{"id":26849,"date":"2022-06-05T08:30:30","date_gmt":"2022-06-05T03:00:30","guid":{"rendered":"https:\/\/mcq-questions.com\/?p=26849"},"modified":"2022-05-23T15:48:04","modified_gmt":"2022-05-23T10:18:04","slug":"ncert-solutions-for-class-10-maths-chapter-3-ex-3-3","status":"publish","type":"post","link":"https:\/\/mcq-questions.com\/ncert-solutions-for-class-10-maths-chapter-3-ex-3-3\/","title":{"rendered":"NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3"},"content":{"rendered":"

These NCERT Solutions for Class 10 Maths<\/a> Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3 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 3.3<\/h2>\n

\"NCERT<\/p>\n

Question 1.
\nSolve the following pair of linear equations by the substitution method,
\n\"NCERT
\nSolution:
\n(i) The pair of linear equation formed are
\nx + y = 14 … (i)
\nx – y = 14 … (ii)
\nWe express x in the term of y from equation (ii) to get
\nx = y + 4
\nNow we substitute this value of x in equation (i) we get
\n(y + 4) + y = 14
\n2y = 14 – 4
\ny = 5
\nPutting this value of equation (ii)
\nx – 5 = 4
\nx = 9
\n\u2234 x = 9, y = 5<\/p>\n

(ii) We have
\ns – t = 3 … (i)
\n\\(\\frac { s }{ 3 }\\) + \\(\\frac { t }{ 2 }\\) = 6 … (ii)
\nWe express s in the terms of t from equation (i)
\ns = 3 + t
\n2t + 6 + 3t = 36
\n5t = 30
\nt = 6
\nPutting this value in equation (i)
\ns – 6 = 3
\ns = 9
\n\u2234 t = 6 and s = 9<\/p>\n

(iii) We have
\n3x – y = 3 … (i)
\n9x – 3y = 9 … (ii)
\nFrom equation (i)
\ny = 3x – 3
\nSubstituting this value in equation (ii) we get
\n9x – 3(3x – 3) = 9
\n9x – 9x + 9 = 0
\n9 = 9
\nThe statement is true for all value of y
\nSo y = 3x – 3
\nWhere x can take any value i.e., infinite many solution.<\/p>\n

(iv) We have
\n0.2x + 0.3y = 1.3 …(i)
\n0.4x + 0.5y = 2.3 …(ii)
\nFrom equation (i) we get
\nx = \\(\\frac{1.3-0.3 y}{0.2}\\)
\nSubstituting this value in equation (ii)
\n0.4\\(\\frac{(1.3-0.3 y)}{0.2}\\) + 0.5y = 2.3
\n2.6 – 0.6y + 0.5y = 2.3
\n– 0.1y = – 0.3
\ny = 3
\nPutting this value in equation (i) we get
\n\u2234 0.2x + 0.3 x 3 = 1.3
\n0.2x = 1.3-0.9 0.2x = 0.4 x= 2
\n\u2234 x = 2 and y = 3<\/p>\n

(v) We have
\n\\(\\sqrt{2x}\\) + \\(\\sqrt{3y}\\)y = 0 … (i)
\n\\(\\sqrt{3x}\\) + \\(\\sqrt{8y}\\)y = 0 … (ii)
\nFrom equation (i) we can get
\nx = \\(\\frac{-\\sqrt{3} y}{\\sqrt{2}}\\)
\nSubstituting this value in equation (ii)
\n\"NCERT
\nPutting this value in equation (i) we get
\n\"NCERT
\nSubstituting this value in equation (i) we get
\n\"NCERT
\nPutting this value if y if equation (i) we get<\/p>\n

\"NCERT<\/p>\n

Question 2.
\nSolve 2x + 3y = 11 and 2x – 4y = – 24 and hence find the value of \u2018m\u2019 for which y = mx + 3.
\nSolution:
\nWe have the two equations
\n2x + 3y = 11 … (i)
\n2x – 4y = – 24 … (ii)
\nFrom equation (ii)
\n2x = 4y – 24
\nx = 2y – 12
\nSubstituting this value in equation (ii)
\n2(2y – 12) + 3y = 11
\n4y – 24 + 3y = 11
\n7y = 35
\ny = 5
\nPutting this value in equation (i)
\n2x + 15 – 11 x = – 2
\nPutting these value in
\ny = mx + 3
\nwe get
\n5 = – 2m + 3
\n2 = – 2m
\nm = – 1
\nHence the value of m = – 1 and x = – 2, y = 5<\/p>\n

Question 3.
\nForm the pair of linear equations for the following problems and find their solution by substitution method.
\n(i) The difference between two numbers is 26 and one number is three times the other. Find them.<\/p>\n

(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.<\/p>\n

(iii) The coach of a cricket team buys 7 bats and 6 balls for \u20b9 3800. Later, she buys 3 bats and 5 balls for \u20b9 1750. Find the cost of each bat and each ball.<\/p>\n

(iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is \u20b9 105 and for a journey of 15 km, the charge paid is \u20b9 155. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25 km?<\/p>\n

(v) A fraction becomes 9\/11, if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and denominator it becomes 5\/6, Find the fraction.<\/p>\n

(vi) Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob\u2019s age was seven times that of his son. What are their present ages?
\nSolution:
\n(i) Let 1st number be x and 2nd number be y.
\nLet x > y
\n1st condition :
\nx – y = 26
\n2nd condition :
\nx = 3y
\nPutting x = 3y in equation (i)
\n3y – y = 26 \u21d2 2y = 26 \u21d2 y = 13
\nFrom (ii)
\nx = 3 x 13 = 39
\n\u2234 One number is 13 and the other number is 39.<\/p>\n

(ii) Let one angle be x and its supplementary angle = y
\nLet x > y
\n1st Condition :
\nx + y = 180\u00b0
\n2nd Condition :
\nx – y = 18\u00b0 \u21d2 X = 18\u00b0 + y
\nFrom equation (ii), putting the value ofx in equation (i),
\n18\u00b0 + y + y = 180\u00b0 \u21d2 18\u00b0 + 2y = 180\u00b0
\n2y = 162\u00b0 \u21d2 y = 81\u00b0
\nFrom (ii) x = 18\u00b0 + 81\u00b0 = 99\u00b0 \u21d2 x = 99\u00b0
\n\u2234 One angle is 81\u00b0 and another angle is 99\u00b0.<\/p>\n

(iii) Let the cost of each bat be x
\nand the cost of each ball by y.
\nAccording to question
\n7x + 6y = 3800 … (i)
\n3x + 5y = 1750 … (ii)
\nFrom equation (i)
\n3\\(\\frac{(3800-6 y)}{7}\\) + 5y = 1750
\n3(3800 – 6y) + 35y = 7 x 1750
\n17y = 12250 – 11400
\ny = 50
\nPutting this value in equation (i)
\n7x + 300 = 3800
\n7x = 3500
\nx = 500
\nHence the rate of each but in \u20b9 500 and the rate of each ball is \u20b9 50<\/p>\n

(iv) Let the fixed charge be x
\nand let the charge per km be y.
\nThen according to question
\nx + 10y = 105 … (i)
\nx + 15 y = 155 … (ii)
\nFrom equation (i)
\nx = 105 – 10y
\nSubstituting this value in equation (ii)
\n105 – 10y + 15y = 155
\n5y = 50
\ny = \u20b9 10 per km.
\nPutting this value in equation (i)
\nx + 100 = 105
\nx = \u20b9 5
\nA person travelling distance of 25 km the charges will be.
\n= x + 25y
\n= 5 + 25 x 10
\n= \u20b9 255<\/p>\n

(v) Let the numerator of the fraction be x
\nand the denominator of the fraction be y.
\nTherefore according to question
\n\"NCERT<\/p>\n

(vi) Let the present age of Jacob be x
\nand the present age of his son be y
\nAccording to question
\n(x + 5) = 3(y + 5)
\nAgain according to question five years ago
\n(x – 5) = 7(y – 5)
\nx – 7y = – 30
\nFrom equation (i) we get
\nx= 3y + 10
\nSubstituting this value in equation (ii)
\n3y + 10 – 7y = – 30
\n– 4y = – 40
\ny = 10
\nPutting this value in equation (i)
\nx – 30 = 10
\nx = 40
\nHence the present age of Jacob is 40 years and the present age of his son is 10 years.<\/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 3.3 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 3.3 Question 1. Solve the following pair of linear equations …<\/p>\n

NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3<\/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 3.3 - 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-3-3\/\" \/>\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 3.3 - 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 3.3 Questions and Answers are prepared by our highly skilled subject experts. 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