{"id":44610,"date":"2022-05-30T11:30:31","date_gmt":"2022-05-30T06:00:31","guid":{"rendered":"https:\/\/cbselibrary.com\/?p=44610"},"modified":"2023-01-25T12:16:36","modified_gmt":"2023-01-25T06:46:36","slug":"ml-aggarwal-class-8-solutions-for-icse-maths-chapter-8-ex-8-1","status":"publish","type":"post","link":"https:\/\/cbselibrary.com\/ml-aggarwal-class-8-solutions-for-icse-maths-chapter-8-ex-8-1\/","title":{"rendered":"ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 8 Simple and Compound Interest Ex 8.1"},"content":{"rendered":"

ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 8 Simple and Compound Interest Ex 8.1<\/h2>\n

Question 1.
\nFind the simple interest on \u20b94000 at 7.5% p.a. for 3 years 3 months. Also find the amount.
\nSolution:
\nHere, P (Principal) = \u20b94000
\nR (rate of interest) = 7.5% p.a. = \\(\\frac{15}{2}\\)% p.a
\nT (time) = 3 years 3 months
\n\"ML
\nAmount = P + I = \u20b914000 + \u20b9975 = \u20b94975<\/p>\n

Question 2.
\nWhat sum of money will yield \u20b9170\u00b710 as simple interest in 2 years 3 months at 6% per annum?
\nSolution:
\nHere, I = \u20b9170\u00b710, T = 2 years 3 months
\n= 2\\(\\frac{3}{12}\\) years = 2\\(\\frac{1}{4}\\) years
\n= \\(\\frac{9}{4}\\) years
\nR = 6%
\n\"ML<\/p>\n

Question 3.
\nFind the rate of interest when \u20b9800 fetches \u20b9130 as simple interest in 2 years 6 months.
\nSolution:
\nHere, P = \u20b9800, I = \u20b9130
\nT = 2 years 6 months = 2\\(\\frac{6}{12}\\) years
\n\"ML
\n= \\(\\frac{13}{2}\\) % p.a.= 6\u00b75% p.a.<\/p>\n

Question 4.
\nFind the time when simple interest on \u20b93\u00b73 lakhs at 6\u00b75% per annum is \u20b975075.
\nSolution:
\nHere, P = 3\u00b73 lakhs = \u20b93\u00b73 \u00d7 100000 = \u20b9330000
\nR = 6\u00b75% per annum
\nI= \u20b975075
\n\"ML<\/p>\n

Question 5.
\nFind the sum of money when
\n(i) simple interest at 7\\(\\frac{1}{4}\\)% p.a. for years is \u20b92356\u00b725
\n(ii) final amount is \u20b9 11300 at 4% p.a. for 3 years 3 months.
\nSolution:
\n(i) Here, I = \u20b92356\u00b725
\n\"ML
\n\"ML<\/p>\n

(ii) Amount (A) = \u20b911300
\nRate (R) = 4% p.a.
\nTime (T) = 3 years 3 months
\n\"ML
\n\"ML
\nHence principal (P) = \u20b910000<\/p>\n

Question 6.
\nHow long will it take a certain sum of money to triple itself at 13\\(\\frac{1}{3}\\)% per annum simple interest?
\nSolution:
\nLet the sum of money = \u20b9 x
\nAmount = 3 \u00d7 \u20b9 x = \u20b9 3x
\nInterest = Amount – Principal = \u20b93x – \u20b9x = \u20b92x
\n\"ML<\/p>\n

Question 7.
\nAt a certain rate of simple interest \u20b94050 amounts to \u20b94576\u00b750 in 2 years. At the same rate of simple interest, how much would \u20b91 lakh amount to in 3 years?
\nSolution:
\nHere, P = \u20b94000, A = \u20b94576\u00b750,
\nT = 2 years
\nI = A – P = \u20b94576\u00b750 – \u20b94050 = \u20b9526.50
\nLet the rate of simple interest be R% per annum
\n\"ML
\n\"ML
\nAmount = Principal + Interest = \u20b9100000 + \u20b919500 = \u20b9119500<\/p>\n

Question 8.
\nWhat sum of money invested at 7.5% p.a. simple interest for 2 years produces twice as much interest as \u20b99600 in 3 years 6 months at 10% p.a. simple interest?
\nSolution:
\nIn first case,
\nPrincipal (P1<\/sub>) = \u20b99600
\nRate (R1<\/sub>) = 10%
\nPeriod (T) = 3 years 6 months = \\(3 \\frac{1}{2}=\\frac{7}{2}\\) years
\nS. Interest = \\(\\frac{P R T}{100}=\\frac{9600 \\times 10 \\times 7}{100 \\times 2}\\) = \u20b93360
\nIn second case,
\nS. Interest = \u20b93360 \u00d7 2 = \u20b96720
\nRate (R) = 7.5% p.a.
\nand Period (T) = 2 years
\n\"ML<\/p>\n

ML Aggarwal Class 8 Solutions for ICSE Maths<\/a><\/h4>\n","protected":false},"excerpt":{"rendered":"

ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 8 Simple and Compound Interest Ex 8.1 Question 1. Find the simple interest on \u20b94000 at 7.5% p.a. for 3 years 3 months. Also find the amount. Solution: Here, P (Principal) = \u20b94000 R (rate of interest) = 7.5% p.a. = % p.a T (time) = … Read more<\/a><\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[3034],"tags":[],"yoast_head":"\nML Aggarwal Class 8 Solutions for ICSE Maths Chapter 8 Simple and Compound Interest Ex 8.1 - CBSE Library<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/cbselibrary.com\/ml-aggarwal-class-8-solutions-for-icse-maths-chapter-8-ex-8-1\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 8 Simple and Compound Interest Ex 8.1\" \/>\n<meta property=\"og:description\" content=\"ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 8 Simple and Compound Interest Ex 8.1 Question 1. Find the simple interest on \u20b94000 at 7.5% p.a. for 3 years 3 months. Also find the amount. Solution: Here, P (Principal) = \u20b94000 R (rate of interest) = 7.5% p.a. = % p.a T (time) = ... 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