Substitute the values into the formula for compound interest.

Formula for compound interest

A = P {(1 + \frac{r}{100})}^{n}

.
A = final amount = 206.46
P = initial amount = 200
r = interest rate = r
n = number of years = 2.

Substitute the values into the formula to form an equation.

200 {(1 + \frac{r}{100})}^{2} = 206.46

[1]

Divide both sides by 200.

{(1 + \frac{r}{100})}^{2} = \frac{206.46}{200} {(1 + \frac{r}{100})}^{2} = 1.0323

Expand the brackets by multiplying every term in the first bracket by every term in the second bracket.

\begin{array}{rcl} (1 + \frac{r}{100}) (1 + \frac{r}{100}) & = & 1.0323 \\ 1 (1 + \frac{r}{100}) + \frac{r}{100} (1 + \frac{r}{100}) & = & 1.0323 \\ 1 + \frac{r}{100} + \frac{r}{100} + {(\frac{r}{100})}^{2} & = & 1.0323 \\ 1 + \frac{2 r}{100} + {(\frac{r}{100})}^{2} & = & 1.0323 \end{array}

[1]

Set equal to zero and rearrange.

\begin{array}{rcl} 1 + \frac{2 r}{100} + {(\frac{r}{100})}^{2} - 1.0323 & = & 0 \\ {(\frac{r}{100})}^{2} + \frac{2 r}{100} - 0.0323 & = & 0 \end{array}

Remove the fractions by multiplying every term by 100² .

\begin{array}{rcl} {(\frac{r}{100})}^{2} {(100)}^{2} + \frac{2 r}{100} {(100)}^{2} - 0.0323 {(100)}^{2} & = & 0 \\ \frac{10000 r^{2}}{10000} + \frac{20000 r}{100} - 323 & = & 0 \end{array}

r^{2} + 200 r - 323 = 0

[1]

Simple & Compound Interest, Growth & Decay (CIE IGCSE Maths: Extended)

Exam Questions