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6.Interest
hard
The principal that amounts to ₹ $4913$ in $3$ years at $6 \frac{1}{4} \%$ per annum compound interest compounded annually (In ₹) is.
A
$3096$
B
$4076$
C
$4085$
D
$4096$
Solution
Let the principal be $P$.
$\therefore$
$P\left(1+\frac{6 \frac{1}{4}}{100}\right)^{3}=4913$
or $P\left(1+\frac{25}{400}\right)^{3}=4913$
or
$P=\frac{4913}{\left(1+\frac{1}{16}\right)^{3}}=\frac{4913 \times 16^{3}}{17^{3}}=₹ 4096$
Standard 13
Quantitative Aptitude