# Exercise 8.3 Comparing Quantities- NCERT Solutions Class 8

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## Chapter 8 Ex.8.3 Question 1

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.

### 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 $$= \rm{Rs}\, 816$$

## Chapter 8 Ex.8.3 Question 2

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.)

### 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 $$=\rm{Rs}\;36659.70$$

## Chapter 8 Ex.8.3 Question 3

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?

### 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

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

## Chapter 8 Ex.8.3 Question 4

I borrowed $$\rm{Rs} \,12000$$ from Jamshed at $$6\%$$ per annum simple interest for $$2$$ years. Had I borrowed this sum at $$6\%$$ per annum compound interest, what extra amount would I have to pay?

### 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}

$$P = \rm{Rs}\,12,000$$

$$N = 2$$ years

$$R = 6\%$$ simple interest

For Compound Interest:

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

$$P = \rm{Rs}\,12,000$$

$$N =2$$ years

$$R = 6\%$$ compounded annually

Steps:

Simple Interest to be paid for $$2$$ years at the rate of $$6\%$$ per annum

\begin{align}{\rm{S}}{\rm{.I}}{\rm{. for \;}}2\;{\rm{years}} &= 2 \times 12000 \times \frac{6}{{100}}\\&= 2 \times 120 \times 6\\&= 1440\end{align}

Compound Interest to be paid for $$2$$ years at the rate of $$6\%$$ per annum

\begin{align}A& = P{\left( {{1 + }\frac{{r}}{{{100}}}} \right)^{n}}\\& = 12000{\left( {{1 + }\frac{{6}}{{{100}}}} \right)^{2}}\\& = 12000{\left( {\frac{{100}}{{100}} + \frac{{6}}{{{100}}}} \right)^{2}}\\& = 12000\left( {\frac{{{106 \times 106}}}{{{100 \times 100}}}} \right)\\& = 12000\left( {\frac{{{11236}}}{{{10000}}}} \right)\\&= {12000} \times {1}{.1236}\\&= 13483.20\end{align}

Compound Interest

\begin{align} &= 13483.20 - 12000\\&= 1483.20\end{align}

Compound Interest - Simple Interest

\begin{align} &= 1483.20 - 1440\\&= 43.20\end{align}

The extra amount that would have been paid $$= \rm{Rs}\,43.20$$

## Chapter 8 Ex.8.3 Question 5

Vasudevan invested $$\rm{Rs}\, 60,000$$ at an interest rate of $$12\%$$ per annum compounded half yearly. What amount would he get

(i) after $$6$$ months?

(ii) after $$1$$ year?

### 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}

$$P =\rm{ Rs}\,60,000$$

$$N =6$$ months and $$1$$ year

$$R =12\%$$ p.a. compounded half yearly

Steps:

For easy calculation of compound interest, we will put Interest Rate as $$6\%$$ half yearly and $$'n'$$ as $$1$$

(i) Compound Interest to be paid for $$6$$ month

\begin{align}A & = P{\left( {{1 + }\frac{{r}}{{{100}}}} \right)^{n}}\\& = 60000{\left( {{1 + }\frac{{6}}{{{100}}}} \right)^{1}}\\& = 60000{\left( {\frac{{100}}{{100}} + \frac{{6}}{{{100}}}} \right)^{1}}\\& = 60000\left( {\frac{{{106}}}{{{100}}}} \right)\\&= {60000} \times {1}{.06}\\&= 63600\end{align}

Compound Interest for $$6$$ month

\begin{align} &= 63600 - 60000\\&= 3600\end{align}

(ii) Compound Interest to be paid for $$12$$ months ($$1$$ year) Compounded half yearly So $$n=2,r=6\%,$$

\begin{align}A &= P{\left( {{1 + }\frac{{r}}{{{100}}}} \right)^{n}}\\&= 60000{\left( {{1 + }\frac{{6}}{{{100}}}} \right)^{2}}\\& = 60000{\left( {\frac{{100}}{{100}} + \frac{{6}}{{{100}}}} \right)^{2}}\\& = 60000{\left( {\frac{{{106}}}{{{100}}}} \right)^2}\\& = 60000\left( {\frac{{{106} \times {106}}}{{{100} \times 100}}} \right)\\& = 60000\left( {\frac{{11236}}{{{100}00}}} \right)\\&= {60000} \times {1}{.1236}\\&= 67416\end{align}

Compound Interest for $$12$$ months

\begin{align} &= 67416 - 60000\\&= 7416\end{align}

The amount that Vasudevan will get after $$6$$ months $$= \rm{Rs}\,63600$$

The amount that Vasudevan will get after $$1$$ year  $$= \rm{Rs}\, 67416$$

## Chapter 8 Ex.8.3 Question 6

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

### 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 $$=\rm{Rs}\, 92,610 − \rm{Rs}\, 92,400 = \rm{Rs}\, 210$$

## Chapter 8 Ex.8.3 Question 7

Maria invested Rs $$8,000$$ in a business. She would be paid interest at $$5\%$$ per annum compounded annually. Find:

(i) The amount credited against her name at the end of the second year

(ii) The interest for the $$3$$rd year.

### Solution

What is known?

Principal, Time Period and Rate of Interest

What is unknown?

Amount and Compound Interest (C.I.)

Reasoning:

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

$$P=\rm{Rs}\, 8,000$$

$$N =2$$ years and $$3$$rd year

$$R= 5\%$$ p.a. compounded annually

Steps:

(i) For calculation of amount credited at the end of second year:

\begin{align}A &= P\left( {{1 + }\frac{{r}}{{{100}}}} \right)^{n} \\ &= 8000\left( {{1 + }\frac{{5}}{{{100}}}} \right)^{2} \\ &= 8000\left( {{1 + }\frac{{1}}{{{20}}}} \right)^{2} \\ &= 8000\left( {\frac{{21}}{{20}}} \right)^2 \\ & = 8000 \times \frac{{21}}{{20}} \times \frac{{21}}{{20}} \\ & = 8000 \times \frac{{441}}{{400}} \\ &= 20 \times 441 \\ & = 8820 \\ \end{align}

(ii) For calculating C.I. for the $$3$$rd year, the principal $$= 8820$$

\begin{align}{\rm{S}}{\rm{.I}}{\rm{.}} &= \frac{P \times R \times T}{100}\\ &= \left( {\frac{{8820 \times 5 \times 1}}{{100}}} \right)\\&= 441\end{align}

The amount credited at the end of the $$2$$nd year $$= \rm{Rs}\, 8,820$$

The interest for the $$3^\rm{rd}$$ year $$= \rm{Rs}\, 441$$

## Chapter 8 Ex.8.3 Question 8

Find the amount and the compound interest on $$\rm{Rs}\, 10,000$$ for (1\begin{align}\frac{1}{2})\end{align} years at $$10\%$$ per annum, compounded half yearly. Would this interest be more than the interest he would get if it was compounded annually?

### 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}

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

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

$$R= 10\%$$ p.a. compounded annually and half-yearly

Steps:

For calculation of C.I. compounded half yearly, we will take Interest rate as $$5\%$$

\begin{align}A &= P\left( {{1 + }\frac{{r}}{{{100}}}} \right)^{n} \\ &= 10000\left( {{1 + }\frac{{5}}{{{100}}}} \right)^{3} \\ &= 10000\left( {{1 + }\frac{{1}}{{{20}}}} \right)^{3} \\ & = 10000\left( {\frac{{21}}{{20}}} \right)^3 \\ &= 10000\times \frac{{21}}{{20}} \times \frac{{21}}{{20}} \times \frac{{21}}{{20}} \\ &= 10000 \times \frac{{9261}}{{8000}} \\ &= {5} \times \frac{{9261}}{4} \\ &= 11576.25 \\ \end{align}

Interest earned at $$10\%$$ p.a. compounded half-yearly $$= 11576.25 \,– \,10000=\rm{Rs}\, 1576.25$$

The amount earned at $$10\%$$ p.a. compounded half-yearly $$= 11576.25$$

The C.I. earned at $$10\%$$ p.a. compounded half-yearly $$= 1576.25$$

The above interest earned being compounded half-yearly would be more than the interest compounded annually since interest compounded half yearly is always more than compounded annually at the same rate of interest.

## Chapter 8 Ex.8.3 Question 9

Find the amount which Ram will get on $$\rm{Rs}\; 4,096$$, if he gave it for $$18$$ months at 12\begin{align}\frac{1}{2}\% \end{align} per annum, interest being compounded half yearly.

### Solution

What is known?

Principal, Time Period and Rate of Interest

What is unknown?

Amount

Reasoning:

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

$$P=\rm{Rs}\,4,096$$

$$N =18$$ months

R =12\begin{align}\frac{1}{2}\% \end{align} p.a. compounded half-yearly

Steps:

For calculation of C.I. compounded half yearly, we will take Interest rate as 6\begin{align}\frac{1}{4}\% \; = \;\frac{{25}}{4}\%\end{align} and  $$'n'$$ as $$3$$ $$(18\div 6= 3)$$

\begin{align}A &= P\left( {{1 + }\frac{{r}}{{{100}}}} \right)^{n} \\ &= 4096\left( {{1 + }\frac{{{25}}}{{{100} \times {4}}}} \right)^{3} \\ & = 4096\left( {{1 + }\frac{{{25}}}{{{400}}}} \right)^{3} \\ & = 4096\left( {\frac{{425}}{{400}}} \right)^3 \\ & = 4096\left( {\frac{{17}}{{16}}} \right)^3 \\ & = 4096 \times \frac{{17}}{{16}} \times \frac{{17}}{{16}} \times \frac{{17}}{{16}} \\ &= 4096 \times \frac{{4913}}{{4096}} \\ &= 4913 \\ \end{align}

The total amount that Ram will get at the end of $$18$$ months $$= \rm{Rs}\, 4,913$$

## Chapter 8 Ex.8.3 Question 10

The population of a place increased to $$\rm{}\,54,000$$ in $$\rm{}\,2003$$ at a rate of $$5\%$$ per annum

(i) find the population in $$2001.$$

(ii) what would be its population in $$2005$$?

### Solution

What is known?

Population in $$2003$$, Time Period and Rate of population growth

What is unknown?

Population in $$2005$$ and $$2001$$

Reasoning:

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

$$P=\rm{}\, 54000$$ in the year $$2003$$

$$N =2$$ years

$$R =5\%$$ p.a. compounded annually

Steps:

(i) Population in the year $$2001$$

\begin{align}A &= P\left( {{1 + }\frac{{r}}{{{100}}}} \right)^{n} \\ 54000 &= P\left( {{1 + }\frac{{5}}{{{100}}}} \right)^2 \\ 54000 &= P\left( {{1 + }\frac{{1}}{{{20}}}} \right)^{2} \\ 54000 &= P\left( {\frac{{21}}{{20}}} \right)^2 \\ 54000 &= P \times \frac{{21}}{{20}} \times \frac{{21}}{{20}} \\ P &= 54000 \times \frac{{400}}{{441}} \\ P &= 48979.6 \\ \end{align}

The population in $$2001$$ $$= \rm{}\,48980$$.

(ii) Population in the year $$2005$$

\begin{align}A& = P\left( {{1 + }\frac{{r}}{{{100}}}} \right)^{n} \\ & = 54000\left( {{1 + }\frac{{5}}{{{100}}}} \right)^{2} \\ & = 54000\left( {{1 + }\frac{{1}}{{{20}}}} \right)^{2} \\ & = 54000\left( {\frac{{21}}{{20}}} \right)^2 \\ &= 54000 \times \frac{{21}}{{20}} \times \frac{{21}}{{20}} \\ & = 54000 \times \frac{{441}}{{400}} \\ &= 135 \times 441 \\ &= 59535 \\ \end{align}

The population in $$2005$$ $$=\rm{} 59535.$$

The population in $$2001$$ $$=\rm{} 48980$$.

The population in $$2005$$ $$= \rm{}\,59535$$.

## Chapter 8 Ex.8.3 Question 11

In a Laboratory, the count of bacteria in a certain experiment was increasing at the rate of $$2.5\%$$ per hour. Find the bacteria at the end of $$2$$ hours if the count was initially $$\rm{}\,5, 06,000$$.

### Solution

What is known?

Original Count, Time Period and Rate of Increase

What is unknown?

The total count after $$2$$ hours

Reasoning:

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

$$P= \rm{}\,5,06,000$$

$$N = 2$$ hours

$$R= 2.5\%$$ hour =\begin{align}\frac{{25}}{{10}} \end{align}hours

Steps:

\begin{align}A &= P\left( {{1 + }\frac{{r}}{{{100}}}} \right)^{n} \\ &= 506000\left( {{1 + }\frac{{{25}}}{{{1000}}}} \right)^{2} \\ &= 506000\left( {{1 + }\frac{{1}}{{{40}}}} \right)^{2} \\ &= 506000\left( {\frac{{41}}{{40}}} \right)^2 \\ &= 506000 \times \frac{{41}}{{40}} \times \frac{{41}}{{40}} \\ &= 506000 \times \frac{{1681}}{{1600}} \\ &= {506000} \times 1.050625 \\ &= 531616 \\ \end{align}

The total count of bacteria after $$2$$ hours $$=\rm{}\, 531616$$

## Chapter 8 Ex.8.3 Question 12

A scooter was bought at $$\rm{Rs}\, 42,000$$. Its value depreciated at the rate of $$8\%$$ per annum. Find its value after one year

### Solution

What is known?

Original Value, Rate of Depreciation

What is unknown?

The value of scooter after $$1$$ year

Reasoning:

Original value of the scooter $$= \rm{Rs}\, 42,000$$

Rate of depreciation $$= 8\%$$

Steps:

The value of the scooter after $$1$$ year

\begin{align} &= 42000 - \left( {42000 \times \frac{8}{{100}}} \right) \\ &= 42000 - \left( {42000 \times \frac{2}{{25}}} \right) \\ &= 42000 - \left( {1680 \times 2} \right) \\ &= 42000 - 3360 \\ &= 38640 \\ \end{align}

The value of the scooter after $$1$$ year ($$8\%$$ depreciation rate) $$= \rm{Rs}\, 38640.$$

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