{"id":9890,"date":"2021-04-12T11:34:04","date_gmt":"2021-04-12T06:04:04","guid":{"rendered":"https:\/\/apboardsolutions.guru\/?p=9890"},"modified":"2021-09-15T11:34:12","modified_gmt":"2021-09-15T06:04:12","slug":"ap-board-7th-class-maths-solutions-chapter-6-ex-6","status":"publish","type":"post","link":"https:\/\/apboardsolutions.guru\/ap-board-7th-class-maths-solutions-chapter-6-ex-6\/","title":{"rendered":"AP Board 7th Class Maths Solutions Chapter 6 Ratio – Applications Ex 6"},"content":{"rendered":"

AP State Syllabus AP Board 7th Class Maths Solutions<\/a> Chapter 6 Ratio – Applications Ex 6 Textbook Questions and Answers.<\/p>\n

AP State Syllabus 7th Class Maths Solutions 6th Lesson Ratio – Applications Exercise 6<\/h2>\n

\"AP<\/p>\n

Question 1.
\nHow long will it take for a sum of \u20b9 12600 invested at 9% per annum beocme to \u20b9 15624?
\nSolution:
\nHereA = \u20b9 15,624 R = 9% T = ?
\nP = \u20b9 12,600
\n\u2234 I = A – P = 15,624 – 12,600 = \u20b9 3024
\nAlso I = \\(\\frac{P R T}{100}\\)
\n\u2234 3024 = \\(\\frac{12,600 \\times 9 \\times \\mathrm{T}}{100}\\)
\n\"AP<\/p>\n

Question 2.
\nAt what rate a sum doubles itself in 8 year 4 months?
\nSolution:
\nGiven that A = double the sum
\nLet the principle be P
\nthen A = 2P
\nT =8 years 4 months = 8y \\(\\frac { 4 }{ 12 }\\) y = 8\\(\\frac { 1 }{ 3 }\\) years = \\(\\frac { 25 }{ 3 }\\)
\nR = R% say
\nWe know that
\nI = \\(\\frac{\\text { PTR }}{100}\\)
\nHere I = A – P
\n= 2P – P = P
\n\u2234 P = \\(\\frac{P \\cdot \\frac{25}{3} \\times R}{100}\\)
\n\u2234 R = 100 x \\(\\frac{3}{25}\\) = 12%<\/p>\n

\"AP<\/p>\n

Question 3.
\nA child friendly bank announces a savings scheme for school children. They will give kiddy banks to children. Children have to keep their savings in it and the bank collects all the money once in a year. To encourage children savings, they give 6% interest if the amount exceeds by \u20b9 10000. and other wise 5%. Find the interest received by a school if they deposit is \u20b9 9000 for one year.
\nSolution:
\nMoney deposited = \u20b9 9000
\nInterest applicable 5% on \u20b9 9000
\n= 5 \u00d7 \\(\\frac{9000}{100}\\) = \u20b9 450<\/p>\n

Question 4.
\nA sum of money invested at 8% per annum for simple interest amounts to \u20b9 12122 in 2 years. What will it amounts to in 2 year 8 months at 9% rate of interest?
\nSolution:
\nFirst part
\nPirnciple = P say
\nR=8%
\nT = 2 years
\nA = \u20b9 12,122
\n\"AP<\/p>\n

Second part
\nP = \u20b910,450
\nR = 9%
\nT = 2 years 8 months, A = ?
\n\"AP<\/p>\n

\"AP<\/p>\n

Question 5.
\nIn 4 years, \u20b9 6500 amounts to \u20b9 8840 at a certain rate of interest. In what time will \u20b9 1600 amounts to 1816 at the same rate?
\nSolution:
\nFirst part
\nT = 4 years
\nP = \u20b9 6500
\nA = \u20b9 8840
\nR = R% = ?
\nI = A – P = 8840 – 6500 = \u20b9 2340
\nBut I = \\(\\frac{P T R}{100}\\)
\n\u2234 2340 = \\(\\frac{6500 \\times 4 \\times R}{100}\\)
\n\u2234 R = \\(\\frac{2340 \\times 10 \\theta}{6500 \\times 4}=\\frac{36}{4}\\) = 9%<\/p>\n

Second part
\nP = \u20b9 1600
\nA = \u20b9 1816
\nT = ?
\nR = 9%
\nI = A – P = 1816 – 1600 = \u20b9 216
\nBut I = \\(\\frac{P T R}{100}\\)
\n\u2234 2340 = \\(\\frac{1600 \\times T \\times 9}{100}\\)
\n\u2234 R = \\(\\frac{216 \\times 100 \\theta}{1600 \\times 9}=\\frac{3}{2}\\) = 1 \\(\\frac { 1 }{ 2 }\\) years.<\/p>\n","protected":false},"excerpt":{"rendered":"

AP State Syllabus AP Board 7th Class Maths Solutions Chapter 6 Ratio – Applications Ex 6 Textbook Questions and Answers. AP State Syllabus 7th Class Maths Solutions 6th Lesson Ratio – Applications Exercise 6 Question 1. How long will it take for a sum of \u20b9 12600 invested at 9% per annum beocme to \u20b9 …<\/p>\n","protected":false},"author":4,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[9],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/9890"}],"collection":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/users\/4"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/comments?post=9890"}],"version-history":[{"count":0,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/9890\/revisions"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/media?parent=9890"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/categories?post=9890"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/tags?post=9890"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}