{"id":8911,"date":"2024-01-07T10:27:01","date_gmt":"2024-01-07T04:57:01","guid":{"rendered":"https:\/\/wordpress-505192-1602719.cloudwaysapps.com\/?p=8911"},"modified":"2024-01-08T17:46:30","modified_gmt":"2024-01-08T12:16:30","slug":"samacheer-kalvi-7th-maths-solutions-term-3-chapter-2-ex-2-4","status":"publish","type":"post","link":"https:\/\/samacheerkalviguru.com\/samacheer-kalvi-7th-maths-solutions-term-3-chapter-2-ex-2-4\/","title":{"rendered":"Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 2 Percentage and Simple Interest Ex 2.4"},"content":{"rendered":"

Students can Download Maths Chapter 2 Percentage and Simple Interest Ex 2.4 Questions and Answers, Notes Pdf, Samacheer Kalvi 7th Maths Book Solutions<\/a> Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.<\/p>\n

Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 2 Percentage and Simple Interest Ex 2.4<\/h2>\n

Question 1.
\nFind the simple interest on \u20b9 35,000 at 9% per annum for 2 years?
\nSolution:
\n\"Samacheer
\nPrincipal P = \u20b9 35,000
\nRate of interest r = 9 % Per annum
\nTime (n) = 2 years
\nSimple Interest I = \\(\\frac { Pnr }{ 100 } \\) = \\(\\frac{35000 \\times 2 \\times 9}{100}\\) = \u20b9 6300
\nSimple intrest I = \u20b9 6300<\/p>\n

\"Samacheer<\/p>\n

Question 2.
\nAravind borrowed a sum of \u20b9 8,000 from Akash at 7% per annum. Find the interest and amount to be paid at the end of two years.
\nSolution:
\nHere Principal P = \u20b9 8,000
\nRate of interest r = 7% Per annum
\nTime (n) = 2 Years
\nSimple Interest (I) = \\(\\frac { Pnr }{ 100 } \\) = \\(\\frac{8000 \\times 2 \\times 7}{100}\\)
\nI = \u20b9 1120
\nAmount = P + I
\nI = \u20b9 8000 + 1120 = 9120
\nInterest to be paid = \u20b9 1,120
\nAmount to be paid = \u20b9 9,120<\/p>\n

Question 3.
\nSheela has paid simple interest on a certain sum for 4 years at 9.5% per annum is \u20b9 21,280. Find the sum.
\nSolution:
\nLet the Principal be \u20b9 P
\nRate of interest r = 9.5% per annum
\nTime (n) = 4 years
\nSimple Interest I = \\(\\frac { Pnr }{ 100 } \\)
\nGiven I = \u20b9 21,280
\n\"Samacheer
\n\u2234 Sum of money Sheela bought = \u20b9 56,000<\/p>\n

\"Samacheer<\/p>\n

Question 4.
\nBasha borrowed \u20b9 8,500 from a bank at a particular rate of simple interest. After 3 years, he paid \u20b9 11,050 to settle his debt. At what rate of interest he borrowed the money?
\nSolution:
\nLet the rate of interest be r% per annum
\nHere Principal P = \u20b9 8,500
\nTime n = 3 years
\nTotal amount paid = \u20b9 11,050
\nA = P + 1 = \u20b9 11,050
\ni.e. 8,500 + 1 = \u20b9 11,050
\nI = \u20b9 11,050 – \u20b9 8,500 = \u20b9 2,550
\n\"Samacheer<\/p>\n

\"Samacheer<\/p>\n

Question 5.
\nIn What time will \u20b9 16,500 amount to \u20b9 22,935 at 13% per annum?
\nSolution:
\nRate of interest r = 13% per annum
\nHere Amount A = \u20b9 22,935
\nPrincipal P = \u20b9 16,500
\nA = P + I
\n22935 = 16,500 + I
\n\u2234 Interest I = 22935 – 16,500 = \u20b9 6,435
\nSimple Interest I = \\(\\frac { pnr }{ 100 } \\)
\n\"Samacheer
\n6435 = \\(\\frac{16500 \\times n \\times 13}{100}\\)
\nn =\\(\\frac{6435 \\times 100}{16500 \\times 13}\\)
\nn = 3 years
\nRequired time n = 3 years<\/p>\n

Question 6.
\nIn what time will \u20b9 17800 amount to \u20b9 19936 at 6% per annum?
\nSolution:
\nLet the require time be n years
\nHere Principal P = \u20b9 17,800
\nRate of interest r = 6% per annum
\nAmount A = \u20b9 19,936
\nA = P + I
\n19936 = 17800 + 1
\n19936 – 17800 = I
\n2136 = I
\nSimple Interest (I) = \\(\\frac { pnr }{ 100 } \\)
\n2136 = \\(\\frac{17800 \\times n \\times 6}{100}\\)
\nn = \\(\\frac{2136 \\times 100}{17800 \\times 6}\\)
\n\"Samacheer
\nn = 2 Years
\nRequired time = 2 years<\/p>\n

\"Samacheer<\/p>\n

Question 7.
\nA sum of \u20b9 48,000 was lent out at simple interest and at the end of 2 years and 3 months the total amount was \u20b9 55,560. Find the rate of interest per year.
\nSolution:
\nGiven Principal P = \u20b9 48,000
\nTime n = 2 years 3 months
\n= 2 + \\(\\frac { 3 }{ 12 } \\) years = 2 + \\(\\frac { 1 }{ 4 } \\) years
\n= \\(\\frac { 8 }{ 4 } \\) + \\(\\frac { 1 }{ 4 } \\) years = \\(\\frac { 9 }{ 4 } \\) years
\nAmount A = \u20b9 55,660
\nA = p + 1
\n55660 = 48000 + I
\nI = 55660 – 48000 = \u20b9 7660
\n\u2234 Interest for \\(\\frac { 9 }{ 4 } \\) years = \u20b9 7660
\nSimple intrest = \\(\\frac { pnr }{ 100 } \\)
\n7660 = 48000 \u00d7 \\(\\frac { 9 }{ 4 } \\) \u00d7 \\(\\frac { r }{ 100 } \\)
\nr = \\(\\frac{7660 \\times 4 \\times 100}{9 \\times 48000}\\) = 7.09 % = 7 %
\nRate of interest = 7 % Per annum<\/p>\n

Question 8.
\nA principal becomes \u20b9 17,000 at the rate of 12% in 3 years. Find the principal.
\nSolution:
\nGiven the Principal becomes \u20b9 17,000
\nLet the principle initially be P
\nRate of Interest r Time = 12 % Per annum
\nTime n = 3 years
\nAccording to the problem given I = 17000 – P = \\(\\frac{P \\times 3 \\times 12}{100}\\)
\n17000 = \\(\\frac { 36 }{ 100 } \\) p + p
\n17000 = p(\\(\\frac { 36 }{ 100 } \\) + 1)
\n17000 = p(\\(\\frac { 136 }{ 100 } \\))
\np = \\(\\frac{17000 \\times 100}{136}\\) = 12,500
\n\u2234 Principal P = \u20b9 12,500<\/p>\n

\"Samacheer<\/p>\n

Objective Type Questions<\/span><\/p>\n

Question 9.
\nThe interest for a principle of? 4,500 which gives an amount of? 5,000 at end of certain period is
\n(i) \u20b9 500
\n(ii) \u20b9 200
\n(iii) 20%
\n(iv) 15%
\nHint: Interest = Amount – Principle = \u20b9 5000 – \u20b9 4500 = \u20b9 500
\nAnswer:
\n(i) \u20b9 500<\/p>\n

Question 10.
\nWhich among the following is the simple interest for the principle of \u20b9 1,000 for one year at the rate of 10% interest per annum?
\n(i) \u20b9 200
\n(ii) \u20b9 10
\n(iii) \u20b9 100
\n(iv) \u20b9 1,000
\nHint: Intrest = \\(\\frac { pnr }{ 100 } \\) = \\(\\frac{1000 \\times 1 \\times 10}{100}\\) = \u20b9 100
\nAnswer:
\n(iii) \u20b9 100<\/p>\n

\"Samacheer<\/p>\n

Question 11.
\nWhich among the following rate of interest yields an interest of \u20b9 200 for the principle of \u20b9 2,000 for one year.
\n(i) 10%
\n(ii) 20%
\n(iii) 5%
\n(iv) 15%
\nHint: r = \\(\\frac{I \\times 100}{P \\times n}\\) = \\(\\frac{200 \\times 100}{2000 \\times 1}\\) = 10 %
\nAnswer:
\n(i) 10%<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can Download Maths Chapter 2 Percentage and Simple Interest Ex 2.4 Questions and Answers, Notes Pdf, Samacheer Kalvi 7th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations. Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 2 Percentage and …<\/p>\n","protected":false},"author":4,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":"","footnotes":""},"categories":[5],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/posts\/8911"}],"collection":[{"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/users\/4"}],"replies":[{"embeddable":true,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/comments?post=8911"}],"version-history":[{"count":1,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/posts\/8911\/revisions"}],"predecessor-version":[{"id":60307,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/posts\/8911\/revisions\/60307"}],"wp:attachment":[{"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/media?parent=8911"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/categories?post=8911"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/tags?post=8911"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}