Interest & Depreciation (Edexcel IGCSE Maths)

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Compound Interest

What is compound interest?

  • Compound interest is where interest is paid on the interest from the previous year (or whatever time frame is being used), as well as on the original amount
  • This is different from simple interest where interest is only paid on the original amount
    • Simple interest goes up by the same amount each time whereas compound interest goes up by an increasing amount each time

How do you work with compound interest?

  • Keep multiplying by the decimal equivalent of the percentage you want (the multiplier, p)
  • A 25% increase (p = 1 + 0.25) each year for 3 years is the same as multiplying by 1.25 × 1.25 × 1.25
    • Using powers, this is the same as × 1.253
  • In general, the multiplier p applied n times gives an overall multiplier of pn
  • If the percentages change varies from year to year, multiply by each one in order
    • a 5% increase one year followed by a 45% increase the next year is 1.05 × 1.45
  • In general, the multiplier p1 followed by the multiplier p2 followed by the multiplier p3... etc gives an overall multiplier of p1p2p3... 

Compound interest formula

  • Alternative method: A formula for the final ("after") amount is  P open parentheses 1 plus r over 100 close parentheses to the power of n space end exponent where...
    • ...P is the original ("before") amount, r is the % increase, and n is the number of years
    • Note that 1 plus r over 100 is the same value as the multiplier

Examiner Tip

  • It is easier to multiply by the decimal equivalent "raised to a power" than to multiply by the decimal equivalent several times in a row

Worked example

Jasmina invests £1200 in a savings account which pays compound interest at the rate of 2% per year for 7 years.

To the nearest pound, what is her investment worth at the end of the 7 years?

We want an increase of 2% per year, this is equivalent to a multiplier of 1.02, or 102% of the original amount

This multiplier is applied 7 times; cross times 1.02 cross times 1.02 cross times 1.02 cross times 1.02 cross times 1.02 cross times 1.02 cross times 1.02 space equals space 1.02 to the power of 7

Therefore the final value after 7 years will be

£ 1200 space cross times space 1.02 to the power of 7 space equals space £ 1378.42

Round to the nearest pound

bold £ bold 1378

Alternative method
Or use the formula for the final amount   P open parentheses 1 plus r over 100 close parentheses to the power of n space end exponent
Substitute P is 1200, r = 2 and n = 7 into the formula 

1200 open parentheses 1 plus 2 over 100 close parentheses to the power of 7

£1378 (to the nearest pound)

Depreciation

What is meant by depreciation?

  • Depreciation is where an item loses value over time
    • For example: cars, game consoles, etc
  • Depreciation is usually calculated as a percentage decrease at the end of each year
    • This works the same as compound interest, but with a percentage decrease

How do I calculate a depreciation?

  • You would calculate the new value after depreciation using the same method as compound interest
    • Identify the multiplier, p (1 - "% as a decimal")
      • 10% depreciation would have a multiplier of p = 1 - 0.1 = 0.9
      • 1% depreciation would have a multiplier of p = 1 - 0.01 = 0.99
    • Raise the multiplier to the power of the number of years (or months etc)
      • p to the power of n
    • Multiply by the starting value
  • New value is A cross times p to the power of n
    • A is the starting value
    • p is the multiplier for the depreciation 
    • is the number of years
  • Alternative method: A formula for the final ("after") amount is P open parentheses 1 minus r over 100 close parentheses to the power of n where...   
    • ...P is the original ("before") amount, r is the % decrease, and n is the number of years
  • If you are asked to find the amount the value has depreciated by:
    • Find the difference between the starting value and the new value

Worked example

Mercy buys a car for £20 000. Each year its value depreciates by 15%.

Find the value of the car after 3 full years.

Identify the multiplier

100% - 15% = 85%

p = 1 - 0.15 = 0.85

Raise to the power of number of years

0.853

Multiply by the starting value

£20 000 × 0.853

= £12 282.50

Alternative method
Or use the formula for the final amount   P open parentheses 1 minus r over 100 close parentheses to the power of n space end exponent
Substitute P is 20 000, r = 15 and n = 3 into the formula 

20 space 00 open parentheses 1 minus 15 over 100 close parentheses cubed

£12 282.50

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Mark

Author: Mark

Expertise: Maths

Mark graduated twice from the University of Oxford: once in 2009 with a First in Mathematics, then again in 2013 with a PhD (DPhil) in Mathematics. He has had nine successful years as a secondary school teacher, specialising in A-Level Further Maths and running extension classes for Oxbridge Maths applicants. Alongside his teaching, he has written five internal textbooks, introduced new spiralling school curriculums and trained other Maths teachers through outreach programmes.