{"id":26268,"date":"2022-06-05T11:00:01","date_gmt":"2022-06-05T05:30:01","guid":{"rendered":"https:\/\/mcq-questions.com\/?p=26268"},"modified":"2022-05-23T15:49:23","modified_gmt":"2022-05-23T10:19:23","slug":"ncert-solutions-for-class-10-maths-chapter-3-ex-3-6","status":"publish","type":"post","link":"https:\/\/mcq-questions.com\/ncert-solutions-for-class-10-maths-chapter-3-ex-3-6\/","title":{"rendered":"NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.6"},"content":{"rendered":"

These NCERT Solutions for Class 10 Maths<\/a> Chapter 3 Pair of Linear Equations in Two Variables Ex 3.6 Questions and Answers are prepared by our highly skilled subject experts.<\/p>\n

NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.6<\/h2>\n

\"NCERT<\/p>\n

Question 1.
\nSolve the following pairs of equations by reducing them to a pair of linear equations:
\n\"NCERT
\nSolution:
\n\"NCERT
\n<\/strong>By cross multiplication method
\n\"NCERT
\n<\/strong>By cross multiplication method
\n\"NCERT
\n<\/strong>By cross multiplication method
\n\"NCERT
\n<\/strong>By cross multiplication method
\n\"NCERT
\n\"NCERT
\n<\/strong>solving for u and v by cross multiplication method:
\n\"NCERT
\n<\/strong>By cross multiplication method:
\n\"NCERT
\n<\/strong>By cross multiplication method:
\n\"NCERT<\/p>\n

(viii) We have
\n\"NCERT
\n(i) and (iii), we get
\nP + q = – \\(\\frac { 3 }{ 4 }\\) … (iii)
\n\\(\\frac { 1 }{ 2 }\\)P – \\(\\frac { 1 }{ 2q }\\) = \\(\\frac { -1 }{ 8 }\\) … (iv)
\nBy eliminating method Adding equations (iii) and (iv) we get
\nq = \\(\\frac { 1 }{ 2 }\\)
\nBut P = \\(\\frac { 1 }{ 3x+y }\\)
\nand q = \\(\\frac { 1 }{ 3x – y }\\)
\n\\(\\frac { 1 }{ 3x+y }\\) = \\(\\frac { 1 }{ 4 }\\)
\nand
\n3x + y = 4 … (v)
\n3x – y = 2 … (vi)
\nBy eliminating method
\nAdding equations (v) and (vi) we get
\n6x = 6
\nx = 1<\/p>\n

\"NCERT<\/p>\n

Question 2.
\nFormulate the following problems as a pair of equations, and hence find their solutions:
\n(i) Ritu can row’ downstream 20 km in 2 hours, and upstream 4 km in 2 hours. Find her speed of rowing in still water and the speed of the current.<\/p>\n

(ii) 2 women and 5 men can together finish an embroidery work in 4 days, while 3 women and 6 men can finish it in 3 days. Find the time taken by 1 woman alone to finish the work, and also that taken by 1 man alone.<\/p>\n

(iii) Roohi travels 300 km to her home partly by train and partly by bus. She takes 4 hours if she travels 60 km by train and the remaining by bus. If she travels 100 km by train and the remaining by bus, she takes 10 minutes longer. Find the speed of the train and the bus separately.
\nSolution:
\n(i) Let Ritu\u2019s speed in still water = x km\/h
\nSpeed of current = y km\/h
\nDuring downstream, speed = (x + y) km\/h
\nDuring upstream, speed = (x – y) km\/h
\nIn the first case
\nThe time taken in hour be f, then
\n\"NCERT
\nBy eliminating method
\nAdding equations (i) and (ii), we get
\n2x = 12
\nx = 6
\nPutting this value in equation (ii), we get
\nWe get
\n– y = 2 – 6
\ny = 4
\nThen x = 6 and y = 4 where x and y are respectively speed (in km\/h) of rowing and current.<\/p>\n

(ii) Let the number of days taken by one woman by n and the number of days taken by one man by m.
\nAccording to question
\n\\(\\frac { 2 }{ n }\\) + \\(\\frac { 5 }{ m }\\) = \\(\\frac { 1 }{ 4 }\\) … (i)
\n\\(\\frac { 3 }{ n }\\) + \\(\\frac { 6 }{ m }\\) = \\(\\frac { 1 }{ 3 }\\) … (i)
\nPutting \\(\\frac { 1 }{ n }\\) = p and \\(\\frac { 1 }{ m }\\) = q in equations (i) and (ii) we get
\n2p + 5q = \\(\\frac { 1 }{ 4 }\\) … (iii)
\n3p + 6q = \\(\\frac { 1 }{ 3 }\\) … (iv)
\nor 8p + 20q – 1 = 0 … (iii)
\n9p + 18q -1 = 0 … (iv)
\nBy cross multiplication method
\n\"NCERT<\/p>\n

(iii) Let the speed of the train be u km\/h
\nand the speed of the bus be v km\/h
\nIn the first case
\n\\(\\frac { 60 }{ u }\\) + \\(\\frac { 240 }{ v }\\) = 4 … (i)
\nIn the second case
\n\\(\\frac { 100 }{ u }\\) + \\(\\frac { 200 }{ v }\\) = \\(\\frac { 25 }{ 6 }\\) … (ii)
\nPutting \\(\\frac { 1 }{ u }\\) = p and \\(\\frac { 1 }{ v }\\) = q in equations (i) and (ii)
\nwe get
\n60p + 240q = 4 … (iii)
\n100p + 200p = \\(\\frac { 25 }{ 6 }\\) … (iv)
\nor 60p + 240p – 4 = 0 … (v)
\n600p + 120q – 25 = 0 … (vi)
\nBy using elemination method multiply equation (ii) by 10 we get
\n600p + 2400 – 40 = 0 … (vii)
\nSubstract equation (iv) from equations (v)
\n1200q = 15
\n1200q = 15
\nq = \\(\\frac { 15 }{ 12000 }\\)
\nq = \\(\\frac { 1 }{ 80 }\\)
\nPutting this value in equations (iv)
\nP = \\(\\frac { 1 }{ 60 }\\)
\nBut p = \\(\\frac { 1 }{ u }\\) and q = \\(\\frac { 1 }{ v }\\)
\n\\(\\frac { 1 }{ u }\\) = \\(\\frac { 1 }{ 60 }\\) and \\(\\frac { 1 }{ v }\\) = \\(\\frac { 1 }{ 80 }\\)
\nu = 60 and v = 80<\/p>\n

\"NCERT<\/p>\n","protected":false},"excerpt":{"rendered":"

These NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.6 Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.6 Question 1. Solve the following pairs of equations by …<\/p>\n

NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.6<\/span> Read More »<\/a><\/p>\n","protected":false},"author":9,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"default","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","theme-transparent-header-meta":"default","adv-header-id-meta":"","stick-header-meta":"default","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","spay_email":""},"categories":[2],"tags":[],"yoast_head":"\nNCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.6 - MCQ Questions<\/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:\/\/mcq-questions.com\/ncert-solutions-for-class-10-maths-chapter-3-ex-3-6\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.6 - MCQ Questions\" \/>\n<meta property=\"og:description\" content=\"These NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.6 Questions and Answers are prepared by our highly skilled subject experts. 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