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Question no.14

**14. **A man invests Rs 1200 for two years at compound interest. After one year the money amounts to Rs 1275. Find the interest for the second year correct to the nearest rupee.

Amount After one year =1275

Let's calculate rate of interest r for first year.

Formula used for comound interest is as follows

$A=P{\left(1+\frac{r}{n}\right)}^{nt}$

In our case ${A}_{1}=1275andP=1200$,

Assuming interest is compounded anually so n=1 and since we are considering first year , t =1

$1275=1200{\left(1+\frac{r}{1}\right)}^{1*1}$

$\frac{1275-1200}{1200}=r$

$r=\frac{75}{1200}=0.0625$

calculating value of A that is ${A}_{2}$ for two years with r=0.625 and t=2 years.

${A}_{2}=1200{\left(1+\frac{0.0625}{1}\right)}^{1*2}\phantom{\rule{0ex}{0ex}}=1200*(1.0625{)}^{2}=1354.6875$

Interest earned in second year is

${A}_{2}-{A}_{1}=1354.6875-1275=79.6875\approx 80$

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