# NCERT Solutions Class 8 Maths Chapter 8 Exercise 8.3

A government organisation called the National Council of Educational Research and Training (NCERT) was created to assist Indian schools in raising the standard of instruction. Since multiple books were being recommended to students in the same standard, the Government of India developed it to maintain the curriculum’s uniformity. The NCERT is in charge of publishing the NCERT textbooks. The primary objectives of NCERT and the units that comprise the organisation are to conduct, promote, and coordinate research in areas related to school education; develop digital multimedia materials; and create and publish model textbooks, supplemental materials, newsletters, journals, and educational kits. NCERT curriculum is the model used by the Central Board of Secondary Education (CBSE) and many other state boards.

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**NCERT Solutions for Class 8 Maths Chapter 8 Comparing Quantities (EX 8.3) Exercise 8.3 **

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**Exercise 8.3**

Comparing amounts refers to the quantitative relationship between two quantities that captures their respective sizes. It is merely a tool for comparing quantities. Numerous techniques, including ratio and proportion, percentage, profit and loss, and simple interest, can be used to compare quantities.

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**Chapter 8 Comparing Quantities of Class 8:-**

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**NCERT Solutions for Class 8**

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**Q.1 **

$\begin{array}{l}\text{Calculate the amount and compound interest on}\\ \left(\mathrm{a}\right)\text{\hspace{0.33em}\u20b9}10,800\mathrm{for}3\text{\hspace{0.33em}}\mathrm{years}\text{\hspace{0.33em}}\mathrm{at}\text{\hspace{0.33em}}12\frac{1}{2}\mathrm{\%}\mathrm{per}\mathrm{annum}\\ \mathrm{compounded}\text{\hspace{0.33em}}\mathrm{annually}.\\ \left(\mathrm{b}\right)\text{\hspace{0.33em}\u20b9}18,000\mathrm{for}2\frac{1}{2}\text{\hspace{0.33em}}\mathrm{years}\text{\hspace{0.33em}}\mathrm{at}\text{\hspace{0.33em}}10\mathrm{\%}\text{\hspace{0.33em}}\mathrm{per}\mathrm{annum}\\ \mathrm{compounded}\text{\hspace{0.33em}}\mathrm{annually}.\\ \left(\mathrm{c}\right)\text{\hspace{0.33em}\u20b9}62,500\mathrm{for}1\frac{1}{2}\text{\hspace{0.33em}}\mathrm{years}\text{\hspace{0.33em}}\mathrm{at}\text{\hspace{0.33em}}8\mathrm{\%}\text{\hspace{0.33em}}\mathrm{per}\mathrm{annum}\\ \mathrm{compounded}\text{\hspace{0.33em}}\mathrm{half}\text{\hspace{0.33em}}\mathrm{yearly}.\\ \left(\mathrm{d}\right)\text{\hspace{0.33em}\u20b9\hspace{0.33em}}8,000\mathrm{for}1\text{\hspace{0.33em}}\mathrm{year}\text{\hspace{0.33em}}\mathrm{at}\text{\hspace{0.33em}}9\mathrm{\%}\text{\hspace{0.33em}}\mathrm{per}\mathrm{annum}\\ \mathrm{compounded}\text{\hspace{0.33em}}\mathrm{half}\text{\hspace{0.33em}}\mathrm{yearly}.\\ \mathrm{}\left(\begin{array}{l}\mathrm{You}\text{\hspace{0.33em}}\mathrm{could}\text{\hspace{0.33em}}\mathrm{use}\text{\hspace{0.33em}}\mathrm{they}\text{\hspace{0.33em}}\mathrm{ear}\text{\hspace{0.33em}}\mathrm{by}\text{\hspace{0.33em}}\mathrm{year}\text{\hspace{0.33em}}\mathrm{calculation}\\ \mathrm{using}\text{\hspace{0.33em}}\mathrm{SI}\text{\hspace{0.33em}}\mathrm{for}\text{\hspace{0.33em}}\mathrm{mulatoverify}\end{array}\right).\\ \left(\mathrm{e}\right)\text{\hspace{0.33em}\u20b9}10,000\mathrm{for}1\text{\hspace{0.33em}}\mathrm{yearat}\text{\hspace{0.33em}}8\mathrm{\%}\text{\hspace{0.33em}}\mathrm{per}\mathrm{annum}\\ \mathrm{compounded}\text{\hspace{0.33em}}\mathrm{half}\text{\hspace{0.33em}}\mathrm{yearly}.\end{array}$

**Ans**

$\begin{array}{l}\left(\mathrm{a}\right)\text{Principal}=\u20b9\text{}10,800\\ \text{}\mathrm{R}=12\frac{1}{2}\mathrm{\%}=\frac{25}{2}\mathrm{\%}\\ \text{}\mathrm{n}=3\\ \therefore \mathrm{Amount}=\mathrm{P}{(1+\frac{\mathrm{R}}{100})}^{\mathrm{n}}\\ =10800{(1+\frac{25}{2\times 100})}^{3}\\ =10800{(1+\frac{25}{200})}^{3}\\ =10800{\left(\frac{225}{200}\right)}^{3}\\ =10800\times \frac{225}{200}\times \frac{225}{200}\times \frac{225}{200}\\ =15377.34\\ \\ \therefore \mathrm{C}.\mathrm{I}=\mathrm{Amount}-\mathrm{Principal}\\ =15377.34-10800\\ =4,577.34\end{array}$

$\begin{array}{l}\left(\mathrm{b}\right)\text{Principal}=\u20b9\text{}18,000\\ \text{}\mathrm{R}=10\mathrm{\%}\\ \text{}\mathrm{n}=2\frac{1}{2}\\ \therefore \mathrm{Amount}=\mathrm{P}{(1+\frac{\mathrm{R}}{100})}^{\mathrm{n}}\\ \text{}=18000{(1+\frac{10}{100})}^{2\frac{1}{2}}\\ \\ \text{Since,the years are in fractional form,so we will calculate}\\ \text{the Amount for the whole part first and then we will calculate}\\ \text{the simple interest for the remaining fractional part.}\\ \therefore \mathrm{Amount}=18000{(1+\frac{10}{100})}^{2}\\ =18000{\left(\frac{11}{10}\right)}^{2}\\ =18000\times \frac{11}{10}\times \frac{11}{10}\\ =\u20b9\text{}21,780\end{array}$

$\begin{array}{l}\text{Now,we will take \u20b9 21,780 as principal and calculate the simple}\\ \text{interest for the next}\frac{1}{2}\mathrm{year}.\\ \therefore \mathrm{S}.\mathrm{I}=\frac{\mathrm{PRT}}{100}\\ =\frac{21780\times 10\times 1}{2\times 100}\\ =\u20b9\text{1089}\\ \\ \therefore \text{Interest for the first 2 years}=\u20b9(\text{2178}0-\text{18}000)=\u20b9\text{378}0\\ \text{And interest for the next}\frac{1}{2}\text{year}=\u20b9\text{1}0\text{89}\\ \therefore \text{Total C}.\text{I}.=\u20b9\text{378}0+\u20b9\text{1}0\text{89}=\u20b9\text{4},\text{869, and}\\ \\ \text{A}=\text{P}+\text{C}.\text{I}.=\u20b9\text{18}000+\u20b9\text{4869}=\u20b9\text{22},\text{869}\end{array}$

$\begin{array}{l}\left(\mathrm{c}\right)\text{Principal}=\u20b9\text{}62,500\\ \text{}\mathrm{R}=8\mathrm{\%}\text{}\mathrm{yearly}=4\mathrm{\%}\text{}\mathrm{half}\text{}\mathrm{yearly}\\ \text{}\mathrm{n}=1\frac{1}{2}=3\text{}\mathrm{half}\text{}\mathrm{years}\\ \therefore \mathrm{Amount}=\mathrm{P}{(1+\frac{\mathrm{R}}{100})}^{\mathrm{n}}\\ \text{}=62500{(1+\frac{4}{100})}^{3}\\ \text{}=62500{\left(\frac{26}{25}\right)}^{3}\\ \text{}=62500\times \frac{26}{25}\times \frac{26}{25}\times \frac{26}{25}\\ \text{}=\u20b9\text{}70,304\\ \\ \therefore \mathrm{C}.\mathrm{I}=\mathrm{A}-\mathrm{P}=\u20b9\text{}70,304-\u20b9\text{}62,500\\ \text{}=\u20b9\text{}7,804\\ \left(\mathrm{d}\right)\text{Principal}=\u20b9\text{}8000\\ \text{}\mathrm{R}=9\mathrm{\%}\text{}\mathrm{yearly}=\frac{9}{2}\mathrm{\%}\text{}\mathrm{half}\text{}\mathrm{yearly}\\ \text{}\mathrm{n}=1=2\text{}\mathrm{half}\text{}\mathrm{years}\\ \therefore \mathrm{Amount}=\mathrm{P}{(1+\frac{\mathrm{R}}{100})}^{\mathrm{n}}\\ \text{}=8000{(1+\frac{9}{200})}^{2}\\ \text{}=8000{\left(\frac{209}{200}\right)}^{2}\\ \text{}=\u20b9\text{}8,736.20\\ \\ \therefore \mathrm{C}.\mathrm{I}=\mathrm{A}-\mathrm{P}=\u20b9\text{}8,736.20-\mathrm{Rs}\text{}8000\\ \text{}=\u20b9\text{}736.20\\ \left(\mathrm{e}\right)\text{}\mathrm{Principal}=\mathrm{Rs}\text{}10,000\\ \text{}\mathrm{R}=8\mathrm{\%}\text{}\mathrm{yearly}=4\mathrm{\%}\text{}\mathrm{half}\text{}\mathrm{yearly}\\ \text{}\mathrm{n}=1=2\text{}\mathrm{half}\text{}\mathrm{years}\\ \therefore \mathrm{Amount}=\mathrm{P}{(1+\frac{\mathrm{R}}{100})}^{\mathrm{n}}\\ \text{}=10,000{(1+\frac{4}{100})}^{2}\\ \text{}=10,000{(1+\frac{1}{25})}^{2}\\ \text{}=10,000{\left(\frac{26}{25}\right)}^{2}\\ \text{}=10,000\times \frac{26}{25}\times \frac{26}{25}\\ \text{}=\u20b9\text{}10,816\\ \\ \therefore \mathrm{C}.\mathrm{I}=\mathrm{A}-\mathrm{P}=\mathrm{Rs}\text{}10,816-\mathrm{Rs}\text{}10,000\\ =\u20b9\text{}816\end{array}$

**Q.2 ****Kamala borrowed ₹ 26,400 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 SI on the 2nd year amount for 4/12 years).**

**Ans**

$\begin{array}{l}\mathrm{Principal}=\u20b9\text{}26,400\\ \text{}\mathrm{R}=15\mathrm{\%}\text{}\mathrm{per}\text{}\mathrm{annum}\\ \text{}\mathrm{n}=2\frac{4}{12}\mathrm{years}\\ \mathrm{Since},\mathrm{the}\text{}\mathrm{years}\text{}\mathrm{are}\text{}\mathrm{in}\text{}\mathrm{fractional}\text{}\mathrm{form},\mathrm{so}\text{}\mathrm{we}\text{}\mathrm{will}\text{}\mathrm{calculate}\\ \mathrm{the}\text{}\mathrm{Amount}\text{}\mathrm{for}\text{}\mathrm{the}\text{}\mathrm{whole}\text{}\mathrm{part}\text{}\mathrm{first}\text{}\mathrm{and}\text{}\mathrm{then}\text{}\mathrm{we}\text{}\mathrm{will}\text{}\mathrm{calculate}\\ \mathrm{the}\text{}\mathrm{simple}\text{}\mathrm{interest}\text{}\mathrm{for}\text{}\mathrm{the}\text{}\mathrm{remaining}\text{}\mathrm{fractional}\text{}\mathrm{part}.\\ \\ \therefore \mathrm{Amount}=26,400{(1+\frac{15}{100})}^{2}\\ =26,400{\left(\frac{23}{20}\right)}^{2}\end{array}$

$\begin{array}{l}=26,400\times \frac{23}{20}\times \frac{23}{20}\\ =\u20b9\text{}34,914\\ \mathrm{Now},\mathrm{we}\text{}\mathrm{will}\text{}\mathrm{take}\text{}\mathrm{Rs}\text{}34,914\text{}\mathrm{as}\text{}\mathrm{principal}\text{}\mathrm{and}\text{}\mathrm{calculate}\text{}\mathrm{the}\text{}\mathrm{simple}\text{}\\ \mathrm{interest}\text{}\mathrm{for}\text{}\mathrm{the}\text{}\mathrm{next}\text{}\frac{1}{3}\mathrm{years}.\\ \therefore \mathrm{S}.\mathrm{I}=\frac{\mathrm{PRT}}{100}\\ =\frac{34,914\times 15\times 1}{3\times 100}\\ =\u20b9\text{}1745.70\\ \\ \therefore \text{}\mathrm{Interest}\text{}\mathrm{for}\text{}\mathrm{the}\text{}\mathrm{first}\text{}2\text{}\mathrm{years}=\u20b9(34,914-26,400)=\u20b9\text{}8,514\\ \mathrm{And}\text{}\mathrm{interest}\text{}\mathrm{for}\text{}\mathrm{the}\text{}\mathrm{next}\text{}\frac{1}{3}\mathrm{year}=\u20b9\text{}1745.70\\ \therefore \text{}\mathrm{Total}\text{}\mathrm{C}.\mathrm{I}.=\u20b9\text{}8,514+\u20b9\text{}1745.70=\u20b9\text{}10,259.70,\text{}\mathrm{and}\\ \\ \mathrm{A}\text{}=\text{}\mathrm{P}\text{}+\text{}\mathrm{C}.\mathrm{I}.=\u20b9\text{}26,400+\u20b9\text{}10,259.70=\u20b9\text{}36,659.70\end{array}$

**Q.3 ****Fabina borrows ₹ 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?**

**Ans**

$\begin{array}{l}\mathrm{Interest}\text{}\mathrm{paid}\text{}\mathrm{by}\text{}\mathrm{Fabina}=\frac{\mathrm{PRT}}{100}\\ =\u20b9\left(\frac{12,500\times 12\times 3}{100}\right)\\ =\u20b9\text{}4,500\\ \\ \mathrm{Amount}\text{}\mathrm{paid}\text{}\mathrm{by}\text{}\mathrm{Radha}=\mathrm{P}{(1+\frac{\mathrm{R}}{100})}^{\mathrm{n}}\\ =\u20b9\text{}12,500{(1+\frac{10}{100})}^{3}\\ \\ =\u20b9\text{}12,500\times {\left(\frac{110}{100}\right)}^{3}\\ =\u20b9\text{}16,637.50\\ \\ \mathrm{C}.\mathrm{I}=\mathrm{A}-\mathrm{P}=\u20b9\text{}16,637.50\text{}-\u20b9\text{}12,500\\ =4,137.50\\ \\ \mathrm{The}\text{}\mathrm{interest}\text{}\mathrm{paid}\text{}\mathrm{by}\text{}\mathrm{Fabina}\text{}\mathrm{is}\text{\u20b9}4,500\text{}\mathrm{and}\text{}\mathrm{by}\text{}\mathrm{Radha}\text{}\mathrm{is}\text{\u20b9}4,137.50.\\ \mathrm{Thus},\text{}\mathrm{Fabina}\text{}\mathrm{pays}\text{}\mathrm{more}\text{}\mathrm{interest}.\\ \mathrm{i}.\mathrm{e}.\text{\u20b9}4500-\u20b9\text{}4137.50=\u20b9\text{}362.50\\ \\ \mathrm{Hence},\text{}\mathrm{Fabina}\text{}\mathrm{will}\text{}\mathrm{have}\text{}\mathrm{to}\text{}\mathrm{pay}\text{\u20b9}362.50\text{}\mathrm{more}.\end{array}$

**Q.4 ****I borrowed ₹ 12,000 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?**

**Ans**

$\begin{array}{l}\text{}\mathrm{Principal}=\u20b9\text{}12,000\\ \text{}\mathrm{R}=6\mathrm{\%}\text{}\mathrm{yearly}\\ \text{}\mathrm{n}=2\text{}\mathrm{years}\\ \mathrm{Simple}\text{}\mathrm{Interest}\text{}=\frac{\mathrm{PRT}}{100}\\ \text{}=\u20b9\left(\frac{12,000\times 6\times 2}{100}\right)\\ \text{}=\u20b9\text{}1,440\\ \therefore \mathrm{Amount}=\mathrm{P}{(1+\frac{\mathrm{R}}{100})}^{\mathrm{n}}\\ \text{}=12,000{(1+\frac{6}{100})}^{2}\\ \text{}=12,000{(1+\frac{3}{50})}^{2}\\ \text{}=12,000{\left(\frac{53}{50}\right)}^{2}\\ \text{}=12,000\times \frac{53}{50}\times \frac{53}{50}\\ \text{}=\u20b9\text{}13,483.20\\ \\ \therefore \mathrm{C}.\mathrm{I}=\mathrm{A}-\mathrm{P}=\u20b9\text{}13,483.20-\u20b9\text{}12,000=\u20b9\text{}1,483.20\\ \text{}\mathrm{C}.\mathrm{I}-\mathrm{S}.\mathrm{I}=\u20b9\text{}1,483.20-\u20b9\text{}1,440=\u20b9\text{}43.20\\ \\ \mathrm{Therefore},\text{}\mathrm{the}\text{}\mathrm{extra}\text{}\mathrm{amount}\text{}\mathrm{to}\text{}\mathrm{be}\text{}\mathrm{paid}\text{}\mathrm{is}\text{\u20b9}43.20.\end{array}$

**Q.5 ****Vasudevan invested ₹ 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?**

**Ans**

$\begin{array}{l}\left(\mathrm{i}\right)\text{}\mathrm{Principal}=\u20b9\text{}60,000\\ \text{}\mathrm{R}=12\mathrm{\%}\text{}\mathrm{yearly}=6\mathrm{\%}\text{}\mathrm{half}\text{}\mathrm{yearly}\\ \text{}\mathrm{n}=6\text{}\mathrm{months}=1\text{}\mathrm{half}\text{}\mathrm{year}\\ \\ \therefore \mathrm{Amount}=\mathrm{P}{(1+\frac{\mathrm{R}}{100})}^{\mathrm{n}}\\ \text{}=60,000{(1+\frac{6}{100})}^{1}\\ \text{}=60,000\left(\frac{106}{100}\right)\\ \text{}=\text{}\u20b9\text{}63,600\\ \left(\mathrm{ii}\right)\text{}\mathrm{Principal}=\u20b9\text{}60,000\\ \text{}\mathrm{R}=12\mathrm{\%}\text{}\mathrm{yearly}=6\mathrm{\%}\text{}\mathrm{half}\text{}\mathrm{yearly}\\ \text{}\mathrm{n}=1\text{}\mathrm{year}=2\text{}\mathrm{half}\text{}\mathrm{years}\\ \\ \therefore \mathrm{Amount}=\mathrm{P}{(1+\frac{\mathrm{R}}{100})}^{\mathrm{n}}\\ \text{}=60,000{(1+\frac{6}{100})}^{2}\\ \text{}=60,000{\left(\frac{106}{100}\right)}^{2}\\ \text{}=\text{}60,000\times \frac{106}{100}\times \frac{106}{100}\\ \text{}=\text{}\u20b9\text{}67,416\end{array}$

**Q.6 ****Arif took a loan of ₹ 80,000 from a bank. If the rate of interest is 10% per annum, find the difference in amounts he would be paying after**

$1\frac{1}{2}$

**years if the interest is**

**(i) compounded annually.**

**(ii) compounded half yearly.**

**Ans**

$\begin{array}{l}\text{(i)}\mathrm{Principal}=\u20b9\text{80},000\\ \mathrm{R}=10\%\text{}\mathrm{per}\text{}\mathrm{annum}\\ \mathrm{n}=1\frac{1}{2}\mathrm{years}\\ \mathrm{Since},\mathrm{the}\text{}\mathrm{years}\text{}\mathrm{are}\text{}\mathrm{in}\text{}\mathrm{fractional}\text{}\mathrm{form},\mathrm{so}\text{}\mathrm{we}\text{}\mathrm{will}\text{}\mathrm{calculate}\\ \mathrm{the}\text{}\mathrm{Amount}\text{}\mathrm{for}\text{}\mathrm{the}\text{}\mathrm{whole}\text{}\mathrm{part}\text{}\mathrm{first}\text{}\mathrm{and}\text{}\mathrm{then}\text{}\mathrm{we}\text{}\mathrm{will}\text{}\mathrm{calculate}\\ \mathrm{the}\text{}\mathrm{simple}\text{}\mathrm{interest}\text{}\mathrm{for}\text{}\mathrm{the}\text{}\mathrm{remaining}\text{}\mathrm{fractional}\text{}\mathrm{part}.\\ \therefore \mathrm{Amount}=80,000{\left(1+\frac{10}{100}\right)}^{1}\\ =80,000\left(1+\frac{1}{10}\right)\\ =80,000{\left(\frac{11}{10}\right)}^{2}\\ =\u20b9\text{}88,000\\ \\ \mathrm{Now},\mathrm{we}\text{}\mathrm{will}\text{}\mathrm{take}\text{}\mathrm{Rs}\text{}88,000\text{}\mathrm{as}\text{}\mathrm{principal}\text{}\mathrm{and}\text{}\mathrm{calculate}\text{}\mathrm{the}\text{}\mathrm{simple}\text{}\\ \mathrm{interest}\text{}\mathrm{for}\text{}\mathrm{the}\text{}\mathrm{next}\text{}\frac{1}{2}\mathrm{year}.\\ \therefore \mathrm{S}.\mathrm{I}=\frac{\mathrm{PRT}}{100}\\ =\frac{88,000\times 10\times 1}{2\times 100}\\ =\u20b9\text{}4,400\\ \text{Interest for the first year}=\text{\u20b9 88}000\text{}-\text{\u20b9 8}0000\text{}=\text{\u20b9 8},000\\ \text{And interest for the next}\frac{1}{2}\text{year}=\text{\u20b9 4},\text{4}00\\ \text{Total C}.\text{I}.\text{}=\text{\u20b9 8}000\text{}+\text{\u20b9 4},\text{4}00=\text{\u20b9 1},\text{24}00\\ \\ \therefore \text{A}=\text{P}+\text{C}.\text{I}.\text{}=\text{\u20b9}\left(\text{8}0000\text{}+\text{124}00\right)\text{}=\text{\u20b9 92},\text{4}00\end{array}$

$\begin{array}{l}\left(\text{ii}\right)\text{Now,the interest is compounded half yearly}.\\ \text{Rate}=\text{1}0\%\text{per annum}=\text{5}\%\text{per half year}\\ \text{n=}1\frac{1}{2}\mathrm{years}=3\text{half years}\\ \therefore \mathrm{Amount}=80,000{\left(1+\frac{5}{100}\right)}^{3}\\ =80,000\left(1+\frac{1}{20}\right)3\\ =80,000{\left(\frac{21}{20}\right)}^{2}\\ =\u20b9\text{92},610\end{array}$

Therefore, the difference between the amounts = ₹ 92,610 − ₹ 92,400 = ₹ 210

**Q.7 ****Maria invested ₹ 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 3rd year.**

**Ans**

$\begin{array}{l}\left(\mathrm{i}\right)\text{}\mathrm{Principal}=\u20b9\text{8},000\\ \text{}\mathrm{R}=5\mathrm{\%}\text{}\mathrm{yearly}\\ \text{}\mathrm{n}=2\text{}\mathrm{years}\\ \\ \therefore \mathrm{Amount}=\mathrm{P}{(1+\frac{\mathrm{R}}{100})}^{\mathrm{n}}\\ \text{}=8,000{(1+\frac{5}{100})}^{2}\\ \text{}=8,000{\left(\frac{21}{20}\right)}^{2}\\ \text{}=\text{8},000\times \frac{21}{20}\times \frac{21}{20}\\ \text{}=\text{}\u20b9\text{}8,820\end{array}$

$\begin{array}{l}\text{(ii)}\mathrm{Now},\text{we have to calculate the interest for the third year.}\\ \text{So we}\mathrm{will}\text{}\mathrm{take}\text{}\mathrm{Rs}\text{8,820}\mathrm{as}\text{}\mathrm{principal}\text{}\mathrm{and}\text{}\mathrm{calculate}\text{}\mathrm{the}\text{}\mathrm{simple}\text{}\\ \mathrm{interest}\text{}\mathrm{for}\text{}\mathrm{the}\text{}\mathrm{next}\text{}\mathrm{year}.\\ \therefore \mathrm{S}.\mathrm{I}=\frac{\mathrm{PRT}}{100}\\ =\frac{8820\times 5\times 1}{100}\\ =\u20b9\text{}441\end{array}$

**Q.8 ****Find the amount and the compound interest on ₹10,000 for years at 10% per annum, compounded half yearly. Would this interest be more than the interest he would get if it was compounded annually?**

**Ans**

$\begin{array}{l}\mathrm{Principal}=\u20b9\text{}10,000\\ \text{}\mathrm{R}=10\mathrm{\%}\text{}\mathrm{yearly}=5\mathrm{\%}\text{}\mathrm{half}\text{}\mathrm{yearly}\\ \text{}\mathrm{n}=1\frac{1}{2}\text{}\mathrm{years}=3\text{}\mathrm{half}\text{}\mathrm{years}\\ \\ \therefore \mathrm{Amount}=\mathrm{P}{(1+\frac{\mathrm{R}}{100})}^{\mathrm{n}}\\ \text{}=10,000{(1+\frac{5}{100})}^{3}\\ \text{}=10,000{(1+\frac{1}{20})}^{3}\\ \text{}=10,000{\left(\frac{21}{20}\right)}^{3}\end{array}$

$\begin{array}{l}\text{}=\text{}60,000\times \frac{21}{20}\times \frac{21}{20}\times \frac{21}{20}\\ \text{}=\text{}\u20b9\text{}11,576.25\\ \\ \mathrm{C}.\mathrm{I}=\mathrm{A}-\mathrm{P}\\ \text{}=\u20b9\text{}11,576.25-\mathrm{Rs}\text{}10,000\\ \text{}=\u20b9\text{}1,576.25\\ \\ \mathrm{Now}\text{we have to calculate the interest compounded annually.}\\ \mathrm{Since},\mathrm{the}\text{}\mathrm{years}\text{}\mathrm{are}\text{}\mathrm{in}\text{}\mathrm{fractional}\text{}\mathrm{form},\mathrm{so}\text{}\mathrm{we}\text{}\mathrm{will}\text{}\mathrm{calculate}\\ \mathrm{the}\text{}\mathrm{Amount}\text{}\mathrm{for}\text{}\mathrm{the}\text{}\mathrm{whole}\text{}\mathrm{part}\text{}\mathrm{first}\text{}\mathrm{and}\text{}\mathrm{then}\text{}\mathrm{we}\text{}\mathrm{will}\text{}\mathrm{calculate}\\ \mathrm{the}\text{}\mathrm{simple}\text{}\mathrm{interest}\text{}\mathrm{for}\text{}\mathrm{the}\text{}\mathrm{remaining}\text{}\mathrm{fractional}\text{}\mathrm{part}.\\ \\ \therefore \mathrm{Amount}=\u20b9\text{}10,000{(1+\frac{10}{100})}^{1}\\ =\u20b9\text{}10,000\left(\frac{11}{10}\right)\\ =11,000\end{array}$ \begin{array}{l}\end{array}

$\begin{array}{l}\mathrm{Now},\mathrm{we}\text{}\mathrm{will}\text{}\mathrm{take}\text{}\u20b911,000\text{}\mathrm{as}\text{}\mathrm{principal}\text{}\mathrm{and}\text{}\mathrm{calculate}\text{}\mathrm{the}\text{}\mathrm{simple}\text{}\\ \mathrm{interest}\text{}\mathrm{for}\text{}\mathrm{the}\text{}\mathrm{next}\text{}\frac{1}{2}\mathrm{years}.\\ \therefore \mathrm{S}.\mathrm{I}=\frac{\mathrm{PRT}}{100}\\ =\frac{11,000\times 10\times 1}{2\times 100}\\ =\u20b9\text{}550\\ \\ \therefore \text{Interest for the first year}=\u20b9\text{11}000-\u20b9\text{1}0000=\u20b9\text{1},000\\ \therefore \text{Total compound interest}=\u20b9\text{1}000+\u20b9\text{55}0=\u20b9\text{1},\text{55}0\\ \\ \text{Yes,the interest would be more when compounded half yearly}\\ \text{than the interest when compounded annually}.\end{array}$

**Q.9 ****Find the amount which Ram will get on ₹ 4096, if he gave it for 18 months at**

$12\frac{1}{2}$

**% 2 per annum, interest being compounded half yearly.**

**Ans**

$\begin{array}{l}\mathrm{Principal}=\u20b9\text{}4,096\\ \text{}\mathrm{R}=12\frac{1}{2}\mathrm{\%}\text{}\mathrm{yearly}=\frac{25}{4}\mathrm{\%}\text{}\mathrm{half}\text{}\mathrm{yearly}\\ \text{}\mathrm{n}=18\text{}\mathrm{months}=3\text{}\mathrm{half}\text{}\mathrm{years}\\ \\ \therefore \mathrm{Amount}=\mathrm{P}{(1+\frac{\mathrm{R}}{100})}^{\mathrm{n}}\\ \text{}=\u20b9\text{}4096{(1+\frac{25}{400})}^{3}\\ \text{}=\u20b9\text{}4096{(1+\frac{1}{16})}^{3}\\ \text{}=\u20b9\text{}4096{\left(\frac{17}{16}\right)}^{3}\\ \text{}=\text{\u20b9}4096\times \frac{17}{16}\times \frac{17}{16}\times \frac{17}{16}\\ \text{}=\text{\u20b9}4,913\\ \\ \text{Therefore},\text{the required amount is \u20b9 4},\text{913}.\end{array}$

**Q.10 ****The population of a place increased to 54,000 in 2003 at a rate of 5% per annum**

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

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

**Ans**

$\begin{array}{l}\text{(i) The population in the year 2003= 54},000\\ \text{}\\ \therefore 54,000=\left(\text{Population in 2001}\right){(1+\frac{5}{100})}^{2}\\ \text{}54,000=\left(\text{Population in 2001}\right){\left(\frac{21}{20}\right)}^{2}\\ \text{Population in 2001}=54,000\times {\left(\frac{20}{21}\right)}^{2}\\ \text{}=48979.59\\ \\ \text{Therefore},\text{the population in the year 2}00\text{1 was approximately 48},\text{98}0.\\ \text{(ii) Population in 2005}=54,000\times {(1+\frac{5}{100})}^{2}\\ =54,000\times {(1+\frac{1}{20})}^{2}\\ =54,000\times {\left(\frac{21}{20}\right)}^{2}\\ =59,535\\ \\ \text{Therefore},\text{the population in the year 2}00\text{5 would be 59},\text{535}.\end{array}$

**Q.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 5, 06,000.**

**Ans**

\begin{array}{l}\text{The initial count of bacteria is given as}5,06,000.\\ \text{Bacteria at the end of 2 hours}=5,06,000{\left(1+\frac{1}{40}\right)}^{2}\\ =5,06,000{\left(\frac{41}{40}\right)}^{2}\text{}\\ =5,06,000\times \frac{41}{40}\times \frac{41}{40}\\ =\text{}531616.25\\ \\ \text{Thus, the count of bacteria at the end of 2 hours will be}5,31,616\text{}\left(\text{approx}.\right).\end{array}

**Q.12 ****A scooter was bought at ₹ 42,000. Its value depreciated at the rate of 8% per annum. Find its value after one year.**

**Ans**

$\begin{array}{l}\mathrm{Principal}=\mathrm{Cost}\text{}\mathrm{price}\text{}\mathrm{of}\text{}\mathrm{the}\text{}\mathrm{scooter}=\u20b9\text{}42,000\\ \mathrm{Depreciation}=8\mathrm{\%}\text{}\mathrm{of}\text{\u20b9}42,000\text{}\\ \text{}=\text{}42000\times \frac{8}{100}=\u20b9\text{}3,360\\ \\ \therefore \mathrm{Value}\text{}\mathrm{after}\text{}1\text{}\mathrm{year}=\u20b9\text{}42000-\u20b9\text{}3360=\u20b9\text{}38,640.\end{array}$

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