Question

if a certain sum of money amounts to rupees 27960 in 3 yrs and amounts to rupees 34560 in 8 yrs at the same rate of simple interest find the sum and rate of interest applicable​

Answer 1

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

A certain sum of money amounts to Rs 27960 in 3 years and amounts to Rs 34560 in 8 years at the same rate of simple interest.

Let assume that

Sum of money invested be Rs P

Simple interest received per annum be SI

Rate of interest be r % per annum.

According to first condition

A sum of money P amounts to Rs 27960 in 3 years.

$\rm \: P + 3SI = 27960 - - - - (1)$

According to second condition

A sum of money P amounts to Rs 34560 in 8 years.

$\rm \: P + 8SI = 34560 - - - - (2)$

On Subtracting equation (1) from equation (2), we get

$\rm \: 5SI = 6600$

$\rm\implies \:SI \: = \: 1320$

On substituting the value of SI in equation (1), we get

$\rm \: P + 3 \times 1320= 27960$

$\rm \: P + 3960= 27960$

$\rm \: P= 27960 - 3960$

$\rm\implies \:P = 24000$

So, we have with us

$\rm \:P \: = \: Rs \: 24000$

$\rm \: SI \: =  \: \: Rs \: 1320$

$\rm \: n \: = \: 1 \: year$

We know,

Simple interest (SI) received on a certain sum of money of Rs P invested at the rate of r % per annum for n years is given by

$\boxed{\sf{  \:SI \: = \frac{P \times r \times n}{100} \: }}$

So, on substituting the values, we

$\rm \: 1320 = \dfrac{24000 \times r \times 1}{100}$

$\rm \: 132 = 24r$

$\rm\implies \:r \: = \: 5.5 \: \% \: per \: annum$

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Additional Information :-

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

$\boxed{\sf{  \:\rm \: Amount \: = \: P {\bigg[1 + \dfrac{r}{100} \bigg]}^{n} }}$

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

$\boxed{\sf{  \:\rm \: Amount \: = \: P {\bigg[1 + \dfrac{r}{200} \bigg]}^{2n} }}$

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

$\boxed{\sf{  \:\rm \: Amount \: = \: P {\bigg[1 + \dfrac{r}{400} \bigg]}^{4n} }}$

Answer 2

Step-by-step explanation:

GIVEN:--

$\text{\(A_{1}=27960 \: \tt \: \: \: T: 3 \) years}$

FIND: P=?, R: ?

SOLVE: S.I.

$\tt=\dfrac{P \times R \times I}{100}$

$\tt 27960-P=\dfrac{P \times R \times 3}{100} \qquad- - - - (1)$

$\tt \: 34560-P=\dfrac{P \times R \times 8}{100} \qquad - - - - - - -(2 )$

$\operatorname{eqn} \: (1) \div(2)$

$\tt \dfrac{27960-P}{34560-P}=\dfrac{3}{8}$

$223680-8 P=103680-3 P$

$\tt P=120000 / s=24000 Rs$

$\tt 27960-24000=\dfrac{24000 \times R \times 3}{100}$

$\tt \R=\dfrac{3960}{240 \times 3}=5.5 \%$