If the difference between the compound interest and simple interest on

1.	If the difference between the compound interest and simple interest on a sum of money is Rs. 32 at a rate of 5% pa for 2 yr. What is SI?
Answer» $\large\underline{\sf{Solution-}}$ GIVEN that, $\sf{\rm :\longmapsto\: Time\:_{(annually)}, \: <klux>N</klux> \: = \: 2\: years}$ $\sf{\rm :\longmapsto\: Rate\:_{(interest)} \: r \: = \: 5\% \: per \: annum}$ Let assume that sum of money invested be Rs P. We know that, Simple Interest (SI) on a CERTAIN sum of money of Rs P invested at the rate of r % per annum for n years is $\red{\rm :\longmapsto\:SI = \dfrac{P \times r \times n}{100}}$ and Compound interest (CI) 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 $\red{\rm :\longmapsto\: CI=P\bigg(1+\dfrac{r}{100}\bigg)^{n}-P}$ Now, According to statement, it is given that, $\rm :\longmapsto\: CI \: - \: SI \: = \: 32$ On substituting the values, we get $\rm :\longmapsto\:P\bigg(1+\dfrac{r}{100}\bigg)^{n} - P - \dfrac{Prn}{100} = 32$ $\rm :\longmapsto\:P\bigg(1+\dfrac{5}{100}\bigg)^{2} - P - \dfrac{P \times 5 \times 2}{100} = 32$ $\rm :\longmapsto\:P\bigg(1+\dfrac{1}{20}\bigg)^{2} - P - \dfrac{P }{10} = 32$ $\rm :\longmapsto\:P\bigg(\dfrac{20 + 1}{20}\bigg)^{2} - P - \dfrac{P }{10} = 32$ $\rm :\longmapsto\:P\bigg(\dfrac{21}{20}\bigg)^{2} - P - \dfrac{P }{10} = 32$ $\rm :\longmapsto\:P\bigg(\dfrac{441}{400}\bigg) - P - \dfrac{P }{10} = 32$ $\rm :\longmapsto\:\dfrac{441P - 400P - 40P}{400} = 32$ $\rm :\longmapsto\:\dfrac{441P - 440P}{400} = 32$ $\rm :\longmapsto\:\dfrac{P}{400} = 32$ $\bf\implies \:P = 12800$ So, Sum of money invested = Rs 12800 Rate of interest, r = 5 % per annum Time, n = 2 years So, $\red{\rm :\longmapsto\:SI = \dfrac{P \times r \times n}{100}}$ On substituting the values, we get $\rm :\longmapsto\:SI = \dfrac{12800 \times 5 \times 2}{100}$ $\bf\implies \:SI \: = \: Rs \: 1280$ Additional Information :- 1. Amount on a certain sum of money of Rs P invested at the rate of r % per annum for n years compounded annually is $\boxed{ \rm{ Amount = P\bigg(1+\dfrac{r}{100}\bigg)^{n}}}$ 2. Amount on a certain sum of money of Rs P invested at the rate of r % per annum for n years compounded semi - annually is $\boxed{ \rm{ Amount = P\bigg(1+\dfrac{r}{200}\bigg)^{2n}}}$ 3. Amount on a certain sum of money of Rs P invested at the rate of r % per annum for n years compounded QUARTERLY is $\boxed{ \rm{ Amount = P\bigg(1+\dfrac{r}{400}\bigg)^{4n}}}$

If the difference between the compound interest and simple interest on a sum of money is Rs. 32 at a rate of 5% pa for 2 yr. What is SI?

Answer»

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

GIVEN that,

$\sf{\rm :\longmapsto\: Time\:_{(annually)}, \: <klux>N</klux> \: = \: 2\: years}$

$\sf{\rm :\longmapsto\: Rate\:_{(interest)} \: r \: = \: 5\% \: per \: annum}$

Let assume that sum of money invested be Rs P.

We know that,

Simple Interest (SI) on a CERTAIN sum of money of Rs P invested at the rate of r % per annum for n years is

$\red{\rm :\longmapsto\:SI = \dfrac{P \times r \times n}{100}}$

and

Compound interest (CI) 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

$\red{\rm :\longmapsto\: CI=P\bigg(1+\dfrac{r}{100}\bigg)^{n}-P}$

Now, According to statement, it is given that,

$\rm :\longmapsto\: CI \: - \: SI \: = \: 32$

On substituting the values, we get

$\rm :\longmapsto\:P\bigg(1+\dfrac{r}{100}\bigg)^{n} - P - \dfrac{Prn}{100} = 32$

$\rm :\longmapsto\:P\bigg(1+\dfrac{5}{100}\bigg)^{2} - P - \dfrac{P \times 5 \times 2}{100} = 32$

$\rm :\longmapsto\:P\bigg(1+\dfrac{1}{20}\bigg)^{2} - P - \dfrac{P }{10} = 32$

$\rm :\longmapsto\:P\bigg(\dfrac{20 + 1}{20}\bigg)^{2} - P - \dfrac{P }{10} = 32$

$\rm :\longmapsto\:P\bigg(\dfrac{21}{20}\bigg)^{2} - P - \dfrac{P }{10} = 32$

$\rm :\longmapsto\:P\bigg(\dfrac{441}{400}\bigg) - P - \dfrac{P }{10} = 32$

$\rm :\longmapsto\:\dfrac{441P - 400P - 40P}{400} = 32$

$\rm :\longmapsto\:\dfrac{441P - 440P}{400} = 32$

$\rm :\longmapsto\:\dfrac{P}{400} = 32$

$\bf\implies \:P = 12800$

So, Sum of money invested = Rs 12800

Rate of interest, r = 5 % per annum

Time, n = 2 years

So,

$\red{\rm :\longmapsto\:SI = \dfrac{P \times r \times n}{100}}$

On substituting the values, we get

$\rm :\longmapsto\:SI = \dfrac{12800 \times 5 \times 2}{100}$

$\bf\implies \:SI \: = \: Rs \: 1280$

Additional Information :-

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

$\boxed{ \rm{ Amount = P\bigg(1+\dfrac{r}{100}\bigg)^{n}}}$

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

$\boxed{ \rm{ Amount = P\bigg(1+\dfrac{r}{200}\bigg)^{2n}}}$

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

$\boxed{ \rm{ Amount = P\bigg(1+\dfrac{r}{400}\bigg)^{4n}}}$