Question

Q5. A sum of money put at 10 % per annum simple interest amounts to Rs 41600 in 3 years. what will it amount to in 2 years at the same rate? *​

Answer 1

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

Given that,

• A sum of money put at 10 % per annum simple interest amounts to Rs 41600 in 3 years.

So, we have

• Amount = Rs 41600

• Time, n = 3 years

• Rate of interest, r = 10 % per annum

Let assume that

• Sum of money deposited be Rs P.

We know,

Amount received on a certain sum of money of Rs P invested at the rate of r % per annum for n years is given by

$\boxed{ \rm{ \:Amount \: = \: P\bigg[1 + \dfrac{rn}{100} \bigg] \: }}$

So, on substituting the values, we get

$\rm \: \:41600 \: = \: P\bigg[1 + \dfrac{10 \times 3}{100} \bigg] \:$

$\rm \: \:41600 \: = \: P\bigg[1 + \dfrac{3}{10} \bigg] \:$

$\rm \: \:41600 \: = \: P\bigg[\dfrac{10 + 3}{10} \bigg] \:$

$\rm \: \:41600 \: = \: P\bigg[\dfrac{13}{10} \bigg] \:$

$\rm \: P = \dfrac{41600 \times 10}{13}$

$\rm \: P = 3200 \times 10$

$\bf\implies \:P \: = \: Rs \: 32000$

Now, we have to find what Rs 32000 amounts to in 2 years at the rate of 10 % per annum.

So, we have now

• Principal, P = Rs 32000

• Rate of interest, r = 10 % per annum

• Time, n = 2 years

So, Amount received in 2 years is

$\rm \: \:Amount \: = \: 32000\bigg[1 + \dfrac{10 \times 2}{100} \bigg] \:$

$\rm \: \:Amount \: = \: 32000\bigg[1 + \dfrac{2}{10} \bigg] \:$

$\rm \: \:Amount \: = \: 32000\bigg[1 + \dfrac{1}{5} \bigg] \:$

$\rm \: \:Amount \: = \: 32000\bigg[\dfrac{5 + 1}{5} \bigg] \:$

$\rm \: \:Amount \: = \: 32000\bigg[\dfrac{6}{5} \bigg] \:$

$\rm \: \:Amount \: = \: 6400 \times 6 \:$

$\bf\implies \:Amount \: = \: Rs \: 38400$

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Additional Information :-

$\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: Formulae}}}} \\ \\ \bigstar \: \bf{P = \dfrac{SI \times 100}{r \times n} }\\ \\ \bigstar \: \bf{r = \dfrac{SI \times 100}{P \times n} }\\ \\ \bigstar \: \bf{n = \dfrac{SI \times 100}{P \times r} }\\ \\ \bigstar \: \bf{Amount = P\bigg(\dfrac{100 + rn}{100} \bigg) }\: \end{array} }}\end{gathered}\end{gathered}\end{gathered}$