NCERT Solutions Class 8 Maths Chapter 8

Q.1 Find the ratio of the following.

(a) Speed of a cycle 15 km per hour to the speed of scooter 30 km per hour.

(b) 5 m to 10 km

(c) 50 paise to ₹ 5

Ans

$\begin{array}{l}\left(\mathrm{a}\right)\mathrm{Ratio}\mathrm{of}\mathrm{the}\mathrm{speed}\mathrm{of}\mathrm{cycle}\mathrm{to}\mathrm{the}\mathrm{speed}\mathrm{of}\mathrm{scooter}=\frac{15}{30}=1:2\\ \left(\mathrm{b}\right)\mathrm{Since}1\mathrm{km}=1000\mathrm{m},\\ \therefore \mathrm{Required}\mathrm{ratio}=\frac{5\text{}\mathrm{m}}{10\mathrm{km}}=\frac{5\text{}\mathrm{m}}{10\times 1000\mathrm{m}}=1:2000\\ \\ \left(\mathrm{c}\right)\mathrm{Since}\text{₹}1=100\mathrm{paise},\\ \therefore \mathrm{Required}\mathrm{ratio}=\frac{50\mathrm{paise}}{\text{₹\hspace{0.17em}}5}=\frac{50\mathrm{paise}}{500\mathrm{paise}}=1:10\end{array}$

Q.2 Convert the following ratios to percentages.

(a) 3 : 4

(b) 2 : 3

Ans

$\begin{array}{l}\left(\text{a}\right)3:4=\frac{3}{4}=\frac{3}{4}\times \frac{100}{100}=\frac{3}{4}\times 100\%=75\%\\ \left(\text{b}\right)2:3=\frac{2}{3}=\frac{2}{3}\times \frac{100}{100}=\frac{2}{3}\times 100\%=\frac{200}{3}\%=66\frac{2}{3}\%\end{array}$

Q.3 72% of 25 students are good in mathematics. How many are not good in mathematics?

Ans

$\begin{array}{l}\text{Given that 72}\mathrm{\%}\text{of 25 students are good in mathematics}.\\ \\ \text{Therefore},\text{Percentage of students who are not good}\\ \text{in mathematics}=(\text{1}00-\text{72})\mathrm{\%}=\text{28}\mathrm{\%}.\\ \\ \mathrm{Thus},\mathrm{the}\mathrm{n}\text{umber of students who are not good in}\\ \text{mathematics}=\frac{28}{100}\times 25=\text{7}\\ \\ \text{Hence},\text{7 students are not good in mathematics}.\end{array}$

Q.4 A football team won 10 matches out of the total number of matches they played. If their win percentage was 40, then how many matches did they play in all?

Ans

$\begin{array}{l}\text{Let the total number of matches played by the team be}\mathrm{x}.\\ \\ \text{It is given that the team won 10 matches and the winning}\\ \text{percentage of the team was 40}\mathrm{\%}\text{.}\\ \\ \text{Therefore,}\\ \frac{40}{100}\times \mathrm{x}=10\\ \Rightarrow \mathrm{x}=10\times \frac{100}{40}\\ \Rightarrow \mathrm{x}=25\\ \\ \text{Hence, the team played 25 matches in all.}\end{array}$

Q.5 If Chameli had ₹ 600 left after spending 75% of her money, how much did she have in the beginning?

Ans

$\begin{array}{l}\text{Let the amount of money which Chameli had in the}\\ \text{beginning be}\mathrm{x}.\\ \text{It is given that after spending}75\mathrm{\%}\text{of ₹}\mathrm{x},\text{she}\\ \text{was left with ₹}600.\\ \text{Therefore},(100-75)\mathrm{\%}\text{of}\mathrm{x}=\text{Rs}600\end{array}$ $\begin{array}{l}\Rightarrow 25\mathrm{\%}\text{of}\mathrm{x}=\text{₹}600\\ \Rightarrow \frac{25}{100}\times \mathrm{x}=\text{₹}600\\ \Rightarrow \mathrm{x}=₹(600\times \frac{100}{25})=₹2400\\ \\ \text{Hence, she had ₹ 2400 in the beginning.}\end{array}$

Q.6 If 60% people in a city like cricket, 30% like football and the remaining like other games, then what per cent of the people like other games? If the total number of people are 50 lakh, find the exact number who like each type of game.

Ans

$\begin{array}{l}\mathrm{Percentage}\mathrm{of}\mathrm{people}\mathrm{who}\mathrm{like}\mathrm{other}\mathrm{games}=(100-60-30)\mathrm{\%}\\ =(100-90)\mathrm{\%}\\ =10\mathrm{\%}\\ \text{It is given that the total num}\mathrm{ber}\mathrm{of}\mathrm{people}=50\mathrm{lakh}\\ \\ \therefore \mathrm{Number}\mathrm{of}\mathrm{people}\mathrm{who}\mathrm{like}\mathrm{cricket}=(\frac{60}{100}\times 50)\mathrm{lakh}=30\text{}\mathrm{lakh}\\ \mathrm{Number}\mathrm{of}\mathrm{people}\mathrm{who}\mathrm{like}\mathrm{football}=(\frac{30}{100}\times 50)\mathrm{lakh}=15\text{}\mathrm{lakh}\\ \mathrm{Number}\mathrm{of}\mathrm{people}\mathrm{who}\mathrm{like}\mathrm{other}\mathrm{games}=(\frac{10}{100}\times 50)\mathrm{lakh}=5\text{}\mathrm{lakh}\end{array}$

Q.7 A man got a 10% increase in his salary. If his new salary is ₹ 1,54,000, find his original salary.

Ans

$\begin{array}{l}\text{Let the original salary be}\mathrm{x}.\\ \text{It is given that the new salary is ₹}1,54,000.\\ \\ \text{Therefore, Original salary + Increment = New salary}\\ \\ \text{But it is given that the increment is}10\mathrm{\%}\text{of the original salary.}\\ \\ \text{Therefore,we have}\\ \mathrm{x}\text{+}\frac{10}{100}\times \mathrm{x}=1,54,000\\ \Rightarrow \frac{110\mathrm{x}}{100}=1,54,000\\ \Rightarrow \mathrm{x}=1,54,000\times \frac{100}{110}\\ \Rightarrow \mathrm{x}=1,40,000\\ \\ \text{Thus, the original salary was ₹}1,40,000.\end{array}$

Q.8 On Sunday 845 people went to the Zoo. On Monday only 169 people went. What is the per cent decrease in the people visiting the Zoo on Monday?

Ans

\begin{array}{l}\text{It is given that, 845 people went to the zoo on Sunday and}\\ \text{169 people went on Monday}.\\ \\ \text{Thus,decrease in the number of people}=\text{845}-\text{169}=\text{676}\\ \\ \text{Percentage decrease}=\left(\frac{\text{Decrease in the number of people}}{\begin{array}{l}\text{Number of people who went to}\\ \text{the zoo on sunday}\end{array}}\times 100\right)\%\\ \\ =\left(\frac{676}{845}\times 100\right)\%\\ \\ =80\%\end{array}

Q.9 A shopkeeper buys 80 articles for ₹ 2,400 and sells them for a profit of 16%. Find the selling price of one article.

Ans

$\begin{array}{l}\mathrm{It}\mathrm{is}\mathrm{given}\mathrm{that}\mathrm{the}\mathrm{shopkeeper}\mathrm{buys}80\mathrm{articles}\mathrm{for}₹2,400.\\ \mathrm{Cost}\mathrm{Price}\mathrm{of}\mathrm{one}\mathrm{article}=₹\frac{2,400}{80}=\text{₹}30\\ \mathrm{Profit}\mathrm{percent}=16\mathrm{\%}\\ \mathrm{Profit}\mathrm{percent}=\frac{\mathrm{Profit}}{\mathrm{Cost}\mathrm{Price}}\times 100\\ 16=\frac{\mathrm{Profit}}{\text{₹}30}\times 100\\ \mathrm{Profit}=₹\frac{16\times 30}{100}=₹4.80\\ \\ \mathrm{Selling}\mathrm{price}\mathrm{of}\mathrm{one}\mathrm{article}=\mathrm{C}.\mathrm{P}.+\mathrm{Profit}\\ =\text{₹}(30+4.80)\\ =₹34.80\end{array}$

Q.10 The cost of an article was ₹ 15,500. ₹ 450 were spent on its repairs. If it is sold for a profit of 15%, find the selling price of the article.

Ans

The cost of an article was ₹ 15,500
The amount spent on its repairs ₹ 450

$\begin{array}{l}\therefore \text{Total cost of an article}=\text{Cost}+\text{Overhead expenses}\\ \\ =\text{₹ 155}00+\text{Rs 45}0\\ \\ =\text{₹ 1595}0\\ \mathrm{Profit}\mathrm{\%}=\frac{\mathrm{Profit}}{\mathrm{C}.\mathrm{P}.}\times 100\\ \Rightarrow 15=\frac{\text{Profit}}{\text{₹}15,950}\times 100\\ \therefore \text{Profit}=₹\left(\frac{15,950\times 15}{100}\right)\\ =₹\text{2392.50}\\ \\ \text{We know, S.P. of an article}=\text{C}.\text{P}.+\text{Profit}\\ \\ =\text{₹}(\text{1595}0+\text{2392}.\text{5}0)\\ \\ =\text{₹ 18,342}.\text{5}0\end{array}$

Q.11 A VCR and TV were bought for ₹ 8,000 each. The shopkeeper made a loss of 4% on the VCR and a profit of 8% on the TV. Find the gain or loss percent on the whole transaction.

Ans

$\begin{array}{l}\mathrm{C}.\text{P}.\text{of a VCR}=\text{₹ 8}000\\ \text{Loss\% = 4}\%\\ \end{array}$ $\begin{array}{l}\text{Let us suppose that C}.\text{P}.\text{is ₹ 1}00\\ \text{Since loss is}4\mathrm{\%}\text{,so S}.\text{P}.\text{is ₹ 96}.\\ \\ \therefore \text{When C}.\text{P}.\text{is Rs 8}000,\text{S}.\text{P}.=₹(\frac{96}{100}\times 8000)=₹7680\\ \\ \text{Now, C}.\text{P}.\text{of a TV}=\text{₹ 8}000\\ \text{The shopkeeper made a profit of 8}\mathrm{\%}\text{on TV}.\\ \\ \text{Let us suppose that C}.\text{P}.\text{of TV is ₹ 1}00\\ \text{Then S}.\text{P}.\text{is ₹ 1}0\text{8}.\\ \text{When C}.\text{P}.\text{is Rs 8}000,\text{S}.\text{P}.=₹(\frac{108}{100}\times 8000)=₹8640\\ \\ \text{To find the profit or loss on whole transaction,we have}\\ \text{to find total C.P.and S.P.}\\ \\ \text{Total S}.\text{P}.=₹\text{768}0+₹\text{864}0=₹\text{1632}0\\ \text{Total C}.\text{P}.=₹\text{8}000+₹\text{8}000=₹\text{16}000\\ \\ \text{Since total S}.\text{P}.\text{is greater than total C}.\text{P}.,\mathrm{therefore},\mathrm{we}\mathrm{have}\text{a profit}.\end{array}$ $\begin{array}{l}\text{Profit}=\text{Total S}.\text{P}-\text{Total C}.\text{P}.\\ =₹\text{1632}0-₹\text{16}000\\ =₹\text{32}0\\ \\ \text{Now,Profit}\mathrm{\%}=\frac{\text{Profit}}{\mathrm{C}.\mathrm{P}.}\times 100\\ =\frac{\text{320}}{16000}\times 100\\ =2\mathrm{\%}\\ \\ \text{Hence},\text{the shopkeeper had a gain of 2}\mathrm{\%}\text{on the whole transaction}.\end{array}$

Q.12 During a sale, a shop offered a discount of 10% on the marked prices of all the items. What would a customer have to pay for a pair of jeans marked at ₹ 1450 and two shirts marked at ₹ 850 each?

Ans

$\begin{array}{l}\text{Total marked price}=₹(\text{1},\text{45}0+\text{2}\times \text{85}0)\\ =₹(\text{1},\text{45}0+\text{1},\text{7}00)\\ =₹\text{3},\text{15}0\\ \text{Given that},\text{discount}\%=\text{1}0\%\\ \therefore \text{Discount}=₹(\text{3},\text{15}0\times \frac{10}{100})=₹\text{315}\\ \text{Also},\text{Discount}=\text{Marked price}-\text{Sale price}\\ \Rightarrow \text{₹ 315}=₹\text{315}0-\text{Sale price}\\ \Rightarrow \text{Sale price}=₹(\text{315}0-\text{315})\\ =₹\text{2835}\\ \\ \text{Thus},\text{the customer will have to pay ₹ 2},\text{835}.\end{array}$

Q.13 A milkman sold two of his buffaloes for ₹ 20,000 each.On one he made a gain of 5% and on the other a loss of 10%. Find his overall gain or loss. (Hint: Find CP of each)

Ans

$\begin{array}{l}\mathrm{Given},\mathrm{S}.\mathrm{P}.\mathrm{of}\mathrm{each}\mathrm{buffalo}=\text{₹}20000\\ \mathrm{Gain}\mathrm{\%}=5\mathrm{\%}\\ \\ \mathrm{This}\mathrm{means}\mathrm{if}\mathrm{C}.\mathrm{P}.\mathrm{is}\text{₹}100,\mathrm{then}\mathrm{S}.\mathrm{P}.\mathrm{is}\text{₹}105.\\ \\ \therefore \mathrm{C}.\mathrm{P}.\mathrm{of}\mathrm{one}\mathrm{buffalo}=₹20000\times \frac{100}{105}=₹19047.62\\ \\ \mathrm{Also},\mathrm{the}\mathrm{second}\mathrm{buffalo}\mathrm{was}\mathrm{sold}\mathrm{at}\mathrm{a}\mathrm{loss}\mathrm{of}10\mathrm{\%}.\\ \\ \mathrm{This}\mathrm{means}\mathrm{if}\mathrm{C}.\mathrm{P}.\mathrm{is}\text{₹}100,\mathrm{then}\mathrm{S}.\mathrm{P}.\mathrm{is}\text{₹}90.\\ \\ \therefore \mathrm{C}.\mathrm{P}.\mathrm{of}\mathrm{other}\mathrm{buffalo}=₹20000\times \frac{100}{90}=₹22222.22\\ \\ \therefore \mathrm{Total}\mathrm{C}.\mathrm{P}.=₹19047.62+₹22222.22=₹41269.84\\ \mathrm{Total}\mathrm{S}.\mathrm{P}.=\text{₹}20000+\text{₹}20000=\text{₹}40000\\ \\ \mathrm{Loss}=₹41269.84-\text{₹}40000=₹1269.84\\ \\ \mathrm{Thus},\mathrm{the}\mathrm{overall}\text{loss}\mathrm{of}\mathrm{the}\mathrm{m}\text{ilkman}\mathrm{was}\text{₹}1,269.84.\end{array}$

Q.14 The price of a TV is ₹ 13,000. The sales tax charged on it is at the rate of 12%. Find the amount that Vinod will have to pay if he buys it.

Ans

$\begin{array}{l}\text{The price of the TV = ₹ 13,000}\\ \text{Sales tax =12\%}\\ \text{i}\text{.e}\text{.= ₹}\left(\frac{12}{100}\text{×13,000}\right)\\ =₹1,560\\ \therefore \text{Required amount}=\text{Cost}+\text{Sales Tax}\\ =\text{₹ 13}000+\text{₹ 156}0\\ =₹\text{1456}0\\ \\ \text{Therefore},\text{Vinod have to pay ₹ 14},\text{56}0\text{for the T}.\text{V}.\end{array}$

Q.15 Arun bought a pair of skates at a sale where the discount given was 20%. If the amount he pays is ₹ 1,600, find the marked price.

Ans

$\begin{array}{l}\text{Let the marked price be}\mathrm{x}.\\ \text{Discount}\mathrm{\%}=\frac{\text{Discount}}{\text{Marked price}}\times 100\\ 20=\frac{\text{Discount}}{\mathrm{x}}\times 100\\ \text{Discount}=\frac{20\mathrm{x}}{100}=\frac{1}{5}\mathrm{x}\\ \text{Also,Discount = Marked price}-\text{Sale price}\\ \frac{1}{5}\mathrm{x}=\mathrm{x}-₹1600\\ \mathrm{x}-\frac{1}{5}\mathrm{x}=₹1600\\ \frac{4}{5}\mathrm{x}=₹1600\\ \mathrm{x}=\text{₹}1600\times \frac{5}{4}=₹2000\\ \\ \therefore \text{The marked price of the scates was ₹ 2000.}\end{array}$

Q.16 I purchased a hair-dryer for ₹ 5,400 including 8% VAT. Find the price before VAT was added.

Ans

$\begin{array}{l}\text{It is given that the price of the hair drier includes VAT}.\\ \\ \text{Let us suppose the price without VAT be ₹ 1}00,\text{then}\\ \text{the price including VAT will be ₹ 1}0\text{8}.\\ \\ \text{When price including VAT is ₹ 1}0\text{8},\text{original price}=₹\text{1}00\\ \text{When price including VAT is ₹ 5400,original price}\\ =₹(\frac{100}{108}\times 5400)\\ =₹5000\\ \\ \text{Thus},\text{the price of the hair}-\text{dryer before the}\\ \text{addition of VAT was ₹ 5},000.\end{array}$

Q.17

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

Ans

$\begin{array}{l}\left(\mathrm{a}\right)\text{Principal}=₹10,800\\ \mathrm{R}=12\frac{1}{2}\mathrm{\%}=\frac{25}{2}\mathrm{\%}\\ \mathrm{n}=3\\ \therefore \mathrm{Amount}=\mathrm{P}{(1+\frac{\mathrm{R}}{100})}^{\mathrm{n}}\\ =10800{(1+\frac{25}{2\times 100})}^3\\ =10800{(1+\frac{25}{200})}^3\\ =10800{\left(\frac{225}{200}\right)}^3\\ =10800\times \frac{225}{200}\times \frac{225}{200}\times \frac{225}{200}\\ =15377.34\\ \\ \therefore \mathrm{C}.\mathrm{I}=\mathrm{Amount}-\mathrm{Principal}\\ =15377.34-10800\\ =4,577.34\end{array}$ $\begin{array}{l}\left(\mathrm{b}\right)\text{Principal}=₹18,000\\ \mathrm{R}=10\mathrm{\%}\\ \mathrm{n}=2\frac{1}{2}\\ \therefore \mathrm{Amount}=\mathrm{P}{(1+\frac{\mathrm{R}}{100})}^{\mathrm{n}}\\ =18000{(1+\frac{10}{100})}^{2\frac{1}{2}}\\ \\ \text{Since,the years are in fractional form,so we will calculate}\\ \text{the Amount for the whole part first and then we will calculate}\\ \text{the simple interest for the remaining fractional part.}\\ \therefore \mathrm{Amount}=18000{(1+\frac{10}{100})}^2\\ =18000{\left(\frac{11}{10}\right)}^2\\ =18000\times \frac{11}{10}\times \frac{11}{10}\\ =₹21,780\end{array}$ $\begin{array}{l}\text{Now,we will take ₹ 21,780 as principal and calculate the simple}\\ \text{interest for the next}\frac{1}{2}\mathrm{year}.\\ \therefore \mathrm{S}.\mathrm{I}=\frac{\mathrm{PRT}}{100}\\ =\frac{21780\times 10\times 1}{2\times 100}\\ =₹\text{1089}\\ \\ \therefore \text{Interest for the first 2 years}=₹(\text{2178}0-\text{18}000)=₹\text{378}0\\ \text{And interest for the next}\frac{1}{2}\text{year}=₹\text{1}0\text{89}\\ \therefore \text{Total C}.\text{I}.=₹\text{378}0+₹\text{1}0\text{89}=₹\text{4},\text{869, and}\\ \\ \text{A}=\text{P}+\text{C}.\text{I}.=₹\text{18}000+₹\text{4869}=₹\text{22},\text{869}\end{array}$ $\begin{array}{l}\left(\mathrm{c}\right)\text{Principal}=₹62,500\\ \mathrm{R}=8\mathrm{\%}\mathrm{yearly}=4\mathrm{\%}\mathrm{half}\mathrm{yearly}\\ \mathrm{n}=1\frac{1}{2}=3\mathrm{half}\mathrm{years}\\ \therefore \mathrm{Amount}=\mathrm{P}{(1+\frac{\mathrm{R}}{100})}^{\mathrm{n}}\\ =62500{(1+\frac{4}{100})}^3\\ =62500{\left(\frac{26}{25}\right)}^3\\ =62500\times \frac{26}{25}\times \frac{26}{25}\times \frac{26}{25}\\ =₹70,304\\ \\ \therefore \mathrm{C}.\mathrm{I}=\mathrm{A}-\mathrm{P}=₹70,304-₹62,500\\ =₹7,804\\ \left(\mathrm{d}\right)\text{Principal}=₹8000\\ \mathrm{R}=9\mathrm{\%}\mathrm{yearly}=\frac{9}{2}\mathrm{\%}\mathrm{half}\mathrm{yearly}\\ \mathrm{n}=1=2\mathrm{half}\mathrm{years}\\ \therefore \mathrm{Amount}=\mathrm{P}{(1+\frac{\mathrm{R}}{100})}^{\mathrm{n}}\\ =8000{(1+\frac{9}{200})}^2\\ =8000{\left(\frac{209}{200}\right)}^2\\ =₹8,736.20\\ \\ \therefore \mathrm{C}.\mathrm{I}=\mathrm{A}-\mathrm{P}=₹8,736.20-\mathrm{Rs}8000\\ =₹736.20\\ \left(\mathrm{e}\right)\mathrm{Principal}=\mathrm{Rs}10,000\\ \mathrm{R}=8\mathrm{\%}\mathrm{yearly}=4\mathrm{\%}\mathrm{half}\mathrm{yearly}\\ \mathrm{n}=1=2\mathrm{half}\mathrm{years}\\ \therefore \mathrm{Amount}=\mathrm{P}{(1+\frac{\mathrm{R}}{100})}^{\mathrm{n}}\\ =10,000{(1+\frac{4}{100})}^2\\ =10,000{(1+\frac{1}{25})}^2\\ =10,000{\left(\frac{26}{25}\right)}^2\\ =10,000\times \frac{26}{25}\times \frac{26}{25}\\ =₹10,816\\ \\ \therefore \mathrm{C}.\mathrm{I}=\mathrm{A}-\mathrm{P}=\mathrm{Rs}10,816-\mathrm{Rs}10,000\\ =₹816\end{array}$

Q.18 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

$\begin{array}{l}\mathrm{Principal}=₹26,400\\ \mathrm{R}=15\mathrm{\%}\mathrm{per}\mathrm{annum}\\ \mathrm{n}=2\frac{4}{12}\mathrm{years}\\ \mathrm{Since},\mathrm{the}\mathrm{years}\mathrm{are}\mathrm{in}\mathrm{fractional}\mathrm{form},\mathrm{so}\mathrm{we}\mathrm{will}\mathrm{calculate}\\ \mathrm{the}\mathrm{Amount}\mathrm{for}\mathrm{the}\mathrm{whole}\mathrm{part}\mathrm{first}\mathrm{and}\mathrm{then}\mathrm{we}\mathrm{will}\mathrm{calculate}\\ \mathrm{the}\mathrm{simple}\mathrm{interest}\mathrm{for}\mathrm{the}\mathrm{remaining}\mathrm{fractional}\mathrm{part}.\\ \\ \therefore \mathrm{Amount}=26,400{(1+\frac{15}{100})}^2\\ =26,400{\left(\frac{23}{20}\right)}^2\end{array}$ $\begin{array}{l}=26,400\times \frac{23}{20}\times \frac{23}{20}\\ =₹34,914\\ \mathrm{Now},\mathrm{we}\mathrm{will}\mathrm{take}\mathrm{Rs}34,914\mathrm{as}\mathrm{principal}\mathrm{and}\mathrm{calculate}\mathrm{the}\mathrm{simple}\\ \mathrm{interest}\mathrm{for}\mathrm{the}\mathrm{next}\frac{1}{3}\mathrm{years}.\\ \therefore \mathrm{S}.\mathrm{I}=\frac{\mathrm{PRT}}{100}\\ =\frac{34,914\times 15\times 1}{3\times 100}\\ =₹1745.70\\ \\ \therefore \mathrm{Interest}\mathrm{for}\mathrm{the}\mathrm{first}2\mathrm{years}=₹(34,914-26,400)=₹8,514\\ \mathrm{And}\mathrm{interest}\mathrm{for}\mathrm{the}\mathrm{next}\frac{1}{3}\mathrm{year}=₹1745.70\\ \therefore \mathrm{Total}\mathrm{C}.\mathrm{I}.=₹8,514+₹1745.70=₹10,259.70,\mathrm{and}\\ \\ \mathrm{A}=\mathrm{P}+\mathrm{C}.\mathrm{I}.=₹26,400+₹10,259.70=₹36,659.70\end{array}$

Q.19 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

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

Q.20 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

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

Q.21 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

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

Q.22 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

$\begin{array}{l}\text{(i)}\mathrm{Principal}=₹\text{80},000\\ \mathrm{R}=10\%\mathrm{per}\mathrm{annum}\\ \mathrm{n}=1\frac{1}{2}\mathrm{years}\\ \mathrm{Since},\mathrm{the}\mathrm{years}\mathrm{are}\mathrm{in}\mathrm{fractional}\mathrm{form},\mathrm{so}\mathrm{we}\mathrm{will}\mathrm{calculate}\\ \mathrm{the}\mathrm{Amount}\mathrm{for}\mathrm{the}\mathrm{whole}\mathrm{part}\mathrm{first}\mathrm{and}\mathrm{then}\mathrm{we}\mathrm{will}\mathrm{calculate}\\ \mathrm{the}\mathrm{simple}\mathrm{interest}\mathrm{for}\mathrm{the}\mathrm{remaining}\mathrm{fractional}\mathrm{part}.\\ \therefore \mathrm{Amount}=80,000{\left(1+\frac{10}{100}\right)}^1\\ =80,000\left(1+\frac{1}{10}\right)\\ =80,000{\left(\frac{11}{10}\right)}^2\\ =₹88,000\\ \\ \mathrm{Now},\mathrm{we}\mathrm{will}\mathrm{take}\mathrm{Rs}88,000\mathrm{as}\mathrm{principal}\mathrm{and}\mathrm{calculate}\mathrm{the}\mathrm{simple}\\ \mathrm{interest}\mathrm{for}\mathrm{the}\mathrm{next}\frac{1}{2}\mathrm{year}.\\ \therefore \mathrm{S}.\mathrm{I}=\frac{\mathrm{PRT}}{100}\\ =\frac{88,000\times 10\times 1}{2\times 100}\\ =₹4,400\\ \text{Interest for the first year}=\text{₹ 88}000-\text{₹ 8}0000=\text{₹ 8},000\\ \text{And interest for the next}\frac{1}{2}\text{year}=\text{₹ 4},\text{4}00\\ \text{Total C}.\text{I}.=\text{₹ 8}000+\text{₹ 4},\text{4}00=\text{₹ 1},\text{24}00\\ \\ \therefore \text{A}=\text{P}+\text{C}.\text{I}.=\text{₹}\left(\text{8}0000+\text{124}00\right)=\text{₹ 92},\text{4}00\end{array}$ $\begin{array}{l}\left(\text{ii}\right)\text{Now,the interest is compounded half yearly}.\\ \text{Rate}=\text{1}0\%\text{per annum}=\text{5}\%\text{per half year}\\ \text{n=}1\frac{1}{2}\mathrm{years}=3\text{half years}\\ \therefore \mathrm{Amount}=80,000{\left(1+\frac{5}{100}\right)}^3\\ =80,000\left(1+\frac{1}{20}\right)3\\ =80,000{\left(\frac{21}{20}\right)}^2\\ =₹\text{92},610\end{array}$

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

Q.23 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

$\begin{array}{l}\left(\mathrm{i}\right)\mathrm{Principal}=₹\text{8},000\\ \mathrm{R}=5\mathrm{\%}\mathrm{yearly}\\ \mathrm{n}=2\mathrm{years}\\ \\ \therefore \mathrm{Amount}=\mathrm{P}{(1+\frac{\mathrm{R}}{100})}^{\mathrm{n}}\\ =8,000{(1+\frac{5}{100})}^2\\ =8,000{\left(\frac{21}{20}\right)}^2\\ =\text{8},000\times \frac{21}{20}\times \frac{21}{20}\\ =₹8,820\end{array}$ $\begin{array}{l}\text{(ii)}\mathrm{Now},\text{we have to calculate the interest for the third year.}\\ \text{So we}\mathrm{will}\mathrm{take}\mathrm{Rs}\text{8,820}\mathrm{as}\mathrm{principal}\mathrm{and}\mathrm{calculate}\mathrm{the}\mathrm{simple}\\ \mathrm{interest}\mathrm{for}\mathrm{the}\mathrm{next}\mathrm{year}.\\ \therefore \mathrm{S}.\mathrm{I}=\frac{\mathrm{PRT}}{100}\\ =\frac{8820\times 5\times 1}{100}\\ =₹441\end{array}$

Q.24 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

$\begin{array}{l}\mathrm{Principal}=₹10,000\\ \mathrm{R}=10\mathrm{\%}\mathrm{yearly}=5\mathrm{\%}\mathrm{half}\mathrm{yearly}\\ \mathrm{n}=1\frac{1}{2}\mathrm{years}=3\mathrm{half}\mathrm{years}\\ \\ \therefore \mathrm{Amount}=\mathrm{P}{(1+\frac{\mathrm{R}}{100})}^{\mathrm{n}}\\ =10,000{(1+\frac{5}{100})}^3\\ =10,000{(1+\frac{1}{20})}^3\\ =10,000{\left(\frac{21}{20}\right)}^3\end{array}$ $\begin{array}{l}=60,000\times \frac{21}{20}\times \frac{21}{20}\times \frac{21}{20}\\ =₹11,576.25\\ \\ \mathrm{C}.\mathrm{I}=\mathrm{A}-\mathrm{P}\\ =₹11,576.25-\mathrm{Rs}10,000\\ =₹1,576.25\\ \\ \mathrm{Now}\text{we have to calculate the interest compounded annually.}\\ \mathrm{Since},\mathrm{the}\mathrm{years}\mathrm{are}\mathrm{in}\mathrm{fractional}\mathrm{form},\mathrm{so}\mathrm{we}\mathrm{will}\mathrm{calculate}\\ \mathrm{the}\mathrm{Amount}\mathrm{for}\mathrm{the}\mathrm{whole}\mathrm{part}\mathrm{first}\mathrm{and}\mathrm{then}\mathrm{we}\mathrm{will}\mathrm{calculate}\\ \mathrm{the}\mathrm{simple}\mathrm{interest}\mathrm{for}\mathrm{the}\mathrm{remaining}\mathrm{fractional}\mathrm{part}.\\ \\ \therefore \mathrm{Amount}=₹10,000{(1+\frac{10}{100})}^1\\ =₹10,000\left(\frac{11}{10}\right)\\ =11,000\end{array}$ \begin{array}{l}\end{array} $\begin{array}{l}\mathrm{Now},\mathrm{we}\mathrm{will}\mathrm{take}₹11,000\mathrm{as}\mathrm{principal}\mathrm{and}\mathrm{calculate}\mathrm{the}\mathrm{simple}\\ \mathrm{interest}\mathrm{for}\mathrm{the}\mathrm{next}\frac{1}{2}\mathrm{years}.\\ \therefore \mathrm{S}.\mathrm{I}=\frac{\mathrm{PRT}}{100}\\ =\frac{11,000\times 10\times 1}{2\times 100}\\ =₹550\\ \\ \therefore \text{Interest for the first year}=₹\text{11}000-₹\text{1}0000=₹\text{1},000\\ \therefore \text{Total compound interest}=₹\text{1}000+₹\text{55}0=₹\text{1},\text{55}0\\ \\ \text{Yes,the interest would be more when compounded half yearly}\\ \text{than the interest when compounded annually}.\end{array}$

Q.25 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

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

Q.26 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

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

Q.27 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\\ =5,06,000\times \frac{41}{40}\times \frac{41}{40}\\ =531616.25\\ \\ \text{Thus, the count of bacteria at the end of 2 hours will be}5,31,616\left(\text{approx}.\right).\end{array}

Q.28 A scooter was bought at ₹ 42,000. Its value depreciated at the rate of 8% per annum. Find its value after one year.

Ans

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