{"id":108665,"date":"2022-02-12T20:21:17","date_gmt":"2022-02-12T14:51:17","guid":{"rendered":"https:\/\/infinitylearn.com\/surge\/?p=108665"},"modified":"2025-06-20T13:12:54","modified_gmt":"2025-06-20T07:42:54","slug":"ncert-solutions-for-class-8-maths-chapter-8-comparing-quantities-ex-8-3","status":"publish","type":"post","link":"https:\/\/infinitylearn.com\/surge\/study-material\/ncert-solutions\/class-8\/maths\/chapter-8\/comparing-quantities\/ex-8-3\/","title":{"rendered":"NCERT Solutions for Class 8 Maths Chapter 8 Comparing Quantities Ex 8.3"},"content":{"rendered":"

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

 <\/p>\n

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

Ex 8.3 Class 8 Maths Question 1.
\nCalculate the amount and compound interest on
\n(a) \u20b9 10,800 for 3 years at 12\\(\\frac { 1 }{ 2 }\\) % per annum compounded annually.
\n(b) \u20b9 18,000 for 2\\(\\frac { 1 }{ 2 }\\) years at 10% per annum compounded annually.
\n(c) \u20b9 62,500 for 1\\(\\frac { 1 }{ 2 }\\) years at 8% per annum compounded half yearly.
\n(d) \u20b9 8,000 for 1 year at 9% per annum compounded half yearly. (You could use the year by year calculation using SI formula to verify).
\n(e) \u20b9 10,000 for 1 year at 8% per annum compounded half yearly.
\nSolution:
\n(a) Given:
\nP = \u20b9 10,800, n = 3 years,
\n\"NCERT
\nCI = A \u2013 P = \u20b9 15,377.35 \u2013 \u20b9 10,800 = \u20b9 4,577.35
\nHence amount = \u20b9 15,377.34 and CI = \u20b9 4,577.34
\n(b) Given: P = \u20b9 18,000, n = 2\\(\\frac { 1 }{ 2 }\\) years = \\(\\frac { 5 }{ 2 }\\) years
\nR = 10% p.a.
\nThe amount for 2\\(\\frac { 1 }{ 2 }\\) years, i.e., 2 years and 6 months can be calculated by first calculating the amount to 2 years using CI formula and then calculating the simple interest by using SI formula.
\nThe amount for 2 years has to be calculated
\n\"NCERT
\nTotal CI = \u20b9 3780 + \u20b9 1089 = \u20b9 4,869
\nAmount = P + I = \u20b9 21,780 + \u20b9 1,089 = \u20b9 22,869
\nHence, the amount = \u20b9 22,869
\nand CI = \u20b9 4,869
\n(c) Given: P = \u20b9 62,500, n = 1\\(\\frac { 1 }{ 2 }\\) years = \\(\\frac { 3 }{ 2 }\\) years per annum compounded half yearly
\n= \\(\\frac { 3 }{ 2 }\\) \u00d7 2 years = 3 half years
\nR = 8% = \\(\\frac { 8 }{ 2 }\\) % = 4% half yearly
\n\"NCERT<\/p>\n

\"NCERT
\nCI = A \u2013 P = \u20b9 70,304 \u2013 \u20b9 62,500 = \u20b9 7,804
\nHence, amount = \u20b9 70304 and CI = \u20b9 7804
\n(d) Given: P = \u20b9 8,000, n = 1 years R = 9% per annum compounded half yearly
\nSince, the interest is compounded half yearly n = 1 \u00d7 2 = 2 half years
\n\"NCERT
\nCI = A \u2013 P = \u20b9 8,736.20 \u2013 \u20b9 8,000 = \u20b9 736.20
\nHence, the amount = \u20b9 8736.20 and CI = \u20b9 736.20
\n(e) Given: P = \u20b9 10,000, n = 1 year and R = 8% pa compounded half yearly
\nSince the interest is compounded half yearly n = 1 \u00d7 2 = 2 half years
\n\"NCERT<\/p>\n

\"NCERT
\nCI = A \u2013 P = \u20b9 10,816 \u2013 \u20b9 10,000 = \u20b9 816
\nHence the amount = \u20b9 10,816 and Cl = \u20b9 816<\/p>\n

Ex 8.3 Class 8 Maths Question 2.
\nKamala borrowed \u20b9 26,400 from a Bank to buy a scooter at a rate of 15% per annum compounded yearly. What amount will she pay at the end of 2 years and 4 months to clear the loan? (Hint: Find amount for 2 years with interest is compounded yearly and then find SI on the 2nd
\nyear amount for \\(\\frac { 4 }{ 12 }\\) years).
\nSolution:
\nGiven:
\nP = \u20b9 26,400
\nR = 15% p.a. compounded yearly
\nn = 2 years and 4 months
\n\"NCERT
\nAmount after 2 years and 4 months = \u20b9 34,914 + \u20b9 1745.70 = \u20b9 36,659.70
\nHence, the amount to be paid by Kamla = \u20b9 36,659.70<\/p>\n

Ex 8.3 Class 8 Maths Question 3.
\nFabina borrows \u20b9 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?
\nSolution:
\nFor Fabina: P = \u20b9 12,500, R = 12% p.a. and n = 3 years
\n\"NCERT<\/p>\n

\"NCERT
\nDifference between the two interests = \u20b9 4500 \u2013 \u20b9 4137.50 = \u20b9 362.50
\nHence, Fabina pays more interest by \u20b9 362.50.<\/p>\n

Ex 8.3 Class 8 Maths Question 4.
\nI borrowed \u20b9 12,000 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?
\nSolution:
\nGiven: P = \u20b9 12,000, R = 6% p.a., n = 2 years
\n\"NCERT<\/p>\n

\"NCERT
\nDifference between two interests = \u20b9 1483.20 \u2013 \u20b9 1440 = \u20b9 43.20
\nHence, the extra amount to be paid = \u20b9 43.20<\/p>\n

Ex 8.3 Class 8 Maths Question 5.
\nVasudevan invested \u20b9 60,000 at an interest rate of 12% per annum compounded half yearly. What amount would he get
\n(i) after 6 months?
\n(ii) after 1 year?
\nSolution:
\n(i) Given: P = \u20b9 60,000, R = 12% p.a. compounded half yearly
\n\"NCERT<\/p>\n

\"NCERT
\nHence, the required amount = \u20b9 67416<\/p>\n

Ex 8.3 Class 8 Maths Question 6.
\nArif took a loan of \u20b9 80,000 from a bank. If the rate of interest is 10% per annum, find the difference in amounts he would be paying after
\n1\\(\\frac { 1 }{ 2 }\\) years if the interest is
\n(i) compounded annually.
\n(ii) compounded half yearly.
\nSolution:
\n(i) Given: P = \u20b9 80,000
\nR = 10% p.a.
\nn = 1\\(\\frac { 1 }{ 2 }\\) years
\nSince the interest is compounded annually
\n\"NCERT<\/p>\n

\"NCERT<\/p>\n

\"NCERT
\nDifference between the amounts = \u20b9 92,610 \u2013 \u20b9 92,400 = \u20b9 210<\/p>\n

Ex 8.3 Class 8 Maths Question 7.
\nMaria invested \u20b9 8,000 in a business. She would be paid interest at 5% per annum compounded annually. Find
\n(i) The amount credited against her name at the end of the second year.
\n(ii) The interest for the third year.
\nSolution:
\n(i) Given: P = \u20b9 8,000, R = 5% p.a.
\nand n = 2 years
\n\"NCERT
\nHence, interest for the third year = \u20b9 441<\/p>\n

Ex 8.3 Class 8 Maths Question 8.
\nFind the amount and the compound interest on \u20b9 10,000 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?
\nSolution:
\nGiven: P = \u20b9 10,000, n = 1\\(\\frac { 1 }{ 2 }\\) years
\nR = 10% per annum
\nSince the interest is compounded half yearly
\n\"NCERT<\/p>\n

\"NCERT<\/p>\n

Total interest = \u20b9 1,000 + \u20b9 550 = \u20b9 1,550
\nDifference between the two interests = \u20b9 1,576.25 \u2013 \u20b9 1,550 = \u20b9 26.25
\nHence, the interest will be \u20b9 26.25 more when compounded half yearly than the interest when compounded annually.<\/p>\n

Ex 8.3 Class 8 Maths Question 9.
\nFind the amount which Ram will get on \u20b9 4,096, if he gave it for 18 months at 12\\(\\frac { 1 }{ 2 }\\) per annum, interest being compounded half yearly.
\nSolution:
\nGiven: P = \u20b9 4,096, R = 12\\(\\frac { 1 }{ 2 }\\) % pa, n = 18 months
\n\"NCERT
\nHence, the required amount = \u20b9 4913<\/p>\n

Ex 8.3 Class 8 Maths Question 10.
\nThe population of a place increased to 54,000 in 2003 at a rate of 5% per annum.
\n(i) Find the population in 2001.
\n(ii) What would be its population in 2005?
\nSolution:
\n(i) Given: Population in 2003 = 54,000
\nRate = 5% pa
\nTime = 2003 \u2013 2001 = 2 years
\nPopulation in 2003 = Population in 2001
\n\"NCERT<\/p>\n

\"NCERT<\/p>\n

Ex 8.3 Class 8 Maths Question 11.
\nIn 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.
\nSolution:
\nGiven: Initial count of bacteria = 5,06,000
\nRate = 2.5% per hour
\nn = 2 hours
\nNumber of bacteria at the end of 2 hours = Number of count of bacteria initially
\n\"NCERT
\nThus, the number of bacteria after two hours = 5,31,616 (approx).<\/p>\n

Ex 8.3 Class 8 Maths Question 12.
\nA scooter was bought at \u20b9 42,000. Its value depreciated at the rate of 8% per annum. Find its value after one year.
\n\"NCERT
\nSolution:
\nGiven: Cost price of the scooter = \u20b9 42,000
\nRate of depreciation = 8% p.a.
\nTime = 1 year
\nFinal value of the scooter
\n\"NCERT
\nHence, the value of scooter after 1 year = \u20b9 38,640.<\/p>\n

\"NCERT<\/p>\n

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\"NCERT<\/p>\n

For more visit NCERT Exemplar Solutions Class 8 Maths Solutions Chapter 12 Introduction To Graphs<\/a><\/p>\n","protected":false},"excerpt":{"rendered":"

NCERT Solutions for Class 8 Maths Chapter 8 Comparing Quantities Ex 8.3   NCERT Solutions for Class 8 Maths Chapter […]<\/p>\n","protected":false},"author":7,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_yoast_wpseo_focuskw":"Class 8 math solutions","_yoast_wpseo_title":"","_yoast_wpseo_metadesc":"Get NCERT Solutions for Class 8 Maths Chapter 8 Comparing Quantities Ex 8.3 at infinity learn.","custom_permalink":"study-material\/ncert-solutions\/class-8\/maths\/chapter-8\/comparing-quantities\/ex-8-3\/"},"categories":[13,19,21],"tags":[],"table_tags":[],"acf":[],"yoast_head":"\nNCERT Solutions for Class 8 Maths Chapter 8 Comparing Quantities Ex 8.3 - Infinity Learn by Sri Chaitanya<\/title>\n<meta name=\"description\" content=\"Get NCERT Solutions for Class 8 Maths Chapter 8 Comparing Quantities Ex 8.3 at infinity learn.\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/infinitylearn.com\/surge\/study-material\/ncert-solutions\/class-8\/maths\/chapter-8\/comparing-quantities\/ex-8-3\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"NCERT Solutions for Class 8 Maths Chapter 8 Comparing Quantities Ex 8.3 - 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