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4.1F Compound Interest and Depreciation: Foundation Tier Practice

EDEXCEL Edexcel 1MA1 · Calculator allowed · about 60 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

Work out 10% of £120, then add it to 120 to find the amount after a 10% increase.

(Total for Question 1 is 1 mark)

A savings account pays 5% compound interest per year. Work out the value of £150 after 2 years.

(Total for Question 2 is 2 marks)

A smartphone cost £240 new. Its value falls by 20% each year. Work out its value after 3 years. Give your answer to 2 decimal places.

(Total for Question 3 is 2 marks)

A jacket now costs £198 after a 10% increase in price. Work out the original price of the jacket before the increase.

(Total for Question 4 is 2 marks)

A pair of trainers cost £400. After 2 years with compound interest they are worth £484. Work out the annual percentage increase.

(Total for Question 5 is 2 marks)

Jamie invests £800 at 6% compound interest per year.
Part (a) Work out the value after 3 years. Part (b) Work out the interest Jamie has earned after 3 years, to the nearest pound.

(a)Work out the value after 3 years.(3)

(b)Work out the interest earned after 3 years, to the nearest pound.(2)

(Total for Question 6 is 5 marks)

A used car is priced at £6000. It depreciates by 12% each year for 2 years. After that it is sold and the sale price is reduced by a further 5% as a one-off discount. Work out the final sale price to the nearest pound.

(Total for Question 7 is 3 marks)

A tablet is now priced at £96 after losing 20% of its value each year for 2 years. Work out the original price of the tablet.

(Total for Question 8 is 2 marks)

Sarah invests £500 at 4% compound interest. Find how many years it takes for her investment to reach £608. Give your answer to the nearest whole year. Use your calculator and show the method.

(Total for Question 9 is 3 marks)

Start with £250. It is increased by 8% and then decreased by 8%. Work out the final amount to 2 decimal places.

(Total for Question 10 is 2 marks)

Find the compound multiplier for 3% compound interest over 4 years (that is, the number you multiply the original amount by). Give your answer to 4 decimal places.

(Total for Question 11 is 1 mark)

Show that a 10% compound increase for 3 years gives an overall increase of approximately 33.1%. You do not need to show every multiplication; give the multiplier and the percentage increase to 1 decimal place.

(Total for Question 12 is 2 marks)

Mark scheme · 4.1F Compound Interest and Depreciation: Foundation Tier Practice

Question 1

B1 £132 cao
Answer:132.00 (or £132) (132 cao). Note: 10% of 120 is 12; 120 + 12 = 132.

Working check: 120 + (10% of 120 = 12) = 132.

Question 2

M1 method: 150 x (1.05)² or 150 x 1.05 x 1.05
A1 £165.38 cao (to 2 dp) or £165.375 oe
Answer:£165.38 (to 2 dp)

Working check: 150 x (1.05)² = 150 x 1.1025 = 165.375 -> £165.38 (rounded to 2 dp).

Question 3

M1 method: 240 x (0.8)³ or 240 x 0.8 x 0.8 x 0.8
A1 £122.88 cao (to 2 dp)
Answer:£122.88

Working check: 240 x 0.8³ = 240 x 0.512 = 122.88.

Question 4

M1 method: divide by 1.10
A1 £180 cao
Answer:£180

Working check: Original = 198 / 1.10 = 180. Check: 180 x 1.10 = 198.

Question 5

M1 method: find factor 484/400 = 1.21, then find annual factor √1.21 or 1.21^1/2
A1 10% cao
Answer:10%

Working check: overall factor = 484/400 = 1.21. Annual factor = √1.21 = 1.1 so rate = 10%.

Question 6

(a) M1 method: 800 x (1.06)³
(a) M1 calculate (1.06)³ or multiply 800 x 1.06 x 1.06 x 1.06
(a) A1 £952.81 cao (to 2 dp)
(a) Answer: £952.81
(b) M1 method: interest = value after 3 years - original amount
(b) A1 £153 cao (nearest pound)
(b) Answer:£153 (nearest pound)

Working check: interest = 952.8128 - 800 = 152.8128 -> £153 (nearest pound).

Question 7

M1 method: 6000 x (0.88)² then x 0.95
M1 correct intermediate calculation 6000 x 0.88² = 6000 x 0.7744 = 4646.4
A1 £4413 cao (to nearest pound 4646.4 x 0.95 = 4413.08 -> 4413)
Answer:£4413 (nearest pound)

Working check: after 2 years = 6000 x 0.88² = 4646.40; after 5% discount = 4646.40 x 0.95 = 4413.08 -> £4413.

Question 8

M1 method: divide by 0.8² or 0.64
A1 £150 cao
Answer:£150

Working check: original = 96 / 0.64 = 150.

Question 9

M1 method: set up (1.04)ⁿ = 608/500 = 1.216
M1 use logs or calculator to find n: n = ln(1.216)/ln(1.04) or equivalent
A1 5 years cao (rounded to nearest whole year)
Answer:5 years (to nearest whole year)

Working check: 608/500 = 1.216. n = ln(1.216)/ln(1.04) = 4.99 -> 5 years (nearest whole year).

Question 10

M1 method: 250 x 1.08 x 0.92
A1 £248.40 cao (to 2 dp)
Answer:£248.40

Working check: 250 x 1.08 = 270.00; 270.00 x 0.92 = 248.40.

Question 11

B1 1.1255 cao (1.03⁴ = 1.12550881 -> 1.1255 to 4 dp)
Answer:1.1255 (to 4 dp)

Working check: (1.03)⁴ = 1.12550881 -> 1.1255 (4 dp).

Question 12

M1 state multiplier (1.1)³ = 1.331
A1 state increase 33.1% awrt or 33.1% cao
Answer:Multiplier = 1.331, overall increase = 33.1%

Working check: (1.1)³ = 1.331 -> increase = 1.331 - 1 = 0.331 = 33.1%.

Compound Interest and Depreciation: Foundation Tier Practice - Worksheets, Questions and Revision