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NCERT Solutions
Class 8
Maths
Chapter 8 Comparing Quantities

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. 

NCERT Solutions for Class 8 Maths Chapter 7 Comparing Quantities

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 km/hr to the speed of scooter which is 30 km/hr=15:30=1:2. (ii) The ratio is 5 m:10 km or 5 m:10×103 m{∵1 km=103 m} or 1:2000. (iii) The ratio is 50 paise : ₹5 or, 50 paise : 500 paise {∵₹1=100 paise } or, 1:10.
  • Convert the following ratios to percentages : (i) 3:4 (ii) 2:3 Sol. (i) 3:4=43​×100100​=13​×10025​=10075​=75% (ii) Ratio =2:3 and percentage =32​×100=6632​%
  • 72% of 25 students are interested in mathematics. How many are not interested in mathematics? Mathematics. ∴ Percentage of students not interested in Mathematics =100%−72%=28% Also, total number of students =25 Hence, number of students not interested in mathematics =10028​×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 , ∴40% of x=10 ⇒10040​×x=10⇒x=40100​×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% ∴% of money left =100%−75%=25% Let total money in the beginning be x ∴25% of x=F600 or, 10025​×x=₹600 or, 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% ∴ People who like other games =(100−60−30)%=10% Total number of people =5000000 ∴ People who like cricket =10060​×5000000=3000000=30 lakh People who like football =10030​×5000000=1500000=15 lakh and, the people who like other games =10010​×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 =2×850=₹1700 ∴ Total cost =1450+1700=₹3150 Discount = 10% Hence, selling price =3150(1−10010​) = ₹ 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% ∴ Amount to be paid = Price of T.V. (1+10012​) =13000(1+10012​)=₹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 ∴ 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. =Rs80100​ When S.P. is ₹ 1600 , M.P. =Rs(80100​×1600)=₹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+1008x​=5400 100108x​=5400 x=1085400×100​=₹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%×x=1239 x+(10018x​)=1239 (100118x​)=1239 x=1239×(118100​) 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 ∴ Increased salary =₹(100+10)=₹110 If increased salary is ₹110, original salary = ₹ 100 If increased salary is ₹ 154000 , original salary =₹(110100​×154000)=₹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 ∴% decrease =845676​×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 =(100100+ Profit %​)×CP =₹(100100+16​×2400)=₹(116×24) = ₹ 2784 ∴ S.P. of one article =₹(802784​) = ₹ 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 ∴ Total cost = Cost + Repairs =₹15950 Profit = 15% ∴ Selling price =10015950(100+15)​ =15950(1+10015​) = ₹ 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% ∴ S.P. =(100100− Loss %​)× C.P. =₹(100100−4​×8000) =₹(96×80)=₹7680 For TV, we have C.P. =₹8000 and profit =8% ∴ S.P. =(100100+ Gain %​)× C.P. =₹(100100+8​×8000) =₹(108×80)=₹8640 ∴ Total S.P. =₹(7680+8640) = ₹ 16320 Total C.P. =₹(8000+8000)=₹16000 ∵ S.P. > C.P. So, there is profit. ∴ Total profit = S.P. - C.P. =₹(16320−16000)=₹320 Hence, profit percent =(CP Profit ​×100)%=(16000320​×100)% =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% ∴ C.P. =100+ Gain %100​× SP ⇒ C.P. =₹(100+5100​×20000) =(105100​×20000)=₹1052000000​ = ₹ 19047.62 For the second buffalo, we have S.P. =₹20000 and loss =10% ∴ C.P. =100− Loss %100​× SP =₹(100−10100​×20000)=₹(90100​×20000) =902000000​=₹22222.22 ∴ Total cost paid in buying the two buffaloes =₹(1052000000​+902000000​) =₹2000000×90×10590+105​ =(2000000×9450195​) = ₹ 41269.84 Total S.P. =₹(20000+20000)=₹40000 ∴ S.P. < C.P. So, there is a loss. ∴ 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 =P×(1+1005​)2 ⇒54000=P×100105​×100105​ ⇒P=105×10554000×100×100​=48979.59 Hence, the population in the year 2001 is 48980 (approx). (ii) Let P= Initial population =54000, i.e., in the year 2003. ∴ Population after 2 years, i.e., in 2005 =P(1+100R​)n=54000×(1+1005​)2 =54000×100105​×100105​=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, P= Original count of bacteria = 506000, Rate of increase =R=2.5% per hour, Time =2 hours. ∴ Bacteria count after 2 hours =506000×(1+1002.5​)2 =506000×100102.5​×100102.5​ = 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, V0​= Initial value =₹42000 R= Rate of depreciation =8% p.a. ∴ Value after 1 year =V0​(1−100R​) =₹[42000×(1−1008​)] =₹(42000×10092​)=₹38640
  • Calculate the amount and compound interest on (i) ₹ 10,800 for 3 years at 1221​% per annum compounded annually. (ii) ₹ 18,000 for 221​ years at 10% per annum compounded annually. (iii) ₹ 62,500 for 121​ 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, P=₹10,800 R=1221​% p.a. =(225​)% per annum and n=3 years. ∴ Amount after 3 years =₹[10800×(1+2×10025​)3] =₹(10800×89​×89​×89​) = ₹ 15377.34 ∴ C.I. =₹(15377.34−10800)  = ₹ 4577.34 (ii) We have, P=₹18,000 R=10% per annum n=2.5 years or 221​ years ∴ Amount A=P(1+100R​)n At the end of 2 years, A=18000(1+10010​)2=₹21780 ∴ Amount after 2.5 years =21780(1+10010×21​​)=₹22869 ∴ Compound interest =A−P =22869−18000=₹4869 (iii) Here, Principal = ₹ 62,500 Rate =8% per annum =4% per half year, Time =121​ years =3 half years ∴ Amount =₹[62500×(1+1004​)3] =(62500×100104​×100104​×100104​) = ₹ 70304 ∴ C. I=₹(70304−62500) = ₹ 7804 (iv) We have, P=₹8000 R=9 % per annum =4.5 % per half year n=1 year = 2 half years So, amount A=P(1+100R​)n =8000(1+1004.5​)2=₹8736.20 Hence, C.I. =A−P=₹736.20 (v) We have, P=₹10000 R=8% per annum =4% per half year n=1 year =2 half years . Thus, amount A=P(1+100R​)n =10000(1+1004​)2=₹10816 ∴ Compound interest =A−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, P=₹26400 R =15% per annum n=2 years 4 months At the end of 2 years, A=P(1+100R​)n A=26400(1+10015​)2=₹34914 Now, P = ₹ 34914 R=15% per annum n=4 months =31​ year At the end of 2 years, 4 months A=(1+100×315×1​)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: P= Rs 12500,R=12% per annum and T= 3 years. Then, S.I. =100P×R×T​=₹(10012500×12×3​) = ₹ 4500 In case of Radha : Amount =₹[12500×(1+10010​)3] =₹(12500×100110​×100110​×100110​) = ₹ 16637.50 ∴ 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, P=₹12000 R=6% per annum n=2 years For simple interest, I1​=100P×R×T​ =10012000×6×2​=₹1440 For compound interest, I2 =P[(1+100R​)n−1] =12000[(1+1006​)2−1]=₹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 ∴ Amount after 6 months = Amount after 1 half-year =₹[60000×(1+1006​)1]=₹(60000×100106​)=₹63600 (ii) Time =1 year =2 half-years ∴ Amount after 1 year =₹[60000×(1+1006​)2]=₹(60000×100106​×100106​)=₹67416
  • 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 121​ years if the interest is - (i) Compounded annually (ii) Compounded half yearly Sol. We have, P=₹80000 R=10% per annum =5% per half year (i) If interest is compounded annually n=121​ years =1 years +21​ year Amount after 1 year A=80000(1+10010​)=₹88000 Amount after 121​ years A=88000(1+10010×21​​)=₹92400 (ii) If interest is compounded half-yearly n=121​ years =3 half years ∴ Amount A=P(1+100R​)n =80000(1+1005​)3=₹92610 ∴ 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 n=2 years. ∴ Amount after 2 years =P(1+100R​)n =₹[8000×(1+1005​)2]=₹(8000×100105​×100105​)=₹8820 (ii) Principal for the 3rd year = ₹ 8820 Interest for the 3rd year =(1008820×5×1​)=₹441
  • Find the amount and compound interest on ₹ 10000 for 121​ 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, P=₹10000 R=10% per annum =5% per half year n=121​ year =3 half years If interest is compounded half yearly A=10000(1+1005​)3=₹11576.25 Interest =11576.25−10000 = ₹ 1576.25 If interest is compounded annually. Amount after 1 year, A=10000(1+10010​)=₹11000 Amount after 121​ year A=11000(1+100×210×1​)=₹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 1221​% per annum, interest being compounded half yearly. Sol. Here, Principal = ₹ 4096, Time =18 months =3 half years Rate =1221​% per annum =(425​)% per half year. ∴ Amount =[4096×(1+4×10025​)3] =(4096×1617​×1617​×1617​) = ₹ 4913

NCERT Solutions for Class 8 Maths Other Chapters:-

Chapter 1: Rational Numbers

Chapter 2: Linear Equations in One Variable

Chapter 3: Understanding Quadrilaterals

Chapter 4: Data Handling

Chapter 5: Squares and Square Roots

Chapter 6: Cubes and Cube Roots

Chapter 7: Comparing Quantities

Chapter 8: Algebraic Expressions and Identities

Chapter 9: Mensuration

Chapter 10: Exponents and Powers

Chapter 11: Direct and Inverse Proportions

Chapter 12: Factorisation

Chapter 13: Introduction to Graphs


CBSE Notes for Class 8 Maths - All Chapters:-

Class 8 Maths Chapter 1 - Rational Numbers Notes

Class 8 Maths Chapter 2 - Linear Equations In One Variable Notes

Class 8 Maths Chapter 3 - Understanding Quadrilaterals Notes

Class 8 Maths Chapter 4 - Data Handling Notes

Class 8 Maths Chapter 5 - Squares and Square Roots Notes

Class 8 Maths Chapter 6 - Cubes and Cube Roots Notes

Class 8 Maths Chapter 7 - Comparing Quantities Notes

Class 8 Maths Chapter 8 - Algebraic Expressions and Identities Notes

Class 8 Maths Chapter 9 - Mensuration Notes

Class 8 Maths Chapter 10 - Exponents and Powers Notes

Class 8 Maths Chapter 11 - Direct and Inverse Proportions Notes

Class 8 Maths Chapter 12 - Factorisation Notes

Class 8 Maths Chapter 13 - Introduction to Graphs Notes

Frequently Asked Questions:

Ratios and proportions are a fundamental basis of comparing quantities. This will help in solving real-life problems, like determining the ratios in recipes, calculating speeds, and executing financial data analysis.

To solve percentage problems, you need to work on converting percentages to fractions and decimals. Also, make sure you understand how to use percentages in order to calculate increases, decreases, and comparison

Interest calculated on the original principal for the entire period is known as simple interest. Compound interest is interest calculated on both the principal and any accumulated interest over periods

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