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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.

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Last updated date: 20th Jun 2024
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Answer
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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.