Repeated Percentage Change: Fluency and Exam Drill - Worksheets, Questions and Revision

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6.4D Repeated Percentage Change: Fluency and Exam Drill

EDEXCEL 1MA1 · Calculator allowed · about 70 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

Write down the single decimal multiplier for a percentage increase of 8%.

(Total for Question 1 is 1 mark)

Write down the single decimal multiplier for a percentage decrease of 15%.

(Total for Question 2 is 1 mark)

Write down the single decimal multiplier for a percentage increase of 3.5%.

(Total for Question 3 is 1 mark)

Write down the single decimal multiplier for a percentage decrease of 22%.

(Total for Question 4 is 1 mark)

A shop increases the price of a jacket by 20%. In a later sale, the new price is decreased by 20%. Explain why the sale price is not equal to the original price, and state whether the sale price is higher or lower than the original price.

(Total for Question 5 is 2 marks)

The price of a laptop is £40. It increases by 5% per year for 2 years. Calculate the new price.

(Total for Question 6 is 2 marks)

A car is worth £12,000. It depreciates by 10% each year. Find its value after 2 years.

(Total for Question 7 is 2 marks)

£2,500 is invested for 3 years at 3% per annum compound interest. Calculate the total value of the investment at the end of 3 years, to the nearest penny.

(Total for Question 8 is 2 marks)

A population of bacteria increases by 20% every hour. The starting population is 500. Find the population after 3 hours.

(Total for Question 9 is 2 marks)

A tool costs £850 new. It depreciates by 12% per year. Find its value after 2 years, to the nearest penny.

(Total for Question 10 is 2 marks)

Write down the single decimal multiplier that is equivalent to two successive percentage increases of 10% and then 5%.

(Total for Question 11 is 2 marks)

A quantity of 60 kg decreases by 8% and then increases by 8%. Calculate the final quantity, correct to 1 decimal place.

(Total for Question 12 is 2 marks)

£3,000 is invested at 2.5% per annum compound interest. Find the value of the investment after 4 years, to the nearest penny.

(Total for Question 13 is 2 marks)

A house is valued at £180,000. Its value increases by 4% each year for 2 years. Find its new value.

(Total for Question 14 is 2 marks)

A phone costs £900 new. It loses 18% of its value each year. Find its value after 2 years, to the nearest penny.

(Total for Question 15 is 2 marks)

Savings of £1,200 earn compound interest at 1.8% per annum. Find the value of the savings after 5 years, to the nearest penny.

(Total for Question 16 is 2 marks)

A car is bought for £18,500. It depreciates by 15% in the first year, and then by 10% in each subsequent year. Calculate its value after 3 years, giving your answer to the nearest pound.

(Total for Question 17 is 3 marks)

Priya invests £4,000 in a savings account paying 2.2% compound interest per annum. Work out how much interest (not the total value) she has earned after 3 years. Give your answer to the nearest penny.

(Total for Question 18 is 3 marks)

A company's profit increased by 10% in Year 1, then decreased by 10% in Year 2.

(a)The starting profit was £180,000. Calculate the profit at the end of Year 2.(2)

(b)State, with a calculation to support your answer, whether the profit has returned to its starting value of £180,000.(2)

(Total for Question 19 is 4 marks)

The value of a rare coin increases by 7% each year. After 2 years its value is £2,289.80. Find its original value.

(Total for Question 20 is 3 marks)

Aisha invests £6,000 for 3 years. Option A pays simple interest at 4% per annum. Option B pays compound interest at 3.6% per annum. Determine which option gives Aisha more money after 3 years, and find the difference between the final amounts, to the nearest penny.

(Total for Question 21 is 4 marks)

A van is bought for £24,000. Three years later it is valued at £14,739. It depreciated by the same percentage each year. Find the annual percentage rate of depreciation.

(Total for Question 22 is 4 marks)

The population of a town is 42,000. The population is increasing by 2% each year.

(a)Show that the population after 2 years, rounded to the nearest 10, is 43,700.(2)

(b)The council needs to plan for when the population first exceeds 46,000. Find the number of complete years after which this happens.(4)

(Total for Question 23 is 6 marks)

An investment of £5,000 grows at 2.5% compound interest per annum. Find the minimum number of complete years for the investment to first exceed £6,000.

(Total for Question 24 is 5 marks)

A quantity increases by x% and is then decreased by the same x%.

(a)Show algebraically that a percentage increase of x% followed by a percentage decrease of x% gives an overall multiplier of 1 - x²/10000, and that this is always an overall decrease of x²/100 %, for any positive value of x.(4)

(b)Find the overall percentage decrease when x = 15.(2)

(Total for Question 25 is 6 marks)

Mark scheme · 6.4D Repeated Percentage Change: Fluency and Exam Drill

Question 1

B1 1.08 oe
Answer: 1.08

Question 2

B1 0.85 oe
Answer: 0.85

Question 3

B1 1.035 oe
Answer: 1.035

Question 4

B1 0.78 oe
Answer: 0.78

Question 5

B1 correct statement that the sale price is lower than the original price
B1 correct reason: the 20% decrease is applied to the increased (larger) price, not the original price, so more is taken off than was added on
Answer: The sale price is lower than the original price, because the second percentage change (the decrease) is applied to the larger, already-increased price, not the original price.

Question 6

M1 40 * 1.05² (or 40 * 1.05 * 1.05) oe
A1 44.10 cao
Answer: £44.10

Question 7

M1 12000 * 0.9² oe
A1 9720 cao
Answer: £9,720

Question 8

M1 2500 * 1.03³ oe
A1 awrt £2731.82
Answer: £2,731.82

Question 9

M1 500 * 1.2³ oe
A1 864 cao
Answer: 864

Question 10

M1 850 * 0.88² oe
A1 awrt £658.24
Answer: £658.24

Question 11

M1 1.10 * 1.05 oe
A1 1.155 cao
Answer: 1.155

Question 12

M1 60 * 0.92 * 1.08 oe
A1 awrt 59.6 kg
Answer: 59.6 kg

Question 13

M1 3000 * 1.025⁴ oe
A1 awrt £3311.44
Answer: £3,311.44

Question 14

M1 180000 * 1.04² oe
A1 194688 cao
Answer: £194,688

Question 15

M1 900 * 0.82² oe
A1 605.16 cao
Answer: £605.16

Question 16

M1 1200 * 1.018⁵ oe
A1 awrt £1311.96
Answer: £1,311.96

Question 17

M1 correct multipliers identified: 0.85 for year 1, 0.90 for years 2 and 3
M1 18500 * 0.85 * 0.9 * 0.9 (or 18500 * 0.85 * 0.9²) oe
A1 awrt £12,737
Answer: £12,737

Question 18

M1 1.022³ (or equivalent repeated multiplication) oe
M1 4000 * 1.022³ (= total value) oe
A1 awrt £269.85
Answer: £269.85

Question 19

(a) M1 180000 * 1.10 * 0.90 oe
(a) A1 178200 cao
(a) Answer: £178,200
(b) B1 correct conclusion: profit has NOT returned to its starting value (it is lower)
(b) B1 correct supporting reason/calculation, e.g. overall multiplier 1.10 * 0.90 = 0.99, or difference of £1,800 (a 1% decrease), because the 10% decrease is taken from the larger, increased profit
(b) Answer: No; the profit is £178,200, which is £1,800 (1%) lower than the original £180,000, because the 10% decrease was applied to the larger, increased profit rather than to the original profit.

Question 20

M1 1.07² (= 1.1449) identified as the 2-year multiplier
M1 2289.80 / 1.1449 oe
A1 £2,000 cao
Answer: £2,000

Question 21

M1 simple interest: 6000 * 0.04 * 3 (= 720), total £6,720 oe
M1 compound interest: 6000 * 1.036³ oe
A1 awrt £6,671.61 for Option B, and correct conclusion that Option A gives more money
A1 awrt £48.39 difference cao ft
Answer: Option A (simple interest) gives more money; Option A = £6,720.00, Option B = £6,671.61, a difference of £48.39.

Question 22

M1 14739 / 24000 (= 0.614125) identified as the 3-year multiplier oe
M1 cube root of 0.614125 oe
A1 multiplier = 0.85
A1 15% cao (decrease)
Answer: 15% per year

Question 23

(a) M1 42000 * 1.02² oe
(a) A1 43696.8 seen, rounding to 43700 (nearest 10)
(a) Answer: 43,700 (nearest 10)
(b) M1 systematic repeated multiplication by 1.02, continuing beyond year 2
(b) M1 correct value for year 4, awrt 45,462 (below 46,000)
(b) A1 correct value for year 5, awrt 46,371 (exceeding 46,000)
(b) A1 5 (years) cao, with year 4 value shown to be below 46,000
(b) Answer: 5 complete years

Question 24

M1 setting up 5000 * 1.025ⁿ > 6000 oe, or an equivalent trial-and-improvement approach
M1 trial for n = 7: 5000 * 1.025⁷ = awrt £5,943.43 (< 6000)
M1 trial for n = 8: 5000 * 1.025⁸ = awrt £6,092.01 (> 6000)
A1 correct comparison showing n = 7 is insufficient and n = 8 is sufficient
A1 8 (years) cao
Answer: 8 years

Question 25

(a) M1 overall multiplier = (1 + x/100) * (1 - x/100) oe
(a) M1 correct expansion using difference of two squares: 1 - (x/100)² oe
(a) A1 1 - x²/10000 cso
(a) B1 correct conclusion that x²/10000 > 0 for all x not equal to 0, so the multiplier is always less than 1, i.e. an overall decrease of x²/100 %
(a) Answer: Overall multiplier = 1 - x²/10000, an overall decrease of x²/100 %, since x²/10000 is always positive for x not equal to 0.
(b) M1 substitute x = 15 into x²/100 oe
(b) A1 2.25% cao
(b) Answer: 2.25%

Question 25

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Repeated Percentage Change: Fluency and Exam Drill - Worksheets, Questions and Revision

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6.4D Repeated Percentage Change: Fluency and Exam Drill

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