Multiple Choice Questions (MCQs) for Class 10 Mathematics: Chapter 03 Pair of Linear Equations in Two Variables
Review structured MCQ sets for Class 10 Mathematics Chapter 03 Pair of Linear Equations in Two Variables. Built according to official CBSE guidelines, these downloadable questions support daily revision and core concept reinforcement.
Practice Chapter 03 Pair of Linear Equations in Two Variables MCQs for Class 10 Mathematics
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Multiple Choice Questions
(a) infinitely many solutions
(b) no solution
(c) exactly one solution
(d) exactly two solutions
Show Answer & Explanation
Answer: (a) infinitely many solutions
Explanation:
1. Writing the second as 6x - 10y + 4 = 0 gives 3/6 = -5/-10 = 2/4 = 1/2. All three ratios are equal, so the lines coincide and there are infinitely many solutions.
Teacher's Note:
1. For a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0: a1/a2 not equal to b1/b2 gives exactly one solution; a1/a2 = b1/b2 not equal to c1/c2 gives no solution; a1/a2 = b1/b2 = c1/c2 gives infinitely many solutions.
(a) 5x + 4y = 4
(b) 5x + 5y = 4
(c) 3x + 4y = 2
(d) 3x + 4y = 5
Show Answer & Explanation
Answer: (d) 3x + 4y = 5
Explanation:
1. From p = 1 - q: 3(1 - q) + 2q = 1, so q = 2 and p = -1. Check: 3(-1) + 4(2) = 5, while 5(-1) + 4(2) = 3 and 5(-1) + 5(2) = 5.
Teacher's Note:
1. A pair of linear equations can be solved by substitution, by elimination (equating coefficients) or by cross multiplication.
(a) 1
(b) 3
(c) 2
(d) 4
Show Answer & Explanation
Answer: (c) 2
Explanation:
1. With x = y, 3x - x = 4 gives x = y = 2. Then 2 + 2 + z = 6, so z = 2.
Teacher's Note:
1. A pair of linear equations can be solved by substitution, by elimination (equating coefficients) or by cross multiplication.
(a) ad - bc > 0
(b) ad - bc = 0
(c) ad - bc < 0
(d) ad + bc = 0
Show Answer & Explanation
Answer: (b) ad - bc = 0
Explanation:
1. Both lines pass through the origin. They are different lines meeting only at the origin unless a/c = b/d, that is ad - bc = 0, in which case they coincide.
Teacher's Note:
1. For a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0: a1/a2 not equal to b1/b2 gives exactly one solution; a1/a2 = b1/b2 not equal to c1/c2 gives no solution; a1/a2 = b1/b2 = c1/c2 gives infinitely many solutions.
(a) \( \frac{6}{5} \)
(b) \( -\frac{6}{5} \)
(c) \( -\frac{5}{6} \)
(d) \( \frac{14}{3} \)
Show Answer & Explanation
Answer: (b) \( -\frac{6}{5} \)
Explanation:
1. For no solution k/2 = 3/(-5), so k = -6/5. Then c1/c2 = (-7)/(-3) = 7/3, which differs from -3/5, so the lines are parallel.
Teacher's Note:
1. For a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0: a1/a2 not equal to b1/b2 gives exactly one solution; a1/a2 = b1/b2 not equal to c1/c2 gives no solution; a1/a2 = b1/b2 = c1/c2 gives infinitely many solutions.
(a) \( x = 1,\ y = 2 \)
(b) \( x = 2,\ y = 1 \)
(c) \( x = 2,\ y = 3 \)
(d) \( x = 3,\ y = 2 \)
Show Answer & Explanation
Answer: (a) \( x = 1,\ y = 2 \)
Explanation:
1. Adding: 66(x + y) = 198, so x + y = 3. Subtracting: 8(x - y) = -8, so x - y = -1. Hence x = 1 and y = 2.
Teacher's Note:
1. A pair of linear equations can be solved by substitution, by elimination (equating coefficients) or by cross multiplication.
(a) \( x = 8,\ y = 6 \)
(b) \( x = 4,\ y = 6 \)
(c) \( x = 5,\ y = 5 \)
(d) \( x = 6,\ y = 4 \)
Show Answer & Explanation
Answer: (d) \( x = 6,\ y = 4 \)
Explanation:
1. Put u = 1/(x+y) and v = 1/(x-y): 25u - 3v = 1 and 40u + 2v = 5. Solving, u = 1/10 and v = 1/2. So x + y = 10 and x - y = 2, giving x = 6 and y = 4.
Teacher's Note:
1. Equations with variables in denominators become linear after substituting new variables for the reciprocals.
(a) 3/2
(b) -6
(c) 6
(d) 10
Show Answer & Explanation
Answer: (c) 6
Explanation:
1. The ratios must be equal: 2/4 = 3/k = 5/10. So 3/k = 1/2 and k = 6.
Teacher's Note:
1. For a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0: a1/a2 not equal to b1/b2 gives exactly one solution; a1/a2 = b1/b2 not equal to c1/c2 gives no solution; a1/a2 = b1/b2 = c1/c2 gives infinitely many solutions.
(a) consistent with infinitely many solutions
(b) inconsistent
(c) consistent with a unique solution
(d) homogeneous
Show Answer & Explanation
Answer: (c) consistent with a unique solution
Explanation:
1. 1/2 is not equal to 2/1, so the lines intersect at one point (x = 2, y = 1). The pair is consistent with a unique solution.
Teacher's Note:
1. For a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0: a1/a2 not equal to b1/b2 gives exactly one solution; a1/a2 = b1/b2 not equal to c1/c2 gives no solution; a1/a2 = b1/b2 = c1/c2 gives infinitely many solutions.
(a) \( y - x \)
(b) \( \frac{xy}{y-x} \)
(c) \( x - y \)
(d) \( \frac{y-x}{xy} \)
Show Answer & Explanation
Answer: (b) \( \frac{xy}{y-x} \)
Explanation:
1. 1/z = 1/x - 1/y = (y - x)/(xy). Taking reciprocals, z = xy/(y - x).
Teacher's Note:
1. Combine fractions over a common denominator and then take the reciprocal.
(a) \( x = 9,\ y = 6 \)
(b) \( x = 6,\ y = 9 \)
(c) \( x = 12,\ y = 0 \)
(d) \( x = 3,\ y = 18 \)
Show Answer & Explanation
Answer: (a) \( x = 9,\ y = 6 \)
Explanation:
1. Multiply the first by 12: 4x + y = 42. Multiply the second by 24: 4x - 3y = 18. Subtracting, 4y = 24, so y = 6 and x = 9.
Teacher's Note:
1. A pair of linear equations can be solved by substitution, by elimination (equating coefficients) or by cross multiplication.
(a) \( x = 0.3,\ y = 0.1 \)
(b) \( x = 0.2,\ y = 0.2 \)
(c) \( x = 0.5,\ y = 0.02 \)
(d) \( x = 0.1,\ y = 0.3 \)
Show Answer & Explanation
Answer: (d) \( x = 0.1,\ y = 0.3 \)
Explanation:
1. Multiply by 100: 20x + 30y = 11 and 70x - 50y = -8. Multiplying the first by 5 and the second by 3 and adding: 310x = 31, so x = 0.1. Then 30y = 11 - 2 = 9 and y = 0.3.
Teacher's Note:
1. A pair of linear equations can be solved by substitution, by elimination (equating coefficients) or by cross multiplication.
(a) \( x = 1.5,\ y = 0.6 \)
(b) \( x = 0.9,\ y = 1.2 \)
(c) \( x = 0.3,\ y = 1.6 \)
(d) \( x = 0.6,\ y = 1.5 \)
Show Answer & Explanation
Answer: (d) \( x = 0.6,\ y = 1.5 \)
Explanation:
1. The second equation gives x + y/3 = 1.1. With the first, x/3 + y = 1.7. Adding gives (4/3)(x + y) = 2.8, so x + y = 2.1. Multiply the first by 3: x + 3y = 5.1; subtracting x + y = 2.1 gives 2y = 3, so y = 1.5 and x = 0.6.
Teacher's Note:
1. A pair of linear equations can be solved by substitution, by elimination (equating coefficients) or by cross multiplication.
(a) \( a = 1,\ b = 5 \)
(b) \( a = 3,\ b = 1 \)
(c) \( a = 5,\ b = 1 \)
(d) \( a = 5,\ b = 2 \)
Show Answer & Explanation
Answer: (c) \( a = 5,\ b = 1 \)
Explanation:
1. Equal ratios: 2/(a - b) = 3/(a + b) = 7/(3a + b - 2). The first equality gives a = 5b. The second equality gives 3(3a + b - 2) = 7(a + b), so 2a - 4b = 6. Then 6b = 6, b = 1 and a = 5.
Teacher's Note:
1. For a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0: a1/a2 not equal to b1/b2 gives exactly one solution; a1/a2 = b1/b2 not equal to c1/c2 gives no solution; a1/a2 = b1/b2 = c1/c2 gives infinitely many solutions.
(a) \( x = -2,\ y = -1 \)
(b) \( x = -1,\ y = -2 \)
(c) \( x = 1,\ y = 2 \)
(d) \( x = 2,\ y = 1 \)
Show Answer & Explanation
Answer: (b) \( x = -1,\ y = -2 \)
Explanation:
1. Adding: 500(x - y) = 500, so x - y = 1. Subtracting the first from the second: 262(x + y) = -786, so x + y = -3. Hence x = -1 and y = -2.
Teacher's Note:
1. A pair of linear equations can be solved by substitution, by elimination (equating coefficients) or by cross multiplication.
(a) \( x = -a,\ y = b \)
(b) \( x = a,\ y = -b \)
(c) \( x = -b,\ y = a \)
(d) \( x = a,\ y = b \)
Show Answer & Explanation
Answer: (a) \( x = -a,\ y = b \)
Explanation:
1. From the second equation bx = ay - 2ab. Putting this in the first: y - 2b - ay/b + a + b = 0, so y(b - a)/b = b - a, giving y = b (for a different from b). Then bx = ab - 2ab = -ab, so x = -a.
Teacher's Note:
1. A pair of linear equations can be solved by substitution, by elimination (equating coefficients) or by cross multiplication.
(a) \( x = \frac23,\ y = \frac25 \)
(b) \( x = \frac25,\ y = \frac23 \)
(c) \( x = \frac32,\ y = \frac52 \)
(d) \( x = \frac13,\ y = \frac15 \)
Show Answer & Explanation
Answer: (a) \( x = \frac23,\ y = \frac25 \)
Explanation:
1. Let u = 1/x and v = 1/y: u/3 + v/5 = 1 and u/5 + v/3 = 17/15. Multiply by 15: 5u + 3v = 15 and 3u + 5v = 17. Solving, u = 3/2 and v = 5/2, so x = 2/3 and y = 2/5.
Teacher's Note:
1. Equations of the form 1/x are linear in u = 1/x.
(a) 4
(b) 3
(c) 12
(d) 6
Show Answer & Explanation
Answer: (d) 6
Explanation:
1. 6/(2k) = 3/6 gives k = 6. Check the constants: (k - 3)/6 = 3/6 = 1/2, which equals the other ratios, so k = 6 works.
Teacher's Note:
1. For a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0: a1/a2 not equal to b1/b2 gives exactly one solution; a1/a2 = b1/b2 not equal to c1/c2 gives no solution; a1/a2 = b1/b2 = c1/c2 gives infinitely many solutions.
(a) 2
(b) 7
(c) 4
(d) 14
Show Answer & Explanation
Answer: (c) 4
Explanation:
1. Since 1/2 = 7/14, coincident lines need 2/k = 1/2, so k = 4.
Teacher's Note:
1. For a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0: a1/a2 not equal to b1/b2 gives exactly one solution; a1/a2 = b1/b2 not equal to c1/c2 gives no solution; a1/a2 = b1/b2 = c1/c2 gives infinitely many solutions.
(a) 6
(b) -6
(c) 3
(d) -3
Show Answer & Explanation
Answer: (b) -6
Explanation:
1. No solution needs C/12 = 3/C, so C = 6 or C = -6, and also 3/C different from 3/6. For C = 6 the lines coincide (6/12 = 3/6 = 3/6); for C = -6, 3/(-6) = -1/2 is different from 1/2. So C = -6.
Teacher's Note:
1. For a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0: a1/a2 not equal to b1/b2 gives exactly one solution; a1/a2 = b1/b2 not equal to c1/c2 gives no solution; a1/a2 = b1/b2 = c1/c2 gives infinitely many solutions.
(a) \( x = a - b,\ y = \frac{2ab}{a+b} \)
(b) \( x = a + b,\ y = -\frac{2ab}{a+b} \)
(c) \( x = a + b,\ y = \frac{2ab}{a+b} \)
(d) \( x = a,\ y = -b \)
Show Answer & Explanation
Answer: (b) \( x = a + b,\ y = -\frac{2ab}{a+b} \)
Explanation:
1. Verified by substitution: x + y = (a^2 + b^2)/(a + b) and x = a + b gives y = -2ab/(a + b). Then (a - b)(a + b) - 2ab = a^2 - b^2 - 2ab, which matches the first equation.
Teacher's Note:
1. A pair of linear equations can be solved by substitution, by elimination (equating coefficients) or by cross multiplication.
(a) \( x = 1,\ y = 2 \)
(b) \( x = 2,\ y = 1 \)
(c) \( x = 3,\ y = 0 \)
(d) \( x = -1,\ y = 4 \)
Show Answer & Explanation
Answer: (a) \( x = 1,\ y = 2 \)
Explanation:
1. Adding: 80(x + y) = 240, so x + y = 3. Subtracting: 6(y - x) = 6, so y - x = 1. Hence x = 1 and y = 2.
Teacher's Note:
1. A pair of linear equations can be solved by substitution, by elimination (equating coefficients) or by cross multiplication.
(a) parallel to the x-axis, 2 units above it
(b) passing through the origin
(c) parallel to the y-axis, 2 units to its left
(d) parallel to the y-axis, 2 units to its right
Show Answer & Explanation
Answer: (d) parallel to the y-axis, 2 units to its right
Explanation:
1. Every point on x = 2 has x-coordinate 2 whatever the y-coordinate. The line is vertical, parallel to the y-axis, at distance 2 on the positive side.
Teacher's Note:
1. Each linear equation is a straight line. Two points fix a line; the solution of the pair is the point where the two lines meet.
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Practice MCQs for Class 10 Mathematics Chapter 03 Pair of Linear Equations in Two Variables
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