Question

What is the difference between the compound interests on $\$400$ for 1 year at 10% per annum compounded yearly and half-yearly?
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Answer 1

$\text{Compound Interest}=P{(1+\frac{r}{100\times n})}^{n\times t}-P$

$\text{Where,}P\text{= Principal amount}=\$400$
$r=\text{rate of interest (\%)}=10$
$t=\text{time period (years)}=1$

When interest is compounded yearly, then $n=1$
${\text{Compound Interest}}_{n=1}=400{(1+\frac{10}{100\times 1})}^{1\times 1}-400$
$=400(1+\frac{10}{10\times 10})-400$
$=400(1+\frac{1}{10})-400$
$=400(\frac{10}{10}+\frac{1}{10})-400$
$=400\left(\frac{10+1}{10}\right)-400$
$=400\times \frac{11}{10}-400$
$=40\times 11-400$
$=440-400$
$=40$

When interest is compounded half-yearly, then $n=2$
${\text{Compound Interest}}_{n=2}=400{(1+\frac{10}{100\times 2})}^{2\times 1}-400$
$=400{(1+\frac{10}{10\times 10\times 2})}^2-400$
$=400{(1+\frac{1}{20})}^2-400$
$=400{(\frac{20}{20}+\frac{1}{20})}^2-400$
$=400{\left(\frac{20+1}{20}\right)}^2-400$
$=400\times {\left(\frac{21}{20}\right)}^2-400$
$=400\times \frac{441}{400}-400$
$=441-400$
$=41$

Difference between the compound interests
$={\text{Compound Interest}}_{n=2}-{\text{Compound Interest}}_{n=1}$
$=41-40$
$=1$

Hence, difference between the compound interests is $\$1.$