Question

Mrs. Jhaluka deposits Rs 1000 every month in a recurring deposit account for a time period of 3 years at 8% interest per annum. What is the matured value?

• A. Rs 20,440
• B. Rs 30,660
• C. Rs 40,440
• D. Rs 54,000

Answer 1

The correct option is C

Rs 40,440

Monthly installment = Rs. 1000

n = 36,

Amount deposited = 1000 $\times$ 36 = Rs. 36000

Maturity value = Rs. 1236

$\therefore$ Interest on his deposit = Rs. (1236 – 1200) = Rs. 36

We know that interest I = $\frac{n(n+1)}{2}\times \frac{(Installment\times rate)}{(100\times 12)}$

$\Rightarrow =\frac{(36\times 37)}{2}\times \frac{(1000\times 8)}{(100\times 12)}$

Interest = Rs 4440

Amount on maturity = Rs (36000 + 4440) = Rs 40,440