CBSE Class 10 Mathematics Pair of Linear Equations in Two Variables MCQs Set 17

Answer: (a) A is true and R is the correct explanation of A

Explanation:
1. In a homogeneous system \( a_1x + b_1y = 0,\ a_2x + b_2y = 0 \) every constant term is zero.
2. Putting x = 0, y = 0 satisfies both equations, so R is true.
3. Because a solution always exists, the system is always consistent, so A is true and R explains A.

Teacher's Note:
1. Both lines of a homogeneous system pass through the origin.
2. So the two lines always meet at least at (0, 0).

Answer: (b) 3

Explanation:
1. Put x = 2, y = 1 in the second equation: \( 2 + k(1) = 5 \).
2. So k = 3.
3. Check with the first equation: 2 − 1 = 1, which is true.

Teacher's Note:
1. A solution of a pair must satisfy both equations.
2. Substitute it into the equation that still contains k.

Answer: (d) \( k \neq \frac{2}{9} \)

Explanation:
1. A unique solution needs \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \).
2. \( \frac{2}{6} \neq \frac{k}{2/3} \Rightarrow \frac{1}{3} \neq \frac{3k}{2} \).
3. \( \Rightarrow k \neq \frac{2}{9} \).

Teacher's Note:
1. Unique solution means the lines intersect, so the ratios of x and y coefficients differ.
2. Equality would give k = 2/9, which must be excluded.

Answer: (a) \( k \neq -3 \)

Explanation:
1. \( \frac{2k}{6} \neq \frac{5}{-5} \) for a unique solution.
2. \( \frac{k}{3} \neq -1 \Rightarrow k \neq -3 \).

Teacher's Note:
1. Write both equations in the form ax + by + c = 0 and compare a1/a2 with b1/b2.
2. The excluded value of k is the one making the ratios equal.

Answer: (a) k = 2

Explanation:
1. Infinite solutions need \( \frac{3}{9} = \frac{4}{12} = \frac{k}{6} \).
2. \( \frac{1}{3} = \frac{k}{6} \Rightarrow k = 2 \).

Teacher's Note:
1. Equal ratios of all three coefficients mean the lines coincide.
2. Here a1/a2 and b1/b2 are already 1/3, so c1/c2 must be 1/3.

Answer: (b) k = 6

Explanation:
1. \( \frac{2}{k} = \frac{5}{15} = \frac{k}{18} \).
2. \( \frac{2}{k} = \frac{1}{3} \Rightarrow k = 6 \).
3. Check: \( \frac{k}{18} = \frac{6}{18} = \frac{1}{3} \), so k = 6 works.

Teacher's Note:
1. Always check the third ratio too.
2. Both conditions must give the same k.

Answer: (b) k = 10

Explanation:
1. No solution needs \( \frac{3}{6} = \frac{5}{k} \neq \frac{3}{8} \).
2. \( \frac{1}{2} = \frac{5}{k} \Rightarrow k = 10 \).
3. For k = 10, \( \frac{5}{10} = \frac{1}{2} \neq \frac{3}{8} \), so the system is inconsistent.

Teacher's Note:
1. No solution means parallel lines.
2. The first two ratios are equal and the third ratio is different.

Answer: (b) k = 14

Explanation:
1. For no solution, \( \frac{3}{12} = \frac{7}{2k} \neq \frac{-k}{-(4k + 1)} \).
2. \( \frac{1}{4} = \frac{7}{2k} \Rightarrow k = 14 \).
3. For k = 14: \( \frac{k}{4k + 1} = \frac{14}{57} \neq \frac{1}{4} \), so the condition holds.

Teacher's Note:
1. Solve the equality of the first two ratios and then confirm the inequality with the third.
2. k = 14 satisfies both parts.

Answer: (c) k ≠ −9

Explanation:
1. Check when the pair is inconsistent: \( \frac{7}{3} = \frac{-3}{k/7} \Rightarrow k = -9 \).
2. Then \( \frac{c_1}{c_2} = \frac{-4}{-4} = 1 \neq \frac{7}{3} \), so k = −9 gives parallel lines (no solution).
3. For every other k the lines intersect, so the pair is consistent only when \( k \neq -9 \).

Teacher's Note:
1. Find the value of k that makes the lines parallel and exclude it.
2. Consistent means at least one solution exists.

Answer: (a) k ≠ 6

Explanation:
1. The pair is inconsistent when \( \frac{k}{3} = \frac{4}{2} \neq \frac{5}{5} \), i.e., k = 6.
2. So it is consistent exactly when \( k \neq 6 \).

Teacher's Note:
1. Parallel lines are the only inconsistent case here.
2. Exclude k = 6.

Answer: (a) k = 1

Explanation:
1. \( \frac{7}{14} = \frac{k}{2} = \frac{k}{k + 1} \).
2. \( \frac{1}{2} = \frac{k}{2} \Rightarrow k = 1 \).
3. Check: \( \frac{k}{k + 1} = \frac{1}{2} \), which is true for k = 1.

Teacher's Note:
1. Equate the first two ratios, then confirm the third.
2. The same k must satisfy all three equalities.

Answer: (b) k = 2

Explanation:
1. \( \frac{13}{39} = \frac{k}{6} = \frac{k}{k + 4} \).
2. \( \frac{1}{3} = \frac{k}{6} \Rightarrow k = 2 \).
3. Check: \( \frac{2}{6} = \frac{1}{3} \), true.

Teacher's Note:
1. The first ratio is 1/3, so k/6 = 1/3.
2. Verify with the constant terms.

Answer: (a) 1

Explanation:
1. From the first equation, x = 3 − y. Substitute into the second: \( 2(3 - y) + 5y = 12 \Rightarrow 3y = 6 \Rightarrow y = 2 \).
2. So x = 3 − 2 = 1, i.e., x₁ = 1.

Teacher's Note:
1. Solve the pair by substitution.
2. Check: 2(1) + 5(2) = 12.

Answer: (b) −1

Explanation:
1. From the second equation, y = 2x + 3. Substitute in the first: \( 3x - 5(2x + 3) + 1 = 0 \Rightarrow -7x - 14 = 0 \Rightarrow x = -2 \).
2. Then y = 2(−2) + 3 = −1.

Teacher's Note:
1. Use substitution.
2. Check in the first equation: 3(−2) − 5(−1) + 1 = 0.

Answer: (a) x = 1, y = 2

Explanation:
1. Multiply the first equation by 7: \( 7x + 14y = 35 \).
2. Subtract the second equation: \( 11y = 22 \Rightarrow y = 2 \).
3. Then \( x = 5 - 2(2) = 1 \).

Teacher's Note:
1. Use elimination.
2. Verify: 7(1) + 3(2) = 13.

Answer: (a) a unique solution

Explanation:
1. \( \frac{a_1}{a_2} = \frac{1}{3} \) and \( \frac{b_1}{b_2} = \frac{2}{12} = \frac{1}{6} \).
2. Since \( \frac{1}{3} \neq \frac{1}{6} \), the lines intersect at exactly one point.

Teacher's Note:
1. Compare only a1/a2 and b1/b2 first.
2. Unequal ratios always give a unique solution.

Answer: (a) 45 years

Explanation:
1. Let the father be x years and the son y years. Then \( x + y = 65 \) and \( 2(x - y) = 50 \Rightarrow x - y = 25 \).
2. Adding: \( 2x = 90 \Rightarrow x = 45 \).

Teacher's Note:
1. Convert each statement into an equation.
2. The son is 20, which checks: 45 + 20 = 65.

Answer: (c) \( \frac{7}{9} \)

Explanation:
1. Let the fraction be \( \frac{x}{y} \). Then \( \frac{x + 1}{y + 1} = \frac{4}{5} \Rightarrow 5x - 4y = -1 \) ...(i) and \( \frac{x - 5}{y - 5} = \frac{1}{2} \Rightarrow 2x - y = 5 \) ...(ii).
2. From (ii), y = 2x − 5. Substitute in (i): \( 5x - 8x + 20 = -1 \Rightarrow x = 7 \), so y = 9.

Teacher's Note:
1. Form two equations from the two conditions.
2. Check: 8/10 = 4/5 and 2/4 = 1/2.

Answer: (b) Rs. 850

Explanation:
1. Let a chair cost Rs. x and a table Rs. y. Then \( 3x + 2y = 1850 \) and \( 5x + 3y = 2850 \).
2. Multiply the first by 3 and the second by 2: \( 9x + 6y = 5550 \) and \( 10x + 6y = 5700 \).
3. Subtract: x = 150. Then \( 2y = 1850 - 450 \), so y = 700.
4. One chair and one table cost 150 + 700 = Rs. 850.

Teacher's Note:
1. Elimination works well because both equations have integer coefficients.
2. The source printed the second total as Rs. 1850, which gives a negative cost, so it is corrected to Rs. 2850.

Answer: (b) 30 years

Explanation:
1. Let the man be x years and the son y years. Then \( x + 6 = 3(y + 6) \Rightarrow x - 3y = 12 \) and \( x - 3 = 9(y - 3) \Rightarrow x - 9y = -24 \).
2. Subtract: \( 6y = 36 \Rightarrow y = 6 \). Then x = 12 + 18 = 30.

Teacher's Note:
1. Use present age = x and y and shift by the stated number of years.
2. Check: 36 = 3 x 12 and 27 = 9 x 3.