Question

A certain sum amounts to $Rs.9440$ in $3$ years and to $Rs.10400$ in $5$ years. Find sum and rate percent.

Answer 1

Let the principal be $P$

The rate of interest is ${}^{\prime }{r}^{\prime }$ per annum at simple interest.

The amount for $3$ years is $Rs.9440$.
The amount for $5$ years is $Rs.10400.$

The amount is given by the formula

Amount, $A=P+S.I$

$=P+\frac{P\times r\times 3}{100}$ [Amount for $3$ years]

$=P(1+\frac{3r}{100})$

$\therefore P(1+\frac{3r}{100})=9440\dots \dots \left(i\right)$

And, Amount for 5 years is given by

$A=P(1+\frac{5r}{100})$

$\therefore P(1+\frac{5r}{100})=10400\dots \dots \left(ii\right)$

On dividing $eq\left(ii\right)$ by $\left(i\right),$ we get

$\frac{P(1+\frac{5r}{100})}{P(1+\frac{3r}{100})}=\frac{10400}{9440}$

$\Rightarrow \frac{(1+0.05r)}{(1+0.03r)}=\frac{520}{472}=\frac{260}{236}=\frac{130}{118}=\frac{65}{59}$

$\Rightarrow 59(1+0.05r)=65(1+0.03r)$

$\Rightarrow 59+2.95r=65+1.95r$

$\Rightarrow 2.95-1.95r=65-59$

$\Rightarrow r=6$

So, the rate of interest is $6\%.$

On putting the value of ${}^{\prime }{r}^{\prime }$ in $eq.\left(i\right),$ we get

$P(1+\frac{3\times 6}{100})=9440$

$\Rightarrow P(1+\frac{9}{50})=9440$

$\Rightarrow P\left(\frac{50+9}{50}\right)=9440$

$\Rightarrow P=9440\times \frac{50}{59}$

$\therefore P=Rs.8000$

Hence, the sum is $Rs.8000.$