Delhi Police Constable Exam Model Paper 8

GIVEN:
C.I. = Rs. 609, Time = 2 years and Rate = 10
FORMULA USED: $\mathit{S}\mathit{I}=\mathit{P}\times \mathit{R}\times \frac{\mathit{T}}{100}$
$\mathit{C}\mathit{I}=\mathit{P}\times \left[{\left(1+\frac{\mathit{R}}{100}\right)}^{\mathit{T}}\mathit{–}1\right]$
P = Principal, R = Rate, T = Time
CALCULATION
:Let the sum be Rs. P
Then,
$\left[\mathit{P}{\left(1+\frac{\mathit{r}}{100}\right)}^2\mathit{–}\mathit{P}\right]=\mathit{C}\mathit{I}$
⇒ $\left[\mathit{P}{\left(1+\frac{10}{100}\right)}^2\mathit{–}\mathit{P}\right]=609$
⇒ 1 $21\frac{\mathit{P}}{100}\mathit{–}\mathit{P}=609$
⇒ $21\frac{\mathit{P}}{100}=609$
⇒ $\mathit{P}=2900$
Now,
Simple interest on the same sum for double the time at half the rate percent per annum
$\mathit{S}\mathit{I}=\frac{\left(2900\times 5\times 4\right)}{100}=\mathit{R}\mathit{s}.580$
∴ required rate of interest is Rs. 580