CBSE Class 10 Mathematics Pair of Linear Equations in Two Variables MCQs Set 18

Answer: (b) 40

Explanation:
1. Let x be the number of right answers and y the number of wrong answers. Then 3x − y = 40 and 4x − 2y = 40.
2. From the first, y = 3x − 40. Substituting: 4x − 6x + 80 = 40, so x = 20 and y = 20.
3. Total number of questions = 20 + 20 = 40.

Teacher's Note:
1. Form two equations from the two marking schemes.
2. The total is x + y, not just x.

Answer: (c) x = 18 m, y = 12 m

Explanation:
1. x − y = 6 and 2(x + y) = 60, that is x + y = 30.
2. Adding: 2x = 36, so x = 18 and y = 12. The dimensions are x = 18 m, y = 12 m.

Teacher's Note:
1. Length is the larger side.
2. Check the perimeter: 2(18 + 12) = 60.

Answer: (a) 1

Explanation:
1. 1024 = 4⁵, so 4^(x + y) = 4⁵ gives x + y = 5.
2. Also 4 = 1024^(1/5), so x − y = 1/5. Two lines with different slopes (x + y = 5 and x − y = 1/5) intersect at exactly one point.
3. Hence the pair has 1 solution (x = 13/5, y = 12/5).

Teacher's Note:
1. Convert the powers to a common base first.
2. Check the ratios of coefficients: 1/1 ≠ 1/(−1).

Answer: (c) −12x + 8y = 7

Explanation:
1. Parallel lines have a₁/a₂ = b₁/b₂ ≠ c₁/c₂.
2. For −12x + 8y = 7: 3/(−12) = −1/4 and (−2)/8 = −1/4, and 5/7 ≠ −1/4 so the lines are parallel. The other options have different ratios of x and y coefficients.

Teacher's Note:
1. Compare the ratio of the x coefficients with the ratio of the y coefficients.
2. Parallel lines have no common point.

Answer: (b) 4

Explanation:
1. Let the number be 10a + b. Then 10a + b = 7(a + b), so 3a = 6b and a = 2b.
2. b can be 1, 2, 3 or 4 (a must be a single digit, b = 0 gives 0), giving the numbers 21, 42, 63 and 84.
3. So there are 4 such numbers.

Teacher's Note:
1. Digits must be single digit numbers: a ≤ 9.
2. Check: 63 ÷ 9 = 7.

Answer: (d) A is false, but R is true.

Explanation:
1. Coincident lines are consistent, but they have infinitely many solutions, not a unique solution. So A is false.
2. Parallel lines never meet, so there is no solution and the pair is inconsistent. R is true.

Teacher's Note:
1. Unique solution means intersecting lines.
2. Coincident lines: infinitely many solutions.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. The line 3x + 4y = 12 cuts the axes at (4, 0) and (0, 3), forming a triangle of area ½ × 4 × 3 = 6. The line 3x + 4y = 6 cuts the axes at (2, 0) and (0, 3/2), forming a triangle of area ½ × 2 × 3/2 = 3/2.
2. The region between the two lines and the axes has area 6 − 3/2 = 9/2, so A is true.
3. R is also a true statement (intersecting lines give a unique solution), but it does not explain this area, so R is not the correct explanation of A.

Teacher's Note:
1. The two lines are parallel, so the area is the difference of two triangles.
2. The reason is a true general statement, unrelated to the area.

Answer: (d) A is false, but R is true.

Explanation:
1. Unique solution needs k/6 ≠ (−1)/(−2) = 1/2, that is k ≠ 3. So for k = 3 there is no unique solution and A is false.
2. R is the correct condition for intersecting lines, so R is true.

Teacher's Note:
1. A unique solution is possible for every k except one value.
2. Compare k/6 with 1/2.

Answer: (a) x + 10y = 75 and x + 15y = 110

Explanation:
1. Charge = fixed charge + (running charge per km) × distance.
2. 10 km: x + 10y = 75. 15 km: x + 15y = 110. So the pair is x + 10y = 75 and x + 15y = 110.

Teacher's Note:
1. Fixed charge has no multiplier; the running charge is multiplied by the distance.
2. Match each distance with its own charge.

Answer: (b) Rs. 355

Explanation:
1. Subtracting the equations: 5y = 35, so y = 7. Then x = 75 − 70 = 5.
2. Charge for 50 km = 5 + 7 × 50 = Rs. 355.

Teacher's Note:
1. Find y first by subtracting, then x.
2. Check: 5 + 70 = 75.

Answer: (b) Rs. 289

Explanation:
1. x + 8y = 91 and x + 14y = 145. Subtracting: 6y = 54, so y = 9 and x = 91 − 72 = 19.
2. Charge for 30 km = 19 + 9 × 30 = Rs. 289.

Teacher's Note:
1. Use the same method as for Situation 1.
2. Check with the first data point: 19 + 72 = 91.

Answer: (b) 96

Explanation:
1. Let x be the correct answers; the wrong ones are 120 − x.
2. x − ¼(120 − x) = 90 gives 1.25x = 120, so x = 96.

Teacher's Note:
1. Marks = correct × 1 − wrong × ¼.
2. Check: 96 − 24/4 = 96 − 6 = 90.

Answer: (b) 70

Explanation:
1. Correct = 80, so wrong = 120 − 80 = 40.
2. Marks = 80 − ¼ × 40 = 80 − 10 = 70.

Teacher's Note:
1. Count the wrong answers first.
2. Each wrong answer costs ¼ mark.

Answer: (c) 100

Explanation:
1. Let x be the correct answers, so 120 − x are wrong.
2. x − ¼(120 − x) = 95 gives 1.25x = 125, so x = 100.

Teacher's Note:
1. Use the same equation with the new total of marks.
2. Check: 100 − 20/4 = 95.

Answer: (b) Rs. 15

Explanation:
1. Let one round of shooting cost Rs. s and one round of bowling Rs. b in the combos. 3s + 2b = 285 and 4s + 5b = 485.
2. Eliminating s: 4 × (first) − 3 × (second): 8b − 15b = 1140 − 1455, so 7b = 315 and b = 45 (and s = 65).
3. In the solo pack one round of bowling costs Rs. 60, so it is 60 − 45 = Rs. 15 more.

Teacher's Note:
1. Use the combo packs to form the equations; the solo pack gives the bowling price Rs. 60.
2. Check: 3 × 65 + 2 × 45 = 285 and 4 × 65 + 5 × 45 = 485.

Answer: (c) intersecting or coincident

Explanation:
1. A consistent pair has at least one solution.
2. So the lines meet at one point (intersecting) or at all points (coincident): intersecting or coincident.

Teacher's Note:
1. Parallel lines are inconsistent.
2. Consistent means that a solution exists.

Answer: (c) 6 and 36

Explanation:
1. Let the son's age be s and the father's age f. f = 6s and f + 4 = 4(s + 4).
2. 6s + 4 = 4s + 16, so s = 6 and f = 36.
3. The present ages are 6 and 36 years.

Teacher's Note:
1. Add 4 years to both ages for the second condition.
2. Check: 36 + 4 = 40 = 4 × 10.

Answer: (c) 3 and 1

Explanation:
1. Adding the equations: 2x = 6, so x = 3. Then y = 4 − 3 = 1.
2. So a = 3 and b = 1.

Teacher's Note:
1. Adding removes y here.
2. Check: 3 − 1 = 2 and 3 + 1 = 4.

Answer: (c) x + 2y = 10

Explanation:
1. Total cost: 25x + 50y = 250.
2. Dividing by 25: x + 2y = 10.

Teacher's Note:
1. Total spent = brownie price × x + wrap price × y.
2. Divide all terms by the common factor 25.

Answer: (b) 18

Explanation:
1. Let the digits be a (tens) and b (units): a + b = 9 and 9(10a + b) = 2(10b + a).
2. 88a = 11b gives b = 8a. With a + b = 9: 9a = 9, so a = 1 and b = 8.
3. The number is 18. Check: 9 × 18 = 162 = 2 × 81.

Teacher's Note:
1. Reversing 10a + b gives 10b + a.
2. Use both conditions to get two equations in a and b.

Answer: (d) parallel

Explanation:
1. a₁/a₂ = 6/2 = 3, b₁/b₂ = −3/−1 = 3 and c₁/c₂ = 10/9.
2. Since a₁/a₂ = b₁/b₂ ≠ c₁/c₂, the lines are parallel.

Teacher's Note:
1. Equal ratios of the x and y coefficients mean parallel or coincident.
2. The third ratio separates parallel from coincident.

Answer: (a) x − y = 3

Explanation:
1. k₃ passes through (3, 0) and (4, 1), so its slope is (1 − 0)/(4 − 3) = 1.
2. Equation: y − 0 = 1 × (x − 3), that is x − y = 3.

Teacher's Note:
1. Two points are enough to write a line.
2. Check both points: 3 − 0 = 3 and 4 − 1 = 3.

Answer: (d) no solution

Explanation:
1. Rewrite the second as 3x + 6y − 1 = 0. Then a₁/a₂ = 1/3, b₁/b₂ = 2/6 = 1/3 and c₁/c₂ = 5/(−1) = −5.
2. Since a₁/a₂ = b₁/b₂ ≠ c₁/c₂, the lines are parallel and there is no solution.

Teacher's Note:
1. Bring both equations into the form ax + by + c = 0 first.
2. A linear pair never has exactly two solutions.

Answer: (a) 9

Explanation:
1. For coincident lines a₁/a₂ = b₁/b₂ = c₁/c₂.
2. 5/15 = 7/21 = 3/k, so 3/k = 1/3 and k = 9.

Teacher's Note:
1. Coincident lines have all three ratios equal.
2. 15x + 21y = 9 is 3 times 5x + 7y = 3.

Answer: (b) −21x + 15y − 32 = 0

Explanation:
1. An inconsistent pair has parallel lines: a₁/a₂ = b₁/b₂ ≠ c₁/c₂.
2. For −21x + 15y − 32 = 0: 7/(−21) = −5/15 = −1/3 but 8/(−32) = −1/4, so the lines are parallel and the pair is inconsistent. 14x − 10y + 16 = 0 gives coincident lines and 7x + 25y − 16 = 0 gives intersecting lines, both consistent.

Teacher's Note:
1. Check all three ratios for every option.
2. Only one option gives parallel lines.

Answer: (c) x − y + 3 = 0; 2x − y − 1 = 0

Explanation:
1. First condition: x + y = 2y − 3, so x − y + 3 = 0.
2. Second condition: x − 1 = (y − 1)/2, so 2x − 2 = y − 1, that is 2x − y − 1 = 0.
3. So the pair is x − y + 3 = 0; 2x − y − 1 = 0 (the fraction is 4/7).

Teacher's Note:
1. Translate each sentence into one equation.
2. Check with 4/7: 4 + 7 = 11 = 14 − 3.

Answer: (b) Step 2

Explanation:
1. Step 1 only writes the given equations.
2. In Step 2, 65x + 65y = 65 gives x + y = 1, not 65. This is the first error.
3. (Steps 3 and 4 are then consistent with the correct x + y = 1: y = 2 and x = −1.)

Teacher's Note:
1. Check each step one by one from the top.
2. Dividing both sides of 65x + 65y = 65 by 65 gives 1.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. Coincident lines need 3/6 = (−2)/(−k) = (−5)/(−10), so 2/k = 1/2 and k = 4. A is true.
2. Coincident lines do have infinitely many solutions, so R is true, but it does not explain why k = 4. So R is not the correct explanation of A.

Teacher's Note:
1. Use equal ratios of all the coefficients.
2. The reason is a definition-type fact; the value of k comes from the ratios.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Subtracting the second equation from the first: 0.001x − 0.001y = 0.001, so x − y = 1. A is true.
2. Check x = 2, y = 1: 2 + 0.999 = 2.999 and 1.998 + 1 = 2.998. R is true, and x − y = 2 − 1 = 1 follows from it, so R explains A.

Teacher's Note:
1. Subtract the equations to see x − y directly.
2. Always verify the solution in both equations.

Answer: (a) Son 10 years, father 30 years

Explanation:
1. Let the son's age be s. Then f = 3s and s + f = 40, so 4s = 40.
2. s = 10 and f = 30. The ages are son 10 years and father 30 years.

Teacher's Note:
1. Substitute f = 3s into the sum.
2. Check: 10 + 30 = 40.

Answer: (a) x = 85°, y = 55°

Explanation:
1. The angle sum of a triangle is 180°, so x + y = 140. Also x − y = 30.
2. Adding: 2x = 170, so x = 85 and y = 55.

Teacher's Note:
1. Use the angle sum property for the first equation.
2. Check: 85 + 55 + 40 = 180.

Answer: (a) 2

Explanation:
1. Adding: 6x = 30, so x = 5. Then y = 13 − 10 = 3.
2. x − y = 5 − 3 = 2.

Teacher's Note:
1. The y terms cancel on adding.
2. Verify in the second equation: 20 − 3 = 17.

Answer: (b) k = 7

Explanation:
1. For infinitely many solutions 2/(k − 1) = 3/(k + 2) = 7/(3k).
2. From the first two: 2(k + 2) = 3(k − 1), so k = 7. Check: 3/9 = 1/3 and 7/21 = 1/3, and 2/6 = 1/3.

Teacher's Note:
1. Equate the ratios pair by pair and then verify the third ratio.
2. The value of k must satisfy all three equalities.

Answer: (a) \( \dfrac{5}{12} \)

Explanation:
1. Let the fraction be x/y. Then (x − 1)/y = 1/3, so 3x − y = 3, and x/(y + 8) = 1/4, so 4x − y = 8.
2. Subtracting: x = 5 and y = 12. The fraction is 5/12.

Teacher's Note:
1. Clear the denominators in both equations.
2. Check: 4/12 = 1/3 and 5/20 = 1/4.

Answer: (a) −1

Explanation:
1. Here a₁/a₂ = λ, b₁/b₂ = 1/λ and c₁/c₂ = λ².
2. No solution needs a₁/a₂ = b₁/b₂ ≠ c₁/c₂. From λ = 1/λ we get λ = ±1.
3. For λ = 1: c₁/c₂ = 1 = a₁/a₂ (the lines coincide). For λ = −1: a₁/a₂ = −1 and c₁/c₂ = 1, which are different, so the lines are parallel. Hence λ = −1.

Teacher's Note:
1. First find the values making the lines parallel or coincident, then test the c-ratio.
2. The equations are −x + y − 1 = 0 and x − y − 1 = 0 for λ = −1: parallel.

Answer: (b) 1

Explanation:
1. From λ = 1/λ we get λ = ±1 and infinitely many solutions need c₁/c₂ to be equal as well.
2. For λ = 1 both equations become x + y − 1 = 0, so the lines coincide and the pair has infinitely many solutions: λ = 1.

Teacher's Note:
1. Put the value back in the equations to see whether they become identical.
2. λ = −1 gives parallel lines, not coincident.

Answer: (c) all real values of λ except 1 and −1

Explanation:
1. A unique solution needs a₁/a₂ ≠ b₁/b₂, that is λ ≠ 1/λ, so λ² ≠ 1 (this also holds for λ = 0, where the equations are y = 0 and x = 1).
2. So the pair has a unique solution for every real λ except 1 and −1.

Teacher's Note:
1. Unique solution means intersecting lines.
2. Exclude only the values that make the ratios equal.

Answer: (c) inconsistent

Explanation:
1. a₁/a₂ = 1/3, b₁/b₂ = −2/−6 = 1/3 and c₁/c₂ = −6/0, which is not equal to 1/3 (the second equation x − 2y = 0 has constant term 0, while the first has −6).
2. The lines x − 2y = 6 and x − 2y = 0 are parallel, so the pair is inconsistent (no solution).

Teacher's Note:
1. 3x − 6y = 0 is the same as x − 2y = 0.
2. Parallel lines never meet.

Answer: (a) (4, 2), (1, 0), (2, 3)

Explanation:
1. Solve the lines in pairs: 2x − 3y = 2 and x + 2y = 8 give (4, 2). 3x − y = 3 and 2x − 3y = 2 give (1, 0). 3x − y = 3 and x + 2y = 8 give (2, 3).
2. The vertices are (4, 2), (1, 0) and (2, 3).

Teacher's Note:
1. Each vertex is the intersection of two of the three lines.
2. Check each point in both of its lines.

Answer: (b) 2

Explanation:
1. Let the digits be a and b. (10a + b) + (10b + a) = 11(a + b) = 66, so a + b = 6.
2. Digits differ by 2: a − b = 2 gives (4, 2), so the number is 42; b − a = 2 gives (2, 4), so the number is 24.
3. So there are 2 such numbers: 42 and 24.

Teacher's Note:
1. The difference of the digits can be either way: a − b = ±2.
2. Check: 42 + 24 = 66.