Worksheets · Foundation and Higher

Original values, profit and simple interest

8 exam-style questions, grades 2 to 7. Worked solutions and the marks are on the last page.

  1. Question 1Non-calculator · 3 marks

    (a) £600 is invested at 3% simple interest per year for 4 years. Work out the total interest earned. (3)

  2. Question 2Calculator · 4 marks

    You may use a calculator.

    (a) Nadia invests £2400 for 5 years at 3%3\% per year simple interest. Work out the total interest she earns. (2)

    (b) A trader buys a lamp for £40 and sells it for £52. Work out the percentage profit. (2)

  3. Question 3Non-calculator · 1 mark

    (a) After a 20% increase, a repair charge is £90. What was the charge before the increase? Select one answer. (1)

    1. £72
    2. £108
    3. £112.50
    4. £75
  4. Question 4Non-calculator · 4 marks

    A trader sells a table for £156, making a profit of 30% of its cost price. Delivery had cost the trader another £12 and was not included in the cost price.

    (a) Find the cost price of the table. (2)

    (b) Find the profit after allowing for the delivery cost. (2)

  5. Question 5Calculator · 4 marks

    You may use a calculator.

    (a) In a sale, the price of a television is reduced by 15%15\%. The sale price is £323. Work out the price before the sale. (2)

    (b) Ravi's rent went up by 4%4\%. His new rent is £832 per month. What was his rent before the increase? (2)

  6. Question 6Non-calculator · 3 marks

    (a) A shop reduces a lamp by 25%, then takes another £9 off the reduced price. The final price is £45. Work out the original price. You must show your working. (3)

  7. Question 7Calculator · 4 marks

    A shop sells jackets for £72 each. At this price the shop makes a profit of 20%20\% on the cost price. In a sale, the shop reduces the selling price by 15%15\%.

    (a) Work out the shop's percentage profit on each jacket sold in the sale. (4)

  8. Question 8Non-calculator · 1 mark

    (a) A price is increased by p%, where p > 0. It is then reduced by q% to return exactly to its original value. Which expression gives q in terms of p? Select one answer. (1)

    1. p
    2. 100p/(100 −- p)
    3. 100p/(100 + p)
    4. p/100

Worked solutions and marks

Question 1

(a) £7272

  1. 600×0.03600\times 0.03
  2. 18×418\times 4
  3. Simple interest uses the original £600 each year.
  4. 600 ×\times 0.03 ×\times 4 = £72 interest.
  • P1 Establishing 600×0.03600\times 0.03 or an equivalent valid method.
  • P1 Establishing 18×418\times 4 or an equivalent valid method.
  • A1 Correct answer: £7272

Question 2

(a) £360

  1. Interest each year: 0.03×2400=£720.03 \times 2400 = £72.
  2. Over 5 years: 5×72=£3605 \times 72 = £360.
  • M1 Finding one year's interest, £72.
  • A1 The correct answer, £360.

(b) 30%30\%

  1. Profit: 52−40=£1252 - 40 = £12.
  2. As a percentage of the cost price:
    1240×100=30%\frac{12}{40} \times 100 = 30\%
  • M1 Writing 1240\frac{12}{40}.
  • A1 The correct answer, 30%30\%.

Question 3

(a) £75

  1. The final £90 is 120% of the original charge.
  2. Original charge = 90 ÷\div 1.2 = £75.
  • B1 Correct answer: £75

Question 4

(a) £120120

  1. The selling price is 130% of the cost price.
    156/1.3156/1.3
  2. Therefore £120120.
  • P1 The selling price is 130% of the cost price.
  • A1 Correct answer: £120120

(b) £2424

  1. Subtract both the table cost and delivery from the income.
    156−120−12156-120-12
  2. Therefore £2424.
  • M1 Subtract both the table cost and delivery from the income.
  • A1 Correct answer: £2424

Question 5

(a) £380

  1. The sale price is 85%85\% of the original price.
  2. 0.85×original=323⇒original=323÷0.85=3800.85 \times \text{original} = 323 \Rightarrow \text{original} = 323 \div 0.85 = 380
  • M1 Recognising that £323 is 85%85\% (writing 0.850.85 or 32385\frac{323}{85}).
  • A1 The correct answer, £380.

(b) £800

  1. £832 is 104%104\% of the old rent.
  2. 832÷1.04=800832 \div 1.04 = 800
  • M1 Writing 832÷1.04832 \div 1.04.
  • A1 The correct answer, £800.

Question 6

(a) £7272

  1. 45+945+9
  2. 54/0.7554/0.75
  3. Undo the £9 reduction first: £45 + £9 = £54.
  4. £54 is 75% of the original price.
  5. Original price = £54 ÷\div 0.75 = £72.
  • P1 Establishing 45+945+9 or an equivalent valid method.
  • P1 Establishing 54/0.7554/0.75 or an equivalent valid method.
  • A1 Correct answer: £7272

Question 7

(a) 2%2\% profit

  1. £72 is 120%120\% of the cost price.
    cost=72÷1.2=60\text{cost} = 72 \div 1.2 = 60
  2. Sale price:
    72×0.85=61.2072 \times 0.85 = 61.20
  3. Profit as a percentage of cost:
    61.20−6060×100=2%\frac{61.20 - 60}{60} \times 100 = 2\%
  • P1 Finding the cost price: 72÷1.2=£6072 \div 1.2 = £60.
  • P1 Finding the sale price: 72×0.85=£61.2072 \times 0.85 = £61.20.
  • P1 Finding the profit £1.20 and dividing by the cost, £60.
  • A1 The correct answer, 2%2\% profit.

Question 8

(a) 100p/(100 + p)

  1. Returning to the original requires (1 + p/100)(1 −- q/100) = 1.
  2. So 1 −- q/100 = 100/(100 + p).
  3. Hence q = 100p/(100 + p).
  • B1 Correct answer: 100p/(100 + p)

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Original values, profit and simple interest

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Independent practice for Pearson Edexcel GCSE Mathematics (1MA1), not endorsed by Pearson.

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