Question

Find the compound interest to the nearest rupee on Rs. $7500$ for $2$ years $4$ months at $12\%$ per annum reckoned annually.

• A. Rs. $2240$
• B. Rs. $2250$
• C. Rs. $2284$
• D. Rs. $2276$

Answer 1

Answer: (D)

Rs. $2276$

$P=7500$,

$r=12$,

$n=2\frac{4}{12}=\frac{7}{3}$

$CI=P\left[\right(1+\frac{r}{100}{)}^n-1]$

$CI=Rs.7500\left[\right(1+\frac{12}{100}{)}^{7/3}-1]$

$=2270.19$

Value to the nearest integer$=Rs.2276$.