Compound Interest (Cambridge International IGCSE Mathematics)

Compound Interest

What is compound interest?

• Compound interest is where interest is calculated on the running total, not just the starting amount

  • This is different from simple interest where interest is only based on the starting amount

• E.g. $ 100 earns 10% interest each year, for 3 years

  • At the end of year 1, 10% of $ 100 is earned

    • The total balance will now be 100+10 = $ 110

  • At the end of year 2, 10% of $ 110 is earned

    • The balance will now be 110+11 = $ 121

  • At the end of year 3, 10% of $ 121 is earned

    • The balance will now be 121+12.1 = $ 133.10

How do I calculate compound interest?

• Compound interest increases an amount by a percentage and then increases the new amount by the same percentage

  • This process repeats each time period (yearly or monthly etc)

• We can use a multiplier to carry out the percentage increase multiple times

  • To increase $ 300 by 5% once, we would find 300×1.05

  • To increase $ 300 by 5%, each year for 2 years, we would find (300×1.05)×1.05

    • This could be rewritten as 300×1.05²

  • To increase $ 300 by 5%, each year for 3 years, we would find ((300×1.05)×1.05)×1.05

    • This could be rewritten as 300×1.05³

• This can be extended to any number of periods that the interest is applied for

  • If $ 2000 is subject to 4% compound interest each year for 12 years
  • Find 2000×1.04¹²,****which is $ 3202.06

• Note that this method calculates the** total balance** at the end of the period, not the interest earned

Compound interest formula

• An alternative method is to use the following formula to calculate the final balance

  • Final balance = $P ( 1 + \frac{r}{100} )^{n }$ where

    • P is the original amount,
    • r is the % increase,
    • and n is the number of years

  • Note that $1+\frac{r}{100}$ is the same value as the multiplier

    • e.g. 1.15 for 15% interest

• This formula is not given in the exam

What if there are different rates of interest?

• Instead of applying the same percentage for $n$ years, banks may offer different rates

  • E.g. $3 \%$ for 2 years and then $2 \%$ for 1 year

• Treat the periods under different rates as separate calculations

• For example if a sum of $500 earns 4% interest for 2 years, and then 2.5% for 1 year

  • For the first period

    • 500×1.04² = 540.80

  • For the second period

    • 540.80 × 1.025 = 554.32

  • So the total amount is now $554.32 at the end of the 3 years

How do I solve reverse compound interest problems?

• You could be told the final balance after compound interest has been applied, and need to find the original amount

  • This could be referred to as a "reverse compound interest" problem

• For example if:

  • The final balance is £432
  • After 20% interest has been applied each year
  • For 3 years

• Using the same method as above, this can be written as an equation:

  • $432=P\times1.20^{3}$ where $P$ is the original amount

  • Solve for $P$,

    • Divide both sides by $1.20^{3}$
    • $\begin{gathered} 432 \div 1.20^{3} & = P \end{gathered}$
    • P equals £ 250

• In general, to find the original amount:

  • Divide the final amount by $m^{n}$ where

    • $m$ is the multiplier for the time period
    • and $n$ is the number of time periods (usually years)

Exam Hint

• Double check if the question uses simple interest or compound interest
• The formula for compound interest is not given in the exam

Worked Example Jasmina invests $1200 in a savings account, which pays compound interest at the rate of 4% per year for 7 years.

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

Answer:

Method 1

We want an increase of 4% per year This is equivalent to a multiplier of 1.04, or 104% of the original amount

This multiplier is applied 7 times$\times1.04\times1.04\times1.04\times1.04\times1.04\times1.04\times1.04=1.04^{7}$

Therefore the final value after 7 years will be

1200 \times 1.04^{7} = \ 1579.118135...$

Round to the nearest dollar

bold $ bold 1579

Method 2 Using the formula for the final amount $P ( 1 + \frac{r}{100} )^{n }$ *Substitute P is 1200, r = 4 and n = 7 into the formula *

$1200 ( 1 + \frac{4}{100} )^{7}$

bold $ bold 1579

Worked Example Martin invested$ n in an account for 3 years at a rate of 3.9%.

At the end of the 3 years the balance of the account is $3925.67.

Find the value of $n$ to the nearest dollar.

Answer:

$n$ is increased by 3.9% for 3 years, resulting in 3925

Write an equation describing this

$n\times1.039^{3}=3925.67$

Divide both sides by $1.039^{3}$

$n=\frac{3925.67}{1.039^{3}}$

Work this out on your calculator

$n=3499.992764...$

Round to the nearest dollar

$n=3500$