{"id":9889,"date":"2023-06-01T11:32:22","date_gmt":"2023-06-01T06:02:22","guid":{"rendered":"https:\/\/apboardsolutions.in\/?p=9889"},"modified":"2023-06-02T18:00:58","modified_gmt":"2023-06-02T12:30:58","slug":"ap-board-7th-class-maths-solutions-chapter-6-ex-5","status":"publish","type":"post","link":"https:\/\/apboardsolutions.in\/ap-board-7th-class-maths-solutions-chapter-6-ex-5\/","title":{"rendered":"AP Board 7th Class Maths Solutions Chapter 6 Ratio – Applications Ex 5"},"content":{"rendered":"

AP State Syllabus AP Board 7th Class Maths Solutions<\/a> Chapter 6 Ratio – Applications Ex 5 Textbook Questions and Answers.<\/p>\n

AP State Syllabus 7th Class Maths Solutions 6th Lesson Ratio – Applications Exercise 5<\/h2>\n

\"AP<\/p>\n

Question 1.
\nA shopkeeper bought a suit case for \u20b9 480 and sold it for \u20b9 540. Find his gain percent?
\nSolution:
\nC.P of the suitcase = \u20b9 480
\nSP of the suitcase = \u20b9 540
\n\u2234 gain=S,P-C.P = 540 – 480 = \u20b9 60
\n\u2234 gain percent = \\(\\frac { 60 }{ 450 }\\) \u00d7 100% = 12.5%<\/p>\n

Question 2.
\nAjay bought a TV for \u20b9 15000 and sold it for \u20b9141 00. Find the loss percent?
\nSolution:
\nC.P of the TV = \u20b9 15,000
\nS.P of the TV = \u20b9 14100
\nAs S.P < C.P
\nLoss = C.P – S.P= 15,000 – 14,100 = \u20b9 900
\n\u2234 Loss percent = \\(\\frac { 900 }{ 15000 }\\) \u00d7 100 = 6%<\/p>\n

\"AP<\/p>\n

Question 3.
\nRamu sold a plot of land for \u20b9 2,40,000 gaining 20%. For how much did he purchase the plot?
\nSolution:
\nLet the cost price of the land = \u20b9 x
\ngain percent = 20% of the C.P
\n\u2234 S.P of the land = 120% of C.P
\nBy problem, 120% of x = 2,40,000
\n\\(\\frac { 120. x }{ 100 }\\) = 2,40,000
\nx = 2.40000 \u00d7 \\(\\frac {100 }{ 120 }\\)
\n= \u20b9 2,00,000<\/p>\n

(OR)<\/p>\n

Let the C.P of the land = \u20b9 x
\ngain percent = 20% of C.P = 20% of x = \\(\\frac{20 \\mathrm{x}}{100}\\)
\n\u2234 S.P of the land = C.P + gain
\n= x + \\(\\frac{20 \\mathrm{x}}{100}\\)
\n= \\(\\frac{100 x+20 x}{100}=\\frac{120 x}{100}\\)<\/p>\n

By problem, \\(\\frac{20 \\mathrm{x}}{100}\\) = 2,40.000
\n\u2234 x = 24000 x \\(\\frac { 100 }{ 120 }\\)
\n= \u20b9 2,00,000<\/p>\n

\"AP<\/p>\n

Question 4.
\nOn selling a mobile for \u20b9 750, a shop keeper looses 10%. For what amount should he sell it to gain 5%?
\nSolution:
\nLet the C.P of the mobile be x
\nLosspercent = 10% of C.P = 10% of x = \\(\\frac{10 \\mathrm{x}}{100}\\)
\n\u2234 Loss = \\(\\frac{10 x}{100}\\)
\n\u2234 S.P = C.P – Loss
\n= \\(x – \\frac{10 x}{100}=\\frac{100 x-10 x}{100}=\\frac{90 x}{100}\\)<\/p>\n

By problem, \\(\\frac{90 \\mathrm{x}}{100}\\) = 750
\nx = 750 \u00d7 \\(\\frac { 100 }{ 90 }\\) = \u20b9 \\(\\frac { 2500 }{ 3 }\\)
\nNow to gain by 5% the S.P = C.P + 5% of C.P
\n\"AP<\/p>\n

Question 5.
\nA farmer sold 2 bullocks for \u20b9 24000 each. On one bullock he gained 25% and on the other he lost 20%. Find his total profit or loss percent?
\nSolution:
\nLet the C.P of first bullock be x
\ngain = 25% of x = 25 \u00d7 \\(\\frac{x}{100}=\\frac{25 x}{100}\\)
\nS.P = C.P + gain = x + \\(\\frac{25 x}{100}=\\frac{100 x+25 x}{100}=\\frac{125 x}{100}\\)
\nSelling price of first bullock 24,000 (given)
\n\\(\\frac{125 x}{100}\\) = 24000
\nx = \\(\\frac{24000}{125}\\) \u00d7 100
\n\u2234 C.P of first bullock = \u20b9 19.200
\nLet the C.P of second bullock be y
\nLoss percent = 20%<\/p>\n

\u2234Loss = \\(\\frac{20 \\times \\mathrm{y}}{100}=\\frac{20 \\mathrm{y}}{100}\\)<\/p>\n

S.P = C.P – loss
\n= \\(y-\\frac{20 y}{100}=\\frac{100 y-20 y}{100}=\\frac{80 y}{100}\\)<\/p>\n

But S.P = \u20b9 24,000 (given)
\n\\(\\frac{80 \\mathrm{y}}{100}\\) = 24,000
\n\u2234 y = 24000 x \\(\\frac{100}{80}\\) = 30,000<\/p>\n

\u2234 C.P of second bullock = \u20b9 30,000
\nC.P of 2 bullocks = \u20b919,200 + \u20b930,000
\n= \u20b9 49,200
\nS.P of 2 bullocks \u20b924,000 \u00d7 2 = \u20b948,000
\nLoss = C.P – S.P = 49,200 – 48,000 = \u20b91200
\nLoss % = \\(\\frac{\\text { loss }}{\\text { CP }}\\) \u00d7 100
\n= \\(\\frac{1200}{49200}\\) \u00d7 100 = 2.4
\n\u2234 Loss% = 2.4%<\/p>\n

\"AP<\/p>\n

Question 6.
\nSravya bought a watch for \u20b9480. She sold it to Ridhi at a gain of 6\\(\\frac { 1 }{ 4 }\\) %. Ridhi sold it to Divya at a gain of 10%. How much did Divya pay for it?
\nSolution:
\nCF of the watch = \u20b9 480
\nGain% = \\(6 \\frac{1}{4} \\%=\\frac{25}{4} \\%\\)
\nGain = 480 \u00d7 \\(\\frac{25}{4}\\) x \\(\\frac{1}{100}\\) = 30
\nC.P of the watch for Ridhi = 480 + 30 = \u20b9 510
\nGain% = 10%
\n\u2234 Gain= 51% \u00d7 \\(\\frac{10}{100}\\) = 51
\nS.P of the watch for Ridhi = 510 + 51 = \u20b9561
\n\u2234 Amount paid for Divya for the watch = \u20b9561<\/p>\n

Question 7.
\nThe marked price of a book is \u20b9225.Thc publisher allows a discount of \u20b910% on it. Find the selling price of it?
\nSolution:
\nThe marked price of the book = \u20b9 225
\nDiscount 10%
\n\u2234 Discount = 225 \u00d7 \\(\\frac{10}{100}\\)= \u20b9 22.5
\nS.P = M.P – discount
\n= 225 – 22.5 = \u20b9202.5
\n\u2234 Selling price of the book = \u20b9 202.5<\/p>\n

\"AP<\/p>\n

Question 8.
\nA carpenter allows 15% discount on his goods. Find the marked price of a chair which is sold by him for \u20b9680?
\nSolution:
\nLet the marked price of chair be \u20b9x
\nDiscount = 15% = x \u00d7 \\(\\frac{15}{100}=\\frac{15 x}{100}\\)
\nS.P = M.P – discount
\n\\(x-\\frac{15 x}{100}=\\frac{100 x-15 x}{100}=\\frac{85 x}{100}\\)
\nS.P = \u20b9 680 (given)
\n\u2234 \\(\\frac{85 \\mathrm{x}}{100}\\) = 680
\nx = \\(\\frac{680 \\times 100}{85}\\) = 800
\nMarked price of the chair = \u20b9 800<\/p>\n

\"AP<\/p>\n

Question 9.
\nA dealer allows a discount of \u20b9 10% and still gains by 10%. What should be the marked price if the cost price is \u20b9900?
\nSolution:
\nGiven that C.P = 900
\nThere is a gain of 10% on C.P
\n\u2234 Gain = 10% of \u20b9 900
\n= \\(\\frac{10 \\times 900}{100}\\) = \u20b9 90
\n\u2234 S.P = C.P . gain
\n= 900 + 90
\nS.P = \u20b9990 …………….. (1)
\nLet the marked price = \u20b9 x
\nDiscount on it = 10% on M.P = \\(\\frac{10 \\mathrm{x}}{100}\\)
\n\u2234 Selling Price = MP – Discount
\n= x – \\(\\frac{10 \\mathrm{x}}{100}\\)
\n= \\(\\frac{100 x-10 x}{100}\\)
\n\u2234 S.P = \\(\\frac{90 \\mathrm{x}}{100}\\) …………….(2)
\nFrom (1) & (2) \\(\\frac{90 \\mathrm{x}}{100}\\) = 990
\n\u2234 x = 990 x \\(\\frac{100}{90}\\)
\n\u2234 M.P = x = \u20b91100<\/p>\n","protected":false},"excerpt":{"rendered":"

AP State Syllabus AP Board 7th Class Maths Solutions Chapter 6 Ratio – Applications Ex 5 Textbook Questions and Answers. AP State Syllabus 7th Class Maths Solutions 6th Lesson Ratio – Applications Exercise 5 Question 1. A shopkeeper bought a suit case for \u20b9 480 and sold it for \u20b9 540. Find his gain percent? … 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":[9],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/9889"}],"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=9889"}],"version-history":[{"count":1,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/9889\/revisions"}],"predecessor-version":[{"id":40948,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/9889\/revisions\/40948"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/media?parent=9889"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/categories?post=9889"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/tags?post=9889"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}