{"id":11567,"date":"2020-10-07T10:57:44","date_gmt":"2020-10-07T05:27:44","guid":{"rendered":"https:\/\/mcq-questions.com\/?p=11567"},"modified":"2022-03-02T11:21:06","modified_gmt":"2022-03-02T05:51:06","slug":"pair-of-linear-equations-in-two-variables-class-10-extra-questions","status":"publish","type":"post","link":"https:\/\/mcq-questions.com\/pair-of-linear-equations-in-two-variables-class-10-extra-questions\/","title":{"rendered":"Pair of Linear Equations in Two Variables Class 10 Extra Questions Maths Chapter 3 with Solutions Answers"},"content":{"rendered":"

Here you will find Pair of Linear Equations in Two Variables Class 10 Extra Questions Maths Chapter 3 with Answers Solutions, Extra Questions for Class 10 Maths<\/a> are solved by experts and will guide students in the right direction.<\/p>\n

Extra Questions for Class 10 Maths Pair of Linear Equations in Two Variables with Answers Solutions<\/h2>\n

Extra Questions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables with Solutions Answers<\/strong><\/p>\n

Pair of Linear Equations in Two Variables Class 10 Extra Questions Objective Type<\/h3>\n

Question 1.
\nIf the pair of linear equations x – y – k = 0 and 6x – 2y – 3 = 0 represent an infinite solution, then the value of k is:
\n(a) k = 1
\n(b) k = 2
\n(c) k = 0
\n(d) No value of k
\nAnswer:
\n(d) No value of k<\/p>\n

Question 2.
\nIf the pair of equations 6x + 5y = 4 and 12x + py = -8 has no solution, then the value of p is:
\n(a) 9
\n(b) 10
\n(c) 7
\n(d) 6
\nAnswer:
\n(b) 10<\/p>\n

\"Pair<\/p>\n

Question 3.
\nThe solution of the equation 2x + 3y = 18 and x – 2y = 2 is
\n(a) x = 2, y = 6
\n(b) x = 6, y = 2
\n(c) x = – 6, y = -2
\n(d) None of these
\nAnswer:
\n(b) x = 6, y = 2<\/p>\n

Question 4.
\nFind the values of x and y in the following equations:
\nx – 3y = 8 and 5x + 3y = 10
\n(a) x = 3, y = – \\(\\frac {5}{3}\\)
\n(b) x = -3, y = \\(\\frac {5}{3}\\)
\n(c) x = -3, y = – \\(\\frac {5}{3}\\)
\n(d) None of these
\nAnswer:
\n(a) x = 3, y = – \\(\\frac {5}{3}\\)<\/p>\n

Pair of Linear Equations in Two Variables Class 10 Extra Questions Very Short Answer Type<\/h3>\n

Question 1.
\nSolve the following pair of linear equations by the substitution method.
\n(i) x + y = 14, x – y = 4
\n(ii) 3x – y = 3, 9x – 3y = 9
\n(iii) \u221a2x + \u221a3y = 0, \u221a3x – \u221a8y = 0
\nSolution:
\n(i) We have,
\nx + y = 14 …..(i)
\nand x – y = 4 ……(ii)
\nFrom Eq. (ii), y = x – 4 …(iii)
\nSubstituting y from Eq. (iii) in Eq. (i), we get
\nx + x – 4 = 14
\n\u21d2 2x = 18.
\n\u21d2 x = 9
\nOn substituting x = 9 in Eq. (iii), we get
\ny = 9 – 4 = 5
\n\u21d2 y = 5
\nx = 9, y = 5<\/p>\n

(ii) We have,
\n3x – y = 3 …(i)
\nand 9x – 3y = 9 …(ii)
\nFrom Eq. (i)y = 3x – 3 …. (iii)
\nOn substituting y from Eq. (iii) in Eq. (ii), we get
\n9x – 3(3x – 3) = 9
\n\u21d2 9 = 9
\nIt is a true statement. Hence, every solution of Eq. (i) is a solution of Eq. (ii) and vice-versa.
\nOn putting x = k in Eq. (i), we get
\n3k – y = 3 \u21d2 y = 3k – 3
\n\u2234 x = k, y = 3k – 3 is a solution for every real k.
\nHence, infinitely many solutions exist.<\/p>\n

(iii) We have,
\n\u221a2x + \u221a3y = 0 …..(i)
\nand \u221a3x – \u221a8y = 0 …(ii)
\nFrom Eq. (ii),
\n\u221a8y = \u221a3x
\n\u21d2 y = \\(\\frac{\\sqrt{3} x}{\\sqrt{8}}\\) …..(iii)
\nOn substituting y from Eq. (iii) in Eq. (i), we get
\n\"Pair-of-Linear-Equations-in-Two-Variables-Class-10-Extra-Questions-Maths-Chapter-3-with-Solutions-Answers-1\"
\n\u21d2 \u221a2 x \u221a8x + 3x = 0
\n\u21d2 \u221a16x + 3x = 0
\n\u21d2 4x + 3x = 0
\n\u21d2 7x = 0
\n\u21d2 x = 0
\nPutting x = 0 in Eq. (iii), y = 0<\/p>\n

\"Pair<\/p>\n

Question 2.
\nForm the pair of linear equations in the following problems and find their solutions (if they exist) by the elimination method
\n(i) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes \\(\\frac {1}{2}\\), if we only add 1 to the denominator. What is the fraction ?
\n(ii) Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu ?
\nSolution:
\n(i) Let the fraction be \\(\\frac {x}{y}\\)
\nAccording to the given conditions,
\n\"Pair-of-Linear-Equations-in-Two-Variables-Class-10-Extra-Questions-Maths-Chapter-3-with-Solutions-Answers-2\"
\n\u21d2 x + 1 = y – 1; 2x = y + 1 ……..(i)
\n\u21d2 x – y = -2 and 2x – y = 1 ……..(ii)
\nOn subtracting Eq. (i) from Eq. (ii),
\n(2x – y) – (x – y) = 1 + 2
\n\u21d2 x = 3
\nOn substituting x = 3 in Eq. (i),
\n3 – y = -2 \u21d2 y = 5
\nHence, the fraction is \\(\\frac {3}{5}\\)<\/p>\n

(ii) Let present age of Nuri = x years. Present age of Sonu = y years
\nAccording to the given conditions,
\nFive years ago,
\nx – 5 = 3(y – 5)
\n\u21d2 x – 3y = -10 …(i)
\nTen years later,
\nx + 10 = 2(y + 10)
\n\u21d2 x – 2y = 10 …(ii)
\nOn subtracting Eq. (i) from Eq. (ii),
\n(x – 2y) – (x – 3y) = 10 + 10
\n\u21d2 – 2y + 3y = 20
\n\u21d2 y = 20
\nFrom Eq. (ii), substituting y = 20, we get
\n\u21d2 x = 2y + 10 = 2 \u00d7 20 + 10
\n\u21d2 x = 50
\nTherefore, present age of Nuri = 50 years and present age of Sonu = 20 years<\/p>\n

\"Pair<\/p>\n

Question 3.
\nFor What value of k, will the pair of linear equations kx + y = k2<\/sup> and x + ky = 1 have infinitely many solutions ?
\nSolution:
\nGiven equation are
\nkx + y = k2<\/sup> ……(i)
\nand x + ky = 1 ……(ii)
\nhave infinitely many solution
\n\"Pair-of-Linear-Equations-in-Two-Variables-Class-10-Extra-Questions-Maths-Chapter-3-with-Solutions-Answers-3\"
\n\u2234 k2<\/sup> = 1 = k \u21d2 \u00b1 1
\nand k3<\/sup> = 1 \u21d2 k = 1
\nHence k = 1<\/p>\n

Pair of Linear Equations in Two Variables Class 10 Extra Questions Short Answer Type<\/h3>\n

Question 1.
\nSolve the following pairs of equations by reducing them to a pair of linear equations:
\n(i) \\(\\frac {4}{x}\\) + 3y =14 and \\(\\frac {3}{x}\\) – 4y = 23
\n(ii) \"Pair-of-Linear-Equations-in-Two-Variables-Class-10-Extra-Questions-Maths-Chapter-3-with-Solutions-Answers-4\"
\n\"Pair-of-Linear-Equations-in-Two-Variables-Class-10-Extra-Questions-Maths-Chapter-3-with-Solutions-Answers-5\"
\nSolution:
\n(i) We have,
\n\\(\\frac {4}{x}\\) + 3y =14 and \\(\\frac {3}{x}\\) – 4y = 23
\nOn putting \\(\\frac {1}{x}\\) = X, we get
\n4x + 3y = 14 …..(i)
\nand 3x – 4y = 23 … (ii)
\nOn multiplying Eq. (i) by 4 and Eq. (ii) by 3 and then adding, we get
\n16X + 9X = 4 \u00d7 14 + 3 \u00d7 23
\n\u21d2 25X = 56 + 69
\n\u21d2 25X = 125 = X = 5
\nThen, \\(\\frac {1}{x}\\) = 5
\n\u21d2 x = \\(\\frac {1}{5}\\) From Eq. (i), substituting x = 5, we get
\n4 \u00d7 5 + 3y = 14 = 3y = 14 – 20
\n\u21d2 3y = -6 = y = -2
\nHence, x = and y = -2
\n\"Pair-of-Linear-Equations-in-Two-Variables-Class-10-Extra-Questions-Maths-Chapter-3-with-Solutions-Answers-6\"
\n\"Pair-of-Linear-Equations-in-Two-Variables-Class-10-Extra-Questions-Maths-Chapter-3-with-Solutions-Answers-7\"
\n\u21d2 x – 1 = 3 and y – 2 = 3
\n\u21d2 x = 3 + 1
\nand y = 3 + 2
\n\u21d2 x = 4 and y = 5<\/p>\n

\"Pair<\/p>\n

Question 2.
\nProve that the pair of linear equation \\(\\frac {22}{7}\\)x + \\(\\frac {22}{7}\\)y = 7 and y = 7 and 9x – 10y = 14 is coinsistent find its solution by method cross-multiplication.
\nSolution:
\nGiven equation are
\n\\(\\frac {22}{7}\\)x + \\(\\frac {22}{7}\\)y = 7
\n\u21d2 3x + 5y = 14 ….(i)
\nand 9x – 10y = 14 …(ii)
\nIf \\(\\frac{a_{1}}{a_{2}} \\neq \\frac{b_{1}}{b_{2}}\\) then equation are coinsistance.
\nHere a1<\/sub> = 3, b1<\/sub> = 5, c1<\/sub> = – 14
\nand a2<\/sub> = 9, b2<\/sub> = -10, c2<\/sub> = -14
\n\"Pair-of-Linear-Equations-in-Two-Variables-Class-10-Extra-Questions-Maths-Chapter-3-with-Solutions-Answers-8\"<\/p>\n

Question 3.
\nSolve the following pair of linear equations.
\n\"Pair-of-Linear-Equations-in-Two-Variables-Class-10-Extra-Questions-Maths-Chapter-3-with-Solutions-Answers-9\"
\nx \u2260 0, y \u2260 0
\nSolution:
\nSuppose \\(\\frac{1}{x-y}\\) = a
\nand \\(\\frac{1}{x+y}\\) = b
\nThe reduced equation will be
\n15a + 22b = 5 ….(i)
\nand 40a + 55b = 13 (ii)
\nMultiply equation (i) by 8 and equation (ii) by 3, we get
\n120a + 176b = 40
\nand 120a + 165b = 39
\non substracting 11b = 1
\n\"Pair-of-Linear-Equations-in-Two-Variables-Class-10-Extra-Questions-Maths-Chapter-3-with-Solutions-Answers-10\"
\n\u2234 x – y = 5 … (iv)
\nSolving Eq. (iii) and Eq. (iv) we get.
\nx = 8 and y = 3<\/p>\n

\"Pair<\/p>\n

Question 4.
\nThe sum of a two-digit number and the number obtained by reversing the digits is 66. If the digits of the number differ by 2. Find the number.
\nSolution:
\nLet unit digit of the number bex and ten’s digit of the number be y
\n\u2234 Number = 10y + x
\nand Number obtain by reversing the digits is 10x + y
\n\u2234 By Ist condition,
\n(10y + x) + (10x + y) = 66
\n\u21d2 11x + 11y = 66
\n\u2234 x + y = 6 ……(i)
\nBy IInd condition, x – y = 2 (ii)
\nOn adding (i) & (ii)
\n2x = 8
\n\u2234 x = 4
\nfrom equation (1), y = 6 – 4 = 2
\nHence, required Number be 2 \u00d7 10 + 4 = 24<\/p>\n

Pair of Linear Equations in Two Variables Class 10 Extra Questions Short Answer Type 2<\/h3>\n

Question 1.
\n7 chairs and 4 tables for a classroom cost \u20b9 7000 while 5 chairs and 6 tables cost \u20b9 5080. Find the cost of each chair and that of each table.
\nSolution:
\nLet the cost of each chair be \u20b9 x and that of each table be \u20b9 y.
\nThen,
\n7x + 4y = 7000 …(i)
\n5x + 3y = 5080 …(ii)
\nOn multiplying (i) by 3, (ii) by 4 and subtracting, we get:
\n(21x – 20x) = (21000 – 20320)
\n\u21d2 x = 680.
\nOn substituting x = 680 in (i), we get:
\n(7000 – 4760) 4y = 7000
\n\u21d2 4y = (7000 – 4760)
\n\u21d2 y = 560.
\n\u2234 cost of each chair = \u20b9 680 and cost of each table = \u20b9 560.<\/p>\n

\"Pair<\/p>\n

Question 2.
\nThe sum of a two-digit number and the number obtained by reversing the order of its digits is 99. If the digits differ by 3, find the number.
\nSolution:
\nLet the tens and units digits of the required number be x and y respectively.
\nThen, the number = (10x + y).
\nThe number obtained on reversing the digits = (10y + x).
\n\u2234 (10y + x) + (10x + y) = 99
\n= 11(x + y) = 99
\n\u21d2 x + y = 9.
\nAlso, (x – y) = \u00b1 3.
\n\u2234 x + y = 9 …..(i)
\nx – y = 3 …..(ii)
\nx + y = 9 ….(iii)
\nx – y = -3 …..(iv)
\nFrom (i) and (ii), we get: x = 6, y = 3.
\nFrom (iii) and (iv), we get: x = 3, y = 6.
\nHence, the required number is 63 or 36.<\/p>\n

\"Pair<\/p>\n

Question 3.
\nThe monthly incomes of A and B are in the ratio 8 : 7 and their expenditures are in the ratio 19 : 16. If each saves \u20b9 5000 per month, find the monthly income of each.
\nSolution:
\nLet the monthly incomes of A and B be \u20b9 8x and \u20b9 7x respectively, and let their expenditures be \u20b9 19y and \u20b9 16y respectively.
\nThen, A’s monthly savings = \u20b9 (8x – 19y).
\nAnd, B’s monthly savings = \u20b9 (7x – 16y).
\nBut, the monthly saving of each is \u20b9 5000.
\n\u2234 8x – 19y = 5000 …..(i)
\n7x – 16y = 5000…..(ii)
\nMultiplying (ii) by 19, (i) by 16 and subtracting the results, we get:
\n(19 \u00d7 7 – 16 \u00d7 8)x = (19 \u00d7 5000 – 16 \u00d7 5000)
\n\u21d2 (133 – 128)x = (19 – 16) 5000
\n\u21d2 5x = 15000
\n\u21d2 x = 3000.
\n\u2234 A’s monthly income = \u20b9 (8x) = \u20b9 (8 \u00d7 3000) = \u20b9 24000.
\nAnd, B’s monthly income = \u20b9 (7x) = \u20b9(7 \u00d7 3000) = \u20b9 21000.<\/p>\n

\"Pair<\/p>\n

Question 4.
\nA two-digit number is such that the product of its digits is 14. If 45 is added to the number, the digits interchange their places. Find the number.
\nSolution:
\nLet the tens and units digits of the required number be x and y respectively.
\nThen, xy = 14.
\nRequired number= (10x + y).
\nNumber obtained on reversing its digits = (10y + x).
\n\u2234 (10x + y) + 45 = (10y + x)
\n\u21d2 9(y – x) = 45
\n\u21d2 y – x = 5 …(i)
\nNow, (y + x)2<\/sup> – (y – x)2<\/sup> = 4xy
\n\"Pair-of-Linear-Equations-in-Two-Variables-Class-10-Extra-Questions-Maths-Chapter-3-with-Solutions-Answers-11\"
\n= y + x = 9 …(ii)
\n[\u2234 digits are never negative)
\n2y = 14 \u21d2 y = 7.
\nPutting y = 7 in (ii), we get:
\n7 + x = 9 \u21d2 x = (9 – 7) = 2.
\nx = 2 and y = 7.
\nHence, the required number is 27.<\/p>\n

\"Pair<\/p>\n

Question 5.
\nThe sum of the numerator and the denominator of a fraction is 12. If the denominator is increased by 3, the fraction becomes \\(\\frac {1}{2}\\). Find the fraction.
\nSolution:
\nLet the required fraction be \\(\\frac {x}{y}\\).
\nThen,
\n\u2234 x + y = 12 ……(i)
\nand \\(\\frac{x}{y+3}\\) = \\(\\frac{1}{2}\\)
\n\u21d2 2x = y + 3
\n\u21d2 2x – y = 3 …(ii)
\nAdding (i) and (ii), we get:
\n3x = 15 x = 5.
\nPutting x = 5 in (i), we get:
\n5 + y = 12 y = (12 – 5) = 7.
\nThus, x = 5 and y = 7.
\nHence, the required fraction is \\(\\frac {5}{7}\\).<\/p>\n

\"Pair<\/p>\n

Question 6.
\nFive year ago, A was thrice as old as B and ten years later A shall be twice as old as B. What are the present ages of A and B ?
\nSolution:
\nLet the present ages of B and A be x years and y years respectively. Then,
\nB’s age 5 years ago = (x – 5 ) years and A’s age 5 years ago = (y – 5) years.
\n\u2234 (y – 5) = 3(x – 5) = 3x – y = 10 …(i)
\nB’s age 10 years hence = (x + 10) years.
\nA’s age 10 years hence = (y + 10) years.
\n\u2234 (y + 10) = 2(x + 10) = 2x – y = -10 (\u00fc)
\nOn subtracting (ii) from (i), we get:
\nx = 20.
\nPutting x = 20 in (i), we get:
\n(3 x 20) – y = 10 = y = (60 – 10) = 50.
\n\u2234 x = 20 and y = 50.
\nHence, B’s present age = 20 years and A’s present age = 50 years.<\/p>\n

\"Pair<\/p>\n

Question 7.
\nIf the length of a rectangle is reduced by 5 units and its breadth is increased by 2 units, then the area of the rectangle is reduced by 80 sq units. However, if we increase its length by 10 units and decrease the breadth by 5 units, its area is increased by 50 sq units. Find the length and breadth of the rectangle.
\nSolution:
\nLet the length and breadth of the rectangle be x units and y units respectively.
\nThen, area of the rectangle = xy sq units
\nCase I: When the length is reduced by 5 units and the breadth is increased by 2 units.
\nThen,new length= (x – 5) units
\nandnew breadth = (y + 2) units.
\n\u2234 new area = (x – 5)(y + 2) sq units.
\n\u2234 xy – (x – 5)(y + 2) = 80
\n\u21d2 5y – 2x = 70…(i)
\nCase II: When the length is increased by 10 units and the breadth is decreased by 5 units.
\nThen, new length= (x + 10) units
\nandnew breadth = (y – 5) units.
\n\u2234 new area = (x + 10)(y – 5) sq units.
\n\u2234 (x + 10)(y – 5) – xy = 50
\n\u21d2 10y – 5x = 100
\n\u21d2 2y – x = 20 …….(ii)
\nOn multiplying (ii) by 2 and subtracting esult from (i), we get:
\ny = 30.
\nPutting y = 30 in (ii), we get:
\n(2 \u00d7 30) – x = 20 \u21d2 60 – x = 20
\n\u21d2 x = (60 – 20) = 40.
\n\u2234 x = 40 and y = 30.
\nHence, length = 40 units and breadth = 30 units.<\/p>\n

\"Pair<\/p>\n

Question 8.
\n8 men and 12 boys can finish a piece of work in 5 days, while 6 men and 8 boys can finish it in 7 days. Find the time taken by 1 man alone and that by 1 boy alone to finish the work…
\nSolution:
\nSuppose 1 man alone can finish the work in x days and 1 boy alone can finish it in y days.
\nThen, 1 man’s 1 day’s work = \\(\\frac {1}{x}\\)
\nAnd, 1 boy’s 1 day’s work = \\(\\frac {1}{y}\\)
\n8 men and 12 boys can finish the work in 5 days
\n\u21d2 (8 men’s 1 day’s work) + (12 boys’ 1 day’s work) = \\(\\frac {1}{5}\\)
\n\u21d2 \\(\\frac {8}{x}\\) + \\(\\frac {12}{y}\\) = \\(\\frac {1}{5}\\),
\n\u21d2 8\u03c5 + 12\u03c5 = \\(\\frac {1}{5}\\),
\n[Where \\(\\frac {1}{x}\\) = \u03c5 and \\(\\frac {1}{y}\\) = \u03c5] …..(i)
\nAgain, 6 men and 8 boys can finish the work in 7 days
\n\u21d2 (6 men’s 1 day’s work) + (8 boys’ 1 day’s work) = \\(\\frac {1}{7}\\)
\n\"Pair-of-Linear-Equations-in-Two-Variables-Class-10-Extra-Questions-Maths-Chapter-3-with-Solutions-Answers-12\"
\n\u2234 one man alone can finish the work in 70 days,
\nand one boy alone can finish the work in 140 days.<\/p>\n

\"Pair<\/p>\n

Question 9.
\nA boat goes 16 km upstream and 24 km downstream in 6 hours. Also, it covers 12 km upstream and 36 km downstream at the same time. Find the speed of the boat in still water and that of the stream.
\nSolution:
\nLet the speed of the boat in still water be x km\/hr and the speed of the stream be y km\/hr.
\nThen, speed upstream = (x – y) km\/hr
\nand speed downstream = (x + y) km\/hr.
\nTime taken to cover 16 km upstream
\n\"Pair-of-Linear-Equations-in-Two-Variables-Class-10-Extra-Questions-Maths-Chapter-3-with-Solutions-Answers-13\"
\n\"Pair-of-Linear-Equations-in-Two-Variables-Class-10-Extra-Questions-Maths-Chapter-3-with-Solutions-Answers-14\"
\nOn adding (v) and (vi), we get: 2x = 16 \u21d2 x = 8.
\nOn subtracting (vi) from (v), we get:
\n2y = 8 y = 4.
\n\u2234 speed of the boat in still water
\n= 8 km\/hr.
\nAnd, speed of the stream = 4 km\/hr.<\/p>\n

\"Pair<\/p>\n

Question 10.
\n90% and 97% pure acid solutions are mixed to obtain 21 litres of 95% pure acid Solution: Find the quantity of each type of acid to be mixed to form the mixture.
\nSolution:
\nLet the given solutions be labelled as A and B respectively.
\nLet x litres of A be mixed with y litres of B. Then,
\nx + y = 21
\nQuantity of acid in x litres of A = (90% of x)
\n\"Pair-of-Linear-Equations-in-Two-Variables-Class-10-Extra-Questions-Maths-Chapter-3-with-Solutions-Answers-15\"
\n\"Pair-of-Linear-Equations-in-Two-Variables-Class-10-Extra-Questions-Maths-Chapter-3-with-Solutions-Answers-16\"
\n\u21d2 90x + 97y – 1995 …….(ii)
\nMultiplying (i) by 90 and subtracting the result from (ii), we get:
\n7y = 105 \u21d2 y = 15.
\nPutting y = 15 in (i), we get:
\nx + 15 = 21 = x = 6.
\n\u2234 x = 6 and y = 15.
\nSo, 6 litres of 90% solution is mixed with 15 litres of 97% Solution.<\/p>\n

\"Pair<\/p>\n

Question 11.
\nOn selling a tea-set at 5% loss and a lemon-set at 15% gain, a shopkeeper gains \u20b9 84. However, if he sells the tea-set at 5% gain and the lemon-set at 10% gain, he gains \u20b9 104. Find the price of the tea-set and that of the lemon-set paid by the shopkeeper.
\nSolution:
\nLet the CP of the tea-set and the lemon-set be \u20b9 x and \u20b9 y respectively.
\n\"Pair-of-Linear-Equations-in-Two-Variables-Class-10-Extra-Questions-Maths-Chapter-3-with-Solutions-Answers-17\"
\nOn adding (i) and (ii), we get:
\n5y = 3760 \u21d2 y = 752.
\nPutting y = 752 in (ii), we get:
\nx + (2 \u00d7 752) = 2080
\nx = (2080 – 1504) = 576.
\n\u2234 x = 576 and y = \u20b9 752.
\nHence, CP of the tea-set = \u20b9 576
\nand CP of the lemon-set = \u20b9 752.<\/p>\n

\"Pair<\/p>\n

Question 12.
\nA chemist has one solution containing 50% acid and a second one containing 25% acid. How much of each should be used to make 10 litres of a 40% acid solution ?
\nSolution:
\nLet x liters of 50% solution be mixed with y liters of 25% Solution. Then,
\nx + y = 10 and 50% of x + 25% of y = 40% of 10
\n\u21d2 x + y 10 and \\(\\frac {50}{100}\\) \u00d7 x + \\(\\frac {25}{100}\\) \u00d7 y = \\(\\frac {40}{100}\\) \u00d7 10
\n\u21d2 x + y 10 and \\(\\frac {x}{2}\\) + \\(\\frac {y}{4}\\) = 4
\n\u21d2 x + y = 10 …(i)
\nand 2x + y = 16 …(ii)
\nOn solving (i) and (ii), we get: x = 6 and y = 4.
\n\u2234 6 litres of 50% solution is to be mixed with 4 litres of 25% Solution.<\/p>\n

\"Pair<\/p>\n

Question 13.
\nIn a \u2206ABC, \u2220C = 3\u2220B = 2(\u2220A + \u2220B). Find the angles.
\nSolution:
\nLet \u2220A = x\u00b0 and \u2220B = y\u00b0. Then,
\n\u2220C = 3\u2220B = (3y)\u00b0.
\nNow, \u2220A + \u2220B + \u2220C = 180\u00b0
\n\u21d2 x + y + 3y = 180\u00b0
\nx + 4y = 180 ……(i)
\nAlso, \u2220C = 2(\u2220A + \u2220B)
\n\u21d2 3y = 2(x + y)
\n\u21d2 2x – y = 0 ……(ii)
\nMultiplying (ii) by 4 and adding the result to (i), we get:
\n9x = 180 \u21d2 x = 20.
\nPutting x = 20 in (i), we get:
\n20 + 4y = 180 \u21d2 4y = 160
\n\u21d2 y = \\(\\frac {160}{4}\\) = 40.
\n\u2234 x = 20 and y = 40.
\n\u2234 \u2220A = 20\u00b0, \u2220B = 40\u00b0
\nand \u2220C = (3 x 40)\u00b0 = 120\u00b0.<\/p>\n

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

Here you will find Pair of Linear Equations in Two Variables Class 10 Extra Questions Maths Chapter 3 with Answers Solutions, Extra Questions for Class 10 Maths are solved by experts and will guide students in the right direction. Extra Questions for Class 10 Maths Pair of Linear Equations in Two Variables with Answers Solutions …<\/p>\n

Pair of Linear Equations in Two Variables Class 10 Extra Questions Maths Chapter 3 with Solutions Answers<\/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":"\nPair of Linear Equations in Two Variables Class 10 Extra Questions Maths Chapter 3 with Solutions Answers - 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\/pair-of-linear-equations-in-two-variables-class-10-extra-questions\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Pair of Linear Equations in Two Variables Class 10 Extra Questions Maths Chapter 3 with Solutions Answers - MCQ Questions\" \/>\n<meta property=\"og:description\" content=\"Here you will find Pair of Linear Equations in Two Variables Class 10 Extra Questions Maths Chapter 3 with Answers Solutions, Extra Questions for Class 10 Maths are solved by experts and will guide students in the right direction. 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