CBSE Class 10 Mathematics Pair of Linear Equations in Two Variables MCQs Set 34

Multiple Choice Questions (MCQs) for Class 10 Mathematics: Chapter 03 Pair of Linear Equations in Two Variables

Access targeted multiple-choice questions for Chapter 03 Pair of Linear Equations in Two Variables designed to align with the latest CBSE academic syllabus for Class 10 Mathematics. These objective practice sets help students evaluate their conceptual understanding and improve exam readiness.

Practice Chapter 03 Pair of Linear Equations in Two Variables MCQs for Class 10 Mathematics

Navigate directly to the 50 objective questions for Chapter 03 Pair of Linear Equations in Two Variables using the digital viewer below. Each practice set includes verified answer keys, allowing students to instantly cross-check their work and identify areas requiring further revision.

Multiple Choice Questions

Question 1: For what value of k does the pair 2x + ky = 1 and 3x − 5y = 7 have a unique solution?
(a) k = −10/3
(b) k ≠ −10/3
(c) k ≠ 10/3
(d) k ≠ −3/10
Show Answer & Explanation

Answer: (b) k ≠ −10/3

Explanation:
1. Unique solution needs 2/3 ≠ k/(−5), so k ≠ −10/3.

Teacher's Note:
1. Form the equations from the data, solve them by substitution, elimination or cross-multiplication, and verify the answer in both equations.
2. For the conditions on k, compare a1/a2, b1/b2 and c1/c2.

Question 2: For what value of k does x − 2y = 3 and 3x + ky = 1 have a unique solution?
(a) k ≠ −6
(b) k = −6
(c) k ≠ 6
(d) k ≠ −3
Show Answer & Explanation

Answer: (a) k ≠ −6

Explanation:
1. Unique solution needs 1/3 ≠ −2/k, i.e. k ≠ −6.

Teacher's Note:
1. Form the equations from the data, solve them by substitution, elimination or cross-multiplication, and verify the answer in both equations.
2. For the conditions on k, compare a1/a2, b1/b2 and c1/c2.

Question 3: For what value of k does 2x + 5y = 7 and 3x − ky = 5 have a unique solution?
(a) k = −15/2
(b) k ≠ 15/2
(c) k ≠ −2/15
(d) k ≠ −15/2
Show Answer & Explanation

Answer: (d) k ≠ −15/2

Explanation:
1. Need 2/3 ≠ 5/(−k), so −2k ≠ 15, i.e. k ≠ −15/2.

Teacher's Note:
1. Form the equations from the data, solve them by substitution, elimination or cross-multiplication, and verify the answer in both equations.
2. For the conditions on k, compare a1/a2, b1/b2 and c1/c2.

Question 4: For which value of k does 5x + 2y = k and 10x + 4y = 3 have infinitely many solutions?
(a) 3
(b) 3/2
(c) 2/3
(d) 6
Show Answer & Explanation

Answer: (b) 3/2

Explanation:
1. 5/10 = 2/4 = k/3 gives k/3 = 1/2, so k = 3/2.

Teacher's Note:
1. Form the equations from the data, solve them by substitution, elimination or cross-multiplication, and verify the answer in both equations.
2. For the conditions on k, compare a1/a2, b1/b2 and c1/c2.

Question 5: For which value of k does kx + 4y = k − 4 and 16x + ky = k have infinitely many solutions?
(a) 8
(b) −8
(c) 4
(d) 16
Show Answer & Explanation

Answer: (a) 8

Explanation:
1. k/16 = 4/k gives k² = 64, so k = ±8. Also 4/k = (k − 4)/k needs k − 4 = 4, so k = 8 (for k = −8 the constants do not match).

Teacher's Note:
1. Form the equations from the data, solve them by substitution, elimination or cross-multiplication, and verify the answer in both equations.
2. For the conditions on k, compare a1/a2, b1/b2 and c1/c2.

Question 6: For which value of k does 8x + 5y = 9 and kx + 10y = 8 have no solution?
(a) 8
(b) −16
(c) 4
(d) 16
Show Answer & Explanation

Answer: (d) 16

Explanation:
1. No solution needs 8/k = 5/10 ≠ 9/8, so k = 16. Then 9/8 ≠ 1/2 holds.

Teacher's Note:
1. Form the equations from the data, solve them by substitution, elimination or cross-multiplication, and verify the answer in both equations.
2. For the conditions on k, compare a1/a2, b1/b2 and c1/c2.

Question 7: For which value of k does 2x + ky = 11 and 5x − 7y = 5 have no solution?
(a) 14/5
(b) −5/14
(c) −14/5
(d) 5/14
Show Answer & Explanation

Answer: (c) −14/5

Explanation:
1. No solution needs 2/5 = k/(−7) ≠ 11/5, so k = −14/5.

Teacher's Note:
1. Form the equations from the data, solve them by substitution, elimination or cross-multiplication, and verify the answer in both equations.
2. For the conditions on k, compare a1/a2, b1/b2 and c1/c2.

Question 8: For which value of k does x − 4y = 6 and 3x + ky = 5 have no solution?
(a) 12
(b) −12
(c) −4
(d) 3
Show Answer & Explanation

Answer: (b) −12

Explanation:
1. No solution needs 1/3 = −4/k ≠ 6/5, so k = −12.

Teacher's Note:
1. Form the equations from the data, solve them by substitution, elimination or cross-multiplication, and verify the answer in both equations.
2. For the conditions on k, compare a1/a2, b1/b2 and c1/c2.

Question 9: Five years hence a father's age will be three times his son's age. Five years ago he was seven times as old as his son. Their present ages are
(a) father 35 years, son 10 years
(b) father 45 years, son 12 years
(c) father 40 years, son 12 years
(d) father 40 years, son 10 years
Show Answer & Explanation

Answer: (d) father 40 years, son 10 years

Explanation:
1. x + 5 = 3(y + 5) gives x − 3y = 10. x − 5 = 7(y − 5) gives x − 7y = −30. Subtracting: 4y = 40, y = 10 and x = 40.

Teacher's Note:
1. Form the equations from the data, solve them by substitution, elimination or cross-multiplication, and verify the answer in both equations.
2. For the conditions on k, compare a1/a2, b1/b2 and c1/c2.

Question 10: A father is 4 times as old as his son. After 5 years his age will be 3 times the age of the son. The present ages are
(a) father 36 years, son 9 years
(b) father 48 years, son 12 years
(c) father 40 years, son 10 years
(d) father 44 years, son 11 years
Show Answer & Explanation

Answer: (c) father 40 years, son 10 years

Explanation:
1. x = 4y and x + 5 = 3(y + 5). So 4y + 5 = 3y + 15 and y = 10, x = 40.

Teacher's Note:
1. Form the equations from the data, solve them by substitution, elimination or cross-multiplication, and verify the answer in both equations.
2. For the conditions on k, compare a1/a2, b1/b2 and c1/c2.

Question 11: A two-digit number is 4 times the sum of its digits. If 18 is added to the number, the digits get reversed. The number is
(a) 42
(b) 24
(c) 36
(d) 48
Show Answer & Explanation

Answer: (b) 24

Explanation:
1. Let the number be 10x + y. 10x + y = 4(x + y) gives 2x = y. Also 10x + y + 18 = 10y + x gives y − x = 2. So x = 2, y = 4 and the number is 24. Check: 24 = 4(2 + 4) and 24 + 18 = 42.

Teacher's Note:
1. Form the equations from the data, solve them by substitution, elimination or cross-multiplication, and verify the answer in both equations.
2. For the conditions on k, compare a1/a2, b1/b2 and c1/c2.

Question 12: A two-digit number is 3 more than 4 times the sum of its digits. If 18 is added, the digits are reversed. The number is
(a) 35
(b) 53
(c) 27
(d) 47
Show Answer & Explanation

Answer: (a) 35

Explanation:
1. 10x + y = 4(x + y) + 3 gives 6x − 3y = 3, i.e. 2x − y = 1. Also y − x = 2. Adding: x = 3, y = 5. The number is 35. Check: 4 × 8 + 3 = 35 and 35 + 18 = 53.

Teacher's Note:
1. Form the equations from the data, solve them by substitution, elimination or cross-multiplication, and verify the answer in both equations.
2. For the conditions on k, compare a1/a2, b1/b2 and c1/c2.

Question 13: The sum of a two-digit number and the number formed by interchanging its digits is 132. If 12 is added to the number, the new number becomes 5 times the sum of its digits. The number is
(a) 84
(b) 57
(c) 39
(d) 48
Show Answer & Explanation

Answer: (d) 48

Explanation:
1. 11(x + y) = 132 gives x + y = 12. 10x + y + 12 = 5(x + y) = 60 gives 10x + y = 48. So x = 4, y = 8 and the number is 48.

Teacher's Note:
1. Form the equations from the data, solve them by substitution, elimination or cross-multiplication, and verify the answer in both equations.
2. For the conditions on k, compare a1/a2, b1/b2 and c1/c2.

Question 14: The sum of numerator and denominator of a fraction is 18. If the denominator is increased by 2, the fraction becomes 1/3. The fraction is
(a) 13/5
(b) 5/13
(c) 6/12
(d) 7/11
Show Answer & Explanation

Answer: (b) 5/13

Explanation:
1. x + y = 18 and x/(y + 2) = 1/3 gives 3x = y + 2. So 3x = 18 − x + 2, hence x = 5 and y = 13.

Teacher's Note:
1. Form the equations from the data, solve them by substitution, elimination or cross-multiplication, and verify the answer in both equations.
2. For the conditions on k, compare a1/a2, b1/b2 and c1/c2.

Question 15: The length of a rectangle exceeds its width by 8 cm and its area is 240 sq. cm. Its dimensions are
(a) length 20 cm, width 12 cm
(b) length 24 cm, width 10 cm
(c) length 18 cm, width 10 cm
(d) length 16 cm, width 15 cm
Show Answer & Explanation

Answer: (a) length 20 cm, width 12 cm

Explanation:
1. Let the width be w, length w + 8. w(w + 8) = 240 gives w² + 8w − 240 = 0, so (w + 20)(w − 12) = 0 and w = 12. Length = 20 cm.

Teacher's Note:
1. Form the equations from the data, solve them by substitution, elimination or cross-multiplication, and verify the answer in both equations.
2. For the conditions on k, compare a1/a2, b1/b2 and c1/c2.

Question 16: The side of a square exceeds the side of another by 4 cm and the sum of their areas is 400 sq. cm. The sides are
(a) 10 cm and 14 cm
(b) 8 cm and 12 cm
(c) 14 cm and 18 cm
(d) 12 cm and 16 cm
Show Answer & Explanation

Answer: (d) 12 cm and 16 cm

Explanation:
1. x² + (x + 4)² = 400 gives x² + 4x − 192 = 0, so x = 12. The sides are 12 cm and 16 cm (squares 144 + 256 = 400).

Teacher's Note:
1. Form the equations from the data, solve them by substitution, elimination or cross-multiplication, and verify the answer in both equations.
2. For the conditions on k, compare a1/a2, b1/b2 and c1/c2.

Question 17: In a triangle the sum of two angles equals the third angle and their difference is 50°. The two smaller angles are
(a) 60° and 10°
(b) 65° and 15°
(c) 70° and 20°
(d) 80° and 30°
Show Answer & Explanation

Answer: (c) 70° and 20°

Explanation:
1. If the sum of two angles equals the third, the third is 90° and the other two add to 90°. With difference 50°: (90 + 50)/2 = 70° and 20°.

Teacher's Note:
1. Form the equations from the data, solve them by substitution, elimination or cross-multiplication, and verify the answer in both equations.
2. For the conditions on k, compare a1/a2, b1/b2 and c1/c2.

Question 18: Two numbers are in the ratio 3 : 5. If 5 is subtracted from each, the new numbers are in the ratio 1 : 2. The numbers are
(a) 12 and 20
(b) 15 and 25
(c) 18 and 30
(d) 9 and 15
Show Answer & Explanation

Answer: (b) 15 and 25

Explanation:
1. Let the numbers be 3m and 5m. (3m − 5)/(5m − 5) = 1/2 gives 6m − 10 = 5m − 5, so m = 5. The numbers are 15 and 25.

Teacher's Note:
1. Form the equations from the data, solve them by substitution, elimination or cross-multiplication, and verify the answer in both equations.
2. For the conditions on k, compare a1/a2, b1/b2 and c1/c2.

Question 19: The ages of Ram and Mohan are in the ratio 2 : 3 and the sum of their ages is 65. The difference of their ages is
(a) 10 years
(b) 15 years
(c) 26 years
(d) 13 years
Show Answer & Explanation

Answer: (d) 13 years

Explanation:
1. Ages: 26 and 39 (2 : 3, sum 65). The difference is 39 − 26 = 13 years.

Teacher's Note:
1. Form the equations from the data, solve them by substitution, elimination or cross-multiplication, and verify the answer in both equations.
2. For the conditions on k, compare a1/a2, b1/b2 and c1/c2.

Question 20: The difference between two numbers is 1365. When the larger is divided by the smaller, the quotient is 6 and the remainder is 15. The numbers are
(a) 1500 and 135
(b) 1600 and 235
(c) 1635 and 270
(d) 1365 and 15
Show Answer & Explanation

Answer: (c) 1635 and 270

Explanation:
1. Larger = 6 × smaller + 15 and larger − smaller = 1365. So 5 × smaller + 15 = 1365, hence smaller = 270 and larger = 1635.

Teacher's Note:
1. Form the equations from the data, solve them by substitution, elimination or cross-multiplication, and verify the answer in both equations.
2. For the conditions on k, compare a1/a2, b1/b2 and c1/c2.

Question 21: The denominator of a fraction is 1 more than its numerator. If 1 is subtracted from both numerator and denominator, the fraction becomes 1/2. The fraction is
(a) 3/4
(b) 2/3
(c) 1/2
(d) 3/2
Show Answer & Explanation

Answer: (b) 2/3

Explanation:
1. y = x + 1 and (x − 1)/(y − 1) = 1/2 gives 2x − 2 = x, so x = 2 and y = 3.

Teacher's Note:
1. Form the equations from the data, solve them by substitution, elimination or cross-multiplication, and verify the answer in both equations.
2. For the conditions on k, compare a1/a2, b1/b2 and c1/c2.

Question 22: 2 tables and 3 chairs together cost Rs. 2000 whereas 3 tables and 2 chairs together cost Rs. 2500. The total cost of 1 table and 5 chairs is
(a) Rs. 1700
(b) Rs. 1500
(c) Rs. 1900
(d) Rs. 1100
Show Answer & Explanation

Answer: (a) Rs. 1700

Explanation:
1. 2x + 3y = 2000 and 3x + 2y = 2500. Adding: 5x + 5y = 4500, so x + y = 900. Subtracting: x − y = 500. So x = 700 and y = 200. Then x + 5y = 700 + 1000 = Rs. 1700.

Teacher's Note:
1. Form the equations from the data, solve them by substitution, elimination or cross-multiplication, and verify the answer in both equations.
2. For the conditions on k, compare a1/a2, b1/b2 and c1/c2.

Question 23: 3 bags and 4 pens together cost Rs. 257 whereas 4 bags and 3 pens together cost Rs. 324. The total cost of 1 bag and 10 pens is
(a) Rs. 135
(b) Rs. 175
(c) Rs. 89
(d) Rs. 155
Show Answer & Explanation

Answer: (d) Rs. 155

Explanation:
1. 3x + 4y = 257 and 4x + 3y = 324. Adding: 7x + 7y = 581, so x + y = 83. Subtracting: x − y = 67. So x = 75 and y = 8. Then x + 10y = 75 + 80 = Rs. 155.

Teacher's Note:
1. Form the equations from the data, solve them by substitution, elimination or cross-multiplication, and verify the answer in both equations.
2. For the conditions on k, compare a1/a2, b1/b2 and c1/c2.

Question 24: Solve: x + y = 5xy and 3x + 2y = 13xy (x, y ≠ 0).
(a) x = 2, y = 3
(b) x = 1/2, y = 1/3
(c) x = 1/3, y = 1/2
(d) x = 5, y = 13
Show Answer & Explanation

Answer: (b) x = 1/2, y = 1/3

Explanation:
1. Divide each equation by xy: 1/y + 1/x = 5 and 3/y + 2/x = 13. With u = 1/x and v = 1/y: u + v = 5 and 2u + 3v = 13. So v = 3, u = 2, giving x = 1/2 and y = 1/3. (x = y = 0 is the trivial solution.)

Teacher's Note:
1. Form the equations from the data, solve them by substitution, elimination or cross-multiplication, and verify the answer in both equations.
2. For the conditions on k, compare a1/a2, b1/b2 and c1/c2.

Question 25: Solve: xy/(x + y) = 6/5 and xy/(y − x) = 6 (xy ≠ 0).
(a) x = 2, y = 3
(b) x = 3, y = 2
(c) x = 6, y = 5
(d) x = 1, y = 6
Show Answer & Explanation

Answer: (a) x = 2, y = 3

Explanation:
1. Taking reciprocals: (x + y)/xy = 5/6 gives 1/y + 1/x = 5/6, and (y − x)/xy = 1/6 gives 1/x − 1/y = 1/6. Adding: 2/x = 1, so x = 2 and 1/y = 5/6 − 1/2 = 1/3, so y = 3.

Teacher's Note:
1. Form the equations from the data, solve them by substitution, elimination or cross-multiplication, and verify the answer in both equations.
2. For the conditions on k, compare a1/a2, b1/b2 and c1/c2.

Question 26: Solve: x + y = a + b and ax − by = a² − b².
(a) x = b, y = a
(b) x = a, y = −b
(c) x = −a, y = b
(d) x = a, y = b
Show Answer & Explanation

Answer: (d) x = a, y = b

Explanation:
1. Multiply the first equation by b: bx + by = ab + b². Adding to ax − by = a² − b²: (a + b)x = a² + ab = a(a + b), so x = a (taking a + b ≠ 0). Then y = a + b − a = b.

Teacher's Note:
1. Form the equations from the data, solve them by substitution, elimination or cross-multiplication, and verify the answer in both equations.
2. For the conditions on k, compare a1/a2, b1/b2 and c1/c2.

Question 27: Solve: x − y = a + b and ax + by = a² − b².
(a) x = a, y = b
(b) x = b, y = −a
(c) x = a, y = −b
(d) x = −a, y = b
Show Answer & Explanation

Answer: (c) x = a, y = −b

Explanation:
1. From the first, x = a + b + y. Then a(a + b + y) + by = a² − b² gives (a + b)y = −ab − b² = −b(a + b), so y = −b (a + b ≠ 0) and x = a.

Teacher's Note:
1. Form the equations from the data, solve them by substitution, elimination or cross-multiplication, and verify the answer in both equations.
2. For the conditions on k, compare a1/a2, b1/b2 and c1/c2.

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