Ex.8.3 Q6 Comparing Quantities Solutions - NCERT Maths Class 8


Question

Arif took a loan of \(\rm{Rs}\,80,000\) from a bank. If the rate of interest is \(10\%\) per annum, find the difference in amounts he would be paying after \(1\begin{align}\frac{1}{2}\end{align}\) years if the interest is

(i) Compounded annually

(ii) Compounded half-yearly

 Video Solution
Comparing Quantities
Ex 8.3 | Question 6

Text Solution

What is known?

Principal, Time Period and Rate of Interest

What is unknown?

Amount and Compound Interest (C.I.)

Reasoning:

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

\(P=\rm{Rs}\, 80,000\)

\(N = 1\begin{align}\frac{1}{2}\end{align}\)years

\(R=10\%\) p.a. compounded half yearly and \(10\%\) p.a. compounded yearly

Steps:

For calculation of Compound Interest (C.I.) compounded annually:

Since ‘\(n\)’ is \(1\begin{align}\frac{{1}}{{2}}\end{align}\)years, amount can be calculated for \(1\) year and having that amount as principal, S.I can be calculated for \(\begin{align}\frac{1}{2}\end{align}\) year, because C.I is only annually.

\[\begin{align} A &= P\left( {{1 + }\frac{{r}}{{{100}}}} \right)^{n}  \\   &= 80000\left( {{1 + }\frac{{{10}}}{{{100}}}} \right)^{1}  \\   &= {80000 \times }\frac{{{11}}}{{{10}}} \\  &=  {80000 \times 1}{.1} \\ 
& = { 88000} \\  \end{align}\]

Amount after \(1\) years \(= \rm{ Rs}\, 88,000\)

Therefore, the principal for the \({1}\begin{align}\frac{{1}}{{2}}\end{align}\)th year \(= \rm{ Rs}\, 88,000\)

S.I. for \( \frac{1}{2} \) years

\[\begin{align} &= \frac{1}{2} \times 88000 \times \frac{{10}}{{100}}\\&= \frac{{8800}}{2}\\&= 4400\end{align}\]

Amount after \( 1\frac{1}{2} \) years

\[\begin{align}  &= 88000 + 4400\\&={\rm{Rs}}\;92400\end{align}\]

Compound Interest after \(1\frac{1}{2}\) years

\(\begin{align}  &= 92400 - 80000\\ &= {\rm{Rs}}\;12400 \dots \left\{\text{For the C.I. to be }\\{\text{charged yearly}} \right\} \end{align}\) 

For calculation of Compound Interest (C.I.) compounded half-yearly, we will consider as rate \(5\%\) p.a. and \(‘n'\) as \(3\)

\[\begin{align}A &= P\left( {{1 + }\frac{{r}}{{{100}}}} \right)^{n}  \\  &= 80000\left( {{1 + }\frac{{5}}{{{100}}}} \right)^{3}  \\ &= 80000\left( {{1 + }\frac{{1}}{{{20}}}} \right)^{3}  \\&= 80000\left( {\frac{{21}}{{20}}} \right)^3  \\ &= 80000 \times \frac{{21}}{{20}} \times \frac{{21}}{{20}} \times \frac{{21}}{{20}} \\ 
&= 80000 \times \frac{{9261}}{{8000}} \\  &=10 \times 9261 \\   &= 92610 \\ \end{align}\]

Compound Interest after \(1\frac{1}{2}\) years

\(\begin{align} & = 92610 - 80000\\&= {\rm{Rs}}\;12610  \dots \left\{ {\text{For the C.I. to be }}\\{\text{charged half - yearly}} \right\}\end{align}\)

Difference in amounts he would be paying \(=\rm{Rs}\, 92,610 − \rm{Rs}\, 92,400 = \rm{Rs}\, 210\)

Learn from the best math teachers and top your exams

  • Live one on one classroom and doubt clearing
  • Practice worksheets in and after class for conceptual clarity
  • Personalized curriculum to keep up with school