Year 11 Exam  >  Year 11 Notes  >  Mathematics for GCSE/IGCSE  >  Compound Interest

Compound Interest | Mathematics for GCSE/IGCSE - Year 11 PDF Download

Compound Interest

  • Compound interest is when interest is paid on both the original amount and the interest accumulated from previous periods.
  • This contrasts with simple interest, where interest is only calculated on the original amount.
    • Simple interest increases by the same amount each period, while compound interest increases by a progressively larger amount each period.

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 pfollowed 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 Compound Interest | Mathematics for GCSE/IGCSE - Year 11where...
  • ...P is the original ("before") amount, r is the % increase, and n is the number of years
  • Note that Compound Interest | Mathematics for GCSE/IGCSE - Year 11 is the same value as the multiplier

Depreciation

  • Depreciation is the process by which an item loses value over time.
  • Examples include cars, game consoles, and other electronics.
  • Depreciation is typically calculated as a percentage decrease at the end of each year.
  • This calculation works similarly to compound interest, but it involves a percentage decrease instead of an increase.

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) pn
    • Multiply by the starting value
  • New value is A x pn
    • A is the starting value
    • p is the multiplier for the depreciation 
    • n is the number of years
  • Alternative method: A formula for the final ("after") amount is Compound Interest | Mathematics for GCSE/IGCSE - Year 11 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
The document Compound Interest | Mathematics for GCSE/IGCSE - Year 11 is a part of the Year 11 Course Mathematics for GCSE/IGCSE.
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