Question

A certain sum of money is invested at the rate of 10% per annum compound interest , the interest compounded annually . If the difference between the interests of third year and first year is Rs. 1,105, find the sum invested.

Answer 1

Rate of interest$=10$%

Difference between interest of $3rd$ and $1st$ year is $Rs.1105$

i.e.,$P{(1+\frac{10}{100})}^3-P(1+\frac{10}{100})=Rs.1105$

$\Rightarrow P[{\left(\frac{11}{10}\right)}^3-\frac{11}{10}]=1105\Rightarrow P[\frac{1331}{1000}-\frac{11}{10}]=1105\Rightarrow P\times \left[\frac{1331-1100}{1000}\right]=1105\Rightarrow P\times \frac{231}{1000}=1105\Rightarrow P=\frac{1105\times 1000}{231}=Rs.4783.5$