Question

A man has a Recurring Deposit Account in a bank for $3^{1/2}years.$ If the rate of interest is 12% per annum and the man gets Rs.2,892 on maturity, find the monthly installment.

Answer 1

Maturity value $=P\times n+P\times \frac{n(n+1)}{2\times 12}\times \frac{r}{100}$

P=monthly deposit

n= number of months $=3\frac{1}{2}years=42$ months

r= rate of interest $12$

$2892=P\times 42+P\times \frac{42(42+1)}{2\times 12}\times \frac{12}{100}$

$2892=42P+P\times \frac{42\times 43}{2\times 100}$

$2892=42P(1+\frac{43}{2\times 100})$

$2892=42P\left(1.215\right)$

$P=Rs.56.67$ is the monthly deposit.