# Ex.8.3 Q2 Comparing Quantities Solution - NCERT Maths Class 8

## Question

Kamala borrowed \(\rm{Rs}\, 26400\) from a Bank to buy a scooter at a rate of \(15\%\) p.a. compounded yearly. What amount will she pay at the end of \(2\) years and \(4\) months to clear the loan?

(**Hint:** Find \(A\) for \(2\) years with interest is compounded yearly and then find S.I. on the \(2\)nd year amount for \(\begin{align}\frac{4}{{12}}\end{align}\) years.)

## Text Solution

**What is known?**

Principal, Time Period and Rate of Interest

**What is unknown?**

Amount and Compound Interest (C.I.)

**Reasoning:**

\({A = P}{\left( {{1 + }\frac{{r}}{{{100}}}} \right)^{\rm{n}}}\)

\(P = \rm{Rs}\, 26400\)

\(N = 2\) years \(4\) months

\(R=15\%\) compounded annually

**Steps:**

First, we will calculate Compound Interest (C.I) for the period of \(2\) years

\[\begin{align}A &= P{\left( {{1 + }\frac{{r}}{{{100}}}} \right)^{n}}\\&= 26400{\left( {{1 + }\frac{{{15}}}{{{100}}}} \right)^{2}}\\& = 26400{\left( {\frac{{100}}{{100}} + \frac{{{15}}}{{{100}}}} \right)^{2}}\\&= 26400 \times \frac{{{115 \times 115}}}{{{100 \times 100}}}\\& = {26400}\left( {\frac{{{23}}}{{{20}}}{ \times }\frac{{{23}}}{{{20}}}} \right)\\&= {26400} \times {1}{.3225}\\

&= 34914\end{align}\]

\[\begin{align}{\rm{C}}{\rm{.I}}. &= A - {\rm{P}}\\&{ = {\rm{34914}} - {\rm{26400}}}\\&{ = {\rm{8514}}}\end{align}\]

Second, we will find Simple Interest (S.I) for the period of \(4\) months

Principal for \(4\) months after C.I. for \(2\) years \(= \rm{Rs}\,34,914\)

S.I. for \(4\) months

\[\begin{align} &= \frac{4}{{12}} \times 34914 \times \frac{{15}}{{100}}\\&= \frac{1}{3} \times 34914 \times \frac{3}{{20}}\\&= \frac{{34914}}{{20}}\\&= 1745.70\end{align}\]

Total interest for \(2\) years \(4\) months

\[\begin{align} &= 8514 + 1745.70\\&= 10259.70\end{align}\]

Total amount for \(2\) years \(4\) months

\[\begin{align} &= 26400 + 10259.70\\&= 36659.70\end{align}\]

The amount Kamala will have to pay after \(2\) years \(4\) months \(=\rm{Rs}\;36659.70\)