Answer

Verified

405.3k+ views

**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.

Recently Updated Pages

Differentiate between Shortterm and Longterm adapt class 1 biology CBSE

How do you find slope point slope slope intercept standard class 12 maths CBSE

How do you find B1 We know that B2B+2I3 class 12 maths CBSE

How do you integrate int dfracxsqrt x2 + 9 dx class 12 maths CBSE

How do you integrate int left dfracx2 1x + 1 right class 12 maths CBSE

How do you find the critical points of yx2sin x on class 12 maths CBSE

Trending doubts

Which are the Top 10 Largest Countries of the World?

Define limiting molar conductivity Why does the conductivity class 12 chemistry CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Difference Between Plant Cell and Animal Cell

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Name 10 Living and Non living things class 9 biology CBSE

The Buddhist universities of Nalanda and Vikramshila class 7 social science CBSE

Write a letter to the principal requesting him to grant class 10 english CBSE