CBSE Class 10 Mathematics Pair of Linear Equations in Two Variables MCQs Set 23

Answer: (a) has a unique solution

Explanation:
1. Write the equations as ax + by − c = 0 and lx + my − n = 0.
2. The ratio of the x-coefficients is \( \frac{a}{l} \) and of the y-coefficients is \( \frac{b}{m} \).
3. am ≠ bl means \( \frac{a}{l} \neq \frac{b}{m} \), so the lines intersect at exactly one point.
4. The system has a unique solution.

Teacher's Note:
1. Unique solution: \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \).
2. Cross-multiplying \( \frac{a}{l} \neq \frac{b}{m} \) gives am ≠ bl.

Answer: (c) 68

Explanation:
1. Let the tens digit be x and the units digit be y, so x + y = 14.
2. The number is 10x + y. Adding 18 reverses the digits: 10x + y + 18 = 10y + x, so 9y − 9x = 18 and y − x = 2.
3. Adding x + y = 14 and y − x = 2 gives 2y = 16, so y = 8 and x = 6.
4. The number is 68.

Teacher's Note:
1. Always write the reversed number as 10y + x.
2. Check: 68 + 18 = 86 and 6 + 8 = 14.

Answer: (b) 5 and 15

Explanation:
1. Let the numbers be x and 3x.
2. After adding 5 to each: \( \frac{x+5}{3x+5} = \frac{1}{2} \), so 2x + 10 = 3x + 5 and x = 5.
3. The numbers are 5 and 15.

Teacher's Note:
1. Use one variable for numbers given in a ratio.
2. Check: 10 : 20 = 1 : 2.

Answer: (b) 15 and 5

Explanation:
1. Let the sister's age be x years. Then Sanya's age is 3x years.
2. Five years hence: 3x + 5 = 2(x + 5), so 3x + 5 = 2x + 10 and x = 5.
3. Sanya's age is 15 years and her sister's age is 5 years.

Teacher's Note:
1. Add the same number of years to both ages.
2. Check: after 5 years the ages are 20 and 10, in ratio 2 : 1.

Answer: (b) 35

Explanation:
1. Let the tens digit be x and the units digit be y, so x + y = 8.
2. Adding 18 reverses the digits: 10x + y + 18 = 10y + x, so y − x = 2.
3. Solving: y = 5 and x = 3. The number is 35.

Teacher's Note:
1. The units digit is larger than the tens digit when adding 18 reverses the digits.
2. Check: 35 + 18 = 53.

Answer: (a) 15 and 10

Explanation:
1. Let the number of Rs. 2 coins be x and of Rs. 5 coins be y. Then x + y = 25 and 2x + 5y = 80.
2. Multiply the first equation by 2: 2x + 2y = 50. Subtract from the second: 3y = 30, so y = 10.
3. Then x = 25 − 10 = 15.
4. She has 15 coins of Rs. 2 and 10 coins of Rs. 5.

Teacher's Note:
1. Form one equation for the number of coins and one for the value.
2. Check: 15 × 2 + 10 × 5 = 30 + 50 = 80.

Answer: (c) A pair of linear equations in two variables cannot have exactly two solutions.

Explanation:
1. Two straight lines meet in no point, exactly one point, or every point (if they coincide). Two distinct points of meeting is impossible, so (c) is true.
2. A linear equation in two variables has infinitely many solutions, so (a) is false.
3. A consistent pair has at least one solution, so (b) is false.
4. Parallel lines have no common point, so (d) is false.

Teacher's Note:
1. Count the possible numbers of solutions: 0, 1 or infinite.
2. Two distinct lines cannot intersect twice.

Answer: (c) The pair of equations 4x − 5y = 8 and 8x − 10y = 3 has a unique solution.

Explanation:
1. For 4x − 5y = 8 and 8x − 10y = 3: \( \frac{a_1}{a_2} = \frac{4}{8} = \frac{1}{2} \), \( \frac{b_1}{b_2} = \frac{-5}{-10} = \frac{1}{2} \) and \( \frac{c_1}{c_2} = \frac{-8}{-3} = \frac{8}{3} \).
2. Since \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \), the lines are parallel and there is no solution. So (c) is false.
3. Statements (a), (b) and (d) are all true.

Teacher's Note:
1. Compare the three ratios a1/a2, b1/b2, c1/c2.
2. Equal first two ratios and a different third ratio mean parallel lines.