CBSE Class 10 Mathematics Pair of Linear Equations in Two Variables MCQs Set 30

Answer: (b) \( x = 3,\ y = 2 \)

Explanation:
1. Adding x − y = 1 and 2x + y = 8: 3x = 9, so x = 3.
2. y = x − 1 = 2.
3. Check: 2(3) + 2 = 8.

Teacher's Note:
1. The solution is the point where the two lines intersect.
2. Check by substituting in both equations.

Answer: (a) 13.5

Explanation:
1. x − y = 1 meets the y-axis at (0, −1); 2x + y = 8 meets it at (0, 8).
2. Base = 8 − (−1) = 9; the lines meet at (3, 2), so height = 3.
3. Area = ½ × 9 × 3 = 13.5 square units.

Teacher's Note:
1. The side on the y-axis is the base.
2. The x-coordinate of the intersection point is the height.

Answer: (d) \( x = a^2,\ y = b^2 \)

Explanation:
1. From the first equation a²/x = b²/y = t.
2. Second equation: b(a²/x) + a(b²/y) = bt + at = a + b, so t = 1.
3. a²/x = 1 gives x = a² and b²/y = 1 gives y = b².

Teacher's Note:
1. Reduce to a simple pair by letting a²/x = b²/y = t.
2. Then solve for t first.

Answer: (c) \( k \neq -4 \)

Explanation:
1. A unique solution needs a₁/a₂ ≠ b₁/b₂, i.e. 2/k ≠ 3/(−6) = −1/2, so k ≠ −4.

Teacher's Note:
1. Unique solution: a₁/a₂ ≠ b₁/b₂.
2. Here b₁/b₂ = −1/2.

Answer: (b) k = 2

Explanation:
1. 2/(k + 2) = 3/6 = 1/2 gives k + 2 = 4, so k = 2.
2. Check 4/(3k + 2) = 4/8 = 1/2. All three ratios agree.

Teacher's Note:
1. Set all three ratios equal.
2. Verify with the third ratio.

Answer: (a) k = −2

Explanation:
1. (3k + 1)(k − 1) = 3(k² + 1) gives −2k − 1 = 3, so k = −2.
2. Then (3k + 1)/(k² + 1) = −5/5 = −1 = 3/(k − 1).
3. c₁/c₂ = 2/5 ≠ −1, so there is no solution.

Teacher's Note:
1. The lines are parallel: a₁/a₂ = b₁/b₂ ≠ c₁/c₂.
2. Check the third ratio after finding k.

Answer: (d) \( x = a + b,\ y = -\frac{2ab}{a+b} \)

Explanation:
1. Take x = a + b, y = −2ab/(a + b).
2. (a + b)(x + y) = (a + b)² − 2ab = a² + b², so equation 2 holds.
3. (a − b)(a + b) + (a + b)y = a² − b² − 2ab = a² − 2ab − b², so equation 1 holds.

Teacher's Note:
1. Use the second equation to eliminate one variable.
2. Verify the answer in both equations.

Answer: (c) \( x = m + n,\ y = m - n \)

Explanation:
1. y = 2m − x, so mx − 2mn + nx = m² + n².
2. (m + n)x = m² + 2mn + n² = (m + n)².
3. x = m + n and y = 2m − (m + n) = m − n.

Teacher's Note:
1. Eliminate y using the second equation.
2. Factor the right-hand side as (m + n)².

Answer: (b) 45

Explanation:
1. ab = 20 and 10a + b + 9 = 10b + a.
2. 9a + 9 = 9b, so b = a + 1.
3. a(a + 1) = 20, so a = 4, b = 5. The number is 45.

Teacher's Note:
1. Write the number as 10a + b.
2. Use both conditions to find the digits.

Answer: (a) 3/4

Explanation:
1. n + d = 7 and 4n = 5d − 8.
2. 4(7 − d) = 5d − 8, so 36 = 9d and d = 4.
3. n = 3. The fraction is 3/4.

Teacher's Note:
1. Form two equations in the numerator and denominator.
2. Substitute one into the other.

Answer: (d) 70°, 53°, 110°, 127°

Explanation:
1. A + C = 180: 2x + 2y + 14 = 180, so x + y = 83.
2. B + D = 180: 4x + y − 2 = 180, so 4x + y = 182.
3. Subtracting: 3x = 99, x = 33; y = 50.
4. A = 70°, B = 53°, C = 110°, D = 127°.

Teacher's Note:
1. Opposite angles of a cyclic quadrilateral are supplementary.
2. Form two equations from A + C and B + D.

Answer: (c) 10/3 km/h and 5 km/h

Explanation:
1. 30/x − 30/y = 3 and 30/y − 15/x = 3/2.
2. Substituting 15/x = (3 + 30/y)/2: 15/y − 3/2 = 3/2, so y = 5.
3. 30/x = 3 + 6 = 9, so x = 10/3 km/h.

Teacher's Note:
1. Use time = distance ÷ speed for both conditions.
2. With doubled speed X's time is 15/x.

Answer: (b) 692

Explanation:
1. a + b + c = 17 and b − (a + c) = 1 give b = 9, a + c = 8.
2. (100a + 10b + c) − (100c + 10b + a) = 99(a − c) = 396, so a − c = 4.
3. a = 6, c = 2. The number is 692.

Teacher's Note:
1. Write the number as 100a + 10b + c.
2. The difference of the number and its reverse is 99(a − c).

Answer: (a) 96

Explanation:
1. Wrong answers = 120 − c.
2. c − (120 − c)/4 = 90.
3. 4c − 120 + c = 360, so c = 96.

Teacher's Note:
1. Wrong answers = 120 − c.
2. Marks = c − ¼(120 − c).

Answer: (d) A in 50 days, B in 75 days

Explanation:
1. Let B's one day's work = b, so A's = 1.5b.
2. 2.5b = 1/30, so b = 1/75.
3. B takes 75 days and A takes 1/(1.5/75) = 50 days.

Teacher's Note:
1. Work done per day adds for people working together.
2. Take B's rate as the unknown.

Answer: (c) 47 or 74

Explanation:
1. (10a + b) + (10b + a) = 11(a + b) = 121, so a + b = 11.
2. |a − b| = 3 gives the digits 7 and 4.
3. The number is 74 or 47.

Teacher's Note:
1. The sum of a number and its reverse is 11(a + b).
2. Both orders of the digits are valid.

Answer: (b) Rs. 210 and Rs. 6

Explanation:
1. F + r = 216.
2. F/2 + r = 327 − 216 = 111.
3. Subtracting: F/2 = 105, so F = 210 and r = 6.

Teacher's Note:
1. Set up the full ticket and the half ticket separately.
2. Subtract the first equation from the second.

Answer: (a) Rs. 5,000 and Rs. 25,000

Explanation:
1. 0.05x + 0.07(30000 − x) = 2000.
2. 2100 − 0.02x = 2000, so x = 5000.
3. So Rs. 5,000 is at 5% and Rs. 25,000 is at 7%.

Teacher's Note:
1. Interest = principal × rate.
2. The two amounts add to 30,000.

Answer: (d) 6 litres of 50% and 4 litres of 25%

Explanation:
1. 0.5x + 0.25(10 − x) = 0.4 × 10 = 4.
2. 0.25x = 1.5, so x = 6.
3. The 25% solution is 10 − 6 = 4 litres.

Teacher's Note:
1. The acid in the mixture equals the acid in the parts.
2. The total volume is 10 litres.

Answer: (c) Rs. 700 and Rs. 500

Explanation:
1. 0.10x + 0.12y = 130 and 0.12x + 0.10y = 134.
2. Adding: 0.22(x + y) = 264, so x + y = 1200.
3. Subtracting: 0.02(y − x) = −4, so y − x = −200 and x = 700, y = 500.

Teacher's Note:
1. Form two equations from the two interest totals.
2. Add and subtract them.

Answer: (b) Father 42 years, son 10 years

Explanation:
1. f − 2 = 5s − 10, so f = 5s − 8.
2. f + 2 = 3s + 6 + 8, so f = 3s + 12.
3. 5s − 8 = 3s + 12, so s = 10 and f = 42.

Teacher's Note:
1. Write each condition as an equation in the present ages.
2. Eliminate f.

Answer: (a) Rs. 400 and Rs. 30

Explanation:
1. x + 20y = 1000 and x + 26y = 1180.
2. 6y = 180, so y = 30.
3. x = 1000 − 600 = 400.

Teacher's Note:
1. Subtract the two equations to remove the fixed charge.
2. Substitute back.

Answer: (d) 35 km/h and 25 km/h

Explanation:
1. Same direction: 8(u − v) = 80, so u − v = 10.
2. Opposite directions: (4/3)(u + v) = 80, so u + v = 60.
3. u = 35 and v = 25.

Teacher's Note:
1. Relative speed is the difference when moving in the same direction and the sum when moving towards each other.
2. 1 hour 20 minutes = 4/3 hours.

Answer: (c) 6 km/h and 2 km/h

Explanation:
1. 12p + 40q = 8 and 16p + 32q = 8, where p = 1/(b − s), q = 1/(b + s).
2. From the second, p = 0.5 − 2q; the first gives 6 + 16q = 8, so q = 1/8 and p = 1/4.
3. b − s = 4 and b + s = 8, so b = 6 and s = 2.

Teacher's Note:
1. Let upstream speed be b − s and downstream speed b + s.
2. Substitute reciprocals to get a linear pair.

Answer: (b) \( x = 2\frac{1}{8},\ y = 1\frac{7}{8} \)

Explanation:
1. x + y = 4.
2. 4(x − y) = 1, so x − y = 1/4.
3. x = (4 + 1/4)/2 = 17/8 = 2 1/8 and y = 15/8 = 1 7/8.

Teacher's Note:
1. Write both sides with base 3.
2. Equate the exponents to get a linear pair.