You can also download Class 10 Maths NCERT Solutions<\/strong><\/a> to help you to revise complete syllabus and score more marks in your examinations.<\/p>\n Question 1. Question 2. Question 3. Question 4. Question 5. Question 6. Question 7. Question 1. Question 2. Question 3. Question 4. Question 1. Question 2. Question 3. Question 4. Question 5. A.T.Q. Question 6. At condition I: At condition II: Question 7. Condition II: Question 8. [Important Tips: If the present age of a person is x years, i J then, n years later his age = (x + n) years whereas m j i years ago his age = (x – m) years.]<\/p>\n Question 9. Question 10. Question 1. Table of solutions for 2x – y + 2 = 0 Question 2. Case I: Case II. Question 3. Question 4. Question 5. Question 1. Question 2. First condition: Second condition: Question 3. Question 4. Multiple Choice Questions<\/span><\/p>\n Choose the correct option out of four given in each of the following:<\/p>\n Question 1. Question 2. Question 3. Question 4. Question 5. Question 6. Question 7. Question 8. Question 9. Question 10. Question 11. Question 12. Question 13. Question 14. Question 15. Fill in the blanks:<\/span><\/p>\n Question 1. Question 2. Question 3. Question 4. Question 5. Question 6. Question 7. Question 8. Question 9. Question 10. Extra Questions for Class 10 Maths Pair of Linear Equations in Two Variables with Answers Extra Questions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables. According to new CBSE Exam Pattern,\u00a0MCQ Questions for Class 10 Maths\u00a0Carries 20 Marks. You can also download Class 10 Maths NCERT Solutions to help you […]<\/p>\n","protected":false},"author":27,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_genesis_hide_title":false,"_genesis_hide_breadcrumbs":false,"_genesis_hide_singular_image":false,"_genesis_hide_footer_widgets":false,"_genesis_custom_body_class":"","_genesis_custom_post_class":"","_genesis_layout":""},"categories":[2],"tags":[],"yoast_head":"\nPair of Linear Equations in Two Variables Class 10 Extra Questions Very Short Answer Type<\/h3>\n
\nFind the value of k for which the following pair of linear equations have infinitely many solutions. 2x + 3y = 7, (k + 1)x + (2k – 1 )y = 4k + 1 [CBSE 2019 Set-B]
\nAnswer:
\nGiven equations: 2x + 3y – 7 = 0
\n(k + 1)x + (2k – 1 )y – (4k + 1) = 0
\nHere, a1<\/sub> = 2, b1<\/sub> = 3, c1<\/sub> = – 7
\na2<\/sub> = (k + 1), b2<\/sub> = (2k – 1), c2<\/sub> = – (4k + 1)
\nFor infinitely many solutions
\n![\"Pair](\"https:\/\/www.learncbse.in\/wp-content\/uploads\/2019\/10\/Pair-of-Linear-Equations-in-Two-Variables-Class-10-Extra-Questions-Maths-Chapter-3-with-Solutions-1.png\")
\n\u2234 4k – 2 = 3k + 3
\nand 12k + 3 = 14k – 7
\n\u21d2 k = 5 ………………. (1)
\n2k = 10
\n\u21d2 k = 5 ………………. (2)
\nUsing (1) and (2) \u21d2 k = 5.<\/p>\n
\nFind the relation between p and y if x = 3 and y = 1 is the solution of the pair of equations x – 4y + p = 0 and 2x + y – q – 2 = 0 [CBSE 2019 (C)]
\nAnswer:
\nSince x = 3, y = 1 is the solution of
\nx – 4y + p = 0 …………….. (i)
\n2x + y – q – 2 = 0 …………… (ii)
\nSo, x = 3, y = 1 must satisfy both (i) and (ii).
\n\u21d2 3 – 4(1) + p = 0 \u21d2 p = 1.
\nand 2(3) + (1) – q – 2 = 0 \u21d2 q = 5
\nSince, 1 = – 4 + 5
\n\u2234 p = 1 = – 4 + 5
\n= q – 4
\n\u21d2 p = q – 4
\nThus, p is lesser than q by 4.<\/p>\n
\nIn Fig., ABCD is a rectangle. Find the values of x and y. [CBSE 2018]
\n![\"Pair](\"https:\/\/www.learncbse.in\/wp-content\/uploads\/2019\/10\/Pair-of-Linear-Equations-in-Two-Variables-Class-10-Extra-Questions-Maths-Chapter-3-with-Solutions-2.png\")
\nAnswer:
\nSince ABCD is a rectangle
\n\u21d2 AB = CD and BC = AD
\nx + y = 30 …………….. (i)
\nx – y = 14 ……………. (ii)
\n(i) + (ii) \u21d2 2x = 44
\n\u21d2 x = 22
\nPlug in x = 22 in (i)
\n\u21d2 22 + y = 30
\n\u21d2 y = 8<\/p>\n
\nName the geometrical figure enclosed by graph of the equations x + 7 = 0, y – 2 = 0 and x – 2 = 0, y + 7 = 0.
\nAnswer:
\nClearly, a square of side 9 units is enclosed by lines.
\n![\"Pair](\"https:\/\/www.learncbse.in\/wp-content\/uploads\/2019\/10\/Pair-of-Linear-Equations-in-Two-Variables-Class-10-Extra-Questions-Maths-Chapter-3-with-Solutions-3.png\")
<\/p>\n
\nDetermine whether the following system of linear equations is inconsistent or not.
\n3x – 5y = 20
\n6x – 10y = -40
\nAnswer:
\nGiven
\n3x – 5y = 20 ……… (i)
\n6x – 10y = – 40 ………… (ii)
\n![\"Pair](\"https:\/\/www.learncbse.in\/wp-content\/uploads\/2019\/10\/Pair-of-Linear-Equations-in-Two-Variables-Class-10-Extra-Questions-Maths-Chapter-3-with-Solutions-4.png\")
\nHence given pair of linear equation are parallel.
\n\u2234 It is inconsistent.<\/p>\n
\nIf 51x + 23y = 116 and 23x + 51y = 106, then find the value of (x – y).
\nAnswer:
\n51x + 23y = 116 ……….. (i)
\n23x + 51y = 106 ………… (ii)
\nSubtracting (ii) from (i)
\n28x – 28y = 10
\n28(x – y) = 10
\n\u21d2 (x – y) = \\(\\frac{10}{28}\\) = \\(\\frac{5}{14}\\)<\/p>\n
\nFor what value of V the point (3, a) lies on the line represented by 2x – 3y = 5?
\nAnswer:
\nSince (3, a) lies on the equation 2x – 3y = 5.
\n\u2234 (3, a) must satisfy this equation
\n\u21d2 2(3) – 3(a) = 5
\n\u21d2 6 – 3a = 5
\n\u21d2 – 3a = 5 – 6 = – 1
\na = \\(\\frac{1}{3}\\)
\n\u2234 a = \\(\\frac{1}{3}\\)<\/p>\nPair of Linear Equations in Two Variables Class 10 Extra Questions Short Answer Type-1<\/h3>\n
\nSolve the following pair of linear equations using elimination method. [CBSE Delhi 2016]
\nx – y + 1 = 0; 4x + 3y – 10 = 0
\nAnswer:
\nx – y + 1 = o …………. (i)
\n4x + 3y – 10 = 0 ……… (ii)
\nEq (i) is multiplied by 4, then we get
\n4x – 4y + 4 = 0
\nBy elimination method.
\n![\"Pair](\"https:\/\/www.learncbse.in\/wp-content\/uploads\/2019\/10\/Pair-of-Linear-Equations-in-Two-Variables-Class-10-Extra-Questions-Maths-Chapter-3-with-Solutions-5.png\")
\n7y = 14
\n\u2234 y = 2
\nPutting y = 2 in (i), we get
\nx = 1
\nHence, x = 1, y = 2<\/p>\n
\nFind the values of k for which the pair of linear equations kx + y = k2<\/sup> and x + ky = 1 have infinitely many solutions. [CBSE Sample Paper 2017]
\nAnswer:
\nWe have
\n![\"Pair](\"https:\/\/www.learncbse.in\/wp-content\/uploads\/2019\/10\/Pair-of-Linear-Equations-in-Two-Variables-Class-10-Extra-Questions-Maths-Chapter-3-with-Solutions-6.png\")
\nIst<\/sup> & 2nd<\/sup> Ratio \u21d2 k2<\/sup> = 1
\n\u21d2 k = \u00b11 ……………. (i)
\n2nd<\/sup> & 3rd<\/sup> Ratio \u21d2 k3<\/sup> = 1
\n\u21d2 k = 1 ………………. (ii)
\n(i) and (ii) \u21d2 k = 1.<\/p>\n
\nFind out whether the following pair of equation is consistent or inconsistant.
\n3x + 2y = 5; 2x – 3y = 7
\nAnswer:
\nHere a1<\/sub> = 3; a2<\/sub> = 2; b1<\/sub> = 2, b2<\/sub> = – 3, c1<\/sub> = 5, c2<\/sub> = 7
\n![\"Pair](\"https:\/\/www.learncbse.in\/wp-content\/uploads\/2019\/10\/Pair-of-Linear-Equations-in-Two-Variables-Class-10-Extra-Questions-Maths-Chapter-3-with-Solutions-7.png\")
\nHence, the equations will have a unique solution.<\/p>\n
\nFor what value of k,
\n2x + 2y + 2 = 0
\n4x + ky + 8 = 0
\nwill have unique solution.
\nAnswer:
\nHere,
\na1<\/sub> = 2, b1<\/sub> = 2, c1<\/sub> = 2
\na2<\/sub> = 4, b2<\/sub> = k, c2<\/sub> = 8
\nFor the given system of linear equations to have a unique solution.
\n\\(\\frac{a_1}{a_2} \\neq \\frac{b_1}{b_2}\\)
\n\u2234 \\(\\frac{2}{4} \\neq \\frac{2}{k}\\)
\n\u21d2 2k \u2260 2 \u00d7 4, or k \u2260 4<\/p>\nPair of Linear Equations in Two Variables Class 10 Extra Questions Short Answer Type-2<\/h3>\n
\nSolve 2x + 3y = 11 and x-2 y = -12 algebraically and hence find the value of m for which y = mx + 3. [C.B.S.E. 2019]
\nAnswer:
\n2x – 3y = 11 …………….. (i)
\nx – 2y = – 12 …………….. (ii)
\n(ii) \u21d2 x = 2y – 12 ……. (iii)
\nSubstitute value of x from (iii) in (i), we get
\n2(2y – 12) + 3y = 11
\n\u21d2 4y – 24 + 3y = 11
\n\u21d2 7y = 35
\n\u21d2 y = 5
\nSubstituting value of y = 5 in equation (iii), we get
\n4 = 2(5) – 12 = 10 – 12 = – 2
\nHence, x = – 2, y = 5 is the required solution
\nNow, 5 = – 2m + 3
\n\u21d2 2m = 3 – 5
\n\u21d2 2m = – 2
\n\u21d2 m = – 1.<\/p>\n
\nRepresent the system of linear equations 3x + y = 5 and 2x + y = 5 graphically. From the graph, find the points where the lines intersect y-axis.
\nAnswer:
\n3x + y – 5 = 0, 2x + y – 5 = 0
\nGiven system of linear equations is:
\n3x + y – 5 = 0 …………… (i)
\n2x + y – 5 = 0 ………….. (ii)
\n(i) \u21d2 y = 5 – 3x …………. (iii)
\nTable of solutions for (iii)
\n![\"Pair](\"https:\/\/www.learncbse.in\/wp-content\/uploads\/2019\/10\/Pair-of-Linear-Equations-in-Two-Variables-Class-10-Extra-Questions-Maths-Chapter-3-with-Solutions-8.png\")
\n(ii) \u21d2 y = 5 – 2x …….. (iv)
\nTable of solutions for (iv)
\n![\"Pair](\"https:\/\/www.learncbse.in\/wp-content\/uploads\/2019\/10\/Pair-of-Linear-Equations-in-Two-Variables-Class-10-Extra-Questions-Maths-Chapter-3-with-Solutions-9.png\")
\nPlot the points and draw the lines passing through them. We observe, the two lines intersect at the point (0, 5) which lies on y-axis. So from graph, we conclude that (0, 5) is the required point where the lines intersect y-axis.
\n![\"Pair](\"https:\/\/www.learncbse.in\/wp-content\/uploads\/2019\/10\/Pair-of-Linear-Equations-in-Two-Variables-Class-10-Extra-Questions-Maths-Chapter-3-with-Solutions-10.png\")
<\/p>\n
\nSolve for x and y: [CBSE 2016]
\nx + 4y = 27xy ………. (i)
\nx + 2 y = 21xy ………… (ii)
\nAnswer:
\nDividing (i) and (ii) by xy. We get
\n![\"Pair](\"https:\/\/www.learncbse.in\/wp-content\/uploads\/2019\/10\/Pair-of-Linear-Equations-in-Two-Variables-Class-10-Extra-Questions-Maths-Chapter-3-with-Solutions-11.png\")
<\/p>\n
\nA fraction becomes \\(\\frac{1}{3}\\) when 1 is subtracted from the numerator and it becomes \\(\\frac{1}{4}\\) when 8 is added to its denominator. Find the fraction.
\nAnswer:
\nLet the fraction be \\(\\frac{x}{y}\\).
\nAccording to given conditions
\n\\(\\frac{x-1}{y}=\\frac{1}{3}\\) and \\(\\frac{x}{y+8}=\\frac{1}{4}\\)
\n\u21d2 3x – 3 = y and 4x = y + 8
\n\u21d2 3x – y – 3 = 0
\nand 4x – y – 8 = 0
\nSubtracting (it) from (i)
\n\u21d2 – x + 5 = 0
\n\u21d2 x = 5
\nPut x = 5 in (i), we get
\n3(5) – y – 3 = 0 \u21d2 y = 15
\n\u21d2 y = 12
\n\u2234 Fraction is \\(\\frac{5}{12}\\).<\/p>\n
\nFind the two numbers whose sum is 75 and difference is 15. [CBSE 2016]
\nAnswer:
\nLet x is the first number and y be the second number.<\/p>\n
\n![\"Pair](\"https:\/\/www.learncbse.in\/wp-content\/uploads\/2019\/10\/Pair-of-Linear-Equations-in-Two-Variables-Class-10-Extra-Questions-Maths-Chapter-3-with-Solutions-12.png\")
\ny = 30
\nPutting the value of y in eq. (i), we get,
\nx + 30 = 75
\nx = 75 – 30
\nx = 45
\n\u2234 x = 45
\n\u2234 Hence the first number = 45 and
\nSecond number = 30<\/p>\n
\nIn a two digit number, the ten’s digit number is three times the unit’s digit. When the number is decreased by 54, the digits are reversed. Find the number.
\nAnswer:
\nLet the digit at units place = x and ten’s place = y<\/p>\n
\ny = 3x ………… (i)
\nNow number obtained after reversing the digit
\n= 10x + y<\/p>\n
\n10y + x – 54 = 10x + y
\n\u21d2 9x – 9y = – 54
\n\u21d2 x – y = – 6 …………. (ii)
\nPut y = 3x in (ii) we get x – 3x = – 6 \u21d2 x = 3
\nNow put x = 3 in (i), we get y = 3 \u00d7 3 = 9
\n\u2234 Number is 9 \u00d7 10 + 3 = 93<\/p>\n
\nWhen the son will be as old as what his father is today their ages will add upto 126 years. When the father was as old as what his son is today, their ages added upto 38 years. Find their present ages.
\nAnswer:
\nLet son’s present age be x years and father’s present age be y years.
\nDifference in father’s and son’s age = (y – x) years
\nIf this difference is added to son’s age, then the son would turn as old as his father is now and at the time the father’s age will be [y + (y – x)] years. Also if we subtract this difference from father’s present age, then father would be as old as his son is today and at that time son\u2019s age will be [x – (y – x)] years.
\nThus, Condition I:
\n\u21d2 [x + (y – x)] + [y + (y – x)] = 126
\n\u21d2 3y – x = 126
\n\u21d2 x – 3y = – 126 ……. (i)<\/p>\n
\n\u21d2 [y – (y – x)] + [x – (y – x)] = 38
\n\u21d2 3x – y = 38 …… (ii)
\n(i) + (ii) \u21d2 4(x – y) = – 88
\n\u21d2 x – y = – 22 ……. (iii)
\n(ii) – (i) \u21d2 2(x + y) = 164
\n\u21d2 x + y = 82 …….. (iv)
\n(iii) + (iv) \u21d2 2x = 60
\n\u21d2 x = 30
\n(iv) – (iii) \u21d2 2 y = 104
\n\u21d2 y = 52
\nHence, present age of father is 52 years and that of son is 30 years.<\/p>\n
\n4 chairs and 3 tables cost \u20b9 2100 and 5 chairs and 2 tables cost \u20b9 1750. Find the cost of one chair and one table separately. [CBSE Delhi 2016]
\nAnswer:
\nLet the cost of one chair = \u20b9 x
\nand and the cost of one table = \u20b9 y
\nAccording to question,
\n4x + 3 y = 2100 ………….. (i)
\n5x + 2y = 1750 ………….. (ii)
\n[4x + 3y = 2100] \u00d7 2 ……………. (iii)
\n[5x + 2y = 1750] \u00d7 3 ……………. (iv)
\nSubtracting (iii) from (iv),
\nwe get
\n7x = 1050
\n\u2234 x = 150
\nPutting the value of x in (i), we get
\n4 \u00d7 150 + 3y = 2100
\n\u2234 3y = 2100 – 600 = 1500
\n\u2234 y = 500
\n\u2234 cost of one chair = \u20b9 150
\nand cost of one table = \u20b9 500<\/p>\n
\nA number consists of two digits. When it is divided by the sum of the digits, the quotient is 6 with no remainder. When the number is diminished by 9, the digit are reversed.
\nAnswer:
\nLet the ten’s and unit’s digit of a number be ‘x’ and y respectively.
\nRequired number = 10 \u00d7 x + 1 \u00d7 y = 10x + y
\nReversed number = 10 \u00d7 y + 1 \u00d7 x = 10y + x
\nAccording to condition I, we have:
\n\u21d2 \\(\\frac{\\text { Number }}{\\text { Sum of digits }}\\) = \\(\\frac{10 x+y}{x+y}=\\frac{6}{1}\\)
\n\u21d2 10x + y = 6x + 6y
\n4x – 5y = 0 ……… (i)
\nAccording to condition II, we have:
\n\u21d2 Reversed number = Number – 9
\n\u21d2 10y + x = (10x + y) – 9
\n\u21d2 9(x – y) = 9 \u21d2 x – y = 1 …….. (ii)
\n(i) \u21d2 x = \\(\\frac{5}{4}\\) y
\nPutting this value of x in (ii), we get
\n\\(\\frac{5}{4}\\)y – y = 1 \u21d2 \\(\\frac{1}{4}\\)y = 1 \u21d2 y = 4
\nPutting y = 4 in (i), we get
\nx = \\(\\frac{5}{4}\\) \u00d7 4 = 5 4
\n\u2234 Required number = 54<\/p>\n
\nSeven times a two digit number is equal to four times the number obtained by reversing the order of the digits. If the difference of the digits is 3, determine the number.
\nAnswer:
\nLet Ten’s and Units digits be y and x respectively.
\n\u2234 Value of Number is 10y + x
\nValue of Number on reversal = 10x + y
\nAccording to as given:
\n7(10y + x) = 4(10x + y)
\n\u21d2 70y + 7x = 40x + 4 y
\n\u21d2 66y = 33x or 2y = x
\nAlso x – y = 3 ………. (ii)
\nPut the value of x in (ii) form (i)
\n2y – y = 3 \u21d2 y = 3
\nPut for y in (z)
\nx = 2(3) = 6
\nHence the number is 36.<\/p>\nPair of Linear Equations in Two Variables Class 10 Extra Questions Long Answer Type 1<\/h3>\n
\nDraw the graph of 2x + y = 6 and 2x – y + 2 = 0. Shade the region bounded by these lines and x-axis. Find the area of the shaded region. [CBSE 2007]
\nAnswer:
\nThe given system of equation is
\n2x + y – 6 = 0 ………. (i)
\n2x – y + 2 = 0 ……… (ii)
\nLet us write three solutions for each equation of the system in a table.
\n(i) \u21d2 y = 6 – 2x
\nTable of solutions for 2x + y – 6 = 0
\n![\"Pair](\"https:\/\/www.learncbse.in\/wp-content\/uploads\/2019\/10\/Pair-of-Linear-Equations-in-Two-Variables-Class-10-Extra-Questions-Maths-Chapter-3-with-Solutions-13.png\")
\nSimilarly (ii) \u21d2 y = 2x + 2<\/p>\n
\n![\"Pair](\"https:\/\/www.learncbse.in\/wp-content\/uploads\/2019\/10\/Pair-of-Linear-Equations-in-Two-Variables-Class-10-Extra-Questions-Maths-Chapter-3-with-Solutions-14.png\")
\nPlotting these points of each table of solutions on the same graph paper and joining them by a ruler, we obtain graph of two lines represented by equation (i) and (ii) respectively as shown in the graph below. Since, the two lines intersect at point P(1, 4). Thus x = 1, y = 4 is the solution of the given system of equations.
\nIn graph, area bounded by the lines and x-axis is \u2206PAB which is shaded.
\n![\"Pair](\"https:\/\/www.learncbse.in\/wp-content\/uploads\/2019\/10\/Pair-of-Linear-Equations-in-Two-Variables-Class-10-Extra-Questions-Maths-Chapter-3-with-Solutions-15.png\")
\nDraw PM \u22a5 x-axis
\nClearly, PM = y-coordinate of P(1, 4)
\n= 4 units
\nAlso AB = 1 + 3 = 4 units
\n\u2234 Area of shaded region
\n= Area of \u2206PAB = \\(\\frac{1}{2}\\) \u00d7 AB \u00d7 PM
\n= \\(\\frac{1}{2}\\) \u00d7 4 \u00d7 4 = 8 sq. units.<\/p>\n
\nA man travel 370 km partly by train and partly by car. If he covers 250 km by train and the rest by car, it takes him 4 hours. But, if he travels 130 km by train and the rest by car, he takes 18 minutes longer. Find the speed of the train and that of the car. [CBSE 2001]
\nAnswer:
\nLet speed of the train= x km\/ hour
\nSpeed of the car = y km\/hour<\/p>\n
\nTotal distance travelled = 370 km
\nDistance travelled by train = 250 km
\nDistance travelled by car = (370 – 250) km = 120 km
\n![\"Pair](\"https:\/\/www.learncbse.in\/wp-content\/uploads\/2019\/10\/Pair-of-Linear-Equations-in-Two-Variables-Class-10-Extra-Questions-Maths-Chapter-3-with-Solutions-16.png\")
\nor 250y + 120x = 4xy
\nor 60x + 125y = 2xy<\/p>\n
\nTotal distance covered = 370 km
\nDistance covered by train = 130 km
\nDistance covered by car = (370 -130) km = 240
\nTime taken by train = \\(\\frac{130}{x}\\) hour
\nTime taken by car = \\(\\frac{240}{y}\\) hour
\nAccording to 2nd condition,
\n\\(\\frac{130}{x}+\\frac{240}{y}\\) = 4 hour 18 minutes
\n\\(\\frac{130}{x}+\\frac{240}{y}=\\frac{43}{10}\\)
\n1300y + 2400x = 43xy
\nor 2400x + 1300y = 43xy
\nMultiplying equation (i) by 40, we get
\n![\"Pair](\"https:\/\/www.learncbse.in\/wp-content\/uploads\/2019\/10\/Pair-of-Linear-Equations-in-Two-Variables-Class-10-Extra-Questions-Maths-Chapter-3-with-Solutions-17.png\")
\nor x = 100
\nFrom (i) and (iii), we get
\n60(100) + 125y = 2(100)y
\n6000 + 125y = 200y
\nor (200 – 125) y = 6000
\nor 75y = 6000
\nor y = \\(\\frac{6000}{75}\\) = 80
\nHence, speed of train and car are 100 km\/hour and 80 km\/hour respectively.<\/p>\n
\nDraw the graphs of the following equations:
\n2x – y = 1, x + 2y = 13
\n(i) Find the solution of the equations from the graph.
\n(ii) Shade the triangular region formed by lines and the y-axis.
\nAnswer:
\n2x – y = 1 …………….. (i)
\nx + 2y = 13 ………………. (ii)
\nLet us draw table of values for (i) and (ii)
\n![\"Pair](\"https:\/\/www.learncbse.in\/wp-content\/uploads\/2019\/10\/Pair-of-Linear-Equations-in-Two-Variables-Class-10-Extra-Questions-Maths-Chapter-3-with-Solutions-18.png\")
\nPlotting these points on the graph paper, we see that the two lines representing equations (i) and (ii) intersect at point (3, 5).
\n(i) Therefore, (3, 5) is solution of given system,
\n(ii) Also, that two lines enclose a triangular region (\u2206ABC) with y-axis which is shaded in graph.<\/p>\n
\nThe taxi charges in a city comprise of a fixed charges together with charge for the distance covered. For a journey of 10 km, the charges paid is \u20b975 and for a journey of 15 km, the charges paid is \u20b9110. What will a person has to pay for travelling a distance of 25 km?
\nAnswer:
\nLet fixed charges of taxi be \u20b9x and charges for covering distance be \u20b9y per km.
\nThen, according to the question, we have
\nx + 10y = 75 ……… (i)
\nand x + 15y = 110
\nSubtracting (i) from (ii), we get
\n5y = 35 \u21d2 y = 35 \u00f7 5 = 7
\nPutting y = 7 in (i), we get
\nx + 10 (7) = 75
\n\u21d2 x = 75 – 70 = 5
\n\u2234 Person will have to pay for travelling a distance of 25 km = x + 25y = 5 + 25(7) = \u20b9180.<\/p>\n
\nSolve the following system by drawing their graph:
\n\\(\\frac{3}{2}\\)x – \\(\\frac{5}{4}\\)y = 6, 6x – 6y = 20.
\nDetermine whether these are consistent, inconsistent or dependent.
\nAnswer:
\n![\"Pair](\"https:\/\/www.learncbse.in\/wp-content\/uploads\/2019\/10\/Pair-of-Linear-Equations-in-Two-Variables-Class-10-Extra-Questions-Maths-Chapter-3-with-Solutions-19.png\")
\nPlotting the points and joining by a ruler in each case. Here, we see that the graph of given equations are parallel lines. The two lines have no point in common. The given system of equations has no solution and is, therefore, inconsistent.
\n![\"Pair](\"https:\/\/www.learncbse.in\/wp-content\/uploads\/2019\/10\/Pair-of-Linear-Equations-in-Two-Variables-Class-10-Extra-Questions-Maths-Chapter-3-with-Solutions-20.png\")
<\/p>\nPair of Linear Equations in Two Variables Class 10 Extra Questions HOTS<\/h3>\n
\nStudents of a class are made to stand in rows. If one student is extra in a row, there would be 2 rows less. If one student is less in a row there would be 3 rows more. Find the number of students in the class. [CBSE 1994]
\nAnswer:
\nLet the number of students in a class = x
\nNumber of rows for students = y
\nNumber of students in one row
\n= \\(\\frac{\\text { Total no. of students }}{\\text { Number of rows }}=\\frac{x}{y}\\)
\nAccording to 1st condition,
\n![\"Pair](\"https:\/\/www.learncbse.in\/wp-content\/uploads\/2019\/10\/Pair-of-Linear-Equations-in-Two-Variables-Class-10-Extra-Questions-Maths-Chapter-3-with-Solutions-21.png\")
\n– 10 + y – 2 = 0
\ny = 12 ………… (iv)
\nFrom (iii) and (iv), we get
\nx = 5 \u00d7 12 = 60
\nHence, number of students in class is 60.<\/p>\n
\nIf 45 is subtracted from twice the greater of two numbers, it results in the other number. If 21 is subtracted from twice the smaller number, it results in greater number. Find the numbers.
\nAnswer:
\nLet x, y (x > y) be the required numbers.<\/p>\n
\nAccording to it:
\n2x – 45 = y
\n\u21d2 2x – y – 45 = 0 …… (i)<\/p>\n
\nAccording to it:
\n2y – 21 = x
\n\u21d2 x – 2y + 21 = 0 ……… (ii)
\nMultiplying (i) by 2 and subtracting from (ii), we get
\n(x – 2y + 21) – 2(2x – y – 45) = 0
\n\u21d2 x – 2y + 21 – 4x + 2y + 90 = 0
\n\u21d2 – 3x + 111 = 0
\n\u21d2 x = \\(\\frac{111}{3}\\) = 37
\nPutting x – 37 in (i), we get
\n2 \u00d7 37 – 45 = y
\n\u21d2 y = 74 – 45
\n\u21d2 y = 29
\n\u2234 The required numbers are 37, 29.<\/p>\n
\nA two digit number is a times the sum of its digits find out how many times the sum of digits equals the number formed by interchanging the digits.
\nAnswer:
\nLet the digit at the unit’s place be x and the digit at the ten’s place by y.
\nTwo digit number = 10y + x
\nAgain, number formed by interchanging the digits = 10x + y
\nAccording to the question,
\n10y + x = \u03b1(x + y) ……….. (i)
\nand 10x + y = k(x + y) ………… (ii)
\nFrom (i), we get
\n(1 – \u03b1)x + (10 – \u03b1)y = 0 ………… (iii)
\nFrom (ii), we get (10 – k)x + (1 – k)y = 0 ………….. (iv)
\n\u2234 \\(\\frac{1-\\alpha}{10-k}=\\frac{10-\\alpha}{1-k}\\)
\n\u21d2(1 – \u03b1) (1 – k) = (10 – \u03b1) (10 – k)
\n\u21d2 1 – k – \u03b1 + \u03b1k = 100 – 10k – 10\u03b1 + \u03b1k
\n\u21d2 9k + 9\u03b1 = 99
\n\u21d2 k + \u03b1 = 11 \u21d2 k = 11 – \u03b1
\nHence, (11 – \u03b1) times sum of digits equal the number obtained by reversing the digits.<\/p>\n
\nA cistern is supplied by three pipes, two of the pipes conveying equal volumes of water. When one of these two pipes and the third pipe are simultaneously opened, the cistern is filled in 12 minutes, but when all the three pipes are opened, the cistern is filled in 7 minutes and 30 seconds. How long does each pipe take separately to fill the cistern?
\nAnswer:
\nLet each of the pipe of equal dimension takes x minutes to fill the whole cistern, and the third pipe takes y minutes.
\nSo, any one of the two equal pipes fills in 1 minute \\(\\frac{1}{x}\\) th of the cistern and the third pipe fills in 1 minute \\(\\frac{1}{y}\\) th of the cistern.
\n\u2234 One of the two equal pipes and the third pipe fill in 1 minute \\(\\left(\\frac{1}{x}+\\frac{1}{y}\\right)\\)th of the cistern.
\nBut, by the condition of the problem, one of the two equal pipes and the third pipe fill in 1 minute \\(\\frac{1}{12}\\)th of the cistern. Then we must have:
\n![\"Pair](\"https:\/\/www.learncbse.in\/wp-content\/uploads\/2019\/10\/Pair-of-Linear-Equations-in-Two-Variables-Class-10-Extra-Questions-Maths-Chapter-3-with-Solutions-22.png\")
\n\u21d2 x = 20, y = 30
\n\u2234 Two pipes of equal dimensions take 20 minutes while third pipe take 30 minutes.<\/p>\n
\nThe system of equations – 3x + 4y = 5 and \\(\\frac{9}{2}\\) – 6y + \\(\\frac{15}{2}\\) = 0 has:
\n(a) Unique Solution
\n(b) No Solution
\n(c) Infinite Solution
\n(d) None of these
\nAnswer:
\n(c) Infinite Solution<\/p>\n
\nA pair of linear equations which have a unique solution x = 2, y = – 3 is:
\n(a) 2x – 3y = – 5, x + y = – 1
\n(b) 2x + 5y + 11 = 0, 4x + 10y + 22 = 0
\n(c) x – 4y – 14 = 0, 5x – y – 13 = 0
\n(d) 2x – y = 1, 3x + 2y = 0
\nAnswer:
\n(c) x – 4y – 14 = 0, 5x – y – 13 = 0<\/p>\n
\nIf a system of pair of linear equations in two unknowns is consistent, then the lines representing the system will be
\n(a) parallel
\n(b) always coincident
\n(c) always intersecting
\n(d) intersecting or coincident
\nAnswer:
\n(d) intersecting or coincident<\/p>\n
\nThe pair of equations x = 0 and y = 0 has
\n(a) one solution
\n(b) two solutions
\n(c) infinitely many solutions
\n(d) no solution
\nAnswer:
\n(a) one solution<\/p>\n
\nA pair of system of equations x = 2, y = -2; x = 3, y = – 3 when represented graphically enclose
\n(a) Square
\n(b) Trapezium
\n(c) Rectangle
\n(d) Triangle
\nAnswer:
\n(c) Rectangle<\/p>\n
\nIf two lines are parallel to each other then the system of equations is
\n(a) consistent
\n(b) inconsistent
\n(c) consistent dependent
\n(d) (a) and (c) both
\nAnswer:
\n(b) inconsistent<\/p>\n
\nIf lq \u2260 mp, then the system of equations lx + my = c, px + qy = k
\n(a) has a unique solution
\n(b) has no solution
\n(c) has infinitely many solutions
\n(d) may or may not have a solution
\nAnswer:
\n(a) has a unique solution<\/p>\n
\nGraphically, the two systems of equations x + 7 = 0, y – 2 = 0 and x-2 = 0, j\/ + 7 = 0 enclose a
\n(a) Square region
\n(b) Rectangular region
\n(c) triangular region
\n(d) Trapezium shaped region
\nAnswer:
\n(a) Square region<\/p>\n
\nIf graph of two lines pass through the same points, then the system of equations representing these lines is
\n(a) Consistent
\n(b) Inconsistent
\n(c) Consistent dependent
\n(d) Inconsistent and dependent
\nAnswer:
\n(c) Consistent dependent<\/p>\n
\nIf x = a, y = b is the solution of the equation x-y = 2 and x + y = 4, then the values of a and b are, respectively
\n(a) 3 and 5
\n(b) 5 and 3
\n(c) 3 and 1
\n(d) – 1 and – 3
\nAnswer:
\n(c) 3 and 1<\/p>\n
\nIf the pair of equations 2x + 3y = 7 and kx + \\(\\frac{9}{2}\\)y = 12 have no solution, then the value of k is:
\n(a) \\(\\frac{2}{3}\\)
\n(b) \\(\\frac{3}{2}\\)
\n(c) 3
\n(d) – 3
\nAnswer:
\n(c) 3<\/p>\n
\n\u20b9 2450 were divided among 65 children. If each girl gets \u20b9 50 and each boy gets \u20b9 30 then the number of girls are
\n(a) 30
\n(b) 40
\n(c) 25
\n(d) 27
\nAnswer:
\n(c) 25<\/p>\n
\nIf 51x + 23y = 116 and 23x + 51y = 106, then value of (x – y) is
\n(a) \\(\\frac{14}{5}\\)
\n(b) \\(\\frac{5}{14}\\)
\n(c) – 5
\n(d) – 14
\nAnswer:
\n(b) \\(\\frac{5}{14}\\)<\/p>\n
\nIf 3x + 2y = 13 and 3x – 2y = 5, then the value of x is:
\n(a) 5
\n(b) 3
\n(c) 7
\n(d) 11
\nAnswer:
\n(b) 3<\/p>\n
\nIf \\(\\frac{x}{2}\\) + y = 0.8 and \\(\\) = 10,
\nthen the value of x + y is:
\n(a) 1
\n(b) 0.6
\n(c) – 0.8
\n(d) 0.5
\nAnswer:
\n(a) 1<\/p>\n
\nIf in a system of equation corresponding coefficients of member equations are proportional then the system has ______________ solution (s).
\nAnswer:
\nInfinite<\/p>\n
\nIf in a system of equation a1<\/sub1<\/sub>y + c1<\/sub> = 0, a2<\/sub>x + b2<\/sub>y + c2<\/sub> = 0, \\(\\frac{a_1}{a_2}=\\frac{b_1}{b_2} \\neq \\frac{c_1}{c_2}\\) then the equation must ___________ at __________ point (s).
\nAnswer:
\nintersect, one<\/p>\n
\nIf in a system of equations a1<\/sub>x + b1<\/sub>y + c1<\/sub> = 0, a2<\/sub>x + b2<\/sub>y + c2<\/sub> = 0, \\(\\frac{a_1}{a_2}=\\frac{b_1}{b_2} \\neq \\frac{c_1}{c_2}\\) then the system of equations is ____________ .
\nAnswer:
\nInconsistent<\/p>\n
\nA pair of linear equations is said to be inconsistent if its graph lines are ____________.
\nAnswer:
\nparallel<\/p>\n
\nA pair of linear equations is said to be ____________ if its graph lines intersect or coincide.
\nAnswer:
\nConsistent<\/p>\n
\nA consistent system of equations where straight lines fall on each other is also called _____________ system of equations.
\nAnswer:
\nDependent<\/p>\n
\nThe pair of linear equations ax + by + c = 0 and lx + my + n = 0 represents two parallel lines if ____________ ____________ .
\nAnswer:
\n\\(\\frac{a}{l}=\\frac{b}{m}=\\frac{c}{n}\\)<\/p>\n
\nSolution of linear equations representing 2x – y = 0, 8x + y = 25 is ____________ .
\nAnswer:
\nx = 25, y = 5<\/p>\n
\nSystem of linear equations representing two numbers whose ratio is 2:3 and on adding 5 to each number the ratio becomes 5 : 7 is ____________ .
\nAnswer:
\n7x – 5y + 10 = 0, 3x + 2y = 0<\/p>\n
\nThe solution of \u221a2x – \u221a5y = 0 and \u221a3x – \u221a7y = 0 is ____________ .
\nAnswer:
\nx = 0, y = 0<\/p>\nExtra Questions for Class 10 Maths<\/a><\/h4>\n
NCERT Solutions for Class 10 Maths<\/a><\/h4>\n","protected":false},"excerpt":{"rendered":"