Pair of Linear Equations in Two Variables — Class 10 MCQs with Answers

Pair of Linear Equations in Two Variables — important questions & MCQs with answers (Class 10 Mathematics)

14 solved questions from this chapter's difficulty levels, each with its answer and explanation. The other 76 are timed and scored when you sign in.

1. Q1Easy

   The pair x + y = 5, 2x + 2y = 10 has:

   A.Unique solution

   B.No solution

   C.Infinitely many solutions✓ Correct

   D.Exactly two solutions

   Answer: C. Infinitely many solutions

   Explanation: Testing a₁/a₂ = b₁/b₂ = c₁/c₂ here gives 1/2 = 1/2 = 5/10, all equal, so the two equations describe the same line, giving infinitely many solutions rather than just one.

2. Q2Easy

   The general form of a linear equation in two variables is:

   A.ax + by + c = 0 where a, b not both zero✓ Correct

   B.ax² + by + c = 0

   C.y = mx only

   D.ax + b = 0

   Answer: A. ax + by + c = 0 where a, b not both zero

   Explanation: A linear equation in two variables is degree-1 in each variable: ax + by + c = 0, with a and b not both zero — unlike ax² + by + c = 0, which is quadratic, not linear.

3. Q3Easy

   Solve: x + y = 5, x − y = 1. Then (x, y) is:

   A.(2, 3)

   B.(3, 2)✓ Correct

   C.(4, 1)

   D.(1, 4)

   Answer: B. (3, 2)

   Explanation: Elimination: add the equations to cancel y, 2x = 6 so x = 3, then y = 5 − 3 = 2 — the pair (3, 2), easy to flip with (2, 3) if x and y are swapped.

4. Q4Easy

   Solve: 2x + y = 7, x + y = 4. Then (x, y) is:

   A.(3, 1)✓ Correct

   B.(1, 3)

   C.(2, 5)

   D.(5, 2)

   Answer: A. (3, 1)

   Explanation: Elimination: subtracting x + y = 4 from 2x + y = 7 cancels y, giving x = 3, then y = 4 − 3 = 1 — the pair (3, 1), not (1, 3) which reverses the order.

5. Q5Easy

   The sum of two numbers is 12 and their difference is 4. The numbers are:

   A.7 and 5

   B.6 and 6

   C.10 and 2

   D.8 and 4✓ Correct

   Answer: D. 8 and 4

   Explanation: Elimination: add x + y = 12 and x − y = 4 to get 2x = 16, so x = 8 and y = 4 — the numbers 8 and 4, not 6 and 6 which ignores the given difference.

6. Q6Easy

   By cross-multiplication, the solution of a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 uses:

   A.x/(b₁c₂ − b₂c₁) = y/(c₁a₂ − c₂a₁) = 1/(a₁b₂ − a₂b₁)✓ Correct

   B.x = b₁c₂

   C.y = a₁c₁

   D.x + y = 0

   Answer: A. x/(b₁c₂ − b₂c₁) = y/(c₁a₂ − c₂a₁) = 1/(a₁b₂ − a₂b₁)

   Explanation: Cross-multiplication solves a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 via x/(b₁c₂ − b₂c₁) = y/(c₁a₂ − c₂a₁) = 1/(a₁b₂ − a₂b₁), reading each term cyclically from the coefficients.

7. Q7Medium

   Lines 2x − y = 1 and 3x + y = 14 intersect at:

   A.(2, 8)

   B.(5, 3)

   C.(3, 5)✓ Correct

   D.(1, 1)

   Answer: C. (3, 5)

   Explanation: Elimination: adding 2x−y=1 and 3x+y=14 cancels y, giving 5x=15 so x=3, then y=2(3)−1=5. Substituting into the wrong equation to find y is the common slip.

8. Q8Medium

   The pair kx + 3y = k − 3 and 12x + ky = k has infinitely many solutions when k equals:

   A.6✓ Correct

   B.−6

   C.±6

   D.3

   Answer: A. 6

   Explanation: Ratios: k/12 = 3/k = (k−3)/k. From k/12 = 3/k: k² = 36 ⇒ k = ±6. Check (k−3)/k: at k = 6: 3/6 = 1/2 = k/12 ✓. At k = −6: −9/−6 = 3/2 ≠ 6/12 = 1/2. So k = 6.

9. Q9Medium

   Solve: 2x + 3y = 11 and 2x − 4y = −24. Then (x, y) is:

   A.(2, −5)

   B.(−2, 5)✓ Correct

   C.(−5, 2)

   D.(5, −2)

   Answer: B. (−2, 5)

   Explanation: Elimination: subtracting 2x−4y=−24 from 2x+3y=11 cancels x, giving 7y=35 so y=5; then 2x=11−15=−4, x=−2. Sign errors while subtracting negatives are the common trap.

10. Q10Medium

    Solve: 0.4x + 0.3y = 1.7 and 0.7x − 0.2y = 0.8.

    A.(x, y) = (2, 3)✓ Correct

    B.(x, y) = (3, 2)

    C.(x, y) = (1, 4.33)

    D.(x, y) = (4, 0.33)

    Answer: A. (x, y) = (2, 3)

    Explanation: Multiply by 10: 4x + 3y = 17, 7x − 2y = 8. Solve: from 2nd, y = (7x−8)/2. Sub: 4x + 3(7x−8)/2 = 17 ⇒ 8x + 21x − 24 = 34 ⇒ 29x = 58 ⇒ x = 2, y = 3.

11. Q11Medium

    The age of a father is 3 years more than 3 times the son's age. After 3 years, father's age will be 10 years more than twice the son's age. Son's current age is:

    A.12

    B.13

    C.10✓ Correct

    D.9

    Answer: C. 10

    Explanation: Model father=3x+3; after 3 years, 3x+6 = 2(x+3)+10 gives x=10. Forgetting to add 3 to both ages before applying the 'twice' condition is the common error.

12. Q12Hard

    For 3x + 2y = 5, 9x + ky = 15 to have infinite solutions, k = ?

    A.4

    B.9

    C.3

    D.6✓ Correct

    Answer: D. 6

    Explanation: Infinite solutions need a₁/a₂ = b₁/b₂ = c₁/c₂: here 3/9 = 5/15 = 1/3, so 2/k must also equal 1/3, giving k = 6 — not k = 9, which only matches the x-ratio.

13. Q13Hard

    For what value of k will the equations kx + 3y − (k − 3) = 0 and 12x + ky − k = 0 have no solution?

    A.k = 6

    B.k = −6✓ Correct

    C.k = ±6

    D.k = 0

    Answer: B. k = −6

    Explanation: No solution ⇒ k/12 = 3/k ≠ (k − 3)/k. From k/12 = 3/k: k² = 36 ⇒ k = ±6. Check inequality: at k = 6, (k − 3)/k = 1/2 = k/12 — equal, fails. At k = −6, (k−3)/k = (−9)/(−6) = 3/2, k/12 = −1/2 ≠ 3/2 ✓. So k = −6.

14. Q14Hard

    Two trains start from stations A and B and travel toward each other at speeds of 50 km/h and 60 km/h respectively. The faster train covers 120 km more than the slower one before they meet. Distance between A and B is:

    A.1500 km

    B.1200 km

    C.1320 km✓ Correct

    D.1100 km

    Answer: C. 1320 km

    Explanation: Both trains run for the same time t; the 10 km/h speed gap explains the extra 120 km, so 10t=120 gives t=12 h. Distance = (50+60)×12 = 1320 km. Using one train's speed alone is the common error.

Pair of Linear Equations in Two Variables — FAQs

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• 01Real Numbers
• 02Polynomials
• 04Quadratic Equations
• 05Arithmetic Progressions
• 06Triangles
• 07Coordinate Geometry

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