Question

man deposited ₹1200 in a recurring deposit account for 1 year at 5% per

annum simple interest. The interest earned by him on maturity is​

Answer 1

Answer:

390,14790

Step-by-step explanation:

P= 1200  

R%=5

N=12

INTEREST=   1200*12*13*5/

                           2400

                   = 390

MV = PN +  I

         1200*12+390

          14400+390

          14790

Answer 2

Given:

Man deposited , P = ₹1200

n = 12 months

r = 5 %

To find :

The interest earned by him on maturity

Formula to be used:

Interest = P × $\frac{n(n + 1)r}{2400 }$

Solution:

Step 1 of 2 :

To find the interest earned by him on maturity,

Interest = P × $\frac{n(n + 1)r}{2400 }$

Interest = 1200 × $\frac{12(12 + 1)5}{2400 }$

Interest = 1200 × $\frac{12(13)5}{2400 }$

Interest = $\frac{12(13)5}{2 }$

Interest = 6 × 13 × 5

Interest = 390

Step 2 of 2 :

Maturity value = Pn + I

Maturity value = 1200 × 12 + 390

Maturity value = 14400 + 390

Maturity value = 14790

Final answer:

The interest earned by him on maturity is​ ₹14790 .