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KS3 · Ratio, Proportion and Rates of Change

KS3.M-R10 Simple and Compound Interest

AQA KS3.M-R10 · Calculators not allowed · about 80 minutes

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

Calculate the simple interest earned on a savings account that holds 500 pounds at 3% per year for 2 years. Give the interest only.

(2)

Work out 4% of 250 pounds. This represents the interest for one year. Give the interest only.

(1)

A loan of 1200 pounds is charged simple interest at 5% per year for 3 years. Calculate the total interest charged and the total amount to repay after 3 years.

(3)

A deposit of 400 pounds is left for 1 year at 10% compound interest. Work out the amount after 1 year.

(2)

Calculate the amount after 2 years when 600 pounds is invested at 5% compound interest per year. Give your answer to 2 decimal places.

(3)

An amount of 250 pounds is invested at 8% compound interest per year. Find the value after 3 years, correct to 2 decimal places.

(3)

A small musical instrument costs 180 pounds. It depreciates by 15% of its value in one year (simple depreciation used here). Work out its value after one year.

(2)

A laptop worth 1200 pounds loses 10% of its value each year compound. Find its value after 2 years.

(3)

If a multiplier is 1.06, state the annual percentage rate of increase.

(2)

An investment of 1000 pounds grows to 1210 pounds after 2 years with compound interest applied annually. Work out the annual interest rate.

(2)

Compare simple and compound interest on 1000 pounds at 5% per year for 3 years. Find the simple interest total and the compound total, then give the difference between them (compound minus simple).

(3)

Hannah deposits 800 pounds into an account paying 6% compound interest. At the start of the second year she adds 200 pounds to the account. Find the total in the account at the end of the second year. Give your answer to 2 decimal places.

(3)

A shirt is reduced by 20% in a sale. Later the price is increased by 25%. If the original price was 150 pounds, what is the final price after both changes?

(2)

A machine is worth 648 pounds after losing 10% of its value each year for 2 years (compound depreciation). Work out the original price before the two years.

(2)

A car loses 18% of its value each year compound. Its value now, after 3 years, is 1976 pounds. Find the price of the car when it was new (value before the 3 years).

(3)

A price is increased by 10% and then decreased by 10%. Work out the overall percentage change in the price.

(2)

A simple interest loan charges a total of 9% interest over 3 years. Work out the annual simple interest rate.

(2)

Sam wants 1200 pounds at the end of 2 years. He deposits 1000 pounds now into an account paying 4% compound interest and plans to add one extra deposit at the start of the second year. How much must he add at the start of the second year so that at the end of year 2 he has 1200 pounds? Give your answer to 2 decimal places.

(4)

Mark scheme · KS3.M-R10 Simple and Compound Interest

Question 1

M1 method: 500 x 0.03 x 2 or 500 x (3% per year for 2 years) shown
A1 30 (pounds) cao
Answer: 30 pounds

Question 2

B1 10 (pounds) cao
Answer: 10 pounds

Question 3

M1 method: 1200 x 0.05 x 3 or equivalent method for simple interest
A1 180 (pounds) interest cao
A1 1380 (pounds) total amount cao
Answer: 180 pounds; 1380 pounds

Question 4

M1 method: use multiplier 1.10 or 400 x 1.10
A1 440 (pounds) cao
Answer: 440 pounds

Question 5

M1 method: use multiplier 1.05 twice or write 600 x 1.05²
M1 correct intermediate multiplication leading to 600 x 1.1025 or equivalent
A1 661.50 (pounds) cao
Answer: 661.50 pounds

Question 6

M1 method: 250 x 1.08³ or use multiplier 1.08 three times
M1 correct intermediate product leading to about 314.928
A1 314.93 (pounds) cao
Answer: 314.93 pounds

Question 7

M1 method: 15% of 180 or 180 x 0.15 found
A1 153 (pounds) final value cao
Answer: 153 pounds

Question 8

M1 method: use multiplier 0.90 twice or 1200 x 0.9²
M1 correct intermediate product 1200 x 0.81 or equivalent
A1 972 (pounds) cao
Answer: 972 pounds

Question 9

M1 method: subtract 1 from multiplier or state 0.06
A1 6% cao
Answer: 6%

Question 10

M1 method: divide 1210 by 1000 to find multiplier per 2 years then take square root or deduce yearly multiplier 1.10
A1 10% per year cao
Answer: 10%

Question 11

M1 method for simple interest: 1000 x 0.05 x 3 = 150
M1 method for compound: 1000 x 1.05³ = 1157.625 or equivalent
A1 difference 7.63 (pounds) cao (1157.63 - 1150.00 rounded to 2 dp)
Answer: Simple interest = 150 pounds; compound total = 1157.63 pounds; difference = 7.63 pounds

Question 12

M1 value after year 1: 800 x 1.06 = 848 shown
M1 add the deposit: 848 + 200 = 1048 and then apply 6% for year 2 or show (848 + 200) x 1.06
A1 1110.88 (pounds) cao
Answer: 1110.88 pounds

Question 13

M1 method: 150 x 0.80 then result x 1.25 or equivalent
A1 150 (pounds) cao
Answer: 150 pounds

Question 14

M1 method: divide 648 by 0.9² or 648 ÷ 0.81
A1 800 (pounds) cao
Answer: 800 pounds

Question 15

M1 method: use multiplier 0.82³ or set original x 0.82³ = 1976
M1 divide: original = 1976 ÷ 0.82³ shown
A1 3584 (pounds) cao
Answer: 3584 pounds

Question 16

M1 method: use multipliers 1.10 x 0.90 = 0.99 shown
A1 1% decrease cao
Answer: 1% decrease

Question 17

M1 method: divide total percentage by number of years 9% ÷ 3
A1 3% per year cao
Answer: 3% per year

Question 18

M1 value after year 1: 1000 x 1.04 = 1040 shown
M1 let extra deposit = x, then (1040 + x) x 1.04 = 1200 or equivalent equation shown
M1 correct method to find pre-final total: 1040 + x = 1200 ÷ 1.04
A1 113.85 (pounds) cao
Answer: 113.85 pounds

Simple and Compound Interest - Worksheets, Questions and Revision