Question

Honey borrowed a certain sum of money from Anu for 2 years at simple interest. Honey then lent this sum to Arushi at the same rate for 2 years at compound interest. At the end of two years she received Rs. as compound interest but paid Rs. 100 as simple interest. Find the sum and rate of interest.

Answer 1

Hint: To solve this question, firstly we will find the relation between P and r using the formula of Simple Interest. Then, using the condition in question we will solve the compound interest putting it equal to P+120, from where we will get the value of r. And then we will calculate the value of P using the relation, we get from using the formula of S.I.

Complete step by step answer:
Let the money borrowed by honey from Anu be initially, Rs. P and rate of interest be r.
Now, we know that Simple Interest that is S.I is calculated by formula,
\[S.I=\dfrac{P\times r\times t}{100}\] , where P is Principal Amount, r is rate of interest and t is time for which money is borrowed.
Here, in question S.I = Rs. 100, t = 2
So, on substituting values we get
\[100=\dfrac{P\times r\times 2}{100}\]
On solving, we get
\[P=\dfrac{5000}{r}\]
Now, we know that compound interest is equals to $A=P{{\left( 1+\dfrac{r}{100} \right)}^{n}}$
Now, as in question it is given that Honey then lent this sum to Arushi at the same rate for 2 years at compound interest and after the end of two years she received Rs. as compound interest but paid Rs. 100 as simple interest.
So, we can say that
$A=P{{\left( 1+\dfrac{r}{100} \right)}^{2}}$ , where A = P + 120
So, $P+120=P{{\left( 1+\dfrac{r}{100} \right)}^{2}}$
On simplification, we get
$120=P{{\left( 1+\dfrac{r}{100} \right)}^{2}}-P$
\[120=P\left[ {{\left( 1+\dfrac{r}{100} \right)}^{2}}-1 \right]\]
We know that, ${{a}^{2}}-{{b}^{2}}=(a+b)(a-b)$ ,
So, \[120=P\left[ \left( 1+\dfrac{r}{100}+1 \right)\left( 1+\dfrac{r}{100}-1 \right) \right]\]
On simplification, we get
\[120=P\left[ \left( \dfrac{r}{100}+2 \right)\left( \dfrac{r}{100} \right) \right]\]
As, we know that \[P=\dfrac{5000}{r}\]
So, \[120=\dfrac{5000}{r}\left( \dfrac{r}{100}+2 \right)\left( \dfrac{r}{100} \right)\]
On simplification, we get
\[120=50\left( \dfrac{r}{100}+2 \right)\]
\[24=10\left( \dfrac{r+200}{100} \right)\]
On solving, we get
\[240=r+200\]
Adding – 200 both sides, we get
\[240-200=r+200-200\]
r = 40%
as, \[P=\dfrac{5000}{r}\]
so, \[P=\dfrac{5000}{40}\]
P = Rs. 125

Hence, sum is Rs. 125 and rate of interest is 40%.

Note: To solve such question, must know the formulas to calculate the values of S.I and compound interest, so formula for Simple interest is \[S.I=\dfrac{P\times r\times t}{100}\] and for compound interest is $A=P{{\left( 1+\dfrac{r}{100} \right)}^{n}}$and remember the formula ${{a}^{2}}-{{b}^{2}}=(a+b)(a-b)$. Also, while solving calculation mistakes must be avoided as this will affect the solution part in the end or also make the solution more complex.