NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities (EX 8.3) Exercise 8.3

Q.1

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

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.2

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

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.3

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

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.4

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

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.5

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

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.6

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

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.7

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

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.8 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.9 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.10 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.11

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

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}