Repeated change Quiz (Grade 6)

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Score: 0 / 10

Repeated change — Notes & Worked Examples Show / Hide

Explanation:

For repeated percentage change (compound interest or depreciation) over n years, use the multiplier formula: Final Amount = Initial Amount × (multiplier)ⁿ.

For an increase of r%, the multiplier is 1 + r100 (e.g. +3.2% → 1.032). For a decrease of r%, the multiplier is 1 − r100 (e.g. −12% → 0.88). If the percentage rate changes from year to year, multiply by each year's multiplier in turn.

Example 1: £4500 is invested at 3.2% compound interest per year for 4 years. Work out the value of the investment after 4 years.

Solution:

Multiplier for +3.2% = 1 + 0.032 = 1.032

Value = 4500 × 1.032⁴

Value = £5104.243...

Answer: £5104.24 (to the nearest penny)

Example 2: A car bought for £18,000 depreciates by 20% in Year 1, then by 12% per year for the next 2 years. Find its value after 3 years.

Solution:

Year 1 multiplier = 0.80; Years 2 & 3 multiplier = 0.88

Value = 18,000 × 0.80 × 0.88²

Value = 14,400 × 0.7744 = £11,151.36

Answer: £11,151.36

One Page Guide (PDF) Repeated change Worksheets

Question 1

[1 mark]

A piece of machinery costs £4000 and depreciates by 10% each year.
Calculate its value after 2 years.

Worked Solution & Answer

4000 × 0.9² = £3240 B1

Answer: £ £3240

Question 2

[1 mark]

£2000 is invested in an account paying 3% compound interest per annum
How much is in the account after 3 years?

Worked Solution & Answer

Final Amount = Initial Amount × (1 ± rate100)ⁿ
£2185.45

Answer: £ £2185.45

Question 3 IGCSE Jun 2023 2F Q24 / 2H Q8

[3 marks]

Charlotte buys a painting for \$680
The value of the painting increases by 4% each year.
Work out the value of the painting at the end of 3 years.
Give your answer correct to the nearest \$

Worked Solution & Answer

680 × 1.04³ = 764.90752 -> $765. M2 A1

Answer: $ 765

Question Paper (PDF) • Mark Scheme (PDF)

Question 4 IGCSE Nov 2023 2F Q23 / 2H Q7

[3 marks]

Hermione buys a boat for \$26 800
The value of the boat depreciates by 8% each year.
Work out the value of the boat at the end of 3 years.
Give your answer correct to the nearest dollar.

Worked Solution & Answer

26 800 × 0.92³ = 20 868.8384 ≈ $20 869. M2 A1

Answer: $ 20 869

Question Paper (PDF) • Mark Scheme (PDF)

Question 5 AQA Jun 2023 3F Q18 / 3H Q5

[3 marks]

The number of hedgehogs in England is expected to **reduce** by 4% each year.
Assume there are now 1 000 000 hedgehogs in England.
Work out the expected number of hedgehogs in England after **five** years.
You **must** show your working.

Answer

Worked Solution & Answer

Multiplier = 1 – 0.04 = 0.96. M2
1 000 000 × 0.96⁵ = 815 372.6976 ≈ 815 373. A1

Answer: Answer 815 373

Question Paper (PDF) • Mark Scheme (PDF)

Question 6 AQA Jun 2025 2H Q16

[3 marks]

The value of a second-hand car decreases by 12.1% per year.

Work out the number of full years until the car loses **more** than half its value.
You **must** show your working.

Answer years

Worked Solution & Answer

Multiplier = 1 – 0.121 = 0.879. M1
0.879⁵ ≈ 0.525 (52.5% remaining). M1
0.879⁶ ≈ 0.461 (46.1% remaining, so lost more than half)
Answer = 6 years. A1

Answer: Answer 6 years

Question Paper (PDF) • Mark Scheme (PDF)

Question 7 OCR Jun 2024 5H Q12

[5 marks]

(a) Sasha invests £1000 at a rate of 5% per year compound interest.
Sasha says:
"After one year, my investment will get £50 in interest and will be worth £1050.
Therefore, after two years, my investment will get another £50 in interest and will be worth £1100."
Is Sasha correct?
Give a reason for your answer.
(b) Sasha buys a car.
The value, £V, of the car after n years is given by the formula
V = a × bⁿ.
The graph shows some information about the value of the car.

(b) Sasha buys a car.
The value, £V, of the car after n years is given by the formula
V = a × bⁿ.
The graph shows some information about the value of the car. [4 marks]

Find the value of a and the value of b.

Worked Solution & Answer

(a)

Compound interest is calculated on the new amount (£1050) in year 2, not the original £1000. C1

Answer (a): No, because in the second year the 5% interest is calculated on £1050 (giving £52.50 interest, total £1102.50).

(b)

When n = 0, V = 8000 ⇒ a = 8000 M3
When n = 1, V = 6400 ⇒ 8000 × b = 6400 ⇒ b = 6400 ÷ 8000 = 0.8. A1

Answer (b): a = 8000, b = 0.8

Question Paper (PDF) • Mark Scheme (PDF)

Question 8 OCR Nov 2017 5H Q11

[5 marks]

Amelia buys a new car.
The expected future value of this car, £V, is given by
V = 16 000 × 0.75^t
where t is the age of the car in complete years.

(a) Write down the value of the car when new. [1 mark]

(b) Write down the annual percentage decrease in the expected value of the car. [1 mark]

(b) Amelia sketches a graph to show the expected value of her car as it gets older. [1 mark]

Explain how you know that Amelia's graph is incorrect.

Worked Solution & Answer

(a)

When t = 0, V = 16 000 × 0.75⁰ = £16 000. B1

Answer (a): £ 16000

(b)

Multiplier is 0.75 = 1 − 0.25, so the annual percentage decrease is 25%. B1

Answer (b): 25 %

(c)

When t = 2: V = 16 000 × 0.75² = 16 000 × (9/16) = £9000. M1 C1

Answer (c): 16000 × 0.75² = 16000 × (3/4) × (3/4) = 12000 × 3/4 = 9000

(b)

Exponential decay produces a decreasing curve rather than a straight line (it decreases by £4000 in year 1 and £3000 in year 2). C1

Answer (b): The graph should be a curve, not a straight line, because the value does not decrease by a constant amount each year.

Question Paper (PDF) • Mark Scheme (PDF)

Question 9 OCR Jun 2022 3F Q24

[6 marks]

1600 fish are released into a new lake which has no fish.
The number of fish is expected to increase by 5% each year.

(a) The table shows the expected number of fish in the lake at the end of 1 year and at the end of 2 years.
Complete the table.
Round your answers to the nearest integer.
@@TABULAR₀@@ [3 marks]

(b) Use the table to draw a suitable graph to show the expected number of fish in the lake. [3 marks]

Worked Solution & Answer

(a)

Year 3: 1764 × 1.05 = 1852.2 ≈ 1852 M2
Year 4: 1852.2 × 1.05 = 1944.81 ≈ 1945. A1

Answer (a): 1852 and 1945

(b)

Plot all 5 points accurately and connect them. M2 A1

Answer (b): Points plotted at (0, 1600), (1, 1680), (2, 1764), (3, 1852), (4, 1945) and joined with a curve/line segments

Question Paper (PDF) • Mark Scheme (PDF)

Question 10 IGCSE Jun 2022 1F Q21 / 1H Q6

[6 marks]

Asha bought an apartment.
The table gives information about the value of apartments, in euros, and the annual service charge band.

In 2021, the value of Asha’s apartment was 634 400 euros.
The value of Asha’s apartment had increased by 4% from its value in 2020

(a) Has the annual service charge band changed for Asha’s apartment?
Show your working clearly. [3 marks]

(b) Work out the total percentage by which the value of the boat had depreciated by the end of the second year after Pam bought the boat. [3 marks]

Worked Solution & Answer

(a)

634 400 ÷ 1.04 = 610 000 euros (600 000 ≤ x < 700 000, still Band B) M2 A1

Answer (a): No, in 2020 the value was 610 000 euros so it stayed in Band B

(b)

1 – 0.85² = 1 – 0.7225 = 0.2775 = 27.75% M2 A1

Answer (b): 27.75 %

Question Paper (PDF) • Mark Scheme (PDF)

Score: 0 / 10 (0 / 36 marks)

Repeated change Worksheets Next: Triple brackets →

Grade 6 Quizzes

14 Topics

6A - Recurring decimals	6B - Harder indices
6C - The product rule	6D - Repeated change
6E - Triple brackets	6F - Parallel and perpendicular
6G - Inequalities on graphs	6H - Similar shapes (area and vol)
6J - Negative scale factors	6K - Circle theorems
6L - Cumulative frequency	6M - Box plots
6N - Combined transformations	6P - Capture recapture