{"id":6565,"date":"2023-05-25T17:06:34","date_gmt":"2023-05-25T11:36:34","guid":{"rendered":"https:\/\/apboardsolutions.in\/?p=6565"},"modified":"2023-05-26T17:09:45","modified_gmt":"2023-05-26T11:39:45","slug":"ap-board-8th-class-maths-solutions-chapter-5-ex-5-3","status":"publish","type":"post","link":"https:\/\/apboardsolutions.in\/ap-board-8th-class-maths-solutions-chapter-5-ex-5-3\/","title":{"rendered":"AP Board 8th Class Maths Solutions Chapter 5 Comparing Quantities Using Proportion Ex 5.3"},"content":{"rendered":"

AP State Syllabus AP Board 8th Class Maths Solutions<\/a> Chapter 5 Comparing Quantities Using Proportion Ex 5.3 Textbook Questions and Answers.<\/p>\n

AP State Syllabus 8th Class Maths Solutions 5th Lesson Comparing Quantities Using Proportion Exercise 5.3<\/h2>\n

\"AP<\/p>\n

Question 1.
\nSudhakar borrows \u20b9 15000 from a bank to renovate his house. He borrows the money at 9% p.a. simple interest over 8 years. What are his monthly repayments\u2019?
\nSolution:
\nP = 15,000
\nR = 9%
\nT = 8 years
\n\"AP
\nA = \u20b9 25800
\n\u2234 His monthly payment = \\(\\frac{25800}{8 \\times 12}\\)
\n= \u20b9268.75
\n\u2234 Monthly he has to pay = \u20b9268.75<\/p>\n

\"AP<\/p>\n

Question 2.
\nA TV was bought at a price of \u20b9 21000. After 1 year the value of the TV was depreciated by 5% (Depreciation means reduction of the value due to use and age of the item). Find the value of the TV after 1 year.
\nSolution:
\nThe C.P. of T.V = \u20b9 21,000.
\nAfter 1 year its value
\n= 21000 – 5% of 21000
\n=21000 – \\(\\frac { 5 }{ 100 }\\) \u00d7 21000
\n= 21000 – 1050
\n= \u20b919,950<\/p>\n

Question 3.
\nFind the amount and the compound interest on \u20b9 8000 at 5% per annum, for 2 years
\ncompounded annually.
\nSolution:
\nP = \u20b98000
\nR = 5%
\nThe interest is compounded every year.
\nThen 2 time periods wII be occurred.
\n\u2234 n = 2
\n\"AP
\n\u2234 Amount (A) = \u20b98820
\nC.I = A – P
\n= 8820 – 8000 = \u20b9 820<\/p>\n

\"AP<\/p>\n

Question 4.
\nFind the amount and the compound interest on \u20b9 6500 for 2 years, compounded annually, the rate of interest being 5% per annum during the first year and 6% per annum during the second year.
\nSolution:
\nP = \u20b9 6500
\nR = 5%
\nT = 1 years
\n\u2234 \\(\\frac{\\mathrm{PTR}}{100}=\\frac{6500 \\times 5 \\times 1}{100}\\) = 325
\n\u2234 A = P + I = 6500 + 325 = 6825
\n\u2234 P = 6825
\n(At the begining of 2,id year A=P)
\nR = 6%
\nT = 1 year
\n\u2234 \\(\\frac{\\mathrm{PTR}}{100}=\\frac{6825 \\times 6 \\times 1}{100}\\) = 409.5
\n\u2234 A = P + I = 6825 + 409.5
\n\u2234 Amount = \u20b9 7234.50
\nC.I. = A – P
\n= \u20b9 7234.50 – 6500
\n= \u20b9734.50<\/p>\n

Question 5.
\nPrathibha borrows \u20b947000 from a fmance company to buy her first car. The rate of simple interest is 17% and she borrows the money over a 5 year period. Find: (a) How much
\namount Prathibha should repay the finance company at the end of five years. (b) her equal
\nmonthly repayments.
\nSolution:
\nP = \u20b9 47000
\nR = 17%
\nT =5 years
\n\u2234 I = \\(\\frac{\\mathrm{PTR}}{100}=\\frac{47000 \\times 5 \\times 17}{100}\\)
\n= \u20b9 39,950<\/p>\n

\"AP<\/p>\n

a) Amount to be paid
\nA = P + I
\n= 47000 + 39,950
\n= 86950
\n\u2234 Amount to be pay = \u20b9 86950<\/p>\n

b) In monthly equal instalments she has to pay
\n\"AP
\n= 149.1
\n= \u20b9 1450 (approx)<\/p>\n

Question 6.
\nThe population of Hyderabad was 68,09,000 in the year 2011. If it increases at the rate of 4.7% per annum. What will be the population at the end of the year 2015.
\nSolution:
\nThe population of Hyderabad
\n= 68,09,000
\nIf every year increase in 4.7%.
\nThen the population of the city in 2015
\n= 68,09,000 ( 1 + \\(\\frac{4.7}{100}\\) )4<\/sup>
\n100 J
\n[ \u2235 P = 6809000, R = 4.7 %, n = 4(2015 -2011)]
\n= 68,09,000 x \\(\\frac{104.7}{100} \\times \\frac{104.7}{100} \\times \\frac{104.7}{100} \\times \\frac{104.7}{100}\\)
\n= 81,82,199<\/p>\n

\"AP<\/p>\n

Question 7.
\nFind Compound interest paid when a sum of \u20b9 10000 is invested for 1 year and 3 months at 8\\(\\frac{1}{2}\\) % per annum compounded annually.
\nSolution:
\nP = \u20b910,000; R = 8\\(\\frac { 1 }{ 2 }\\) % = \\(\\frac { 17 }{ 2 }\\)%
\nT = 1 year
\n\"AP
\n= 50 \u00d7 17 = 850
\n\u2234 I = \u20b9 850
\n\u2234 A = P + I = 10,000 + 850
\nA = 10,850
\n\u2234 P = 10,850; R = \\(\\frac { 17 }{ 2 }\\)% % ; T = 3 months
\n\"AP
\n= \u20b9 230.50
\n\u2234 Compound Interest
\n= 850 + 230.50
\n= \u20b9 1080.50<\/p>\n

Question 8.
\nArif took a loan of \u20b9 80,000 from a bank. If the rate of interest is 10% per annum, find the
\ndifference in amounts he would be paying after 1\\(\\frac{1}{2}\\) years, if the interest is (i) compounded annually (ii) compounded half yearly.
\nSolution:
\nP = \u20b9 80,000; R = 10%;
\nT = 1 year
\n\u2234 \\(\\frac{\\mathrm{PTR}}{100}\\) = \\(\\frac{80000 \\times 10 \\times 1}{100}\\)
\n= \u20b98000
\n\u2234 A = P + I = 80000 + 8000
\n= \u20b9 88,000<\/p>\n

Interest on 6 months :
\nP = 88,000 ; R = 10% ; T = 6 Months
\n= \\(\\frac { 1 }{ 2 }\\) year
\n\"AP<\/p>\n

i) The amount to be paid after 1 year 6 months = P + I
\n= 88000 + 4400
\nA1<\/sub> = \u20b9 92,400<\/p>\n

ii) He has to pay compounded on
\nevery 6 months in 1 \\(\\frac { 1 }{ 2 }\\) years
\n\u2234 3 time periods will be occurred.
\n\u2234 n = 3
\nR = \\(\\frac { 10 }{ 2 }\\) = 5% P = \u20b9 80,000
\n\"AP
\n\"AP
\nA2<\/sub> = \u20b9 92610
\n\u2234 Difference between the amounts = A2<\/sub> – A1<\/sub> = 92610 – 92400 = \u20b9 210<\/p>\n

\"AP<\/p>\n

Question 9.
\nI borrowed \u20b9 12000 from Prasad at 6\u00b0\/o per annum simple interest for 2 years. Had
\nI borrowed this sum at 6% per annum compounded annually, what extra amount would
\nI have to pay9
\nSolution:
\nSum borrowed from Prasad
\nP = \u20b9 12000
\nT = 2 years;
\nR = 6%
\n\"AP
\n= \u20b9144O
\nA = P + I
\nA1<\/sub> = P + I = 12000 + 1440
\n= \u20b913440
\n12000 + 1440 , = \u20b9 13440
\n\u2234 He has to pay the amount after 2 years at the rate of 6% on C.I.
\nP = \u20b912,000; R = 6%; n = 2 years
\n\"AP
\nA2<\/sub> = \u20b913483.2
\n\u2234 The difference between the C.I and S.I = 13483.2 – 13440
\n= \u20b9 43.20<\/p>\n

Question 10.
\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:
\nNo. of bacteria in a laboratory = 5,06,000
\nIf they are increased at the rate of 2.5% per hour then their number after 2 hours
\n\"AP<\/p>\n

\"AP<\/p>\n

Question 11.
\nKamala borrowed \u20b9 26400 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?
\nSolution:
\nKanala borrowed from bank = \u20b9 26400
\nRah of interest (R) =15%
\nn = 2 years
\n\"AP
\nAfter 4 rpnths the amount will be \u20b9 34914
\n\u2234 P = 34914; R = 15%; T = 4 months
\nT = \\(\\frac { 4 }{ 12 }\\) year
\n= \\(\\frac { 1 }{ 3 }\\) year
\n\u2234 \\(I=\\frac{P T R}{100}=\\frac{34914 \\times 15 \\times \\frac{1}{3}}{100}\\)
\n= \u20b91745.7
\n\u2234 Kamala has to pay the amount after 2 years and 4 months to the bank = 34914 + 1745.7
\n= \u20b936659.7<\/p>\n

\"AP<\/p>\n

Question 12.
\nBharathi borrows an amount of \u20b9 12500 at 12% per annum for3 years at a simple interest and Madhuri borrows the same amount for the same time period at 10% per annum, compounded annually. Who pays more interest and by how much?
\nSolution:
\nBharathi borrowed the sum
\nP = \u20b912500
\nR = 12%
\nT = 3 years
\nS. I (I) = \\(\\frac { PTR }{ 100 }\\)
\n= \\(\\frac{12500 \\times 12 \\times 3}{100}\\)
\n= 125 \u00d7 36
\n= 4500
\nAfter 3 years she has to pay
\n(A1<\/sub>)= P + I
\n= 12500 + 4500 .
\nA1<\/sub> = \u20b917,000
\nMadhuri has to pay the amount on
\n\"AP
\n\"AP
\nA2<\/sub> = 16637.5
\n\u2234 A1<\/sub> > A2<\/sub>
\nA1<\/sub> – A2<\/sub> = 17000 – 16637.5
\n= \u20b9 362.5
\n\u2234 Bharathi has to pay \u20b9 362.5 more than Madhuri.<\/p>\n

\"AP<\/p>\n

Question 13.
\nMachinery worth \u20b9 10000 depreciated by 5%. Find its value after 1 year.
\nSolution:
\nThe value of machinery after 1 year on 5% depreciation
\n\"AP
\n= 95 \u00d7 100
\n= \u20b9 9500<\/p>\n

Question 14.
\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 of a city = 12,00,000 If its population increases at the rate of 4%, then the population after 2 years
\n\"AP
\n= 120 \u00d7 104 \u00d7 104
\n= 12,97,920<\/p>\n

\"AP<\/p>\n

Question 15.
\nCalculate compound interest on \u20b9 1000 over a period of 1 year at 10% per annum, if interest is compounded quarterly?
\nSolution:
\ncompounded quarterly then 4 time periods will be there in 1 year.
\n\u2234 n = 4
\nC.I. on \u20b9 1000 over a period of 1 year at
\n10% per annum A = P (1 + \\(\\frac{\\mathrm{R}}{100}\\) )n<\/sup>
\nP = 1000; n = 4; R = \\(\\frac{10}{4}=\\frac{5}{2}\\) %
\n\"AP
\n= \u20b9 1103.81
\nA = \u20b9 1103.81
\nC.I. for 1 year
\n= 1103.81 – 1000
\n= \u20b9 10.81<\/p>\n","protected":false},"excerpt":{"rendered":"

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 5 Comparing Quantities Using Proportion Ex 5.3 Textbook Questions and Answers. AP State Syllabus 8th Class Maths Solutions 5th Lesson Comparing Quantities Using Proportion Exercise 5.3 Question 1. Sudhakar borrows \u20b9 15000 from a bank to renovate his house. He borrows the money at 9% … Read more<\/a><\/p>\n","protected":false},"author":4,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[6],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/6565"}],"collection":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/users\/4"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/comments?post=6565"}],"version-history":[{"count":1,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/6565\/revisions"}],"predecessor-version":[{"id":40670,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/6565\/revisions\/40670"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/media?parent=6565"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/categories?post=6565"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/tags?post=6565"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}