Math, asked by anees9, 6 days ago

Question 17. Aryan has a recurring deposit account in a bank. If he deposits 2000 per month and the bank pays interest at the rate of 10% per annum, find the total time for which he must hold the account to get 83100 at the time of maturity

Answers

Answered by mathdude500
2

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

Given that,

Aryan has a recurring deposit account in a bank. If he deposits 2000 per month and the bank pays interest at the rate of 10% per annum.

So, we have

Amount deposited every month, P = Rs 2000

Rate of interest, r = 10 % per annum

Maturity Value, MV = Rs 83100

Let assume that number of instâllments be n.

We know,

Maturity Value (MV) on a instâllment of Rs P invested per month at the rate of r % per annum for n months is

\bold{ \pink {\rm :\longmapsto\:\boxed{\text{MV} =\text{nP} +  \text{P} \times \dfrac{ \text{n(n + 1)}}{24} \times \dfrac{ \text{r}}{100} }}}

So, on substituting the values, we get

\rm :\longmapsto\:\text{83100} =\text{2000n} +  \text{2000} \times \dfrac{ \text{n(n + 1)}}{24} \times \dfrac{ \text{10}}{100}

\rm :\longmapsto\:\text{83100} =\text{2000n} +   \dfrac{ \text{25n(n + 1)}}{3}

\rm :\longmapsto\:\text{83100} =\text{2000n} +   \dfrac{ {25 {n}^{2}  + 25n}}{3}

\rm :\longmapsto\:\text{83100} = \dfrac{ {25 {n}^{2}  + 25n + 6000n}}{3}

\rm :\longmapsto\:\text{83100} = \dfrac{ {25 {n}^{2}  + 6025n}}{3}

On dividing both sides by 25, we get

\rm :\longmapsto\:\text{3325} = \dfrac{ { {n}^{2}  + 241n}}{3}

\rm :\longmapsto\: {n}^{2}  + 241n - 9972 = 0

\rm :\longmapsto\: {n}^{2}  + 277n - 36n - 9972 = 0

\rm :\longmapsto\:n(n + 277) - 36(n + 277) = 0

\rm :\longmapsto\:(n + 277)(n  - 36) = 0

\bf\implies \:n = 36

So, Total time period is 3 years.

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More to know

Interest received on maturity on a instâllment of Rs P invested per month at the rate of r % per annum for n months is

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

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