NCERT Solutions Class 7 Maths Chapter 8

Q.1 

$\begin{array}{l}\mathrm{Find}\mathrm{the}\mathrm{ratio}\mathrm{of}:\\ \left(\mathrm{a}\right)\mathrm{Rs}5\mathrm{to}50\mathrm{paise}\left(\mathrm{b}\right)15\mathrm{kg}\mathrm{to}210\mathrm{g}\\ \left(\mathrm{c}\right)9\mathrm{m}\mathrm{to}27\mathrm{cm}\left(\mathrm{d}\right)30\mathrm{days}\mathrm{to}36\mathrm{hour}\end{array}$

Ans

\begin{array}{l}\left(\text{a}\right)\text{Rs 5 to 5}0\text{paise}\\ 1\text{ruppe}=\text{100 paise}\\ \text{So,}\text{5 rupee}=\text{500 paise}\\ \text{So,}\frac{\text{Rs}5}{50\text{paise}}=\frac{500}{50}=\frac{10}{1}\\ \text{Thus, the required ratio is}\boxed{\text{10:1}}.\\ \left(\text{b}\right)\text{15 kg to 21}0\text{g}\\ \text{1 kg}=\text{1000 g}\\ \text{So, 15 kg}=\text{15000 g}\\ \text{So,}\frac{15\text{kg}}{210\text{g}}=\frac{15000}{210}=\frac{500}{7}\\ \text{Thus, the required ratio is}\boxed{\text{500:7}}.\\ \left(\text{c}\right)\text{9 m to 27 cm}\\ \text{1 m}=\text{100 cm}\\ \text{So, 9 m}=\text{900 cm}\\ \text{So,}\frac{9\text{m}}{27\text{cm}}=\frac{900}{27}=\frac{100}{3}\\ \text{Thus, the required ratio is}\boxed{\text{100:3}}.\\ \left(\text{d}\right)\text{3}0\text{days to 36 hour}\\ \text{1 day}=\text{24 hours}\\ \text{So, 30 days}=\text{720 hours}\\ \text{So,}\frac{30\text{days}}{36\text{hour}}=\frac{720}{36}=\frac{20}{1}\\ \text{Thus, the required ratio is}\boxed{\text{20:1}}.\end{array}

Q.2 

$\begin{array}{l}\mathrm{In}\mathrm{a}\mathrm{computer}\mathrm{lab},\mathrm{there}\mathrm{are}3\mathrm{computers}\mathrm{for}\mathrm{every}6\\ \mathrm{students}.\mathrm{How}\mathrm{many}\mathrm{computers}\mathrm{will}\mathrm{be}\mathrm{needed}\mathrm{for}24\\ \mathrm{students}?\end{array}$

Ans

\begin{array}{l}\text{For 6 students, number of computers required}=\text{3}\\ \text{So, for 1 student, number of computers required}=\frac{3}{6}=\frac{1}{2}\\ \text{Thus, for 24 students, number of computers required}=\frac{1}{2}\times 24\\ =\boxed{12}\\ \text{Therefore, 12 computers are needed for 24 students}\text{.}\end{array}

Q.3

$\begin{array}{l}\mathrm{Population}\mathrm{of}\mathrm{Rajasthan}=570\mathrm{lakhs}\mathrm{and}\\ \mathrm{population}\mathrm{of}\mathrm{UP}=1660\mathrm{lakh}\\ \mathrm{Area}\mathrm{of}\mathrm{Rajasthan}=\; 3\mathrm{lakh}{\mathrm{km}}^{\mathrm{2}}\mathrm{and}\\ \mathrm{area}\mathrm{of}\mathrm{UP}=\; 2\mathrm{lakh}{\mathrm{km}}^{\mathrm{2}}\mathrm{.}\\ \left(\mathrm{i}\right)\mathrm{How}\mathrm{many}\mathrm{people}\mathrm{are}\mathrm{there}\mathrm{per}{\mathrm{km}}^{\mathrm{2}}\mathrm{in}\mathrm{both}\mathrm{these}\mathrm{States}?\\ \left(\mathrm{ii}\right)\mathrm{Which}\mathrm{State}\mathrm{is}\mathrm{less}\mathrm{populated}?\end{array}$

Ans

\begin{array}{l}{\text{(i) Population of Rajasthan in 3 lakh km}}^{\text{2}}=\text{570 lakhs}\\ {\text{Population of Rajasthan in 1 lakh km}}^{\text{2}}=\frac{\text{570}}{3}\\ =\boxed{190}\\ {\text{Population of UP in 2 lakh km}}^{\text{2}}=\text{1660 lakhs}\\ {\text{Population of UP in 1 lakh km}}^{\text{2}}=\frac{1660}{2}\\ =\boxed{830}\\ \text{(ii)}\\ \text{From above data, clearly Rajasthan is less populated}\text{.}\end{array}

Q.4

$\begin{array}{l}\mathrm{Convert}\mathrm{the}\mathrm{given}\mathrm{fractional}\mathrm{numbers}\mathrm{to}\mathrm{per}\mathrm{cents}.\\ \left(\mathrm{a}\right)\frac{\mathrm{1}}{\mathrm{9}}\left(\mathrm{b}\right)\frac{\mathrm{5}}{\mathrm{4}}\left(\mathrm{c}\right)\frac{\mathrm{3}}{\mathrm{40}}\left(\mathrm{d}\right)\frac{\mathrm{2}}{\mathrm{7}}\end{array}$

Ans

\begin{array}{l}\text{(a})\frac{1}{8}\\ =\frac{1}{8}\times \frac{100}{100}=\frac{1}{8}\times 100\%=12.5\%\\ \left(\text{b}\right)\frac{5}{4}\\ =\frac{5}{4}\times \frac{100}{100}=\frac{5}{4}\times 100\%=125\%\\ \left(\text{c}\right)\frac{3}{40}\\ =\frac{3}{40}\times \frac{100}{100}=\frac{3}{40}\times 100\%=7.5\%\\ \left(\text{d}\right)\frac{2}{7}\\ =\frac{2}{7}\times \frac{100}{100}=\frac{2}{7}\times 100\%=28\frac{4}{7}\%\end{array}

Q.5 Convert the given decimal fractions to percents.
a. 0.65 b. 2.1 c. 0.02 d. 12.35

Ans

\begin{array}{l}\left(\text{a}\right)0.\text{65}\\ =0.65\times 100\%=\frac{65\times 100}{100}\%=65\%\\ \left(\text{b}\right)\text{2}.\text{1}\\ =2.1\times 100\%=\frac{21\times 100}{10}\%=210\%\\ \left(\text{c}\right)0.0\text{2}\\ =0.02\times 100\%=\frac{2\times 100}{100}\%=2\%\\ \left(\text{d}\right)\text{12}.\text{35}\\ =12.35\times 100\%=\frac{1235\times 100}{100}\%=1235\%\end{array}

Q.6 Estimate what part of the figures is coloured and hence find the percent which is coloured.

Ans

\begin{array}{l}\text{(i)}\\ \text{From the figure, we observe that 1 part out of 4 is shaded,}\\ \text{so its}\frac{1}{4}.\\ \frac{1}{4}\times \frac{100}{100}=\frac{1}{4}\times 100\%=25\%\\ \text{(ii)}\\ \text{From the figure, we observe that 3 part out of 5 is shaded,}\\ \text{so its}\frac{3}{5}.\\ \frac{3}{5}\times \frac{100}{100}=\frac{3}{5}\times 100\%=60\%\\ \text{(iii)}\\ \text{From the figure, we observe that 3 part out of 8 is shaded,}\\ \text{so its}\frac{3}{8}.\\ \frac{3}{8}\times \frac{100}{100}=\frac{3}{8}\times 100\%=37.5\%\end{array}

Q.7

$\begin{array}{l}\mathrm{Find}:\\ \left(\mathrm{a}\right)15\%\mathrm{of}250\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\left(\mathrm{b}\right)1\%\mathrm{of}1\mathrm{hour}\\ \left(\mathrm{c}\right)20\%\mathrm{of}\mathrm{Rs}2500\; \hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\left(\mathrm{d}\right)75\%\mathrm{of}1\mathrm{k}\end{array}$

Ans

\begin{array}{l}\left(\text{a}\right)\text{15}\%\text{of 25}0\\ =\frac{15}{100}\times 250=\frac{75}{2}=37.5\\ \left(\text{b}\right)\text{1}\%\text{of 1 hour}\\ \text{1 hour}=\text{60 minutes}\text{.}\\ \text{So, 1}\%\text{of 1 hour}\\ =\frac{1}{100}\times 60=\frac{6}{10}=\frac{3}{5}\\ \left(\text{c}\right)\text{2}0\%\text{of Rs 25}00\\ =\frac{20}{100}\times 2500=20\times 25=500\\ \left(\text{d}\right)\text{75}\%\text{of 1 kg}\\ \text{1 kg}=\text{1000 g}\text{.}\\ \text{So, 75}\%\text{of 1 kg}\\ =\frac{75}{100}\times 1000=75\times 10=750\end{array}

Q.8

$\begin{array}{l}\mathrm{Find}\mathrm{the}\mathrm{whole}\mathrm{quantity}\mathrm{if}\\ \left(\mathrm{a}\right)5\%\mathrm{of}\mathrm{it}\mathrm{is}600.\left(\mathrm{b}\right)12\%\mathrm{of}\mathrm{it}\mathrm{is}₹1080.\\ \left(\mathrm{c}\right)40\%\mathrm{of}\mathrm{it}\mathrm{is}500\mathrm{km}.\left(\mathrm{d}\right)70\%\mathrm{of}\mathrm{it}\mathrm{is}14\mathrm{minutes}\\ \left(\mathrm{e}\right)8\%\mathrm{of}\mathrm{it}\mathrm{is}40\mathrm{litre}\end{array}$

Ans

\begin{array}{l}\left(\text{a}\right)\text{5}\%\text{of it is 6}00\text{is same as 5}\%\text{of x is 6}00.\\ \frac{5}{100}\times x=600\\ x=\frac{600\times 100}{5}=12000\\ \left(\text{b}\right)\text{12}\%\text{of it is}₹\text{1}0\text{8}0\text{is same as 12}\%\text{of x is}₹1080.\\ \frac{12}{100}\times x=1080\\ x=\frac{1080\times 100}{12}=₹9000\\ \left(\text{c}\right)\text{4}0\%\text{of it is 5}00\text{km is same as 40}\%\text{of x is}500\text{km}.\\ \frac{40}{100}\times x=500\\ x=\frac{500\times 100}{40}=1250\\ \left(\text{d}\right)\text{7}0\%\text{of it is 14 minutes is same as 70}\%\text{of x is}14\text{minutes}\\ \frac{70}{100}\times x=14\\ x=\frac{14\times 100}{70}=20\\ \left(\text{e}\right)\text{8}\%\text{of it is 4}0\text{litres is same as 8}\%\text{of x is}40\text{litres}\\ \frac{8}{100}\times x=40\\ x=\frac{40\times 100}{8}=500\end{array}

Q.9

$\begin{array}{l}\mathrm{Convert}\mathrm{given}\mathrm{per}\mathrm{cents}\mathrm{to}\mathrm{decimal}\mathrm{fractions}\mathrm{and}\mathrm{also}\mathrm{to}\\ \mathrm{fractions}\mathrm{in}\mathrm{simplest}\mathrm{forms}:\\ \left(\mathrm{a}\right)\mathrm{25\%}\left(\mathrm{b}\right)150\%\left(\mathrm{c}\right)\mathrm{20\%}\left(\mathrm{d}\right)\mathrm{5\%}\end{array}$

Ans

\begin{array}{l}\left(\text{a}\right)\text{25}\%\\ =\frac{25}{100}=\frac{1}{4}=0.25\\ \left(\text{b}\right)\text{15}0\%\\ =\frac{150}{100}=\frac{15}{10}=\frac{3}{2}=1.5\\ \left(\text{c}\right)\text{2}0\%\\ =\frac{20}{100}=\frac{2}{10}=\frac{1}{5}=0.2\\ \left(\text{d}\right)\text{5}\%\\ =\frac{5}{100}=\frac{1}{20}=0.05\end{array}

Q.10

$\begin{array}{l}\mathrm{In}\mathrm{a}\mathrm{city},\; 30\%\mathrm{are}\mathrm{females},\; 40\%\mathrm{are}\mathrm{males}\mathrm{and}\mathrm{remaining}\\ \mathrm{are}\mathrm{children}.\mathrm{What}\mathrm{percent}\mathrm{are}\mathrm{children}?\end{array}$

Ans

\begin{array}{l}\text{Children}=\left(100-\left(30+40\right)\right)\%\\ =\left(100-70\right)\%\\ =30\%\end{array}

Q.11

$\begin{array}{l}\mathrm{Out}\mathrm{of}15,000\mathrm{voters}\mathrm{in}\mathrm{a}\mathrm{constituency},\; 60\%\mathrm{voted}.\\ \mathrm{Find}\mathrm{the}\mathrm{percentage}\mathrm{of}\mathrm{voters}\mathrm{who}\mathrm{did}\mathrm{not}\mathrm{vote}.\mathrm{Can}\mathrm{you}\\ \mathrm{now}\mathrm{find}\mathrm{how}\mathrm{many}\mathrm{actually}\mathrm{did}\mathrm{not}\mathrm{vote}?\end{array}$

Ans

\begin{array}{l}\text{Percentage of voters who voted}=60\%\\ \text{Percentage of voters who did not vote}=\text{100\%-60\%}\\ =\text{40\%}\\ \text{Number of voters who did not vote}=\text{40\% of 15,000}\\ =\frac{40}{100}\times 15000\\ =6000\end{array}

Q.12

$\begin{array}{l}\mathrm{Meeta}\mathrm{saves}₹400\mathrm{from}\mathrm{her}\mathrm{salary}.\mathrm{If}\mathrm{this}\mathrm{is}10\%\mathrm{of}\mathrm{her}\\ \mathrm{salary}.\mathrm{What}\mathrm{is}\mathrm{her}\mathrm{salay}?\end{array}$

Ans

\begin{array}{l}\text{Let her salary be x}\text{.}\\ \text{Then according to the question, we have}\\ \text{10\% of x}=\text{400}\\ \frac{10}{100}\times x=400\\ x=\frac{400\times 100}{10}\\ =4000\\ \text{Therefore Meeta’s salary is ₹ 4000}\end{array}

Q.13

$\begin{array}{l}\mathrm{A}\mathrm{local}\mathrm{cricket}\mathrm{team}\mathrm{played}20\mathrm{matches}\mathrm{in}\mathrm{one}\mathrm{season}.\\ \mathrm{It}\mathrm{won}25\%\mathrm{of}\mathrm{them}.\mathrm{How}\mathrm{many}\mathrm{matches}\mathrm{did}\mathrm{they}\mathrm{win}?\end{array}$

Ans

\begin{array}{l}\text{Since team played 20 matches,}\\ \text{So, number of matches won}=\text{25\% of 20}\\ =\frac{25}{100}\times 20\\ =\frac{500}{100}\\ =5\\ \text{So, number they won 5 matchces}\end{array}

Q.14

$\begin{array}{l}\mathrm{Tell}\mathrm{what}\mathrm{is}\mathrm{the}\mathrm{profit}\mathrm{or}\mathrm{loss}\mathrm{in}\mathrm{the}\mathrm{following}\mathrm{transactions}.\\ \mathrm{Also}\mathrm{find}\mathrm{profit}\mathrm{per}\mathrm{cent}\mathrm{or}\mathrm{loss}\mathrm{per}\mathrm{cent}\mathrm{in}\mathrm{each}\mathrm{case}.\\ \left(\mathrm{a}\right)\mathrm{Gardening}\mathrm{shears}\mathrm{bought}\mathrm{for}₹250\mathrm{and}\mathrm{sold}\mathrm{for}₹325.\\ \left(\mathrm{b}\right)\mathrm{A}\mathrm{refrigerater}\mathrm{bought}\mathrm{for}₹12,000\mathrm{and}\mathrm{sold}\mathrm{at}₹13,500.\\ \left(\mathrm{c}\right)\mathrm{A}\mathrm{cupboard}\mathrm{bought}\mathrm{for}₹2,500\mathrm{and}\mathrm{sold}\mathrm{at}₹3,000.\\ \left(\mathrm{d}\right)\mathrm{A}\mathrm{skirt}\mathrm{bought}\mathrm{for}₹250\mathrm{and}\mathrm{sold}\mathrm{at}₹150.\end{array}$

Ans

\begin{array}{l}\text{(a) Cost price}=\text{₹ 250}\\ \text{Selling price}=\text{₹ 325}\\ \text{Profit}=\text{₹}\text{325}-\text{₹}\text{250}\\ =\text{₹}\text{75}\\ \text{Profit\%}=\frac{\text{Profit}}{\text{Cost Price}}\times 100\\ =\frac{75}{250}\times 100\\ =30\%\\ \text{(b) Cost price}=\text{₹ 12000}\\ \text{Selling price}=\text{₹ 13,500}\\ \text{Profit}=\text{₹}\text{13500}-\text{₹}\text{1200}\\ =\text{₹}\text{1500}\\ \text{Profit\%}=\frac{\text{Profit}}{\text{Cost Price}}\times 100\\ =\frac{1500}{12000}\times 100\\ =12.5\%\\ \text{(c) Cost price}=\text{₹ 2500}\\ \text{Selling price}=\text{₹ 3,000}\\ \text{Profit}=\text{₹}\text{3000}-\text{₹}\text{2500}\\ =\text{₹}\text{500}\\ \text{Profit\%}=\frac{\text{Profit}}{\text{Cost Price}}\times 100\\ =\frac{500}{2500}\times 100\\ =20\%\\ \text{(d) Cost price}=\text{₹ 250}\\ \text{Selling price}=\text{₹ 150}\\ \text{Loss}=\text{₹}\text{250}-\text{₹}1\text{50}\\ =\text{₹}\text{1000}\\ \text{Loss\%}=\frac{\text{Loss}}{\text{Cost Price}}\times 100\\ =\frac{100}{250}\times 100\\ =40\%\end{array}

Q.15

$\begin{array}{l}\mathrm{Convert}\mathrm{each}\mathrm{part}\mathrm{of}\mathrm{the}\mathrm{ratio}\mathrm{to}\mathrm{percentage}:\\ \left(\mathrm{a}\right)3:1\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\left(\mathrm{b}\right)2:3:5\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\left(\mathrm{c}\right)1:4\; \hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\left(\mathrm{d}\right)\mathrm{1:\; 2:5.}\end{array}$

Ans

\begin{array}{l}\left(\text{a}\right)\text{3}:\text{1}\\ \text{Here total parts}=3+1=4\\ 1^{\text{st}}\text{part}=\frac{3}{4}=\frac{3}{4}\times 100\%=75\%\\ 2^{\text{nd}}\text{part}=\frac{1}{4}=\frac{1}{4}\times 100\%=25\%\\ \left(\text{b}\right)\text{2}:\text{3}:\text{5}\\ \text{Here total parts}=2+3+5=10\\ 1^{\text{st}}\text{part}=\frac{2}{10}=\frac{2}{10}\times 100\%=20\%\\ 2^{\text{nd}}\text{part}=\frac{3}{10}=\frac{3}{10}\times 100\%=30\%\\ 3^{\text{rd}}\text{part}=\frac{5}{10}=\frac{5}{10}\times 100\%=50\%\\ \left(\text{c}\right)\text{1}:\text{4}\\ \text{Here total parts}=1+4=5\\ 1^{\text{st}}\text{part}=\frac{1}{5}=\frac{1}{5}\times 100\%=20\%\\ 2^{\text{nd}}\text{part}=\frac{4}{5}=\frac{4}{5}\times 100\%=80\%\\ \left(\text{d}\right)\text{1}:\text{2}:\text{5}\\ \text{Here total parts}=1+2+5=8\\ 1^{\text{st}}\text{part}=\frac{1}{8}=\frac{1}{8}\times 100\%=12.5\%\\ 2^{\text{nd}}\text{part}=\frac{2}{8}=\frac{2}{8}\times 100\%=25\%\\ 3^{\text{rd}}\text{part}=\frac{5}{8}=\frac{5}{8}\times 100\%=62.5\%\end{array}

Q.16

$\begin{array}{l}\mathrm{The}\mathrm{population}\mathrm{of}\mathrm{a}\mathrm{city}\mathrm{decreased}\mathrm{from}25,000\mathrm{to}24,500.\\ \mathrm{Find}\mathrm{the}\mathrm{percentage}\mathrm{decrease}.\end{array}$

Ans

\begin{array}{l}\text{Initial population}=2500\text{0}\\ \text{Final populaton}=24500\\ \text{Decrease}=25000-2\text{4500}\\ =500\\ \text{Decrease}\%=\frac{500}{25000}\times 100\\ =2\%\end{array}

Q.17

$\begin{array}{l}\mathrm{Arun}\mathrm{bought}\mathrm{a}\mathrm{car}\mathrm{for}₹3,50,000.\mathrm{The}\mathrm{next}\mathrm{year},\mathrm{the}\mathrm{price}\\ \mathrm{went}\mathrm{up}\mathrm{to}₹3,70,000.\mathrm{What}\mathrm{was}\mathrm{the}\mathrm{Percentage}\mathrm{of}\mathrm{price}\\ \mathrm{increase}?\end{array}$

Ans

\begin{array}{l}\text{Initial Price}=\text{₹ 3,50,000}\\ \text{Final Price}=\text{₹ 3,70,000}\\ \text{Increase}=\text{₹}\text{3,70,000}-\text{₹}\text{3,50,000}\\ =\text{₹}\text{20,000}\\ \text{Increase\%}=\frac{20,000}{3,50,000}\times 100\\ =5\frac{5}{7}\%\end{array}

Q.18

$\begin{array}{l}\mathrm{I}\mathrm{buy}\mathrm{a}\mathrm{T}.\mathrm{V}.\mathrm{for}₹\mathrm{10,000}\mathrm{and}\mathrm{sell}\mathrm{it}\mathrm{at}\mathrm{a}\mathrm{profit}\mathrm{of}20\%.\mathrm{How}\\ \mathrm{much}\mathrm{money}\mathrm{do}\mathrm{I}\mathrm{get}\mathrm{for}\mathrm{it}?\end{array}$

Ans

\begin{array}{l}\text{We know that}\\ \text{Profit\%}=\frac{\text{Profit}}{\text{Cost price}}\times 100\\ \text{So,}\\ 20=\frac{\text{Profit}}{10,000}\times 100\\ =\frac{\text{Profit}}{100\times \cancel{100}}\times \cancel{100}\\ \text{Profit}=\text{20}\times \text{100}\\ =\text{₹}\text{2000}\\ \text{Now,}\\ \text{Profit}=\text{Selling price}-\text{Cost price}\\ \text{₹}\text{2000}=\text{Selling price}-\text{₹}\text{10000}\\ \text{Selling price}=\text{₹}\text{10000}+\text{₹}\text{2000}\\ =\text{₹}\text{12000}\end{array}

Q.19

$\begin{array}{l}\mathrm{Juhi}\mathrm{sells}\mathrm{a}\mathrm{washing}\mathrm{machine}\mathrm{for}₹13,500.\mathrm{She}\mathrm{loses}20\%\mathrm{in}\\ \mathrm{the}\mathrm{bargain}.\mathrm{What}\mathrm{was}\mathrm{the}\mathrm{price}\mathrm{at}\mathrm{which}\mathrm{she}\mathrm{bought}\mathrm{it}?\end{array}$

Ans

\begin{array}{l}\text{Selling price}=\text{₹ 135}00\\ \text{Loss}\%=\text{2}0\%\\ \text{Let the cost price be}x.\\ \therefore \text{Loss}=\text{2}0\%\text{of}x\\ \text{Cost price}-\text{Loss}=\text{Selling price}\\ \text{x}-20\%\text{of x}=\text{₹}\text{13500}\\ \text{x}-\frac{20}{100}\times x=13500\\ \frac{100x-20x}{100}=13500\\ \frac{80x}{100}=13500\\ x=\frac{13500\times 100}{80}\\ =\text{₹}16875\end{array}

Q.20

$\begin{array}{l}\left(\mathrm{i}\right)\mathrm{Chalk}\mathrm{contains}\mathrm{calcium},\mathrm{carbon}\mathrm{and}\mathrm{oxygen}\mathrm{in}\mathrm{the}\mathrm{ratio}\\ 10:3:12.\mathrm{Find}\mathrm{the}\mathrm{percentage}\mathrm{of}\mathrm{carbon}\mathrm{in}\mathrm{chalk}.\\ \left(\mathrm{ii}\right)\mathrm{If}\mathrm{in}\mathrm{a}\mathrm{stick}\mathrm{of}\mathrm{chalk},\mathrm{carbon}\mathrm{is}3\mathrm{g},\mathrm{what}\mathrm{is}\mathrm{the}\mathrm{weight}\\ \mathrm{of}\mathrm{the}\mathrm{chalk}\mathrm{stick}.\end{array}$

Ans

\begin{array}{l}\text{(i) Given ratio}=\text{10:3:12}\\ \text{Total}=\text{10}+\text{3}+\text{12}\\ =\text{25}\\ \text{Percentage of Carbon}=\frac{3}{25}\times 100\\ =12\%\\ \text{(ii) Let the weight of the stick be x g}\\ \text{So, 12\% of x}=\text{3}\\ \frac{12}{100}\times x=3\\ x=\frac{300}{12}\\ =25\text{g}\end{array}

Q.21 Amina buys a book for ₹ 275 and sells it at a loss of 15%. How much does she sell it for?

Ans

\begin{array}{l}\text{Cost price of book}=\text{₹ 275}\\ \text{Loss\%}=\text{15}\\ \text{We have,}\\ \text{Loss\%}=\frac{\text{Loss}}{\text{Cost price}}\times 100\\ 15=\frac{\text{Loss}}{275}\times 100\\ \text{Loss}=\frac{15\times 275}{100}\\ =41.25\\ \text{Selling price}=\text{Cost price}-\text{Loss}\\ =\text{₹}\text{275}-\text{₹}\text{41}\text{.25}\\ =\text{₹ 233}\text{.75}\end{array}

Q.22 Find the amount to be paid at the end of 3 years in each case:
(a) Principal = ₹1,200 at 12% p.a.
(b) Principal = ₹7,500 at 5% p.a.

Ans

\begin{array}{l}\text{(a)}\\ \text{Principal (P)}=\text{₹ 1200}\\ \text{Rate (R)}=\text{12\%}\\ \text{Time (T)}=3\text{years}\\ \text{S}\text{.I}\text{.}=\frac{\text{P}\times \text{R}\times \text{T}}{100}\\ =\frac{1200\times 12\times 3}{100}\\ =\text{₹}432\\ \text{Amount}=\text{P}+\text{S}\text{.I}\text{.}\\ =\text{₹}\text{1200}+\text{₹}\text{432}=\text{₹ 1632}\\ \text{(b)}\\ \text{Principal (P)}=\text{₹ 7500}\\ \text{Rate (R)}=\text{5\%}\\ \text{Time (T)}=\text{3 years}\\ \text{S}\text{.I}\text{.}=\frac{\text{P}\times \text{R}\times \text{T}}{100}\\ =\frac{7500\times 5\times 3}{100}\\ =\text{₹}1125\\ \text{Amount}=\text{P}+\text{S}\text{.I}\text{.}\\ =\text{₹}\text{7500}+\text{₹}\text{1125}\\ =\text{₹ 8625}\end{array}

Q.23 What rate gives ₹280 as interest on a sum of ₹56000 in 2 years?

Ans

$\begin{array}{l}\mathrm{We}\mathrm{know}\mathrm{that}\\ \mathrm{S}.\mathrm{I}=\frac{\mathrm{P}\times \mathrm{R}\times \mathrm{T}}{100}\\ 280=\frac{\mathrm{56000}\times \mathrm{R}\times 2}{100}\\ \mathrm{R}=\frac{280\times 100}{56000\times 2}\\ =0.25\\ \mathrm{Therefore},\mathrm{the}\mathrm{rate}\mathrm{is}0.25\%.\end{array}$

Q.24

$\begin{array}{l}\mathrm{If}\mathrm{Meena}\mathrm{gives}\mathrm{an}\mathrm{interest}\mathrm{of}₹45\mathrm{for}\mathrm{one}\mathrm{year}\mathrm{at}9\%\\ \mathrm{rate}\mathrm{p}.\mathrm{a}.\mathrm{What}\mathrm{is}\mathrm{the}\mathrm{sum}\mathrm{she}\mathrm{has}\mathrm{borrowed}?\end{array}$

Ans

\begin{array}{l}\text{We know that}\\ \text{S}\text{.I}=\frac{\text{P}\times \text{R}\times \text{T}}{100}\\ \text{So, we get}\\ \text{45=}\frac{\text{P×9×1}}{\text{100}}\\ \text{P=}\frac{\text{45×100}}{\text{9}}\\ \text{=₹}\text{500}\\ \text{Therefore, she borrowed ₹ 500}\text{.}\end{array}