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GCSE · Ratio, Proportion and Rates of Change

4.1D Compound Interest and Depreciation: Fluency and Exam Drill

EDEXCEL 1MA1 · Calculator allowed · about 75 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

Write down the single decimal multiplier for each of the following percentage changes.

(a)an increase of 3%(1)

(b)a decrease of 5%(1)

(c)an increase of 12%(1)

(d)a decrease of 40%(1)

(Total for Question 1 is 4 marks)

Leah invests 400 pounds in a savings account that pays compound interest at a rate of 5% per year. Work out the value of her investment at the end of 2 years.

(Total for Question 2 is 2 marks)

A car is valued at 9000 pounds. It depreciates in value by 10% each year. Work out the value of the car after 2 years.

(Total for Question 3 is 2 marks)

250 pounds is invested for 3 years at a rate of 2% per year compound interest. Work out the value of the investment at the end of 3 years, giving your answer to the nearest penny.

(Total for Question 4 is 2 marks)

A boat is bought for 15000 pounds. It depreciates in value by 8% per year. Work out its value after 2 years, giving your answer to the nearest pound.

(Total for Question 5 is 2 marks)

Omar invests 1200 pounds at a rate of 3.5% per year compound interest for 4 years. Calculate the total interest earned, giving your answer to the nearest penny.

(Total for Question 6 is 3 marks)

180 pounds is invested for 2 years at a rate of 12.5% per year compound interest. Work out the value of the investment at the end of 2 years, giving your answer to the nearest penny.

(Total for Question 7 is 2 marks)

A tractor is bought for 45000 pounds. It depreciates in value by 18% per year. Work out its value after 2 years, giving your answer to the nearest pound.

(Total for Question 8 is 2 marks)

Kwame invests 3500 pounds for 2 years in an account paying 1.6% per year compound interest. Work out the value of his investment at the end of 2 years, giving your answer to the nearest penny.

(Total for Question 9 is 2 marks)

A laptop costs 900 pounds when new. It depreciates in value by 25% in the first year, and then by 10% per year for each of the next 2 years. Work out its value after 3 years.

(Total for Question 10 is 3 marks)

A piece of jewellery is bought for 2400 pounds. It depreciates in value by 6% per year. Work out its value after 3 years, giving your answer to the nearest penny.

(Total for Question 11 is 2 marks)

600 pounds is invested at a rate of 4% per year compound interest for 5 years. Calculate the total interest earned, giving your answer to the nearest penny.

(Total for Question 12 is 2 marks)

5000 pounds is invested for 2 years at a rate of 2.4% per year compound interest. Work out the value of the investment at the end of 2 years, giving your answer to the nearest penny.

(Total for Question 13 is 2 marks)

Fatima invests 2000 pounds for 4 years at a rate of 3% per year compound interest. James invests 2000 pounds for 4 years in an account that pays simple interest of 70 pounds per year. After 4 years, who has more money, and by how much? You must show your working.

(Total for Question 14 is 4 marks)

A courier company buys a delivery van for 18000 pounds. The van depreciates in value by 20% in its first year, and then by 12% per year for each year after that. The company will replace the van once its value falls below 9000 pounds. After how many years will the van need to be replaced? You must show your working.

(Total for Question 15 is 4 marks)

A car has depreciated in value by 15% over the past year and is now worth 10200 pounds. Work out the value of the car before it depreciated.

(Total for Question 16 is 2 marks)

A savings account pays compound interest at a rate of 2% in the first year, and 3.5% per year for each year after that. Sarah invests 4000 pounds. Work out the value of her investment after 3 years, giving your answer to the nearest penny.

(Total for Question 17 is 3 marks)

Priya wants to invest 6000 pounds for 3 years. Bank A offers 2.8% per year compound interest. Bank B offers a fixed bonus of 550 pounds at the end of the 3 years, with no interest paid before then. Which bank should Priya choose to get the most money back after 3 years? You must show your working.

(Total for Question 18 is 4 marks)

A vintage guitar was bought for 800 pounds. Three years later, having appreciated at a constant rate of compound interest each year, it was valued at 926.10 pounds. Work out the annual rate of appreciation.

(Total for Question 19 is 4 marks)

A population of bacteria increases by 8% every hour. At 09:00 there are 250 bacteria. Work out the number of bacteria at 12:00 (3 hours later), giving your answer to the nearest whole number.

(Total for Question 20 is 3 marks)

A car is bought for 24000 pounds. It depreciates in value by r% per year for each of the first two years, and then by (r + 5)% in the third year. After 3 years, the car is worth 13872 pounds. Given that r is a positive whole number, find the value of r. You must show your working.

(Total for Question 21 is 5 marks)

A sum of money, P, is invested for 2 years at r% per year compound interest, where r is not equal to 0. Prove that the total interest earned over the 2 years is not equal to twice the interest earned in the first year alone.

(Total for Question 22 is 4 marks)

Mark scheme · 4.1D Compound Interest and Depreciation: Fluency and Exam Drill

Question 1

(a) B1 1.03 oe
(a) Answer: 1.03
(b) B1 0.95 oe
(b) Answer: 0.95
(c) B1 1.12 oe
(c) Answer: 1.12
(d) B1 0.60 oe
(d) Answer: 0.60

Question 2

M1 400 x 1.05² oe, or a correct year-by-year method
A1 441 (pounds) cao
Answer: 441 pounds

Question 3

M1 9000 x 0.9² oe, or a correct year-by-year method
A1 7290 (pounds) cao
Answer: 7290 pounds

Question 4

M1 250 x 1.02³ oe, or a correct year-by-year method
A1 265.30 (pounds) awrt
Answer: 265.30 pounds

Question 5

M1 15000 x 0.92² oe, or a correct year-by-year method
A1 12696 (pounds) cao
Answer: 12696 pounds

Question 6

M1 1200 x 1.035⁴ oe, or a correct year-by-year method, to find the total value
A1 1377.03 (pounds) awrt, the total value after 4 years
A1 177.03 (pounds) awrt for the interest earned, ft their total value minus 1200
Answer: 177.03 pounds

Question 7

M1 180 x 1.125² oe, or a correct year-by-year method
A1 227.81 (pounds) awrt
Answer: 227.81 pounds

Question 8

M1 45000 x 0.82² oe, or a correct year-by-year method
A1 30258 (pounds) cao
Answer: 30258 pounds

Question 9

M1 3500 x 1.016² oe, or a correct year-by-year method
A1 3612.90 (pounds) awrt
Answer: 3612.90 pounds

Question 10

M1 900 x 0.75 (= 675) for the value after year 1
M1 correct method applying 0.9 for each of the following 2 years
A1 546.75 (pounds) cao
Answer: 546.75 pounds

Question 11

M1 2400 x 0.94³ oe, or a correct year-by-year method
A1 1993.40 (pounds) awrt
Answer: 1993.40 pounds

Question 12

M1 600 x 1.04⁵ oe, or a correct year-by-year method, to find the total value
A1 129.99 (pounds) awrt for the interest earned
Answer: 129.99 pounds

Question 13

M1 5000 x 1.024² oe, or a correct year-by-year method
A1 5242.88 (pounds) cao
Answer: 5242.88 pounds

Question 14

M1 2000 x 1.03⁴ oe, or a correct year-by-year method, to find Fatima's total
A1 2251.02 (pounds) awrt for Fatima's total
B1 2280 (pounds) for James's total (2000 + 4 x 70)
A1 ft correct conclusion: James has more money, by 28.98 (pounds) awrt
Answer: James has more money, by 28.98 pounds.

Question 15

M1 18000 x 0.8 (= 14400) for the value after year 1
M1 correct method repeatedly applying 0.88 for the following years
A1 value after year 4 = 9813.20 awrt, correctly shown to still be above 9000
A1 correct final answer: 5 years, with value after year 5 = 8635.61 awrt shown to be below 9000
Answer: 5 years

Question 16

M1 10200 / 0.85 oe
A1 12000 (pounds) cao
Answer: 12000 pounds

Question 17

M1 4000 x 1.02 (= 4080) for the value after year 1
M1 correct method applying 1.035 for each of the following 2 years
A1 4370.60 (pounds) awrt
Answer: 4370.60 pounds

Question 18

M1 6000 x 1.028³ oe, or a correct year-by-year method, for Bank A's total
A1 6518.24 (pounds) awrt for Bank A's total
B1 6550 (pounds) for Bank B's total (6000 + 550)
A1 ft correct conclusion: Bank B, since 6550 > 6518.24 (a difference of 31.76 awrt)
Answer: Priya should choose Bank B, as it gives 6550 pounds compared to Bank A's 6518.24 pounds (a difference of 31.76 pounds).

Question 19

M1 926.10 / 800 (= 1.157625) oe
M1 finds the cube root of their 1.157625 (dependent on the previous method mark)
A1 multiplier = 1.05 oe
A1 rate = 5% cao
Answer: 5%

Question 20

M1 250 x 1.08³ oe, or a correct repeated multiplication method
A1 314.928 awrt (accept awrt 314.9)
A1 315 cao, correctly rounded to the nearest whole number
Answer: 315 bacteria

Question 21

M1 sets up a correct expression for the value after 3 years, e.g. 24000 x (1 - r/100)² x (1 - (r+5)/100)
M1 sets this expression equal to 13872, or forms the equivalent ratio 13872 / 24000 = 0.578
M1 trials at least one value of r using a correct method, e.g. r = 10 gives 24000 x 0.9² x 0.85 = 16524 (too high, so r > 10)
A1 a further correct trial narrowing the range, e.g. r = 20 gives 24000 x 0.8² x 0.75 = 11520 (too low, so 10 < r < 20)
A1 r = 15 cao, with a valid check shown (24000 x 0.85² x 0.80 = 13872)
Answer: r = 15

Question 22

M1 correct expression for the interest earned in year 1, e.g. P x r / 100
M1 correct expanded expression for the value after 2 years, e.g. P(1 + r/100)² = P + 2Pr/100 + P(r/100)²
A1 correctly identifies the total interest after 2 years as 2Pr/100 + P(r/100)², and compares it with 2 x (Pr/100) = 2Pr/100
B1 correct conclusion that the difference between the two is P(r/100)², which is non-zero whenever r is not equal to 0 (and P is not equal to 0), so the two amounts of interest can never be equal; QED oe
Answer: Proved: the interest after 2 years exceeds twice the year-1 interest by P(r/100)², which is non-zero whenever r is not equal to 0, so the two can never be equal.

Compound Interest and Depreciation: Fluency and Exam Drill - Worksheets, Questions and Revision