{"id":27619,"date":"2021-07-01T12:56:09","date_gmt":"2021-07-01T07:26:09","guid":{"rendered":"https:\/\/ncertsolutions.guru\/?p=27619"},"modified":"2021-07-01T12:58:13","modified_gmt":"2021-07-01T07:28:13","slug":"ncert-solutions-for-class-8-maths-chapter-8-ex-8-3","status":"publish","type":"post","link":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-8-maths-chapter-8-ex-8-3\/","title":{"rendered":"NCERT Solutions for Class 8 Maths Chapter 8 Comparing Quantities Ex 8.3"},"content":{"rendered":"

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

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.
\n(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) Here P = \u20b9 10800, n = 3 years
\n\"NCERT
\nCompound interest = \u20b9 15377.34 – \u20b9 10,800 = \u20b9 4577.34
\nAmount = \u20b9 15377.34
\nCompound interest = \u20b9 4577.34<\/p>\n

(b) HereP = \u20b9 18,000, n = 2\\(\\frac{1}{2}\\), R = 10% p.a
\nFor first 2 years
\n\"NCERT
\nCI = A – P = 21,780 – 18000 = \u20b9 3780
\nFor the last \\(\\frac{1}{2}\\) year, principal become \u20b9 21,780
\nSo, Interest = \\(\\frac{21780 \\times 10 \\times 1}{2 \\times 100}\\) = \u20b9 1089
\n\u2234 Total interest = 3780 + 1089 = \u20b9 4869<\/p>\n

\"NCERT<\/p>\n

(c) Here P = \u20b9 62,500, R = \\(\\frac{8}{2}\\) = 4% per half year
\nn = \\(\\frac{3}{2}\\) \u00d7 2 = 3 years
\n\"NCERT
\nA = \u20b9 4 \u00d7 26 \u00d7 26 \u00d7 26 = \u20b9 70304
\nCompound interest = Amount – Principal
\n= \u20b9 70304 – \u20b9 62500
\n= \u20b9 7804<\/p>\n

(d) Here P = \u20b9 8000, n = 2 (2 half year),
\nR = \\(\\frac{9}{2}\\) % (half yearly)
\n\"NCERT
\nCompound interest = \u20b9 8736.2 – \u20b9 8000 = \u20b9 736.20<\/p>\n

(e) Here P = \u20b9 10,000; n = 2 (2 half year);
\nR = \\(\\frac{8}{2}\\) % = 4% (half yearly)
\n\"NCERT
\nCompound Interest = \u20b9 10816 – \u20b9 10,000 = \u20b9 816<\/p>\n

\"NCERT<\/p>\n

Question 2.
\nKamala borrowed \u20b9 26,400 from a Bank to buy a scooter at a rate 15% p.a. compounded yearly. What amount will she pay at the end of 2 years and 4 months to clear the loan?
\n(Hint: Find A for 2 years with interest is compounded yearly and then find SI on the 2nd year amount for \\(\\frac{4}{12}\\) years).
\nSolution:
\nHere P = \u20b9 26,400, R% = 15% p.a, n = 2 years
\n\"NCERT
\nSimple interest on \u20b9 34914 at 15% p.a for 4 months (\\(\\frac{1}{3}\\) years)
\n= \u20b9 \\(\\frac{34914 \\times 15 \\times 1}{100 \\times 3}\\)
\n= \u20b9 1745.70
\nRequiredamount = \u20b9 34914 + \u20b9 1745.70 = \u20b9 36,659.70<\/p>\n

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,
\nP = \u20b9 12,500, R% = 12%, T = 3 years
\nSimple interest = \\(\\frac{\\text { PRT }}{100}\\)
\n= \u20b9 \\(\\frac{12500 \\times 12 \\times 3}{100}\\)
\n= \u20b9 4500
\nFor Radha,
\nP = \u20b9 12500, R = 10% p.a. n = 3 years
\n\"NCERT
\nCompound interest = Amount – Principal
\n= \u20b9 16,637.50 – \u20b9 12500
\n= \u20b9 4137.50
\nDifference between C.I and S.I = \u20b9 4500 – \u20b9 4137.50 = \u20b9 362.5
\nFabina pays more by \u20b9 362.50<\/p>\n

\"NCERT<\/p>\n

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:
\nSimple interest
\nHere P = \u20b9 12000, R% = 6%, T = 2 years
\nI = \\(\\frac{\\text { PRT }}{100}\\)
\n= \u20b9 \\(\\frac{12000 \\times 6 \\times 2}{100}\\)
\n= \u20b9 1440
\nCompound interest
\nHere P = \u20b9 12000, R = 6%, n = 2 years
\n\"NCERT
\nC.I. = \u20b9 13,483.20 – \u20b9 12000 = \u20b9 1483.20
\n\u2234 Excess amount = \u20b9 1483.20 – \u20b9 1440 = \u20b9 43.20
\nI would have to pay to him an excess amount of \u20b9 43.20.<\/p>\n

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) After 6 months
\nHere, P = \u20b9 60,000, R = 12% per annum = \\(\\frac {1}{2}\\) \u00d7 12 = 6%, n = 1 (one half year)
\n\"NCERT
\nHe would get \u20b9 63,600 after 6 months<\/p>\n

(ii) After one year
\nHere P = \u20b9 60,000, R = \\(\\frac{12}{2}\\) = 6% per half year
\nn = (1 \u00d7 2) = 2 half years
\n\"NCERT<\/p>\n

\"NCERT
\nHence, he would get \u20b9 67,416 after one year<\/p>\n

\"NCERT<\/p>\n

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 1\\(\\frac{1}{2}\\) years if the interest is
\n(i) compounded annually
\n(ii) compounded half-yearly
\nSolution:
\n(i) Compounded annually
\nHere P = \u20b9 80,000, R = 10% p.a, n = 1 year
\n\"NCERT
\nTotal interest = \u20b9 8000 + \u20b9 4400 = \u20b9 12400
\n\u2234 Required amount = \u20b9 80000 + \u20b9 12400 = \u20b9 92400<\/p>\n

(ii) Compound half yearly
\nHere P = \u20b9 80,000
\nR = \\(\\frac{10}{2}\\)% = 5% per half year
\nn = 1\\(\\frac{1}{2}\\) \u00d7 2 years = 3 half years
\n\"NCERT
\n= \\(\\frac{80000 \\times 105 \\times 105 \\times 105}{100 \\times 100 \\times 100}\\)
\n= \u20b9 92610
\n\u2234 The required amount = \u20b9 92610
\nDifference in amount = \u20b9 92610 – \u20b9 92400 = \u20b9 210
\nHence, difference in amount is \u20b9 210.<\/p>\n

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 3rd year.
\nSolution:
\n(i) Here P = \u20b9 8000, R = 5%, n = 2 years
\n\"NCERT
\nThe amount credited at the end of 2nd year = \u20b9 8820<\/p>\n

(ii) Here P = \u20b9 8820, R = 5%, n = 1 year
\nNote: To find the interest for the 3rd year
\n(Find the S.I. for one year)
\nI = \\(\\frac{\\text { PRT }}{100}\\)
\n= \\(\\frac{8820 \\times 5 \\times 1}{100}\\)
\n= \u20b9 441
\n\u2234 Interest for the 3rd year is \u20b9 441<\/p>\n

\"NCERT<\/p>\n

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:
\nIf compounded half-yearly
\nHere P = \u20b9 10,000
\nR = \\(\\frac{10}{2}\\) = 5% per half year
\n\"NCERT
\nC.I = A – P
\n= \u20b9 11576.25 – \u20b9 10,000
\n= \u20b9 1576.25
\n\u20b9 1576.25 is the required C.I, when compounded half-yearly.<\/p>\n

If compounded annually
\nHere, P = \u20b9 10,000, R = 10%, n = 1 year
\n\"NCERT
\nTotal compound Interest = \u20b9 (11000 – 10,000) + 550
\n= \u20b9 1000 + \u20b9 550
\n= \u20b9 1550
\nWhen compounded half-yearly, the compound interest = \u20b9 1576.25
\nWhen compounded annually, the compound interest = \u20b9 1550
\nHence, the interest when compounded half-yearly would be more than the interest when compounded annually.<\/p>\n

Question 9.
\nFind the amount which Ram will get on \u20b9 4096, if he gave it for 18 months at 12\\(\\frac{1}{2}\\)% per annum, interest being compounded half-yearly.
\nSolution:
\n\"NCERT
\nHence, the required amount \u20b9 4913.<\/p>\n

\"NCERT<\/p>\n

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:
\nLet the population in 2001 be ‘x’
\nR = 5%, n = 2years (2003 – 2001)
\nPopulation in 2003 is 54,000
\n\"NCERT
\n= 48979.59
\n= 48980 (approx).
\nHence, the population in 2001 was about 48980<\/p>\n

(ii) Here P = 54,000, R = 5% p.a., n = 2 years (2003 – 2005 = 2)
\n\"NCERT<\/p>\n

\"NCERT
\nHence, the population in 2005 would be 59,535.<\/p>\n

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:
\nHere P = 5,06,000, R = 2.5% per hour, n = 2 hours
\n\"NCERT
\n= 531,616.25
\n= 531616 (approx).
\nNumber of bacteria at the end of 2 hours = 531616 (approx.)<\/p>\n

\"NCERT<\/p>\n

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.
\nSolution:
\nHere P = \u20b9 42,000 R = 8% p.a, n = 1 year
\nWhen the value depreciated
\n\"NCERT
\nValue of the scooter after one year = \u20b9 38,640<\/p>\n","protected":false},"excerpt":{"rendered":"

These NCERT Solutions for Class 8 Maths Chapter 8 Comparing Quantities Ex 8.3 Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 8 Maths Chapter 8 Comparing Quantities Exercise 8.3 Question 1. Calculate the amount and compound interest on (a) \u20b9 10,800 for 3 years at 12% per annum …<\/p>\n

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NCERT Solutions for Class 8 Maths Chapter 8 Comparing Quantities Exercise 8.3 Question 1. 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