# Ex.8.3 Q3 Comparing Quantities Solution - NCERT Maths Class 8

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## Question

Fabina borrows $$\rm{Rs}\,12,500$$ at $$12\%$$ per annum for $$3$$ years at simple interest and Radha borrows the same amount for the same time period at $$10\%$$ per annum, compounded annually. Who pays more interest and by how much?

Video Solution
Comparing Quantities
Ex 8.3 | Question 3

## Text Solution

What is known?

Principal, Time Period and Rate of Interest

What is unknown?

Simple Interest and Compound Interest (C.I)

Reasoning:

For Simple Interest:

\begin{align}{A = }\frac{{{\rm{P \times R \times T}}}}{{100}}\end{align}

$$P = \rm{Rs}\, 12,500$$

$$N=3$$  years

$$R=12\%$$ simple interest

For Compound Interest:

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

$$P = \rm{Rs}\, 12,500$$

$$N=3$$  years

$$R=10\%$$ compounded annually

Steps:

Simple Interest paid by Fabina for $$3$$ years at the rate of $$12\%$$ per annum

\begin{align}{\rm{S}}{\rm{.I}}{\rm{. for\; }}3\;{\rm{years}} &= 3 \times 12500 \times \frac{{12}}{{100}}\\&= 3 \times 125 \times 12\\&= 4500\end{align}

Amount paid by Radha for $$3$$ years at the rate of $$10\%$$ p.a. compounded annually

\begin{align}A& = P{\left( {{1 + }\frac{{r}}{{{100}}}} \right)^{n}}\\ &= 12500{\left( {{1 + }\frac{{{10}}}{{{100}}}} \right)^{3}}\\ &= 12500{\left( { \frac{{11}}{{{10}}}} \right)^{3}}\\& = 12500\left( {\frac{{{11 \times 11 \times 11}}}{{{10 \times 10 \times 10}}}} \right)\\ &= 12500\left( {\frac{{{1331}}}{{{1000}}}} \right)\\&= {12500} \times {1}{.331}\\&= 16637.50\end{align}

Compound Interest

\begin{align} &= 16637.50 - 12500\\&= 4137.50\end{align}

Since $$4500 > 4137.50$$, Fabina paid more interest than Radha

$$= 4500 \,– 4137.50 \\ = \rm Rs \;362.50$$