Question

Mrs Goswami deposits ₹1000 per month in a recurring deposit account for 3 yrs at 8% interest per annum. Find the matured value..​

Answer 1

Given :

The deposited amount per month = P=  Rs 1000

The time period for deposition = T = 3 years = 12 × 3 = 36 months

The rate of interest = R = 8%

To Find :

The matured value

Solution :

∵ Interest = Principal × $\dfrac{n(n+1)}{2}$ × $\dfrac{R \times T}{100}$               where n = number of month

So, Interest = P × $\dfrac{n(n+1)}{2}$ × $\dfrac{R \times 1}{12 \times 100}$

Or,               = Rs 1000 × $\dfrac{36(36+1)}{2}$ × $\dfrac{8 \times 1}{12 \times 100}$

Or,              = Rs 1000 × $\dfrac{36 \times 37}{2}$ × $\dfrac{8 \times 1}{12 \times 100}$

Or,              = Rs 4440

So, Interest = Rs 4440

Now,

Maturity value = money deposit × number of months + Interest

                        = Rs 1000 × 36 + Rs 4440

                        = Rs 36000 + Rs 4440

                       = Rs 40440

Hence, The matured value after 3 years of deposit is Rs 40440  . Answer

Answer 2

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