Question

if the difference between the compound intrest and simple intrest on a certain sum of money for 2 years at 25/2% per annum is ₹150.the sum is​

Answer 1

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

Given that,

The difference between the compound intrest and simple intrest on a certain sum of money for 2 years at 25/2% per annum is ₹ 150.

So,

Rate of interest, r = 12.5 % per annum

Time, n = 2 years

Let assume that, Sum of money invested = ₹ p.

Now, we know that

Simple interest on a certain sum of money of Rs p invested at the rate of r % per annum for n years is

$\red{\boxed{ \tt{ \: S.I. \: = \: \frac{p \times r \times n}{100} \: }}}$

Also,

Compound interest on a certain sum of money of Rs p invested at the rate of r % per annum compounded annually for n years is

$\red{\boxed{ \tt{ \: C.I. = p {\bigg[1 + \dfrac{r}{100} \bigg]}^{n} - p \: }}}$

According to statement, It is given that

$\green{\bf :\longmapsto\: \: C.I. - \: S.I. \: = \: 150 \: }$

$\rm :\longmapsto\:p {\bigg[1 + \dfrac{r}{100} \bigg]}^{n} - p - \dfrac{prn}{100} = 150$

On substituting the values of n and r, we get

$\rm :\longmapsto\:p {\bigg[1 + \dfrac{12.5}{100} \bigg]}^{2} - p - \dfrac{p \times 12.5 \times 2}{100} = 150$

$\rm :\longmapsto\:p {\bigg[1 + \dfrac{125}{1000} \bigg]}^{2} - p - \dfrac{p \times 125 \times 2}{1000} = 150$

$\rm :\longmapsto\:p {\bigg[1 + \dfrac{1}{8} \bigg]}^{2} - p - \dfrac{p \times 1 \times 2}{8} = 150$

$\rm :\longmapsto\:p {\bigg[ \dfrac{8 + 1}{8} \bigg]}^{2} - p - \dfrac{p }{4} = 150$

$\rm :\longmapsto\:p {\bigg[ \dfrac{9}{8} \bigg]}^{2} - p - \dfrac{p }{4} = 150$

$\rm :\longmapsto\:\dfrac{81}{64}p - p - \dfrac{p }{4} = 150$

$\rm :\longmapsto\:\dfrac{81p - 64p - 16p}{64}= 150$

$\rm :\longmapsto\:\dfrac{81p - 80p}{64}= 150$

$\rm :\longmapsto\:\dfrac{p}{64}= 150$

$\bf\implies \:\boxed{ \tt{ \: p \: = \: 9600 \: }}$

Hence,

Sum invested = ₹ 9600

More to know :-

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

$\boxed{ \tt{ \: Amount \: = \: p \: {\bigg[1 + \dfrac{r}{100} \bigg]}^{n} \: }}$

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

$\boxed{ \tt{ \: Amount \: = \: p \: {\bigg[1 + \dfrac{r}{200} \bigg]}^{2n} \: }}$

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

$\boxed{ \tt{ \: Amount \: = \: p \: {\bigg[1 + \dfrac{r}{400} \bigg]}^{4n} \: }}$

4. Amount on a certain sum of money of Rs p invested at the rate of r % per annum compounded monthly for n years is

$\boxed{ \tt{ \: Amount \: = \: p \: {\bigg[1 + \dfrac{r}{1200} \bigg]}^{12n} \: }}$

Answer 2

Answer:

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

Given that,

The difference between the compound intrest and simple intrest on a certain sum of money for 2 years at 25/2% per annum is ₹ 150.

So,

Rate of interest, r = 12.5 % per annum

Time, n = 2 years

Let assume that, Sum of money invested = ₹ p.

Now, we know that

Simple interest on a certain sum of money of Rs p invested at the rate of r % per annum for n years is

$\red{\boxed{ \tt{ \: S.I. \: = \: \frac{p \times r \times n}{100} \: }}}$

Also,

Compound interest on a certain sum of money of Rs p invested at the rate of r % per annum compounded annually for n years is

$\red{\boxed{ \tt{ \: C.I. = p {\bigg[1 + \dfrac{r}{100} \bigg]}^{n} - p \: }}}$

According to statement, It is given that

$\green{\bf :\longmapsto\: \: C.I. - \: S.I. \: = \: 150 \: }$

$\rm :\longmapsto\:p {\bigg[1 + \dfrac{r}{100} \bigg]}^{n} - p - \dfrac{prn}{100} = 150$

On substituting the values of n and r, we get

$\rm :\longmapsto\:p {\bigg[1 + \dfrac{12.5}{100} \bigg]}^{2} - p - \dfrac{p \times 12.5 \times 2}{100} = 150$

$\rm :\longmapsto\:p {\bigg[1 + \dfrac{125}{1000} \bigg]}^{2} - p - \dfrac{p \times 125 \times 2}{1000} = 150$

$\rm :\longmapsto\:p {\bigg[1 + \dfrac{1}{8} \bigg]}^{2} - p - \dfrac{p \times 1 \times 2}{8} = 150$

$\rm :\longmapsto\:p {\bigg[ \dfrac{8 + 1}{8} \bigg]}^{2} - p - \dfrac{p }{4} = 150$

$\rm :\longmapsto\:p {\bigg[ \dfrac{9}{8} \bigg]}^{2} - p - \dfrac{p }{4} = 150$

$\rm :\longmapsto\:\dfrac{81}{64}p - p - \dfrac{p }{4} = 150$

$\rm :\longmapsto\:\dfrac{81p - 64p - 16p}{64}= 150$

$\rm :\longmapsto\:\dfrac{81p - 80p}{64}= 150$

$\rm :\longmapsto\:\dfrac{p}{64}= 150$

$\bf\implies \:\boxed{ \tt{ \: p \: = \: 9600 \: }}$

Hence,

Sum invested = ₹ 9600

More to know :-

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

$\boxed{ \tt{ \: Amount \: = \: p \: {\bigg[1 + \dfrac{r}{100} \bigg]}^{n} \: }}$

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

$\boxed{ \tt{ \: Amount \: = \: p \: {\bigg[1 + \dfrac{r}{200} \bigg]}^{2n} \: }}$

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

$\boxed{ \tt{ \: Amount \: = \: p \: {\bigg[1 + \dfrac{r}{400} \bigg]}^{4n} \: }}$

4. Amount on a certain sum of money of Rs p invested at the rate of r % per annum compounded monthly for n years is

$\boxed{ \tt{ \: Amount \: = \: p \: {\bigg[1 + \dfrac{r}{1200} \bigg]}^{12n} \: }}$