Question

Vikram borrowed Rs. $20,000$ for $4\frac{1}{2}$ years at $10\%$ per annum, compound annually. How much compound interest would he pay at the end of $4\frac{1}{2}$ years?

• A. $30711.22$
• B. $20711.22$
• C. $40711.22$
• D. $10711.22$

Answer 1

Answer: (D)

$10711.22$

We know the formula,
$A=P{(1+\frac{r}{100})}^t$
Where,
$A=$ total amount
$P=$ principal or amount of money deposited,
$r=$ annual interest rate
$n=$ number of times compounded per year
$t=$ time in years
Given:
$P=$ Rs. $20000,r=10\%$ and $t=$ $4\frac{1}{2}$ years
$A=20000{(1+\frac{10}{100})}^{4.5}$
$A=20000\times {1.1}^{4.5}$
$A=20000\times 1.535561$
$A=$ Rs. $30711.22$
To find interest we use formula $A=P+I$, since $A=30711.22$ and $P=20000$ we have:
$A=P+I$
$30711.22=20000+I$
$I=30711.22-20000=10711.22$
Interest, I $=$ Rs. $10711.22$