CBSE Class 10 Mathematics Pair of Linear Equations in Two Variables MCQs Set 20

Answer: (c) \( 4, -2 \)

Explanation:
1. Label the equations \( 2x + 3y = 2 \) (i) and \( x - 2y = 8 \) (ii).
2. Multiply (i) by 2 and (ii) by 3: \( 4x + 6y = 4 \) and \( 3x - 6y = 24 \).
3. Add them: \( 7x = 28 \), so \( x = 4 \).
4. Put \( x = 4 \) in (ii): \( 4 - 2y = 8 \), so \( y = -2 \). The answer is option (c).

Teacher's Note:
1. Elimination works best when you can make the coefficient of one variable equal and opposite.
2. Always check the answer by putting it back in both equations: 2(4) + 3(-2) = 2 and 4 - 2(-2) = 8.

Answer: (b) \( -\frac{11}{8}, \frac{7}{4} \)

Explanation:
1. Write the equations as \( 2x + y = -1 \) (i) and \( 2x - 3y = -8 \) (ii).
2. Subtract (ii) from (i): \( 4y = 7 \), so \( y = \frac{7}{4} \).
3. Put this in (i): \( 2x = -1 - \frac{7}{4} = -\frac{11}{4} \), so \( x = -\frac{11}{8} \). The answer is option (b).

Teacher's Note:
1. When the x-coefficients are already equal, simply subtract the equations.
2. Do not drop the negative sign of x while writing the final answer.

Answer: (a) 20 cm, 20 cm, 25 cm

Explanation:
1. Let each equal side be \( a \) cm. Then the unequal side is \( (a + 5) \) cm.
2. Perimeter: \( a + a + (a + 5) = 65 \), so \( 3a = 60 \) and \( a = 20 \).
3. The sides are 20 cm, 20 cm and 25 cm (a valid triangle, since 20 + 20 > 25). The answer is option (a).

Teacher's Note:
1. Choose one variable for the equal sides and express the third side in terms of it.
2. After finding the sides, check the triangle inequality.

Answer: (d) 2

Explanation:
1. Write the second equation as \( 2x + ky - 3 = 0 \), so \( a_1 = 1, b_1 = 1, c_1 = -4 \) and \( a_2 = 2, b_2 = k, c_2 = -3 \).
2. For no solution, \( \frac{1}{2} = \frac{1}{k} \ne \frac{-4}{-3} \).
3. \( \frac{1}{2} = \frac{1}{k} \) gives \( k = 2 \), and \( \frac{1}{2} \ne \frac{4}{3} \) holds. The answer is option (d).

Teacher's Note:
1. If k is not 2, the lines meet at one point, so the system has a unique solution.
2. Only the single value k = 2 gives parallel lines.

Answer: (c) one unique solution

Explanation:
1. Here \( a_1 = 2, b_1 = 3, a_2 = 3, b_2 = 4 \).
2. \( \frac{a_1}{a_2} = \frac{2}{3} \) and \( \frac{b_1}{b_2} = \frac{3}{4} \), and \( \frac{2}{3} \ne \frac{3}{4} \).
3. So the lines intersect at exactly one point and the pair has one unique solution: option (c).

Teacher's Note:
1. Different ratios of the coefficients of x and y always mean intersecting lines.
2. A pair of linear equations can never have exactly two solutions.

Answer: (c) no solution

Explanation:
1. A system is called inconsistent when no pair of values satisfies both equations.
2. Graphically the two lines are parallel and never meet, so the system has no solution: option (c).

Teacher's Note:
1. Consistent means at least one solution exists; inconsistent means none.
2. Remember: a unique solution or infinitely many solutions are both consistent cases.

Answer: (c) parallel

Explanation:
1. An inconsistent pair has no common solution, so the lines never meet.
2. Lines that never meet are parallel: option (c).

Teacher's Note:
1. Link the algebra to the graph: no solution corresponds to parallel lines.
2. Coincident lines give infinitely many solutions and intersecting lines give exactly one.

Answer: (d) infinite solutions

Explanation:
1. \( \frac{a_1}{a_2} = \frac{3}{6} = \frac{1}{2}, \quad \frac{b_1}{b_2} = \frac{1}{2}, \quad \frac{c_1}{c_2} = \frac{-4}{-8} = \frac{1}{2} \).
2. All three ratios are equal, so the lines coincide and there are infinitely many solutions: option (d).
3. The points (1, 1) and (0, 4) are only two of the infinitely many solutions, so neither (a) nor (b) is a unique solution.

Teacher's Note:
1. The second equation is just twice the first.
2. Equal ratios for all three terms mean coincident lines.

Answer: (c) 5

Explanation:
1. Here \( a_1 = 3, b_1 = -k, c_1 = -20 \) and \( a_2 = 6, b_2 = -10, c_2 = 40 \).
2. For no solution, \( \frac{3}{6} = \frac{-k}{-10} \ne \frac{-20}{40} \).
3. \( \frac{1}{2} = \frac{k}{10} \) gives \( k = 5 \), and \( \frac{1}{2} \ne -\frac{1}{2} \) holds. The answer is option (c).

Teacher's Note:
1. Reduce each ratio to its simplest form before comparing.
2. Check that the constant-term ratio is different.

Answer: (a) 5, 1

Explanation:
1. Write the first equation as \( 2x + 3y - 11 = 0 \). For infinitely many solutions \( \frac{2}{m + n} = \frac{3}{2m - n} = \frac{-11}{-33} = \frac{1}{3} \).
2. So \( m + n = 6 \) (i) and \( 2m - n = 9 \) (ii).
3. Adding (i) and (ii): \( 3m = 15 \), so \( m = 5 \) and \( n = 1 \). The answer is option (a).

Teacher's Note:
1. Use the ratio of the constants first; it gives a simple value such as 1/3.
2. Then form two linear equations in m and n and solve them.

Answer: (d) no solution

Explanation:
1. \( a_1 = 1, b_1 = 2, c_1 = 5 \) and \( a_2 = -3, b_2 = -6, c_2 = 1 \).
2. \( \frac{a_1}{a_2} = -\frac{1}{3}, \quad \frac{b_1}{b_2} = -\frac{1}{3}, \quad \frac{c_1}{c_2} = 5 \).
3. \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \ne \frac{c_1}{c_2} \), so the lines are parallel and the pair has no solution: option (d).

Teacher's Note:
1. Compare the first two ratios; if they match, look at the third ratio.
2. Matching first two ratios and a different third ratio means parallel lines.

Answer: (c) intersecting or coincident

Explanation:
1. A consistent pair has at least one common solution.
2. It has either exactly one solution (intersecting lines) or infinitely many solutions (coincident lines): option (c).

Teacher's Note:
1. Consistent includes both a unique solution and infinitely many solutions.
2. Parallel lines are the inconsistent case.