NCERT Solutions for Class 8 Maths Chapter 7 Comparing Quantities

Struggling with a chapter comparing quantities in Class 8 Maths? Look no further! There are so many things that need to be compared every day; hence, it is one of the key historical events that led to the existence of mathematics.

ALLEN's comprehensive guide, packed with NCERT solutions for Chapter 8 Maths , offers step-by-step explanations of problems to solidify your understanding. Master concepts like ratios, proportions, percentages, profit, loss, discount, simple interest, and compound interest. Boost your marks and build a strong foundation in mathematics with our expert-crafted resources.

1.0NCERT Solutions Class 8 Chapter 7 - Comparing Quantities Free PDF Download

Students can download NCERT Solutions Class 8 Mathematics Chapter 7 from the below table in PDF format. 

Key Concepts in Class 8 Maths Chapter 7: Comparing Quantities

This chapter covers several concepts designed to help students understand real-life applications of mathematics. Some of the main topics include:

• Ratios and Proportions: Discuss the concept of how to compare two quantities using ratios and proportions.
• Percentage: Students need to be able to express amounts as percentages and to convert between percentages, fractions, and decimals.
• Profit and Loss: Understand the computation of profit and loss in solving word problems.
• Simple and Compound Interest: Some important concepts of simple interest and compound interest are learned by step-by-step instructions to calculate the same.

2.0NCERT Questions with Solutions for Class 8 Maths Chapter 7 - Detailed Solutions

Exercise : 7.1

• Find the ratio of the following : (i) Speed of a cycle is 15 km per hour to the speed of scooter 30 km per hour. (ii) 5 m to 10 km (iii) 50 paise to ₹ 5 Sol. (i) Speed of cycle which is $15\mathrm{km}/\mathrm{hr}$ to the speed of scooter which is $30\mathrm{km}/\mathrm{hr}=15:30=1:2$. (ii) The ratio is $5\mathrm{m}:10\mathrm{km}$ or $5\mathrm{m}:10\times 10^3\mathrm{m}\left\{\because 1\mathrm{km}=10^3\mathrm{m}\right\}$ or $1:2000$. (iii) The ratio is 50 paise : ₹5 or, 50 paise : 500 paise $\{\because ₹1=100\text{ paise }\}$ or, $1:10$.
• Convert the following ratios to percentages : (i) $3:4$ (ii) $2:3$ Sol. (i) $3:4=\frac{3}{4}\times \frac{100}{100}=\frac{3}{1}\times \frac{25}{100}=\frac{75}{100}=75\%$ (ii) Ratio $=2:3$ and percentage $=\frac{2}{3}\times 100=66\frac{2}{3}\%$
• $72\%$ of 25 students are interested in mathematics. How many are not interested in mathematics? Mathematics. $\therefore$ Percentage of students not interested in Mathematics $=100\%-72\%=28\%$ Also, total number of students $=25$ Hence, number of students not interested in mathematics $=\frac{28}{100}\times 25=7$ students
• A football team won 10 matches out of the total number of matches they played. If their winning percentage was 40, then how many matches did they play in all? Sol. Let x games be played in all. Since $40\%$ of the total games is given as 10 , $\therefore 40\%$ of $x=10$ $\Rightarrow \frac{40}{100}\times \mathrm{x}=10\Rightarrow \mathrm{x}=\frac{100}{40}\times 10=25$ Hence, the total games played were 25.
• If Chameli had ₹ 600 left after spending $75\%$ of her money, how much did she have in the beginning? Sol. We know, money left with Chameli $=₹600$ $\%$ of money she spent $=75\%$ $\therefore \%$ of money left $=100\%-75\%=25\%$ Let total money in the beginning be x $\therefore 25\%$ of $\mathrm{x}=\mathrm{F}600$ or, $\frac{25}{100}\times \mathrm{x}=₹600$ or, $\mathrm{x}=₹2400$
• 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 the other games? If the total number of people are 50 lakh, find the exact number of people who like each type of game. Sol. We have, People who like cricket $=60\%$ People who like football $=30\%$ $\therefore$ People who like other games $=(100-60-30)\%=10\%$ Total number of people $=5000000$ $\therefore$ People who like cricket $=\frac{60}{100}\times 5000000=3000000=30$ lakh People who like football $=\frac{30}{100}\times 5000000=1500000=15$ lakh and, the people who like other games $=\frac{10}{100}\times 5000000=500000=5$ lakh

Exercise: 7.2

• 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? Sol. We have, Cost of pair of jeans = ₹ 1450 Cost of two shirts $=2\times 850=₹1700$ $\therefore$ Total cost $=1450+1700=₹3150$ Discount = $10\%$ Hence, selling price $=3150\left(1-\frac{10}{100}\right)$ = ₹ 2835
• The price of a TV is ₹ 13000 . 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. Sol. Price of TV = ₹ 13000 Sales tax = 12% $\therefore$ Amount to be paid $=$ Price of T.V. $\left(1+\frac{12}{100}\right)$ $=13000\left(1+\frac{12}{100}\right)=₹14560$
• Arun bought a pair of skates in a sale where the discount given was $20\%$. If the amount he paid was $₹1600$, find the marked price. Sol. Let the marked price of a pair of skates be ₹ 100 . Discount $=20\%$ of M.P. $=20\%$ of $₹100=₹20$ $\therefore$ S.P. $=$ M.P. - Discount $=₹(100-20)$ = ₹ 80 Now, when S.P. is ₹ 80, M.P. $=₹100$ When S.P. is Re 1, M.P. $=\mathrm{Rs}\frac{100}{80}$ When S.P. is ₹ 1600 , M.P. $=\mathrm{Rs}\left(\frac{100}{80}\times 1600\right)=₹2000$ Hence, the marked price of the pair of skates is ₹ 2000 .
• I purchased a hair dryer for ₹ 5400 including 8% VAT. Find the price before VAT was added. Sol. Let the price before VAT be ₹x Price after including VAT $=₹5400$ $x+8\%$ of $x=5400$ $x+\frac{8x}{100}=5400$ $\frac{108\mathrm{x}}{100}=5400$ $x=\frac{5400\times 100}{108}=₹5000$
• An article was purchased for ₹ 1239 including GST of $18\%$. Find the price of the article before GST was added? Sol. Price of an article with GST = Rs 1239 GST% = 18% Let the initial price of the article be Rs x GST $=18\%$ of the initial price of article So, the initial price of article + GST $=1239$ $x+18\%\times x=1239$ $x+\left(\frac{18x}{100}\right)=1239$ $\left(\frac{118x}{100}\right)=1239$ $x=1239\times \left(\frac{100}{118}\right)$ $\mathrm{x}=$ Rs 1050
• A man got a $10\%$ increase in his salary. If his new salary is $₹1,54,000$, find his original salary. Sol. Let the salary before increment be ₹ 100 . Then, increase in salary $=₹10$ $\therefore$ Increased salary $=₹(100+10)=₹110$ If increased salary is $₹110$, original salary = ₹ 100 If increased salary is ₹ 154000 , original salary $=₹\left(\frac{100}{110}\times 154000\right)=₹140000$ Hence, the salary of the man before increment was ₹ 140000 .
• On Sunday 845 people went to the Zoo. On Monday only 169 people went. What is the percent decrease in the people visiting the Zoo on Monday? Sol. People went to zoo on Sunday $=845$ People went to zoo on Monday $=169$ Decrease in people $=845-169=676$ $\therefore \%$ decrease $=\frac{676}{845}\times 100=80\%$
• A shopkeeper buys 80 articles for ₹ 2400 and sells them for a profit of $16\%$. Find the selling price of one article. Sol. C.P. of 80 articles $=₹2400$ Profit $=16\%$ S.P. of 80 articles $=\left(\frac{100+\text{ Profit }\%}{100}\right)\times \mathrm{CP}$ $=₹\left(\frac{100+16}{100}\times 2400\right)=₹(116\times 24)$ = ₹ 2784 $\therefore$ S.P. of one article $=₹\left(\frac{2784}{80}\right)$ = ₹ 34.80
• The cost of an article is ₹ 15,500 , ₹ 450 were spent on its repairs. If it is sold for a profit of $15\%$, find the selling price of the article. Sol. Cost of article $=₹15500$ Repairs $=₹450$ $\therefore$ Total cost $=$ Cost + Repairs $=₹15950$ Profit = 15% $\therefore$ Selling price $=\frac{15950(100+15)}{100}$ $=15950\left(1+\frac{15}{100}\right)$ = ₹ 18342.50
• A VCR and a TV were bought for ₹ 8,000 each. The shopkeeper incurred a loss of $4\%$ on the VCR and earned a profit of $8\%$ on the TV. Find the gain or loss percent on the whole transaction. Sol. For VCR, we have C.P. $=₹8000$ and loss $=4\%$ $\therefore$ S.P. $=\left(\frac{100-\text{ Loss }\%}{100}\right)\times$ C.P. $=₹\left(\frac{100-4}{100}\times 8000\right)$ $=₹(96\times 80)=₹7680$ For TV, we have C.P. $=₹8000$ and profit $=8\%$ $\therefore$ S.P. $=\left(\frac{100+\text{ Gain }\%}{100}\right)\times$ C.P. $=₹\left(\frac{100+8}{100}\times 8000\right)$ $=₹(108\times 80)=₹8640$ $\therefore$ Total S.P. $=₹(7680+8640)$ = ₹ 16320 Total C.P. $=₹(8000+8000)=₹16000$ $\because$ S.P. > C.P. So, there is profit. $\therefore$ Total profit $=$ S.P. - C.P. $=₹(16320-16000)=₹320$ Hence, profit percent $=\left(\frac{\text{ Profit }}{\mathrm{CP}}\times 100\right)\%=\left(\frac{320}{16000}\times 100\right)\%$ $=2\%$
• 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. Sol. For the first buffalo, we have S.P. $=₹20000$ and gain $=5\%$ $\therefore$ C.P. $=\frac{100}{100+\text{ Gain }\%}\times$ SP $\Rightarrow$ C.P. $=₹\left(\frac{100}{100+5}\times 20000\right)$ $=\left(\frac{100}{105}\times 20000\right)=₹\frac{2000000}{105}$ = ₹ 19047.62 For the second buffalo, we have S.P. $=₹20000$ and loss $=10\%$ $\therefore$ C.P. $=\frac{100}{100-\text{ Loss }\%}\times$ SP $=₹\left(\frac{100}{100-10}\times 20000\right)=₹\left(\frac{100}{90}\times 20000\right)$ $=\frac{2000000}{90}=₹22222.22$ $\therefore$ Total cost paid in buying the two buffaloes $=₹\left(\frac{2000000}{105}+\frac{2000000}{90}\right)$ $=₹2000000\times \frac{90+105}{90\times 105}$ $=\left(2000000\times \frac{195}{9450}\right)$ = ₹ 41269.84 Total S.P. $=₹(20000+20000)=₹40000$ $\therefore$ S.P. < C.P. So, there is a loss. $\therefore$ Total loss = C.P. - S.P. $=₹(41269.84-40000)=₹1269.84$

Exercise : 7.3

• The population of a place increased to 54000 in 2003 at a rate of $5\%$ per annum (i) Find the population in 2001. (ii) What would be its population in 2005? Sol. (i) Let $P$ be the population in 2001 , i.e., 2 years ago Then, Present population $=\mathrm{P}\times {\left(1+\frac{5}{100}\right)}^2$ $\Rightarrow 54000=\mathrm{P}\times \frac{105}{100}\times \frac{105}{100}$ $\Rightarrow \mathrm{P}=\frac{54000\times 100\times 100}{105\times 105}=48979.59$ Hence, the population in the year 2001 is 48980 (approx). (ii) Let $\mathrm{P}=$ Initial population $=54000$, i.e., in the year 2003. $\therefore$ Population after 2 years, i.e., in 2005 $=\mathrm{P}{\left(1+\frac{\mathrm{R}}{100}\right)}^{\mathrm{n}}=54000\times {\left(1+\frac{5}{100}\right)}^2$ $=54000\times \frac{105}{100}\times \frac{105}{100}=59535$
• 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. Sol. We have, $\mathrm{P}=$ Original count of bacteria $=$ 506000, Rate of increase $=\mathrm{R}=2.5\%$ per hour, Time $=2$ hours. $\therefore$ Bacteria count after 2 hours $=506000\times {\left(1+\frac{2.5}{100}\right)}^2$ $=506000\times \frac{102.5}{100}\times \frac{102.5}{100}$ = 531616.25 = 531616 (approx.)
• A scooter was bought at $₹42000$. Its value depreciated at the rate of $8\%$ per annum. find its value after one year. Sol. We have, ${\mathrm{V}}_0=$ Initial value $=₹42000$ $R=$ Rate of depreciation $=8\%$ p.a. $\therefore$ Value after 1 year $={\mathrm{V}}_0\left(1-\frac{\mathrm{R}}{100}\right)$ $=₹\left[42000\times \left(1-\frac{8}{100}\right)\right]$ $=₹\left(42000\times \frac{92}{100}\right)=₹38640$
• Calculate the amount and compound interest on (i) ₹ 10,800 for 3 years at $12\frac{1}{2}\%$ per annum compounded annually. (ii) ₹ 18,000 for $2\frac{1}{2}$ years at $10\%$ per annum compounded annually. (iii) ₹ 62,500 for $1\frac{1}{2}$ years at $8\%$ per annum compounded half yearly. (iv) ₹ 8,000 for 1 year at $9\%$ per annum compounded half yearly. (You could use the year by year calculation using S.I. formula to verify). (v) ₹ 10,000 for 1 year at $8\%$ per annum compounded half yearly. Sol. (i) We have, $\mathrm{P}=₹10,800$ $\mathrm{R}=12\frac{1}{2}\%$ p.a. $=\left(\frac{25}{2}\right)\%$ per annum and $\mathrm{n}=3$ years. $\therefore$ Amount after 3 years $=₹\left[10800\times {\left(1+\frac{25}{2\times 100}\right)}^3\right]$ $=₹\left(10800\times \frac{9}{8}\times \frac{9}{8}\times \frac{9}{8}\right)$ = ₹ 15377.34 $\therefore$ C.I. $=₹(15377.34-10800)$ $\text{ = ₹ }4577.34$ (ii) We have, $\mathrm{P}=₹18,000$ $\mathrm{R}=10\%$ per annum $\mathrm{n}=2.5$ years or $2\frac{1}{2}$ years $\therefore$ Amount $\mathrm{A}=\mathrm{P}{\left(1+\frac{\mathrm{R}}{100}\right)}^{\mathrm{n}}$ At the end of 2 years, $A=18000{\left(1+\frac{10}{100}\right)}^2=₹21780$ $\therefore$ Amount after 2.5 years $=21780\left(1+\frac{10\times \frac{1}{2}}{100}\right)=₹22869$ $\therefore$ Compound interest $=\mathrm{A}-\mathrm{P}$ $=22869-18000=₹4869$ (iii) Here, Principal = ₹ 62,500 Rate $=8\%$ per annum $=4\%$ per half year, Time $=1\frac{1}{2}$ years $=3$ half years $\therefore$ Amount $=₹\left[62500\times {\left(1+\frac{4}{100}\right)}^3\right]$ $=\left(62500\times \frac{104}{100}\times \frac{104}{100}\times \frac{104}{100}\right)$ = ₹ 70304 $\therefore$ C. $I=₹(70304-62500)$ = ₹ 7804 (iv) We have, $\mathrm{P}=₹8000$ $\mathrm{R}=9$ % per annum $=4.5$ % per half year $\mathrm{n}=1$ year = 2 half years So, amount $A=P{\left(1+\frac{R}{100}\right)}^n$ $=8000{\left(1+\frac{4.5}{100}\right)}^2=₹8736.20$ Hence, C.I. $=\mathrm{A}-\mathrm{P}=₹736.20$ (v) We have, $\mathrm{P}=₹10000$ $\mathrm{R}=8\%$ per annum $=4\%$ per half year $\mathrm{n}=1$ year $=2$ half years . Thus, amount $A=P{\left(1+\frac{R}{100}\right)}^n$ $=10000{\left(1+\frac{4}{100}\right)}^2=₹10816$ $\therefore$ Compound interest $=\mathrm{A}-\mathrm{P}=₹816$
• Kamla borrowed ₹ 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? Sol. We have, $\mathrm{P}=₹26400$ R $=15\%$ per annum $\mathrm{n}=2$ years 4 months At the end of 2 years, $A=P{\left(1+\frac{R}{100}\right)}^n$ $A=26400{\left(1+\frac{15}{100}\right)}^2=₹34914$ Now, P = ₹ 34914 $\mathrm{R}=15\%$ per annum $\mathrm{n}=4$ months $=\frac{1}{3}$ year At the end of 2 years, 4 months $A=\left(1+\frac{15\times 1}{100\times 3}\right)34914=₹36659.70$
• Fabina borrows ₹ 12500 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? Sol. In case of Fabina: $\mathrm{P}=$ Rs $12500,\mathrm{R}=12\%$ per annum and $\mathrm{T}=$ 3 years. Then, S.I. $=\frac{\mathrm{P}\times \mathrm{R}\times \mathrm{T}}{100}=₹\left(\frac{12500\times 12\times 3}{100}\right)$ = ₹ 4500 In case of Radha : Amount $=₹\left[12500\times {\left(1+\frac{10}{100}\right)}^3\right]$ $=₹\left(12500\times \frac{110}{100}\times \frac{110}{100}\times \frac{110}{100}\right)$ = ₹ 16637.50 $\therefore$ Compound interest $=₹$ (16637.50 12500) $=₹4137.50$ $₹(4500-4137.50)=₹362.50$ Hence, Fabina pays 362.50 more as interest as compared to Radha.
• I borrowed ₹ 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? Sol. We have, $\mathrm{P}=₹12000$ $\mathrm{R}=6\%$ per annum $\mathrm{n}=2$ years For simple interest, $I_1=\frac{P\times R\times T}{100}$ $=\frac{12000\times 6\times 2}{100}=₹1440$ For compound interest, I2 $=\mathrm{P}\left[{\left(1+\frac{\mathrm{R}}{100}\right)}^{\mathrm{n}}-1\right]$ $=12000\left[{\left(1+\frac{6}{100}\right)}^2-1\right]=₹1483.20$ So, he would have to pay $1483.20-1440$ = ₹ 43.20
• 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. Sol. Here, Principal $=₹60000$, Rate $=12\%$ per annum $=6\%$ per half year. (i) Time $=6$ months $=1$ half-year $\therefore$ Amount after 6 months = Amount after 1 half-year $$\begin{gathered} =₹\left[60000\times {\left(1+\frac{6}{100}\right)}^1\right] \\ =₹\left(60000\times \frac{106}{100}\right)=₹63600 \end{gathered}$$ (ii) Time $=1$ year $=2$ half-years $\therefore$ Amount after 1 year $$\begin{gathered} =₹\left[60000\times {\left(1+\frac{6}{100}\right)}^2\right] \\ =₹\left(60000\times \frac{106}{100}\times \frac{106}{100}\right)=₹67416 \end{gathered}$$
• 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 Sol. We have, $\mathrm{P}=₹80000$ $\mathrm{R}=10\%$ per annum $=5\%$ per half year (i) If interest is compounded annually $\mathrm{n}=1\frac{1}{2}$ years $=1$ years $+\frac{1}{2}$ year Amount after 1 year $A=80000\left(1+\frac{10}{100}\right)=₹88000$ Amount after $1\frac{1}{2}$ years $A=88000\left(1+\frac{10\times \frac{1}{2}}{100}\right)=₹92400$ (ii) If interest is compounded half-yearly $\mathrm{n}=1\frac{1}{2}$ years $=3$ half years $\therefore$ Amount $\mathrm{A}=\mathrm{P}{\left(1+\frac{\mathrm{R}}{100}\right)}^{\mathrm{n}}$ $=80000{\left(1+\frac{5}{100}\right)}^3=₹92610$ $\therefore$ Difference in amounts $=92610-92400=₹210$
• 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. Sol. (i) Here, $P=₹8000,R=5\%$ per annum and $\mathrm{n}=2$ years. $\therefore$ Amount after 2 years $=\mathrm{P}{\left(1+\frac{\mathrm{R}}{100}\right)}^{\mathrm{n}}$ $$\begin{gathered} =₹\left[8000\times {\left(1+\frac{5}{100}\right)}^2\right] \\ =₹\left(8000\times \frac{105}{100}\times \frac{105}{100}\right) \\ =₹8820 \end{gathered}$$ (ii) Principal for the 3rd year = ₹ 8820 Interest for the 3rd year $$\begin{gathered} =\left(\frac{8820\times 5\times 1}{100}\right) \\ =₹441 \end{gathered}$$
• Find the amount and compound interest on ₹ 10000 for $1\frac{1}{2}$ years at $10\%$ per annum, compounded half yearly. Would this interest be more than the interest he would get if it was compounded annually? Sol. We have, $\mathrm{P}=₹10000$ $\mathrm{R}=10\%$ per annum $=5\%$ per half year $\mathrm{n}=1\frac{1}{2}$ year $=3$ half years If interest is compounded half yearly $A=10000{\left(1+\frac{5}{100}\right)}^3=₹11576.25$ Interest $=11576.25-10000$ = ₹ 1576.25 If interest is compounded annually. Amount after 1 year, $A=10000\left(1+\frac{10}{100}\right)=₹11000$ Amount after $1\frac{1}{2}$ year $A=11000\left(1+\frac{10\times 1}{100\times 2}\right)=₹11550$ Interest $=11550-10000=₹1550$ Thus, more interest would be generated if interest is calculated half yearly.
• Find the amount which Ram will get on ₹ 4096, if he gave it for 18 months at $12\frac{1}{2}\%$ per annum, interest being compounded half yearly. Sol. Here, Principal = ₹ 4096, Time $=18$ months $=3$ half years Rate $=12\frac{1}{2}\%$ per annum $=\left(\frac{25}{4}\right)\%$ per half year. $\therefore$ Amount $=\left[4096\times {\left(1+\frac{25}{4\times 100}\right)}^3\right]$ $=\left(4096\times \frac{17}{16}\times \frac{17}{16}\times \frac{17}{16}\right)$ = ₹ 4913