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

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## 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.)

Video Solution
Comparing Quantities
Ex 8.3 | Question 2

## 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 $$=\rm{Rs}\;36659.70$$

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