{"id":26825,"date":"2021-06-28T13:41:32","date_gmt":"2021-06-28T08:11:32","guid":{"rendered":"https:\/\/ncertsolutions.guru\/?p=26825"},"modified":"2021-06-28T13:44:03","modified_gmt":"2021-06-28T08:14:03","slug":"ncert-solutions-for-class-10-maths-chapter-3-ex-3-2","status":"publish","type":"post","link":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-10-maths-chapter-3-ex-3-2\/","title":{"rendered":"NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2"},"content":{"rendered":"

These 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>\n

NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.2<\/h2>\n

\"NCERT<\/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\"NCERT<\/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\"NCERT
\nAgain from equation (ii)
\n\"NCERT
\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

\"NCERT<\/p>\n

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>\n

Question 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\"NCERT
\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\"NCERT
\nSo, the given pair of linear equations consistent.<\/p>\n

\"NCERT<\/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\"NCERT
\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\"NCERT
\nTherefore, the given linear equations are inconsistent.<\/p>\n

\"NCERT<\/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

\"NCERT<\/p>\n

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\"NCERT
\none such equation can be 5x + 4y + 1 = 0 (Try forming other equations. How many such equation can be ?)<\/p>\n

\"NCERT
\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\"NCERT
\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

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NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.2 Question 1. 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