $24. A website charges a fee of 5% of the cost of each ticket. Arshad buys 2 tickets in the stalls and 3 tickets in the circle. Work out the total cost of Arshad’s tickets.

Show solution
  1. Cost of the tickets: 2 × $30 + 3 × $24 = $60 + $72 = $132.
  2. Fee = 5% of $132 = 132 × 0.05 = $6.60.
  3. Total = $132 + $6.60 = $138.60
Answer $138.60
242 marksExtended

Oct/Nov 2025 · Paper 2 · 0580/22/O/N/25 · Q4

A computer costs $560. In a sale, this cost is reduced by 20%. Find the cost of the computer in the sale.

Show solution
  1. Find 20% of $560: 10% = 56, so 20% = 2 × 56 = 112.
  2. Sale price = 560 − 112 = $448
  3. Tip: You can also multiply by 0.8, because 100% − 20% = 80%. 560 × 0.8 = 448.
Answer $448
254 marksExtended

Oct/Nov 2025 · Paper 4 · 0580/41/O/N/25 · Q11

In a sale, the prices of coats are reduced by 15%.

(a) The original price of a coat is $60. Calculate the sale price.

(b) The sale price of a different coat is $58.14. Calculate the original price of this coat.

Show solution
Part (a)
  1. A 15% reduction leaves 100% − 15% = 85% of the price.
  2. 85% as a decimal is 0.85 (the multiplier).
  3. Sale price = 60 × 0.85 = $51
Answer $51
Part (b)
  1. The sale price $58.14 is 85% of the original price.
  2. Original × 0.85 = 58.14
  3. Original = 58.14 ÷ 0.85 = $68.40
  4. Check: 68.40 × 0.85 = 58.14 ✓
Answer $68.40
263 marksExtended

May/June 2025 · Paper 2 · 0580/21/M/J/25 · Q3

Sally invests $1500 at 3% per year simple interest. Work out the total value of her investment at the end of 6 years.

Show solution
  1. Simple interest is the same every year. Interest for one year = 3% of 1500 = 1500 × 0.03 = $45.
  2. Interest for 6 years = 6 × 45 = $270.
  3. Total value = 1500 + 270 = $1770
Answer $1770

Question papers used on this page

All percentages questions above, listed by 0580 question paper code.

QP codeFile namePaperQuestions
0580/12/M/J/260580_s26_qp_12May/June 2026 · Paper 1 (Core)Q2(b)
0580/13/M/J/260580_s26_qp_13May/June 2026 · Paper 1 (Core)Q15
0580/23/M/J/260580_s26_qp_23May/June 2026 · Paper 2 (Extended)Q1
0580/31/M/J/260580_s26_qp_31May/June 2026 · Paper 3 (Core)Q4, Q22
0580/32/M/J/260580_s26_qp_32May/June 2026 · Paper 3 (Core)Q10
0580/33/M/J/260580_s26_qp_33May/June 2026 · Paper 3 (Core)Q16
0580/42/M/J/260580_s26_qp_42May/June 2026 · Paper 4 (Extended)Q4, Q10
0580/43/M/J/260580_s26_qp_43May/June 2026 · Paper 4 (Extended)Q2, Q19(a)
0580/22/F/M/260580_m26_qp_22Feb/Mar 2026 · Paper 2 (Extended)Q5(b)
0580/32/F/M/260580_m26_qp_32Feb/Mar 2026 · Paper 3 (Core)Q22, Q24
0580/11/O/N/250580_w25_qp_11Oct/Nov 2025 · Paper 1 (Core)Q9
0580/13/O/N/250580_w25_qp_13Oct/Nov 2025 · Paper 1 (Core)Q14
0580/22/O/N/250580_w25_qp_22Oct/Nov 2025 · Paper 2 (Extended)Q4
0580/31/O/N/250580_w25_qp_31Oct/Nov 2025 · Paper 3 (Core)Q11, Q19
0580/32/O/N/250580_w25_qp_32Oct/Nov 2025 · Paper 3 (Core)Q16, Q28
0580/33/O/N/250580_w25_qp_33Oct/Nov 2025 · Paper 3 (Core)Q11, Q23(a)
0580/41/O/N/250580_w25_qp_41Oct/Nov 2025 · Paper 4 (Extended)Q11
0580/13/M/J/250580_s25_qp_13May/June 2025 · Paper 1 (Core)Q16
0580/21/M/J/250580_s25_qp_21May/June 2025 · Paper 2 (Extended)Q3

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