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>\nNCERT 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 \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 \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
<\/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 \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 \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 \nCompound Interest = \u20b9 10816 – \u20b9 10,000 = \u20b9 816<\/p>\n
<\/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 \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 \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
<\/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 \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 \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 <\/p>\n
\nHence, he would get \u20b9 67,416 after one year<\/p>\n
<\/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 \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 \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 \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
<\/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 \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 \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 \nHence, the required amount \u20b9 4913.<\/p>\n
<\/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 \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 <\/p>\n
\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 \n= 531,616.25 \n= 531616 (approx). \nNumber of bacteria at the end of 2 hours = 531616 (approx.)<\/p>\n
<\/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 \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
NCERT Solutions for Class 8 Maths Chapter 8 Comparing Quantities Ex 8.3<\/span> Read More »<\/a><\/p>\n","protected":false},"author":9,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"default","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","theme-transparent-header-meta":"default","adv-header-id-meta":"","stick-header-meta":"default","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","spay_email":""},"categories":[7],"tags":[],"yoast_head":"\nNCERT Solutions for Class 8 Maths Chapter 8 Comparing Quantities Ex 8.3 - MCQ Questions<\/title>\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n\t \n\t \n\t \n