NCERT Solutions for Class 10 Maths<\/a> Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2 Questions and Answers are prepared by our highly skilled subject experts.<\/p>\nNCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.2<\/h2>\n <\/p>\n
Question 1. \nFrom the pair of linear equations in the following problems and find their solutions graphically. \n(i) 10 students of Class X took part in mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz.<\/p>\n
(ii) 5 pencils and 7 pens together cost \u20b9 50, whereas 7 pencils and 5 pens together cost \u20b9 45. Find the cost of one pencil and one pen. \nSolution: \n(i) Let the number of boys participating in mathematics quiz is x and the number of girls participating in mathematics quiz is y. \nTherefore the equations form in this situations are \n <\/p>\n
(ii) Let the cost of 1 pencil is \u20b9 x \nand the cost of 1 pen is \u20b9 y \nTherefore the linear equations on the given situations are \n5x + 7y = 50 … (i) \nand 7x + 5y = 46 … (ii) \n \nAgain from equation (ii) \n \n7x + 5y = 46 \n\u2234 x = \\(\\frac{46-5 y}{7}\\) \nMultiply equation. (1) by 7 and (ii) by 5 the get \n35x + 49y = 350 … (i) \n35x + 25y = 230 …(ii) \nSubstracting equation (ii) from equation (i) we get \n24y = 120 \ny = 5 \nPutting the value of y in equation (j) \nWe get \n5x + 35 = 50 \n5x = 15 \nx = 5 \nCost of one pencil = \u20b9 3 \nCost of one pen = \u20b9 5<\/p>\n
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Question 2. \nOn comparing the ratios \\(\\frac{a_{1}}{a_{2}}, \\frac{b_{1}}{b_{2}}\\) and \\(\\frac{c_{1}}{c_{2}}\\) find out whether the lines representing the following pairs of linear equations intersect at a point, parallel or coincident. \n(i) 5x – 4y + 8 = 0 \n7x + 6y-9 = 0<\/p>\n
(ii) 9x + 3y + 12 = 0 \n18x + 6y + 24 = 0<\/p>\n
(iii) 6x – 3y + 10 = 0 \n2x – y + 9 = 0 \nSolution: \n(i) We have, \n5x – 4y + 8 = 0 \nand 7x + 6y – 9 = 0 \nHere, a1<\/sub> = 5, b1<\/sub> = – 4, c1<\/sub> = 8 \nand a2<\/sub> = 7, b2<\/sub> = 6 and c2<\/sub> = – 9 \nTherefore, \\(\\frac{5}{7} \\neq \\frac{-4}{6}\\) \nSo, \\(\\frac{a_{1}}{a_{2}} \\neq \\frac{b_{1}}{b_{2}}\\) \nTherefore the following pair of linear equations intersect each other at point.<\/p>\n(ii) We have, \n9x + 3y + 12 = 0 \n18x + 6y + 24 = 0 \nHere, a1<\/sub> = 9, b1<\/sub> = 3, c1<\/sub> = 12 \nand a2<\/sub> = 18, b2<\/sub> = 6 and c2<\/sub> = 24 \nTherefore, \\(\\frac{9}{18}=\\frac{3}{6}=\\frac{12}{24}=\\frac{1}{2}\\) \nSo, \\(\\frac{a_{1}}{a_{2}}=\\frac{b_{1}}{b_{2}}=\\frac{c_{1}}{c_{2}}\\) \nTherefore the lines representing the given pair of linear equations are coincident.<\/p>\n(iii) We have given, \n6x – 3y + 10 = 0 \n2x – y + 9 = 0 \nHere, a1<\/sub> = 6, b1<\/sub> = – 3, c1<\/sub> = 10 \nand a2<\/sub> = 2, b2<\/sub> = – 1 and c2<\/sub> = 9 \nTherefore, \n\\(\\frac{6}{2}=\\frac{-3}{-1} \\neq \\frac{10}{9}\\) \nSo, \\(\\frac{a_{1}}{a_{2}}=\\frac{b_{1}}{b_{2}} \\neq \\frac{c_{1}}{c_{2}}\\) \nTherefore the lines representing the given pair of linear equations are coincident.<\/p>\nQuestion 3. \nOn comparing the ratios \\(\\frac{a_{1}}{a_{2}}, \\frac{b_{1}}{b_{2}}\\) and \\(\\frac{c_{1}}{c_{2}}\\), find out whether the following pair of linear equations are consistent or inconsistent. \n(i) 3% + 2y = 5 \n2x – 3y = 7<\/p>\n
(ii) 2x – 3y = 8 \n4x – 6y = 9<\/p>\n
(iii) \\(\\frac { 3 }{ 2 }\\)x + \\(\\frac { 5 }{ 3 }\\)y = 7 \n9x – 10y = 14<\/p>\n
(iv) 5x – 3y = 11 \n– 10x + 6y = – 22<\/p>\n
(v) \\(\\frac { 4 }{ 3 }\\)x + 2y = 8 \n2x + 3y = 12 \nSolution: \n(i) We have given, \n3x + 2y = 5 \n2x – 3y = 7 \nWe can also write \n3x + 2y – 5 = 0 \n2x – 3y – 7 = 0 \nHere, a1<\/sub> = 3, b1<\/sub> = 2, c1<\/sub> = – 5 \nand a2<\/sub> = 2, b2<\/sub> = – 3 and c2<\/sub> = – 7 \nSo, \\(\\frac { 3 }{ 2 }\\) \u2260 \\(\\frac { 2 }{ – 3 }\\) \nTherefore, \\(\\frac{a_{1}}{a_{2}} \\neq \\frac{b_{1}}{b_{2}}\\) \nSo the given pair of linear equations are consistent.<\/p>\n(ii) We have given, \n2x – 3y = 8 \n4x – 4y = 9 \nWe can also write \n2x – 3y – 8 = 0 \n4x – 6y – 3 = 0 \nHere, a1<\/sub> = 2, b1<\/sub> = – 3, c1<\/sub> = – 8 \nand a2<\/sub> = 4, b2<\/sub> = – 6 and c2<\/sub> = – 3 \nSo, \\(\\frac{2}{4}=\\frac{-3}{-6} \\neq \\frac{-8}{-9}\\) \nTherefore, \nSo, the given pair of linear equations are inconsistent.<\/p>\n(iii) We have given, \n\\(\\frac { 3 }{ 2 }\\)x + \\(\\frac { 5 }{ 3 }\\)y = 7 \n9x – 10y = 14 \nWe can also write, \n\\(\\frac { 3 }{ 2 }\\)x + \\(\\frac { 5 }{ 3 }\\)y – 7 = 0 \n9x – 10y – 14 = 0 \nHere, a1<\/sub> = \\(\\frac { 3 }{ 2 }\\), b1<\/sub> = \\(\\frac { 5 }{ 3 }\\), c1<\/sub> = – 7 \nand a2<\/sub> = 9, b2<\/sub> = – 10 and c2<\/sub> = – 14 \n \nSo, the given pair of linear equations consistent.<\/p>\n(iv) We have given, \n5x – 3y = 11 \n– 10x + 6y = – 22 \nWe can also write, \n5x – 3y – 11 = 0 \n– 10x + 6y + 22 = 0 \nSo, \\(\\frac { 5 }{ -10 }\\) = \\(\\frac { – 3 }{ 6 }\\) = \\(\\frac { -11 }{ 22 }\\) \nHere, a1<\/sub> = 5, b1<\/sub> = – 3, c1<\/sub> = – 11 \nand a2<\/sub> = – 10, b2<\/sub> = 6 and c2<\/sub> = 22 \nTherefore, \\(\\frac{a_{1}}{a_{2}}=\\frac{b_{1}}{b_{2}}=\\frac{c_{1}}{c_{2}}\\) \nSo, the given pair of linear equations consistent.<\/p>\n(v) We have given \n\\(\\frac { 4 }{ 3 }\\)x + 2y = 8 \n2x + 3y = 12 \nWe can also write, \n\\(\\frac { 4 }{ 3 }\\)x + 2y – 8 = 0 \n2x + 3y – 12 = 0 \nHere, a1<\/sub> = \\(\\frac { 4 }{ 3 }\\), b1<\/sub> = 2, c1<\/sub> = 8 \nand a2<\/sub> = 2, b2<\/sub> = 3 and c2<\/sub> = – 12 \n \nSo, the given pair of linear equations consistent.<\/p>\n <\/p>\n
Question 4. \nWhich of the following pairs of linear equations are consistent inconsistent ? If consistent, obtain the solution graphically: \n(i) x + y = 5 \n2x + 2y = 10<\/p>\n
(ii) x – y = 8 \n3x – 3y = 16<\/p>\n
(iii) 2x + y – 6 = 0 \n4x – 2y – 4 = 0<\/p>\n
(iv) 2x – 2y – 2 = 0 \n4x – 4y – 5 = 0 \nSolution: \nWe have given \nx + y = 5 \n2x + 2y = 10 \nWe can also write, \nx + y – 5 = 0 \n2x + 2y – 10 = 0 \nHere, a1<\/sub> = 1, b1<\/sub> = 1, c1<\/sub> = – 5 \nand a2<\/sub> = 2, b2<\/sub> = 2 and c2<\/sub> = – 10 \nSo, \\(\\frac { 1 }{ 2 }\\) = \\(\\frac { 1 }{ 2 }\\) = \\(\\frac { – 5 }{ – 10 }\\) \nTherefore, \\(\\frac{a_{1}}{a_{2}}=\\frac{b_{1}}{b_{2}}=\\frac{c_{1}}{c_{2}}\\) \nSo, the given pair of linear equations are consistent from equation (i) we have, \n \nSo the given pair of linear equations are inconsistent.<\/p>\n(iii) We have given \n2x + y -6 = 0 … (i) \n4x – 2y – 4 = 0 … (ii) \nHere, a1<\/sub> = 2, b1<\/sub> = 1, c1<\/sub> = – 6 \nand a2<\/sub> = 4, b2<\/sub> = – 2 and c2<\/sub> = – 4 \nTherefore, \\(\\frac{2}{4} \\neq \\frac{1}{-2}\\) \nSo, \\(\\frac{a_{1}}{a_{2}} \\neq \\frac{b_{1}}{b_{2}}\\) \nTherefore, the given linear equations are consistent From equation (i) \n \nTherefore, the given linear equations are inconsistent.<\/p>\n <\/p>\n
Question 5. \nHalf the perimeter of rectangular garden, whose length is 4 m more than its width, is 36 m. Find the dimensions of the garden. \nSolution: \nLet the length of the rectangular garden by y \nand the breadth of the rectangular garden by y \nAccording to question \nx = y + 4 \nx – y = 4 … (i) \nPerimeter of rectangular garden = 72 \n2(x + y)= 72 \n2x + 2y = 72 … (ii) \nMultiplying equation (i) and (ii) and adding both equations, we get \n2x = 40 \nx = 20 m \nPutting this value in equation (i) we get \n20 – y = 4 \ny = 16 m \nHence the length of garden is 20 m and the width of garden is 16 m.<\/p>\n
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Question 6. \nGiven the linear equation 2x + 3y – 8 = 0 write another linear equaton in two variables such that the geometrical representation of the pair so formed is (i) intersecting lines (ii) parallel lines, (iii) coincident lines. \nSolution: \nGiven equation is 2x + 3y – 8 = 0 i.e., a1<\/sub> = 2, b1<\/sub> = 3, c1<\/sub> = 8 \n(i) \n \none such equation can be 5x + 4y + 1 = 0 (Try forming other equations. How many such equation can be ?)<\/p>\n \nOne such equation can be 2x + 3y + 5 = 0 (Try forming other equations. How many such equation can be ?]<\/p>\n
(iii) Try yourself.<\/p>\n
Question 7. \nDraw the graphs of the equation x- y +1 = of the vertices of the triangle formed by these line \nSolution: \n \nVerticles of the triangle are (-1, 0) (4, 0) and (2, 3).<\/p>\n","protected":false},"excerpt":{"rendered":"
These NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2 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 3.2 Question 1. From the pair of linear equations in …<\/p>\n
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2<\/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 3.2 - MCQ Questions<\/title>\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n\t \n\t \n\t \n