{"id":27661,"date":"2023-03-06T15:38:00","date_gmt":"2023-03-06T10:08:00","guid":{"rendered":"https:\/\/ncertsolutions.guru\/?p=27661"},"modified":"2023-03-06T17:05:55","modified_gmt":"2023-03-06T11:35:55","slug":"ncert-solutions-for-class-8-maths-chapter-8-intext-questions","status":"publish","type":"post","link":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-8-maths-chapter-8-intext-questions\/","title":{"rendered":"NCERT Solutions for Class 8 Maths Chapter 8 Comparing Quantities InText Questions"},"content":{"rendered":"

These NCERT Solutions for Class 8 Maths<\/a> Chapter 8 Comparing Quantities InText Questions and Answers are prepared by our highly skilled subject experts.<\/p>\n

NCERT Solutions for Class 8 Maths Chapter 8 Comparing Quantities InText Questions<\/h2>\n

Try These (Page No. 119)<\/span><\/p>\n

Question 1.
\nIn a primary school, the parents were asked about the number of hours they spend per day in helping their children to do homework.
\nThere were 90 parents who helped for \\(\\frac{1}{2}\\) hour to 1\\(\\frac{1}{2}\\) hours. The distribution of parents according to the time for which, they said they helped is given in the following figure: 20% helped for more than 1\\(\\frac{1}{2}\\) hours per day; 30% helped for \\(\\frac{1}{2}\\) hour to 1\\(\\frac{1}{2}\\) hours; 50% did not help at all.
\nUsing this, answer the following:
\n(i) How many parents were surveyed?
\n(ii) How many said that they did not help?
\n(iii) How many said that they helped for man than 1\\(\\frac{1}{2}\\) hours?
\n\"NCERT
\nSolution:
\n(i) Since, 30% of total surveyed parents helped their children for \u201c\\(\\frac{1}{2}\\) hours to 1\\(\\frac{1}{2}\\) hours\u201d. And 90 parents helped their children for \u201c\\(\\frac{1}{2}\\) hours to 1\\(\\frac{1}{2}\\) hours\u201d.
\n30% [ surveyed parents] = 90
\nor \\(\\frac{30}{100}\\) \u00d7 [surveyed parents] = 90
\nor surveyed parents = 90 \u00d7 100\/30
\n= 3 \u00d7 100
\n= 300<\/p>\n

\"NCERT<\/p>\n

(ii) Since, 50% of surveyed parents did not help their children.
\nNumber of parents who did not help = 50% of surveyed parents
\n= 50% of 300
\n= 50\/100 \u00d7 300
\n= 150<\/p>\n

(iii) Since, 20% of surveyed parents help their children for more than 1\\(\\frac{1}{2}\\) hours, i.e. 20% of surveyed parents help for more than 1\\(\\frac{1}{2}\\) hours.
\nNumber of parents who helped for more than 1\\(\\frac{1}{2}\\) hours = 20% of 300
\n= 20\/100 \u00d7 300
\n= 20 \u00d7 3
\n= 60
\nNote: \u2018of\u2019 means multiplication<\/p>\n

Try These (Page No. 121)<\/span><\/p>\n

Question 2.
\nA shop gives a 20% discount. What would the sale price of each of these be?
\n(a) A dress marked at \u20b9 120
\n(b) A pair of shoes marked at \u20b9 750
\n(c) A bag marked at \u20b9 250
\nSolution:
\n(a) Marked price of the dress = \u20b9 120
\nDiscount rate = 20%
\nDiscount = 20% of \u20b9 120
\n= \u20b9 \\(\\frac{20}{100}\\) \u00d7 120
\n= \u20b9 2 \u00d7 12
\n= \u20b9 24
\nSale price of the dress = [Marked Price] – [Discount]
\n= \u20b9 120 – \u20b9 24
\n= \u20b9 96<\/p>\n

(b) Marked price of the pair of shoes = \u20b9 750
\nDiscount rate = 20%
\nDiscount = 20% of \u20b9 750
\n= \u20b9 \\(\\frac{20}{100}\\) \u00d7 750
\n= \u20b9 2 \u00d7 75
\n= \u20b9 150
\nNow, Sale price of the pair of shoes = [Marked Price] – [Discount]
\n= \u20b9 750 – \u20b9 150
\n= \u20b9 600<\/p>\n

(c) Marked price of the bag = \u20b9 250
\nDiscount rate = 20%
\nDiscount = 20% of \u20b9 250
\n= \u20b9 \\(\\frac{20}{100}\\) \u00d7 250
\n= \u20b9 2 \u00d7 25
\n= \u20b9 50
\nSale price of the bag = [Marked price] – [ Discount]
\n= \u20b9 250 – \u20b9 50
\n= \u20b9 200<\/p>\n

\"NCERT<\/p>\n

Question 3.
\nA table marked at \u20b9 15,000 is available for \u20b9 14,000. Find the discount given and the discount percent.
\nSolution:
\nThe marked price of the table = \u20b9 15000
\nSale price of the table = \u20b9 14400
\nDiscount = [Marked Price] – [Sale price]
\n= \u20b9 15000 – \u20b9 14400
\n= \u20b9 600
\nDiscount percent = Discount\/Marked Price \u00d7 100<\/p>\n

Question 4.
\nAn almirah is sold at \u20b9 5,225 after allowing a discount of 5%. Find its marked price.
\nSolution:
\nSale price of the almirah = \u20b9 5255
\nDiscount rate = 5%
\nSince, Discount = 5% of marked price
\nMarked Price – Discount = Sale price
\nor [Marked Price] – \\(\\frac{5}{100}\\) \u00d7 [Marked Price] = Sale price
\nor [Marked Price] \u00d7 \\(\\frac{95}{100}\\) = \u20b9 5255
\nor [Marked Price] = \u20b9 5255 \u00d7 \\(\\frac{95}{100}\\) = \u20b9 5500<\/p>\n

Try These (Page No. 123)<\/span><\/p>\n

Question 5.
\nFind selling price (SP) if a profit of 5% is made on
\n(a) a cycle of \u20b9 700 with \u20b9 50 as overhead charges.
\n(b) a lawnmower bought at \u20b9 1150 with \u20b9 50 as transportation charges.
\n(c) a fan bought for \u20b9 560 and expenses of \u20b9 40 made on its repairs.
\nSolution:
\n(a) Total cost price = \u20b9 700 + \u20b9 50 (overhead expenses) = \u20b9 750
\nprofit = 5% of \u20b9 750
\n= \u20b9 \\(\\frac{5}{100}\\) \u00d7 \u20b9 780
\n= \u20b9 5 \u00d7 \\(\\frac{15}{100}\\)
\n= \u20b9 \\(\\frac{75}{100}\\)
\n= \u20b9 37.50
\nNow, SP = CP + profit
\n= \u20b9 750 + \u20b9 37.50
\n= \u20b9 787.50<\/p>\n

\"NCERT<\/p>\n

(b) Total cost price = \u20b9 1150 + \u20b9 50 (Overhead expenses) = \u20b9 1200
\nProfit = 5% of 1200
\n= \\(\\frac{5}{100}\\) \u00d7 \u20b9 1200
\n= \u20b9 5 \u00d7 12
\n= \u20b9 60
\nNow, SP = CP + Profit
\n= \u20b9 1200 + \u20b9 60
\n= \u20b9 1260<\/p>\n

(c) Total cost price = \u20b9 560 + \u20b9 40 (Overhead expenses) = \u20b9 600
\nProfit = 5% of \u20b9 600 = \\(\\frac{5}{100}\\) \u00d7 \u20b9 600
\n= \u20b9 5 \u00d7 6
\n= \u20b9 30
\nNow, SP = CP + profit
\n= \u20b9 600 + \u20b9 30
\n= \u20b9 630<\/p>\n

Try These (Page No. 123)<\/span><\/p>\n

Question 6.
\nA shopkeeper bought two TV sets at \u20b9 10,000 each. He sold one at a profit of 10% and the other at a loss of 10%. Find whether he made an overall profit or loss.
\nSolution:
\n(a) Marked price of the dress = \u20b9 120
\nDiscount rate = 20%
\nDiscount = 20% of \u20b9 120
\n= \u20b9 \\(\\frac{20}{100}\\) \u00d7 120
\n= \u20b9 2 \u00d7 12
\n= \u20b9 24
\nSale price of the dress = [Marked Price] – [Discount]
\n= \u20b9 120 – \u20b9 24
\n= \u20b9 96<\/p>\n

(b) Marked price of the pair of shoes = \u20b9 750
\nDiscount rate = 20%
\nDiscount = 20% of \u20b9 750
\n= \u20b9 \\(\\frac{20}{100}\\) \u00d7 750
\n= \u20b9 2 \u00d7 75
\n= \u20b9 150
\nNow, Sale price of the pair of shoes = [Marked Price] – [Discount]
\n= \u20b9 750 – \u20b9 150
\n= \u20b9 600<\/p>\n

\"NCERT<\/p>\n

(c) Marked price of the bag = \u20b9 250
\nDiscount rate = 20%
\nDiscount = 20% of \u20b9 250
\n= \u20b9 \\(\\frac{20}{100}\\) \u00d7 250
\n= \u20b9 2 \u00d7 25
\n= \u20b9 50
\nSale price of the bag = [Marked price] – [Discount]
\n= \u20b9 250 – \u20b9 50
\n= \u20b9 200<\/p>\n

Try These (Page No. 126)<\/span><\/p>\n

Question 7.
\nFind the interest and amount to be paid on \u20b9 15000 at 5% per annum after 2 years.
\nSolution:
\nHere P = \u20b9 15000, R% = 5%, T = 2 years
\nSimple Interest = \\(\\frac{\\text { PRT }}{100}\\)
\n= \\(\\frac{15000 \\times 5 \\times 2}{100}\\)
\n= \u20b9 1500
\nAmount = \u20b9 15000 + \u20b9 1500 = \u20b9 16,500<\/p>\n

Try These (Page No. 129)<\/span><\/p>\n

Question 8.
\nFind Cl on a sum of \u20b9 8000 for 2 years at 5% per annum compounded annually.
\nSolution:
\nWe have
\nP = \u20b9 8000, R = 55 p.a., T = 2 years
\n\u2234 \\(A=P\\left[1+\\frac{R}{100}\\right]^{n}\\)
\n= \u20b9 8000 \\(\\left[1+\\frac{5}{100}\\right]^{2}\\)
\n= \u20b9 \\(8000\\left[\\frac{21}{20}\\right]^{2}\\)
\n= \u20b9 \\(8000 \\times \\frac{21}{20} \\times \\frac{21}{20}\\)
\n= \u20b9 (20 \u00d7 21 \u00d7 21)
\n= \u20b9 8820
\nNow, compound interest = A – P
\n= \u20b9 8820 – \u20b9 8000
\n= \u20b9 820<\/p>\n

\"NCERT<\/p>\n

Try These (Page No. 130)<\/span><\/p>\n

Question 9.
\nFind the time period and rate for each.
\n1. A sum taken for 1\\(\\frac{1}{2}\\) years at 8% per annum is compounded half-yearly.
\n2. A sum taken for 2 years at 4% per annum compounded half-yearly.
\nSolution:
\n1. We have interest rate 8% per annum 1\\(\\frac{1}{2}\\) year.
\nIt is compounded half-yearly.
\nTime period (n) = 2(1\\(\\frac{1}{2}\\)) = 3 half yearly
\nRate (R) = \\(\\frac{1}{2}\\) (8%) = 4% per half yearly.<\/p>\n

2. We have an interest rate of 4% per annum for the year
\nTime period (n) = 2(2) = 4 half-yearly
\nRate (R) = \\(\\frac{1}{2}\\) (4%) = 2% per half yearly.<\/p>\n

Try These (Page No. 131)<\/span><\/p>\n

Question 10.
\nFind the amount to be paid
\n1. At the end of 2 years on \u20b9 2,400 at 5% per annum compounded annually.
\n2. At the end of 1 year on \u20b9 1,800 at 8% per annum compounded quarterly.
\nSolution:
\n1. We have:
\nP = \u20b9 2400, R = 5% p.a., T = 2 years
\n\u2234 Interest is compounded annually i.e, n = 2
\n\"NCERT<\/p>\n

2. Here, interest compounded quarterly.
\nR = 8%, p.a. = \\(\\frac{8}{4}\\), i.e., 2% per quarter
\nT = 1 year = 4 \u00d7 1, i.e., 4 quarters on
\nn = 4
\n\"NCERT<\/p>\n

Try These (Page No. 133)<\/span><\/p>\n

Question 11.
\nA machinery worth \u20b9 10,500 depreciated by 5%. Find its value after one year?
\nSolution:
\nHere, P = 10, 500, R = 5% p.a.
\nT = 1 year, n = 1
\n\u2234 A = \\(\\mathrm{P}\\left[1+\\frac{5}{100}\\right]^{1}\\)
\n[\u2234 Depreciation is there, r = -5%]
\n= \u20b9 \\(10500\\left[1+\\frac{5}{100}\\right]^{1}\\)
\n= \u20b9 10500 \u00d7 \\(\\frac{19}{20}\\)
\n= \u20b9 525 \u00d7 19
\n= \u20b9 9975
\nThus, machinery value after 1 year = \u20b9 9975
\nNote: For depreciation, we use the formula as A = \\(\\mathrm{P}\\left[1-\\frac{\\mathrm{R}}{100}\\right]^{\\mathrm{n}}\\)<\/p>\n

\"NCERT<\/p>\n

Question 12.
\nFind the population of a city after 2 years, which is at present 12 lakh, if the rate of increase is 4%.
\nSolution:
\nPresent population, P = 12 lakh
\nRate of increase, R = 4% p.a.
\nTime, T = 2 years
\nn = 2
\nPopulation after 2 years = \\(\\mathrm{P}\\left[1+\\frac{\\mathrm{R}}{100}\\right]^{\\mathrm{n}}\\)
\n= 12 lakh \\(\\left[1+\\frac{4}{100}\\right]^{2}\\)
\n= 12 lakh \\(\\left[\\frac{26}{25}\\right]^{2}\\)
\n= 12,00, 000 \u00d7 \\(\\frac{26}{25} \\times \\frac{26}{25}\\)
\n= 1920 \u00d7 26 \u00d7 26
\n= 12,97,920
\nThus, the population of the town will be 12,97,920 after 2 years.<\/p>\n","protected":false},"excerpt":{"rendered":"

These NCERT Solutions for Class 8 Maths Chapter 8 Comparing Quantities InText Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 8 Maths Chapter 8 Comparing Quantities InText Questions Try These (Page No. 119) Question 1. In a primary school, the parents were asked about the number of hours …<\/p>\n

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NCERT Solutions for Class 8 Maths Chapter 8 Comparing Quantities InText Questions Try These (Page No. 119) Question 1. 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