# Ex.8.3 Q1 Comparing Quantities Solution - NCERT Maths Class 8

## Question

Calculate the amount and compound interest on

(i) $$\rm{Rs}\, 10,800$$ for $$3$$ years at \begin{align}12\frac{{1}}{{2}}\%\end{align} per annum compounded annually.

(ii) $$\rm{Rs}\, 18,000$$ for \begin{align} 2\frac{{1}}{{2}}\end{align} years at $$10\%$$ per annum compounded annually.

(iii) $$\rm{Rs}\, 62,500$$ for \begin{align} 1\frac{{1}}{{2}}\end{align} years at $$8\%$$ per annum compounded half yearly.

(iv) $$\rm{Rs}\, 8,000$$ for $$1$$ year at $$9\%$$ per annum compounded half yearly. (You could use the year by year calculation using SI formula to verify).

(v) $$\rm{Rs}\, 10,000$$ for $$1$$ year at $$8\%$$ per annum compounded half yearly.

Video Solution
Comparing Quantities
Ex 8.3 | Question 1

## 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}

$$P = \rm{Rs}\, 10800$$
$$N = 3\;\rm{years}$$
\begin{align}R = 12\frac{1}{2} \%= \frac{25}{2}\%\end{align}  compounded annually

Steps:

(i)

\begin{align}A &= P{\left( {1 + }\frac{r}{{100}} \right)^{n}}\\&= 10800{\left({1 + }\frac{{25}}{{2 \times 100}} \right)^{3}}\\&= 10800{\left( {\frac{{225}}{{200}}} \right)^{3}}\\ &= 10800 \times \frac{{225 \times 225 \times 225}}{{200 \times 200 \times 200}}\\& = 15377.34\end{align}

\begin{align}{\rm{C}}{\rm{.I}}. &= A - P\\&= {\rm{15377}}{\rm{.34}} - {\rm{10800}}\\ &= {\rm{4577}}{\rm{.34}}\end{align}

Amount $$= \rm{Rs}\, 15377.34$$

Compound Interest $$=\rm{ Rs}\, 4577.34$$

(ii)

$$P= \rm{Rs}\, 18000$$

\begin{align} n= 2\frac{{1}}{{2}}\end{align}years

$$\rm{R}= 10\%$$ p.a compounded annually

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

Since $$‘n’$$ is \begin{align} 2\frac{{1}}{{2}}\end{align}years, amount can be calculated for $$2$$ years and having 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}}\\ &= 18000{\left( {{1 + }\frac{{{10}}}{{{100}}}} \right)^{2}}\\ &= 18000 \times \frac{{{11 \times 11}}}{{{10 \times 10}}}\\ &= 21780\end{align}

Amount after $$2$$ years $$=\rm{ Rs}\, 21780$$

\begin{align}{\rm{S}}{\rm{.I}}{\rm{. for }}\frac{1}{2}{\rm{ years}} &= \frac{1}{2} \times 21780 \times \frac{{10}}{{100}}\\ &= 1089\end{align}

Amount after $$2\frac{1}{2}$$ years

\begin{align} &= 21780 + 1089\\ &= {\rm{Rs}}\;22869\end{align}

Compound Interest after $$2\frac{1}{2}$$ years

\begin{align} &= 22869 - 18000\\ &= {\rm{Rs}}\;4869\end{align}

(iii)

$$P=\rm{ Rs}\, 62,500$$

\begin{align} n= 1\frac{{1}}{{2}}\end{align} years

$$R= 8\%$$ p.a. compounded half yearly

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

There are $$3$$ half years in \begin{align} 1\frac{{1}}{{2}}\end{align} years. Therefore, compounding has to be done $$3$$ times and rate of interest will be $$4\%$$

\begin{align}A &= P{\left( {{1 + }\frac{{r}}{{{100}}}} \right)^{n}}\\&= 62500{\left( {{1 + }\frac{4}{{{100}}}} \right)^3}\\&= 62500 \times \frac{{{104 \times 104} \times {104}}}{{{100 \times 100} \times {100}}}\\&= 70304\end{align}

C.I. after $$1\frac{1}{2}$$ years ( $$8\%$$ p.a. interest half yearly )

\begin{align} &= 70304 - 62500\\&= 7804\end{align}

Amount after $$1\frac{1}{2}$$years ( $$8\%$$ p.a. interest half yearly) $$= 70304$$

Amount after \begin{align} 1\frac{{1}}{{2}}\end{align} years $$= \rm{Rs}\, 70304$$

Compound Interest after\begin{align} 1\frac{{1}}{{2}}\end{align} years $$= \rm{Rs}\, 7804$$

(iv)

$$P= \rm{Rs}\, 8000$$

$$n= 1$$ year

$$R= 9\%$$ p.a. compounded half yearly

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

S.I. for $$1^{\rm st}$$  $$6$$ months

\begin{align} &= \frac{1}{2} \times 8000 \times \frac{9}{{100}}\\ &= 40 \times 9\\ &= 360\end{align}

Amount after $$1^{\text{st}} 6$$ months including Simple Interest

\begin{align} &= 8000 + 360\\ &= {\rm{Rs}}\;8360\end{align}

Principal for $$2$$nd $$6$$ months $$= \rm{Rs}\, 8360$$

S.I. for $$2^{\text{nd}} 6$$ months

\begin{align} &= \frac{1}{2} \times 8360 \times \frac{9}{{100}}\\ &= \frac{{418 \times 9}}{{10}}\\ &= 376.20\end{align}

C.I. after $$1$$ year ( $$9\%$$  p.a. interest half yearly )

\begin{align} &= 360 + 376.20\\ &= 736.20\end{align}

Amount after $$1$$ year ( $$9\%$$  p.a. interest half yearly )

\begin{align} &= 8000 + 736.20\\ &= 8736.20\end{align}

Amount after $$1$$ year $$=\rm{ Rs}\, 8736.20$$

Compound Interest after $$1$$ year $$= \rm{Rs}\, 736.20$$

(v)

$$P= \rm{Rs}\, 10,000$$

$$n=1$$ year

$$R= 8\%$$ p.a. compounded half yearly

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

There are $$2$$ half years in $$1$$ years. Therefore, compounding has to be done $$2$$ times and rate of interest will be $$4\%$$

\begin{align}A &= P{\left( {{1 + }\frac{{r}}{{{100}}}} \right)^{n}}\\& = 10000{\left( {{1 + }\frac{4}{{{100}}}} \right)^2}\\& = 10000 \times \frac{{{104 \times 104}}}{{{100 \times 100}}}\\ &= 10816\end{align}

C.I. after $$1$$ year ($$8\%$$  p.a. interest half yearly )

\begin{align} &= 10816 + 10000\\ &= 816\end{align}

Amount after $$1$$ year ( $$8\%$$  p.a. interest half yearly) $$= 10816$$

Amount after $$1$$ year $$= \rm{Rs}\, 10816$$

Compound Interest after $$1$$ year $$= \rm{Rs}\, 816$$

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