Power of Maths Paper 1 OL

Power of Maths: Paper 1 – Section 5 WORKED EXAMPLE

Calculating depreciation

A car bought for €26 000 depreciates by 18% every year. What is its value after 3 years?

At the end of the fi rst year, the value = 82% of €26 000 = €26 000 × 0·82 = €21 320

At the end of the second year, the value = 82% of €21 320 = €21 320 × 0⋅82

= €26 000 × (0⋅82) × (0⋅82) = €26 000 × (0⋅82)2 = €17 482⋅40

At the end of the third year, the value = 82% of €17 482⋅40

= €17 482⋅40 × (0⋅82) = €26 000 × (0⋅82)2 × 0⋅82 = €26 000 × (0⋅82)3 = €14 335⋅57

The general formula for calculating depreciation is given in the Formulae and Tables book:

F = P(1 – i)t

F = fi nal value P = principal ( present value) t = number of depreciation periods i = rate of depreciation as a decimal or fraction

EXAMPLE 12

A washing machine is bought for €580. It depreciates at 15% p.a. What is its value after 5 years? By how much has it depreciated in 5 years?

Solution P = €580 t = 5 years r = 15%

F = P ( 1 − r

____ 100

= 580(0·85)5 = €257⋅35

It has depreciated by €580 – €257⋅35 = €322⋅65.

186 ) t = 580

( 1 − 15 ____

100 ) 5 EXAMPLE 13

Joe bought a scooter for €1500. After 3 years he estimated its value at €768. Calculate the percentage rate of depreciation p.a. using the reducing balance method.

Solution F = €768,

F = P (1 − i)t

768 = 1500(1 – i)3 768

_____ 1500 = 64

1 − i =

i = 1 − 4 __

( 64 ____

125 )

5 = 1 __

_ 3

1

= 4 __

5 5 = 0·2

r = 20% ∴ 20% is the percentage rate of depreciation per annum.

____ 125 = (1 − i ) 3

P = €1500, t = 3

This pattern can be generalised into the depreciation formula:

F = P ( 1 – r

____ 100

) t = P (1 – i) t

F is the fi nal value of the item. P is the initial value of the item.

r is the percentage rate of depreciation per depreciation period. t is the number of depreciation periods. i = r

____ 100 is the decimal rate of depreciation.

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