Question

2. A man oppened a R.D in a bank and deposits Rs. 1000 per month for 5/4 years. If he receives Rs. 500 as interest, then find the rate of interest p.a is *​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

Amount deposited per month, P = Rs 1000

Time = 5/4 years = 5/4 × 12 = 15 months

Number of instâllment, n = 15

Interest received on maturity, I = Rs 500

Let assume that Rate of interest be r % per annum.

We know,

Interest (I) received on a certain sum of money of Rs P deposited every month at the rate of r % per annum for n months is

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

So, on substituting the values, we get

$\rm :\longmapsto\:\text{500} = \text{1000} \times \dfrac{ \text{15(15 + 1)}}{24} \times \dfrac{ \text{r}}{100}$

$\rm :\longmapsto\:\text{500} = \text{10} \times \dfrac{ {15 \times 16}}{24} \times r$

$\rm :\longmapsto\:\text{50} = \dfrac{ {15 \times 2}}{3} \times r$

$\rm :\longmapsto\:\text{50} = 10 \times r$

$\bf\implies \:r \: = \: 5 \: \% \: per \: annum$

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Additional Information

Maturity Value (MV) received on a certain sum of money of Rs P deposited every month at the rate of r % per annum for n months is

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