NCERT Solutions for Class 8 Maths Chapter 8 Comparing Quantities
Ananya Gupta27 Jun, 2024
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NCERT Solutions for Class 8 Maths Chapter 8: NCERT Solutions for Class 8 Maths Chapter 8, "Comparing Quantities," help students understand how to compare different amounts. This chapter covers important topics like percentages, discounts, profit and loss, simple and compound interest, and ratios and proportions.
The solutions explain everything step-by-step, making it easy for students to solve problems and apply these concepts in real life. By using these solutions, students can improve their math skills, understand how to compare quantities better, and do well in their exams.NCERT Solutions for Class 8 Maths Chapter 8 Overview
The NCERT Solutions for Class 8 Maths Chapter 8, "Comparing Quantities," prepared by the subject experts of Physics Wallah, provide clear and easy-to-follow explanations for various mathematical concepts. Each solution is designed to help students understand and solve problems step-by-step. With these expertly prepared solutions, students can improve their math skills, gain confidence in comparing different quantities, and perform better in their exams.NCERT Solutions for Class 8 Maths Chapter 8 PDF
The NCERT Solutions for Class 8 Maths Chapter 8, "Comparing Quantities," are available in a PDF format through the link provided below. These solutions help students understand and solve problems effectively, enhancing their math skills and confidence. Accessing this PDF can assist students in studying efficiently and performing better in their exams.NCERT Solutions for Class 8 Maths Chapter 8 PDF
NCERT Solutions for Class 8 Maths Chapter 8 Comparing Quantities
Here we have provided NCERT Solutions for Class 8 Maths Chapter 8 Comparing Quantities for the ease of students so that they can prepare better for their exams.NCERT Solutions for Class 8 Maths Chapter 8 Comparing Quantities Exercise 8.1 Page No: 119
1. Find the ratio of the following:
(a) Speed of a cycle 15 km per hour to the speed of a scooter 30 km per hour.
(b) 5 m to 10 km
(c) 50 paise to ₹ 5
Solution:
a) Ratio of the speed of the cycle to the speed of the scooter = 15/30 = ½ = 1:2 b) Since 1 km = 1000 m 5m/10 km = 5 m/(10 x 1000)m = 5/10000 = 1/2000 = 1:2000 The required ratio is 1:2000 c) Since, ₹1 = 100 paise 50 paise/₹5 = 50/(5 x 100) = 50/500 = 1/10 = 1:10 The required ratio is 1:102. Convert the following ratio to percentages:
a) 3:4
b) 2:3
Solution:
a) 3:4 = ¾ = ¾ x 100% = 0.75 x 100% = 75% b) 2:3 = 2/3 = 2/3 x 100% = 0.666 x 100% = 66.66% = 66⅔%3. 72% of 25 students are good in mathematics. How many are not good in mathematics?
Solution:
It’s given that 72% of 25 students are good in mathematics So, the percentage of students who are not good in mathematics = (100 – 72)% = 28% Here, the number of students who are good in mathematics = 72/100 x 25 = 18 Thus, the number of students who are not good in mathematics = 25 – 18 = 7 [Also, 28% of 25 = 28/100 x 25 = 7] Therefore, 7 students are not good in mathematics.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?
Solution:
Let the total number of matches played by the team be x. Given that the team won 10 matches and the winning percentage of the team was 40%. ⇒ 40/100 × x = 10 40x = 10 × 100 40x = 1000 x = 1000/40 = 100/4 = 25 Therefore, the team played 25 matches.5. If Chameli had ₹600 left after spending 75% of her money, how much did she have in the beginning?
Solution:
Let the amount of money which Chameli had, in the beginning, be x Given that, after spending 75% of ₹x, she was left with ₹600 So, (100 – 75)% of x = ₹600 Or, 25% of x = ₹600 25/100 × x = ₹600 x = ₹600 × 4 = ₹2400 Therefore, Chameli had ₹2400 in the beginning.6. If 60% of people in the 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 is 50 lakhs, find the exact number who like each type of game.
Solution:
Percentage of people who like other games = (100 – 60 – 30)% = (100 – 90)% = 10% Total number of people = 50 lakhs So, Number of people who like cricket = 60/100 x 50 = 30 lakhs Number of people who like football = 30/100 x 50 = 15 lakhs Number of people who like other games = 10/100 x 50 = 5 lakhsNCERT Solutions for Class 8 Maths Chapter 8 Comparing Quantities Exercise 8.2 Page No: 125
1. A man got a 10% increase in his salary. If his new salary is ₹1,54,000, find his original salary.
Solution:
Let the original salary be x Given that, the new salary is ₹1,54,000 Original salary + Increment = New salary Given that the increment is 10% of the original salary So, (x + 10/100 × x) = 154000 x + x/10 = 154000 11x/10 = 154000 x = 154000 × 10/11 = 140000 Therefore, the original salary was ₹1,40,000.2. On Sunday, 845 people went to the zoo. On Monday, only 169 people went. What is the per cent decrease in the number of people visiting the zoo on Monday?
Solution:
Given that on Sunday, 845 people went to the zoo, and on Monday, 169 people went to the zoo. Decrease in the number of people = 845 – 169 = 676 Thus, Percentage decrease = (Decrease in the number of people/Number of people who went to the zoo on Sunday) x 100% = (676/845 x 100)% = 80%3. A shopkeeper buys 80 articles for ₹ 2,400 and sells them for a profit of 16%. Find the selling price of one article.
Solution:
Given that the shopkeeper buys 80 articles for ₹ 2,400 Cost of one article = 2400/80 = ₹ 30 Profit percentage = 16% Profit percentage = Profit/C.P. x 100 16 = Profit/30 x 100 Profit = (16 x 30)/100 = ₹ 4.8 Therefore, the selling price of one article = C.P. + Profit = ₹ (30 + 4.80) = ₹ 34.804. The cost of an article was ₹ 15,500. ₹ 450 was spent on its repairs. If it is sold for a profit of 15%, find the selling price of the article.
Solution:
The total cost of an article = Cost + Overhead expenses = ₹15500 + ₹450 = ₹15950 Profit percentage = 15% Profit percentage = Profit/C.P. x 100 15 = Profit/15950 x 100 Profit = (15 x 15950)/100 = 2392.50 Therefore, the selling price of the article = C.P. + Profit = ₹(15950 + 2392.50) = ₹18342.505. 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 per cent on the whole transaction.
Solution:
C.P. of a VCR = ₹ 8000 The shopkeeper made a loss of 4 % on VCR This means if C.P. is ₹ 100, then S.P. is ₹ 96. When C.P. is ₹ 8000, S.P. = (96/100 x 8000) = ₹ 7680 C.P. of a TV = ₹ 8000 The shopkeeper made a profit of 8 % on TV. This means that if C.P. is ₹ 100, then S.P. is ₹ 108. When C.P. is ₹ 8000, S.P. = (108/100 x 8000) = ₹ 8640 Total S.P. = ₹ 7680 + ₹ 8640 = ₹ 16320 Total C.P. = ₹ 8000 + ₹ 8000 = ₹ 16000 Since, total S.P.> total C.P. ⇒ profit Profit = ₹ 16320 − ₹ 16000 = ₹ 320 Profit % on the whole transaction = Profit/Total CP x 100 = 320/16000 x 100 = 2% Therefore, the shopkeeper had a gain of 2% on the whole transaction.6. 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?
Solution:
Total marked price = ₹ (1,450 + 2 × 850) = ₹ (1,450 +1,700) = ₹ 3,150 Given that, the discount percentage = 10% Discount = ₹ (10/100 x 3150) = ₹ 315 Also, Discount = Marked price − Sale price ₹ 315 = ₹ 3150 − Sale price ∴ Sale price = ₹ (3150 − 315) = ₹ 2835 Therefore, the customer will have to pay ₹ 2,835.7. 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 the C.P. of each)
Solution:
S.P. of each buffalo = ₹ 20,000 The milkman made a gain of 5% while selling one buffalo This means if C.P. is ₹ 100, then S.P. is ₹ 105. C.P. of one buffalo = 100/105 × 20000 = ₹ 19,047.62 Also, the second buffalo was sold at a loss of 10% This means if C.P. is ₹ 100, then S.P. is ₹ 90 ∴ C.P. of other buffalo = 100/90 × 20000 = ₹ 22222.22 Total C.P. = ₹ 19047.62 + ₹ 22222.22 = ₹ 41269.84 Total S.P. = ₹ 20000 + ₹ 20000 = ₹ 40000 Loss = ₹ 41269.84 − ₹ 40000 = ₹ 1269.84 Therefore, the overall loss of milkman was ₹ 1,269.848. 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.
Solution:
On ₹ 100, the tax to be paid = ₹ 12 Here, on ₹ 13000, the tax to be paid will be = 12/100 × 13000 = ₹ 1560 Required amount = Cost + Sales Tax = ₹ 13000 + ₹ 1560 = ₹ 14560 Therefore, Vinod will have to pay ₹ 14,560 for the TV.9. 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.
Solution:
Let the marked price be x Discount percent = Discount/Marked Price x 100 20 = Discount/x × 100 Discount = 20/100 × x = x/5 Also, Discount = Marked price – Sale price x/5 = x – ₹ 1600 x – x/5 = 1600 4x/5 = 1600 x = 1600 x 5/4 = 2000 Therefore, the marked price was ₹ 2000.10. I purchased a hair dryer for ₹ 5,400, including 8% VAT. Find the price before VAT was added.
Solution:
The price includes VAT So, 8% VAT means that if the price without VAT is ₹ 100, Then, the price including VAT will be ₹ 108 When price including VAT is ₹ 108, original price = ₹ 100 When price including VAT is ₹ 5400, original price = ₹ (100/108 × 5400) = ₹ 5000 Therefore, the price of the hair dryer before the addition of VAT was ₹ 5,000.Exercise 8.3 Page No: 133
1. Calculate the amount and compound interest on
(a) ₹ 10,800 for 3 years at 12½ % per annum compounded annually.
Solution:
Principal (P) = ₹ 10,800 Rate (R) = 12½ % = 25/2 % (annual) Number of years (n) = 3 Amount (A) = P(1 + R/100) n = 10800(1 + 25/200) 3 = 10800(225/200) 3 = 15377.34375 = ₹ 15377.34 (approximately) C.I. = A – P = ₹ (15377.34 – 10800) = ₹ 4,577.34(b) ₹ 18000 for 2½ years at 10% per annum compounded annually.
Solution:
Principal (P) = ₹ 18,000 Rate (R) = 10% annual Number of years (n) = 2½ The amount for 2 years and 6 months can be calculated by calculating the amount for 2 years using the compound interest formula, then calculating the simple interest for 6 months on the amount obtained at the end of 2 years. First, the amount for 2 years has to be calculated Amount, A = P(1 + R/100) n = 18000(1 + 1/10) 2 = 18000(11/10) 2 = ₹ 21780 By taking ₹ 21780 as principal, the S.I. for the next ½ year will be calculated S.I. = (21780 x ½ x 10)/100 = ₹ 1089 Hence, the interest for the first 2 years = ₹ (21780 – 18000) = ₹ 3780 And, interest for the next ½ year = ₹ 1089 Total C.I. = ₹ 3780 + ₹ 1089 = ₹ 4,869 Therefore, Amount, A = P + C.I. = ₹ 18000 + ₹ 4869 = ₹ 22,869(c) ₹ 62500 for 1½ years at 8% per annum compounded half yearly.
Solution:
Principal (P) = ₹ 62,500 Rate = 8% per annum or 4% per half-year Number of years = 1½ There will be 3 half-years in 1½ years Amount, A = P(1 + R/100) n = 62500(1 + 4/100) 3 = 62500(104/100) 3 = 62500(26/25) 3 = ₹ 70304 C.I. = A – P = ₹ 70304 – ₹ 62500 = ₹ 7,804(d) ₹ 8000 for 1 year at 9% per annum compound half yearly.
(You can use the year-by-year calculation using S.I. formula to verify)
Solution:
Principal (P) = ₹ 8000 Rate of interest = 9% per annum or 9/2% per half-year Number of years = 1 year There will be 2 half-years in 1 year Amount, A = P(1 + R/100) n = 8000(1 + 9/200) 2 = 8000(209/200) 2 = 8736.20 C.I. = A – P = ₹ 8736.20 – ₹ 8000 = ₹ 736.20(e) ₹ 10000 for 1 year at 8% per annum compounded half yearly.
Solution:
Principal (P) = ₹ 10,000 Rate = 8% per annum or 4% per half-year Number of years = 1 year There are 2 half-years in 1 year Amount, A = P(1 + R/100) n = 10000(1 + 4/100) 2 = 10000(1 + 1/25) 2 = 10000(26/25) 2 = ₹ 10816 C.I. = A – P = ₹ 10816 – ₹ 10000 = ₹ 8162. Kamala 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?
(Hint: Find A for 2 years with interest compounded yearly and then find S.I. on the 2nd year amount for 4/12 years.)
Solution:
Principal (P) = ₹ 26,400 Rate (R) = 15% per annum Number of years (n) = 2 4/12 The amount for 2 years and 4 months can be calculated by first calculating the amount for 2 years using the compound interest formula, then calculating the simple interest for 4 months on the amount obtained at the end of 2 years. First, the amount for 2 years has to be calculated Amount, A = P(1 + R/100) n = 26400(1 + 15/100) 2 = 26400(1 + 3/20) 2 = 26400(23/20) 2 = ₹ 34914 By taking ₹ 34,914 as principal, the S.I. for the next 1/3 years will be calculated S.I. = (34914 × 1/3 x 15)/100 = ₹ 1745.70 Interest for the first two years = ₹ (34914 – 26400) = ₹ 8,514 And interest for the next 1/3 year = ₹ 1,745.70 Total C.I. = ₹ (8514 + ₹ 1745.70) = ₹ 10,259.70 Amount = P + C.I. = ₹ 26400 + ₹ 10259.70 = ₹ 36,659.703. 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?
Solution:
Interest paid by Fabina = (P x R x T)/100 = (12500 x 12 x 3)/100 = 4500 Amount paid by Radha at the end of 3 years = A = P(1 + R/100) n A = 12500(1 + 10/100) 3 = 12500(110/100) 3 = ₹ 16637.50 C.I. = A – P = ₹ 16637.50 – ₹ 12500 = ₹ 4,137.50 The interest paid by Fabina is ₹ 4,500 and by Radha is ₹ 4,137.50 Thus, Fabina pays more interest ₹ 4500 − ₹ 4137.50 = ₹ 362.50 Hence, Fabina will have to pay ₹ 362.50 more.4. 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?
Solution:
P = ₹ 12000 R = 6% per annum T = 2 years S.I. = (P x R x T)/100 = (12000 x 6 x 2)/100 = ₹ 1440 To find the compound interest, the amount (A) has to be calculated Amount, A = P(1 + R/100) n = 12000(1 + 6/100) 2 = 12000(106/100) 2 = 12000(53/50) 2 = ₹ 13483.20 ∴ C.I. = A − P = ₹ 13483.20 − ₹ 12000 = ₹ 1,483.20 C.I. − S.I. = ₹ 1,483.20 − ₹ 1,440 = ₹ 43.20 Therefore, the extra amount to be paid is ₹ 43.20.5. Vasudevan invested ₹ 60000 at an interest rate of 12% per annum compounded half yearly. What amount would he get
(i) after 6 months?
(ii) after 1 year?
Solution:
(i) P = ₹ 60,000 Rate = 12% per annum = 6% per half-year n = 6 months = 1 half-year Amount, A = P(1 + R/100) n = 60000(1 + 6/100) 1 = 60000(106/100) = 60000(53/50) = ₹ 63600 (ii) There are 2 half-years in 1 year So, n = 2 Amount, A = P(1 + R/100) n = 60000(1 + 6/100) 2 = 60000(106/100) 2 = 60000(53/50) 2 = ₹ 674166. 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 1½ years if the interest is
(i) Compounded annually
(ii) Compounded half yearly
Solution:
(i) P = ₹ 80,000 R = 10% per annum n = 1½ years The amount for 1 year and 6 months can be calculated by first calculating the amount for 1 year using the compound interest formula, then calculating the simple interest for 6 months on the amount obtained at the end of 1 year. First, the amount for 1 year has to be calculated Amount, A = P(1 + R/100) n = 80000(1 + 10/100) 1 = 80000 x 11/100 = ₹ 88000 By taking ₹ 88,000 as principal, the S.I. for the next ½ year will be calculated as S.I. = (P x R x T)/100 = (88000 x 10 x ½)/100 = ₹ 4400 Interest for the first year = ₹ 88000 – ₹ 80000 = ₹ 8000 And interest for the next ½ year = ₹ 4,400 Total C.I. = ₹ 8,000 + ₹ 4,400 = ₹ 12,400 A = P + C.I.= ₹ (80000 + 12400) = ₹ 92,400 (ii) The interest is compounded half yearly Rate = 10% per annum = 5% per half-year There will be three half-years in 1½ years Amount, A = P(1 + R/100) n = 80000(1 + 5/100) 3 = 80000(105/100) 3 = ₹ 92610 Thus, the difference between the amounts = ₹ 92,610 – ₹ 92,400 = ₹ 2107. 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 3 rd year
Solution:
(i) P = ₹ 8,000 R = 5% per annum n = 2 years Amount, A = P(1 + R/100) n = 8000(1 + 5/100) 2 = 8000(105/100) 2 = ₹ 8820 (ii) The interest for the next year, i.e. the third year, has to be calculated. By taking ₹ 8,820 as principal, the S.I. for the next year will be calculated. S.I. = (P x R x T)/100 = (8820 x 5 x 1)/100 = ₹ 4418. Find the amount and the compound interest on ₹ 10,000 for 1½ years at 10% per annum, compounded half yearly. Would this interest be more than the interest he would get if it was compounded annually?
Solution:
P = ₹ 10,000 Rate = 10% per annum = 5% per half-year n = 1½ years There will 3 half-years in 1½ years Amount, A = P(1 + R/100) n = 10000(1 + 5/100) 3 = 10000(105/100) 3 = ₹ 11576.25 C.I. = A − P = ₹ 11576.25 − ₹ 10000 = ₹ 1,576.25 The amount for 1 year and 6 months can be calculated by first calculating the amount for 1 year using the compound interest formula, then calculating the simple interest for 6 months on the amount obtained at the end of 1 year. Amount, A = P(1 + R/100) n = 10000(1 + 10/100) 1 = 10000(110/100) = ₹ 11000 By taking ₹ 11,000 as the principal, the S.I. for the next ½ year will be calculated as S.I. = (P x R x T)/100 = (11000 x 10 x ½)/100 = ₹ 550 So, the interest for the first year = ₹ 11000 − ₹ 10000 = ₹ 1,000 Hence, Total compound interest = ₹ 1000 + ₹ 550 = ₹ 1,550 So the difference between two interests = 1576.25 – 1550 = 26.25 Therefore, the interest would be 26.25 more when compounded half yearly than the interest when compounded annually.9. Find the amount which Ram will get on ₹ 4,096, if he gave it for 18 months at 12½ per annum, interest being compounded half-yearly.
Solution:
P = ₹ 4,096 R = 12½ per annum = 25/2 per annum = 25/4 per half-year n = 18 months There will be 3 half-years in 18 months Therefore, amount A = P(1 + R/100) n = 4096(1 + 25/(4 x 100)) 3 = 4096 x (1 + 1/16) 3 = 4096 x (17/16) 3 = ₹ 4913 Therefore, the required amount is ₹ 4,913.10. 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?
Solution:
(i) It’s given that population in the year 2003 = 54,000 54,000 = (Population in 2001) (1 + 5/100) 2 54,000 = (Population in 2001) (105/100) 2 Population in 2001 = 54000 x (100/105) 2 = 48979.59 Therefore, the population in the year 2001 was approximately 48,980 (ii) Population in 2005 = 54000(1 + 5/100) 2 = 54000(105/100) 2 = 54000(21/20) 2 = 59535 Therefore, the population in the year 2005 would be 59,535.11. 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.
Solution:
The initial count of bacteria is given as 5,06,000 Bacteria at the end of 2 hours = 506000(1 + 2.5/100) 2 = 506000(1 + 1/40) 2 = 506000(41/40) 2 = 531616.25 Therefore, the count of bacteria at the end of 2 hours will be 5,31,616 (approx.).12. A scooter was bought at ₹ 42,000. Its value depreciated at the rate of 8% per annum. Find its value after one year.
Solution:
Principal = Cost price of the scooter = ₹ 42,000 Depreciation = 8% of ₹ 42,000 per year = (P x R x T)/100 = (42000 x 8 x 1)/100 = ₹ 3360 Thus, the value after 1 year = ₹ 42000 − ₹ 3360 = ₹ 38,640.NCERT Solutions for Class 8 Maths Chapter 8 – Comparing Quantities Summary
Chapter 8 of NCERT Class 8 Maths, "Comparing Quantities," deals with the concepts of ratios and percentages, along with other important topics. The major concepts covered in this chapter include:8.1 Recalling Ratios and Percentages: This section reviews the basic concepts of ratios and percentages.
8.2 Finding the Increase or Decrease Percent: Learn how to calculate the percentage increase or decrease in quantities.
8.3 Finding Discounts: This part covers how to calculate discounts on prices.
- 8.3.1 Estimation in Percentages: Learn methods to estimate values using percentages.
8.4 Prices Related to Buying and Selling (Profit and Loss): Understand how to calculate profit and loss in buying and selling transactions.
- 8.4.1 Finding Cost Price/Selling Price, Profit %, and Loss %: This section explains how to determine the cost price, selling price, and the percentages of profit and loss.
8.5 Sales Tax/Value Added Tax: Learn about sales tax and value-added tax and how they are calculated.
8.6 Compound Interest: Introduction to the concept of compound interest.
8.7 Deducing a Formula for Compound Interest: This part explains how to derive the formula for calculating compound interest.
8.8 Rate Compounded Annually or Half-Yearly (Semi-annually): Learn about interest rates that are compounded annually or semi-annually.
8.9 Applications of Compound Interest Formula: This section covers various applications of the compound interest formula in real-life scenarios.
Benefits of NCERT Solutions for Class 8 Maths Chapter 8
- Clear Understanding: These solutions provide clear and detailed explanations of concepts such as ratios, percentages, discounts, profit and loss, sales tax, and compound interest, making it easier for students to grasp these important topics.
- Step-by-Step Solutions: Each problem is solved step-by-step, which helps students understand the methods and procedures involved in solving various types of mathematical problems related to comparing quantities.
- Exam Preparation: These solutions are aligned with the NCERT curriculum and help students prepare thoroughly for exams by covering all the key topics and providing practice problems.
- Concept Clarity: The solutions help in reinforcing concepts learned in class, ensuring that students have a solid understanding of the material.
- Confidence Building: By working through these solutions and understanding how to solve problems on their own, students can build confidence in their mathematical abilities.
- Accessible Learning: The solutions are available in a PDF format, making it easy for students to access and study them anytime, anywhere.
NCERT Solutions for Class 8 Maths Chapter 8 FAQs
What is the concept of ratio in Chapter 8?
How is percentage different from a ratio?
What are sales tax and value-added tax (VAT)?
Why is understanding ratios and percentages important?
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