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6.Interest
hard
A sum of money lent at compound interest amounts to ₹ $1460$ in $2$ years and to ₹ $1606$ in $3$ years. The rate of interest (In $\%$) per annum is
A
$12$
B
$11$
C
$10.5$
D
$10$
Solution
(d) Let the amount be $A$ and the rate of interest be $r \%$ Then,
$A\left(1+\frac{r}{100}\right)^{2}=1460$
$A\left(1+\frac{r}{100}\right)^{3}=1606$
On dividing eqn. (1) by eqn. (2), we get
$1+\frac{r}{100}=\frac{1606}{1460}$
$\Rightarrow \frac{r}{100}=\frac{1606}{1460}-1=\frac{146}{1460}=\frac{1}{10}$
$\therefore r=\frac{100}{10} \%=10 \%$
Standard 13
Quantitative Aptitude
