# Question no.1414.  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.

Principal Amount=1200
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 ,
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.

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

Interest earned in second year is

• 0
calculate the rate of interest using the formula
A = P( 1 + R/100) n

it will be 9*51/2/20

• 1
after you have manipulated this calculate the S.I for the first year
the amount you get after the first year will be the principal for the second year
again calculate the amount at the end of second year
but the problem occurs in 51/2
so instead of that take 51/2 as 2.2