Question

Saransh deposits Rs.200 per month for 36 months in a bank's recurring deposit account. If the bank pays interest at the rate of 11% per annum, find the amount he gets on maturity.​

Answer 1

Given:–

• Saransh deposited or Principal = Rs.200
• n = 36 months
• Rate of interest = 11%

To find:–

• Amount he will get on maturity

Formulas used:–

⭐ Computing maturity value

• $\bold{ \pink {\boxed{\text{I} = \text{P} \times \dfrac{ \text{n(n + 1)}}{2 \times 12} \times \dfrac{ \text{r}}{100} }}}$

where,

• I = Interest

• P = Principal

• n = number of months

• r = Rate

Step by step explaination:-

If a sum of Rs.P is deposited every month in a bank for n number of months. If rate of interest is r% that is for per year, therefore the Interest on whole deposit is calculated by the above given formula.

$\mathfrak{Therefore,\: evaluating \:values \:in \:the \:given\: formula}$

$\: \: \: :\implies \: \: \: \: \: \: \: \text{I} = 200 \times \dfrac{36(36 + 1)}{2 \times 12} \times \dfrac{11}{100}$

$\underline\bold{ \mathfrak{solving..}}$

$\: \: \: :\implies \: \: \: \: \: \: \: \text{I} = 200 \times \dfrac{36 \times 37}{2 \times 12} \times \dfrac{11}{100}$

$\: \: \: :\implies \: \: \: \: \: \: \: \text{I} = \cancel{200} \times \dfrac{36 \times 37}{ \cancel2 \times \cancel{12}} \times \dfrac{11}{100}$

$\: \: \: :\implies \: \: \: \: \: \: \: \text{I} = \text{rs.1221}$

$\underline\bold{ \mathfrak{Total\: sum \:deposited\: in\: n\: months..}}$

→ Sum deposited every month × number of months

→ P × n

→ Rs.200 × Rs.36

→ Rs.7200

$\underline\bold{ \mathfrak{Amount \:Saransh\: will\: receive\: at \:maturity \:time:-}}$

→ Rs.7200 + Rs.1221

→ Rs.8421

Conclusion:–

He will get amount Rs.8421 on maturity