{"id":27868,"date":"2021-07-02T10:17:29","date_gmt":"2021-07-02T04:47:29","guid":{"rendered":"https:\/\/ncertsolutions.guru\/?p=27868"},"modified":"2021-07-02T10:17:29","modified_gmt":"2021-07-02T04:47:29","slug":"ncert-solutions-for-class-7-maths-chapter-8-ex-8-3","status":"publish","type":"post","link":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-7-maths-chapter-8-ex-8-3\/","title":{"rendered":"NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.3"},"content":{"rendered":"

These NCERT Solutions for Class 7 Maths<\/a> Chapter 8 Comparing Quantities Ex 8.3 Questions and Answers are prepared by our highly skilled subject experts.<\/p>\n

NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities Exercise 8.3<\/h2>\n

Question 1.
\nTell what is the profit or loss in the following transactions. Also find profit per cent or loss per cent \u00a1n each case,
\n(a) Gardening shears bought for 250 and sold for \u20b9 325 .
\n(b) A refrigerator bought for \u20b9 12,000 and sold at 13,500.
\n(c) A cupboard bought for \u20b9 2,500 and sold at 3,000.
\n(d) A skirt bought for \u20b9 250 and sold at \u20b9 150.
\nAnswer:
\n(a) Cost Price (CP) = \u20b9 250
\nSelling Price = \u20b9 325
\nProfit = S.P – C.P
\n= \u20b9 325 – 250
\n= \u20b9 75
\nProfit% = \\(\\frac{75}{250}\\) x 100%
\n= 3 x 100%
\n= 30%
\nProfit = \u20b9 75; Profit % = 30%.<\/p>\n

\"NCERT<\/p>\n

(b) Cost Price (CP) = \u20b9 12,000
\nSelling Price (SP) = \u20b9 13,500
\nProfit = SP – CP
\n= \u20b9 13500 – \u20b9 12000
\n= \u20b9 1500
\nProfit% = \\(\\frac{\\text { Profit }}{\\text { C.P. }}\\) x 100
\n= \\(\\frac{1500}{12000} \\times\\) x 100%
\n= \\(\\frac{150}{12} \\%\\)
\n= 12.5%
\nProfit = \u20b9 1500 and Profit% = 12.5%.<\/p>\n

(c) Cost Price (CP) = \u20b9 2500
\nSelling Price (SP) = \u20b9 3000
\nProfit = SP – CP
\n= \u20b9 3000 – 2500
\n= \u20b9 500
\nProfit% = \\(\\frac{\\text { Profit }}{\\text { C.P. }}\\) x 100
\n= \\(\\frac{500}{2500}\\)
\n= \\(\\frac{5}{25}\\) x 100
\n= 5 x 4 = 20%
\nProfit = \u20b9 500
\nand Profit% = 20%.<\/p>\n

(d) Cost Price (CP) = \u20b9 250
\nSelling Price (SP) = \u20b9 150
\nLoss = CP – S.P
\n= \u20b9 250 – 150
\n= \u20b9 1oo
\nLoss% = \\(\\begin{aligned}
\n&\\text { Loss } \\\\
\n&\\hline \\text { C.P. }
\n\\end{aligned}\\) x 100
\n= \u20b9 \\(\\frac{100}{250}\\) x 100%
\n= \\(\\frac{10}{25}\\) x 100%
\n= 10 x 4 = 40%
\nLoss = \u20b9 100; loss% = 40%.<\/p>\n

\"NCERT<\/p>\n

Question 2.
\nConvert each part of the ratio to percentage:
\n(a) 3 : 1
\n(b) 2 : 3 :5
\n(c) 1 : 4
\n(d) 1 : 2 : 5
\nAnswer:
\n(a) 3:1
\nTotal parts of the ratio = 3 + 1 = 4
\nPercentage of the first part
\n= \\(\\frac { 3 }{ 4 }\\) x 100% = 3 x 25% = 75%
\nPercentage of the second part
\n= \\(\\frac { 1 }{ 4 }\\) x 100 = 25%<\/p>\n

(b) 2:3:5
\nTotal parts of the ratio
\n= 2 + 3 + 5 = 10
\nPercentage of the first part
\n= \\(\\frac { 2 }{ 10 }\\) x 100%
\n= 2 x 10% = 20%
\nPercentage of the second part
\n= \\(\\frac { 3 }{ 10 }\\) x 100% = 3 x 10% = 30%
\nPercentage of the third part
\n= \\(\\frac { 5 }{ 10 }\\) x 100% = 5 x 10% = 50%
\n1st part = 20%, 2nd part = 30%; 3rd part = 50%<\/p>\n

(c) 1:4
\nTotal parts of the ratio = 1+4 = 5
\nPercentage of the first part
\n= \\(\\frac { 1 }{ 5 }\\) x 100% = 20%
\nPercentage of the second part
\n= \\(\\frac { 4 }{ 5 }\\) x 100% = 80%
\nPercentage of first part = 20%
\nPercentage of second part = 80%<\/p>\n

\"NCERT<\/p>\n

(d) 1:2:5
\nTotal parts of the ratio = 1 + 2 + 5 = 8
\nPercentage of the first part
\n= \\(\\frac { 1 }{ 8 }\\) x 100% = \\(\\frac { 25 }{ 2 }\\)% = 12\\(\\frac { 1 }{ 2 }\\) % or 12.5%
\nPercentage of the second part
\n= \\(\\frac { 2 }{ 8 }\\) x 100% = 25%
\nPercentage of the third part
\n= \\(\\frac { 5 }{ 8 }\\) x 100% = \\(\\frac { 125 }{ 2 }\\) %
\n= 62\\(\\frac { 1 }{ 2 }\\)% or 62.5%
\nPercentage of the first part = 12\\(\\frac { 1 }{ 2 }\\) %
\nPercentage of the second part = 25%
\nPercentage of the third part = 62\\(\\frac { 1 }{ 2 }\\) %
\n= \\(\\frac { 2 }{ 35 }\\) x 100%
\n= \\(\\frac{2 \\times 20}{7} \\%=\\frac{40}{7} \\%=5 \\frac{5}{7} \\%\\)
\nIncrease percentage = 5\\(\\frac{5}{7}\\)<\/p>\n

Question 3.
\nThe population of a city decreased from 25,000 to 24,500. Find the percentage decrease.
\nAnswer:
\nInitial population = 25000.
\nDecreased population = 24500
\nDecrease in population = 25000 – 24500 = 500
\nPercentage of decrease = \\(\\frac{\\text { Decrease in population }}{\\text { Initial population }}\\) x 100
\n= \\(\\frac{500}{25000}\\) x 100% = \\(\\frac{50 \\times 1}{25}\\) = 2%<\/p>\n

Question 4.
\nArun bought a car for \u20b9 3,50,000. The next year, the price went upto \u20b9 3,70,000. What was the percentage of price increase?
\nAnswer:
\nInitial cost = \u20b9 3,50,000
\nIncreased cost = \u20b9 3,70,000
\nIncrease in price = \u20b9 3,70,000 – 3,50,000
\n= \u20b9 20,000
\nIncrease% = \\(\\frac{\\text { Increase in amount }}{\\text { Initial cost }}\\) x 100
\n= \\(\\frac{20,000}{3,50,000}\\) x 100%
\n= \\(\\frac{2}{35}\\) x 100%
\n= \\(\\frac{2 \\times 20}{7} \\%=\\frac{40}{7} \\%=5 \\frac{5}{7} \\%\\)
\nIncrease percentage = 5\\(\\frac{5}{7} \\%\\)<\/p>\n

\"NCERT<\/p>\n

Question 5.
\nI buy a T.V. for \u20b9 10,000 and sell it at a profit of 20%. How much money do I get for it?
\nAnswer:
\nCost price of the T.V = \u20b9 10,000
\nProfit = 20% of C.P = \\(\\frac{20}{100}\\) x 10,000
\n= \\(\\frac { 1 }{ 5 }\\) x 10000 = \u20b9 2000 5
\nSelling price of the T.V = \u20b9 10,000 + \u20b9 2000 = \u20b9 12,000
\nHence, I get \u20b9 12,000<\/p>\n

Question 6.
\nJuhi sells a washing machine for \u20b9 13,500. She loses 20% in the bargain. What was the price at which she bought it?
\nAnswer:
\nLet the cost price of the washing machine be Y
\nSelling price of the washing machine = \u20b9 13,500
\nLoss = 20% of C.P 20
\n= \\(\\frac{20}{100}\\) x x
\n= \\(\\frac{1}{5}\\) x x = \\(\\frac{x}{5}\\)
\nSelling price = Cost price – Loss
\n13500 = x – \\(\\frac{x}{5}\\)
\n= \\(\\frac{5 x-x}{5}=\\frac{4 x}{5}\\)
\n\\(\\frac{4 \\mathrm{x}}{5}\\) = 13500
\nx = \\(\\frac{13500 \\times 5}{4}\\)
\n= \u20b9 3,375 x 5 = \u20b9 16,875
\nShe bought it for \u20b9 16,875.<\/p>\n

Question 7.
\n(i) Chalk contains calcium, carbon and oxygen in the ratio 10 : 3 : 12. Find the percentage of carbon in chalk.
\n(ii) If in a stick of chalk, carbon is 3 g, what is the weight of the chalk stick?
\nAnswer:
\n(i) The ratio of calcium : carbon :
\noxygen = 10 : 3 : 12
\nTotal ratios = 10 + 3 + 12 = 25
\nPercentage of carbon in chalk
\n= \\(\\frac { 3 }{ 25 }\\) x 100% = 3 x 4% = 12%<\/p>\n

(ii) Let the weight of the stick of chalk be ‘x’ g
\n12% of the chalk mixture is carbon 12% of x = 3
\n\\(\\frac{12}{100}\\) x x = 3
\nx = \\(\\frac{3 \\times 100}{12}=\\frac{100}{4}\\) = 25g
\nWeight of the chalk stick = 25 g<\/p>\n

\"NCERT<\/p>\n

Question 8.
\nAmina buys a book for \u20b9 275 and sells it at a loss of 15%. How much does she sell it for?
\nAnswer:
\nCost price of the book = \u20b9 275
\nLoss = 15% of C.P
\nS.P of the book = \u20b9 275 – \u20b9 41.25
\n= \\(\\frac{15}{100}\\) x 275 = \u20b9 233.75
\n= \u20b9 \\(\\frac{15}{4}\\) x 11
\nS.P of the book = \u20b9 233.75
\n= \u20b9 41.25<\/p>\n

Question 9.
\nFind the amount to be paid at the end of 3 years in each case:
\n(a) Principal = \u20b9 1200 at 12% p.a.
\n(b) Principal = \u20b9 7500 at 5% p.a.
\nAnswer:
\n(a) Principal (P) = \u20b9 1200
\nRate (R) = 12%
\nTime (T) = 3 years
\nT ,T. P x R x T
\nInterest (I) = \\(\\frac{\\mathrm{P} \\times \\mathrm{R} \\times \\mathrm{T}}{100}\\)
\n= \u20b9 \\(\\frac{1200 \\times 12 \\times 3}{100}\\)
\n= \u20b9 12 x 12 x 3
\n= \u20b9 432
\nAmount = Principal + Interest
\n= \u20b9 1200 + 432
\n= \u20b9 1632
\nAmount to be paid at the end of 3 years = \u20b9 1632<\/p>\n

\"NCERT<\/p>\n

(b) Principal (P) = \u20b9 7500
\nRate (R) = 5% p.a.
\nTime (T) = 3 years
\nInterest = \\(\\frac{\\mathrm{P} \\times \\mathrm{R} \\times \\mathrm{T}}{100}\\)
\n= \u20b9 \\(\\frac{7500 \\times 5 \\times 3}{100}\\)
\n= \u20b9 75 x 5 x 3
\n= \u20b9 1125
\nAmount= Principal + Interest
\n= \u20b9 7500+ \u20b9 1125 = \u20b9 8625
\nAmount to be paid = \u20b9 8625<\/p>\n

Question 10.
\nWhat rate gives \u20b9 280 as interest on a sum of \u20b9 56,000 is 2 years?
\nAnswer:
\nPrincipal (P) = \u20b9 56000
\n280 = 560 x R x 2
\nRate (R) = R
\nR = \\(\\frac{280}{560 \\times 2}\\) %
\nTime (T) = 2 years
\nInterest (I) = \u20b9 280
\nR = \\(\\frac{1}{2 \\times 2} \\%=\\frac{1}{4} \\%\\)
\nor 0.25%
\nInterest (I) = \\(\\frac{\\mathrm{P} \\times \\mathrm{R} \\times \\mathrm{T}}{100}\\)
\nThus, rate of interest = \\(\\frac{1}{4}\\) % or 0.25%
\n280 = \\(\\frac{56000 \\times \\mathrm{R} \\times 2}{100}\\)<\/p>\n

Question 11.
\nIf Meena gives an interest of \u20b9 45 for one year at 9% rate p.a. What is the sum she has borrowed?
\nAnswer:
\nLet Principal be \u2018P\u2019
\nRate (R) = 9% p.a.
\n45 = \\(\\frac{\\mathrm{P} \\times 9 \\times 1}{100}\\)
\nInterest (I) = \u20b9 45
\nP x 9 = 45 x 100
\nTime (T) = 1 year
\nP = \\(\\frac{45 \\times 100}{9}\\)
\nSimple interest = \\(\\frac{\\mathrm{P} \\times \\mathrm{R} \\times \\mathrm{T}}{100}\\)
\n= \u20b9 500
\nThe sum borrowed is \u20b9 500<\/p>\n","protected":false},"excerpt":{"rendered":"

These NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.3 Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities Exercise 8.3 Question 1. Tell what is the profit or loss in the following transactions. Also find profit per cent or …<\/p>\n

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NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities Exercise 8.3 Question 1. Tell what is the profit or loss in the following transactions. 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