CBSE Class 8 Mathematics Linear Equations in Two Variables MCQs Set 38

Practice MCQs for Class 8 Mathematics Linear Equations in Two Variables

Access targeted multiple-choice questions for Linear Equations in Two Variables designed to align with the latest CBSE academic syllabus for Class 8 Mathematics. These objective practice sets help students evaluate their conceptual understanding and improve exam readiness.

Access Linear Equations in Two Variables Questions and Solutions

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Multiple Choice Questions

Question 1: The pair of equations \( 3x-5y=20\text{ and }6x-10y=40 \) has:
(a) exactly two solutions
(b) infinitely many solutions
(c) a unique solution
(d) no solution
Show Answer & Explanation

Answer: (b) infinitely many solutions

Explanation:
1. Compare a₁/a₂, b₁/b₂ and c₁/c₂.
2. 3/6 = -5/-10 = 20/40, all ratios equal, so the lines coincide.

Teacher's Note:
1. Use elimination or substitution and check the answer in both equations.
2. A solution must satisfy both equations at the same time.

Question 2: The pair of equations \( x-2y=8\text{ and }5x-10y=10 \) has:
(a) infinitely many solutions
(b) no solution
(c) exactly two solutions
(d) a unique solution
Show Answer & Explanation

Answer: (b) no solution

Explanation:
1. Compare a₁/a₂, b₁/b₂ and c₁/c₂.
2. 1/5 = -2/-10 = 1/5 but this differs from 8/10, so the lines are parallel and distinct.

Teacher's Note:
1. Use elimination or substitution and check the answer in both equations.
2. A solution must satisfy both equations at the same time.

Question 3: In the equation \( x-\frac{y}{5}-10=0 \) written as ax + by + c = 0, the values of a, b, c are:
(a) \( a=1,\ b=-\frac{1}{5},\ c=-10 \)
(b) \( a=-10,\ b=-\frac{1}{5},\ c=1 \)
(c) \( a=1,\ b=\frac{1}{5},\ c=-10 \)
(d) \( a=1,\ b=-5,\ c=-10 \)
Show Answer & Explanation

Answer: (a) \( a=1,\ b=-\frac{1}{5},\ c=-10 \)

Explanation:
1. The equation already has the form 1·x + (-1/5)·y + (-10) = 0.
2. So a = 1, b = -1/5, c = -10.

Teacher's Note:
1. Read off the coefficient of x, the coefficient of y and the constant term with signs.
2. Do not change the signs.

Question 4: In the equation \( 3x+2=0 \) written as ax + by + c = 0, the values of a, b, c are:
(a) \( a=3,\ b=2,\ c=0 \)
(b) \( a=0,\ b=3,\ c=2 \)
(c) \( a=3,\ b=1,\ c=2 \)
(d) \( a=3,\ b=0,\ c=2 \)
Show Answer & Explanation

Answer: (d) \( a=3,\ b=0,\ c=2 \)

Explanation:
1. There is no y term, so b = 0.
2. 3x + 0·y + 2 = 0 gives a = 3, b = 0, c = 2.

Teacher's Note:
1. A missing variable has coefficient zero.
2. The graph of this equation is a vertical line x = -2/3.

Question 5: Does the point (2, 3) lie on the graph of \( y-2x=3 \)?
(a) No, because it lies on the x-axis
(b) No, because 3 - 4 = -1 is not 3
(c) Yes, because 2 + 3 = 5
(d) Yes, because 3 - 4 = 3
Show Answer & Explanation

Answer: (b) No, because 3 - 4 = -1 is not 3

Explanation:
1. Substitute x = 2, y = 3: 3 - 2(2) = -1.
2. -1 is not equal to 3, so the point is not on the line.

Teacher's Note:
1. A point lies on a line only if it satisfies the equation.
2. The point (1, 5) would lie on this line.

Question 6: Express x in terms of y from \( 7x-3y=15 \). At y = -5 the line meets the y-axis because:
(a) \( x=\frac{3y-15}{7} \) and x = 0 when y = 5
(b) \( x=\frac{15-3y}{7} \) and x = 0 when y = -5
(c) \( x=\frac{15+3y}{7} \) and x = 6 when y = -5
(d) \( x=\frac{15+3y}{7} \) and x = 0 when y = -5
Show Answer & Explanation

Answer: (d) \( x=\frac{15+3y}{7} \) and x = 0 when y = -5

Explanation:
1. 7x = 15 + 3y, so x = (15 + 3y)/7.
2. At y = -5: x = (15 - 15)/7 = 0, so the point (0, -5) is on the y-axis.

Teacher's Note:
1. Make x the subject by moving 3y to the other side.
2. Points on the y-axis have x = 0.

Question 7: The line \( 7x+y=-2 \) meets the x-axis at the point:
(a) \( \left(\frac{2}{7},\ 0\right) \)
(b) \( \left(-\frac{2}{7},\ 0\right) \)
(c) \( (0,-2) \)
(d) \( (-2,0) \)
Show Answer & Explanation

Answer: (b) \( \left(-\frac{2}{7},\ 0\right) \)

Explanation:
1. On the x-axis y = 0, so 7x = -2 and x = -2/7.
2. The point (2, 1) does not satisfy the equation since 14 + 1 = 15, not -2.

Teacher's Note:
1. Put y = 0 to find the x-intercept.
2. Put x = 0 to find the y-intercept.

Question 8: If the line \( 2x-ky=9 \) passes through the point (-1, -1), then k equals:
(a) 7
(b) -11
(c) 11
(d) 9
Show Answer & Explanation

Answer: (c) 11

Explanation:
1. Substitute x = -1, y = -1: -2 + k = 9.
2. So k = 11.

Teacher's Note:
1. Replace x and y by the coordinates of the point.
2. Check: 2(-1) - 11(-1) = 9.

Question 9: Is \( x=-6,\ y=-3 \) a solution of \( 2(x-1)-5y=1 \)?
(a) Yes, 2(-7) - 15 = 1
(b) No, 2(-7) + 15 = 29
(c) Yes, 2(-7) + 15 = 1
(d) No, 2(-5) + 15 = 5
Show Answer & Explanation

Answer: (c) Yes, 2(-7) + 15 = 1

Explanation:
1. 2(-6 - 1) - 5(-3) = -14 + 15 = 1.
2. The left side equals the right side, so it is a solution (and the equation has infinitely many solutions).

Teacher's Note:
1. A linear equation in two variables has infinitely many solutions.
2. Substitute both values.

Question 10: From the equation \( 2x+3y=6 \), the value of x when y = 4 is:
(a) -9
(b) 9
(c) 3
(d) -3
Show Answer & Explanation

Answer: (d) -3

Explanation:
1. 2x + 12 = 6, so 2x = -6.
2. x = -3.

Teacher's Note:
1. Substitute y = 4 and solve for x.
2. Check: 2(-3) + 12 = 6.

Question 11: Which of the following points satisfies \( 2x+3y=1 \)?
(a) (5, 3)
(b) (-5, -3)
(c) (2, 1)
(d) (5, -3)
Show Answer & Explanation

Answer: (d) (5, -3)

Explanation:
1. 2(5) + 3(-3) = 10 - 9 = 1.
2. The other points give 19, -19 and 7.

Teacher's Note:
1. Substitute each point.
2. Only one point satisfies the equation.

Question 12: Which point lies on the graph of \( 2(x+3)-3(y+1)=0 \)?
(a) (6, 3)
(b) (0, 0)
(c) (3, 1)
(d) (3, 3)
Show Answer & Explanation

Answer: (d) (3, 3)

Explanation:
1. The equation simplifies to 2x - 3y + 3 = 0.
2. For (3, 3): 6 - 9 + 3 = 0. The others give 8, 6 and 3.

Teacher's Note:
1. Expand the brackets first.
2. Check each point in 2x - 3y + 3.

Question 13: In \( \triangle ABC \), angle A = x°, angle B = (3x - 2)° and angle C = y°, with C - B = 9°. The three angles are:
(a) 30°, 70°, 80°
(b) 20°, 58°, 67°
(c) 25°, 73°, 82°
(d) 25°, 75°, 80°
Show Answer & Explanation

Answer: (c) 25°, 73°, 82°

Explanation:
1. C - B = 9 gives y = 3x + 7.
2. Angle sum: x + (3x - 2) + (3x + 7) = 180, so 7x = 175 and x = 25.
3. B = 73° and C = 82°; check 25 + 73 + 82 = 180.

Teacher's Note:
1. Use the angle sum property and the given difference.
2. Form two equations in x and y.

Question 14: In a cyclic quadrilateral ABCD, A = (2x + 4)°, B = (y + 3)°, C = (2y + 10)° and D = (4x - 5)°. The four angles A, B, C, D are:
(a) 66°, 53°, 114°, 127°
(b) 74°, 60°, 106°, 120°
(c) 70°, 53°, 110°, 127°
(d) 70°, 110°, 53°, 127°
Show Answer & Explanation

Answer: (c) 70°, 53°, 110°, 127°

Explanation:
1. Opposite angles of a cyclic quadrilateral add to 180°: (2x + 4) + (2y + 10) = 180 and (y + 3) + (4x - 5) = 180.
2. These give x + y = 83 and 4x + y = 182, so x = 33 and y = 50.
3. A = 70°, B = 53°, C = 110°, D = 127° (sum 360°).

Teacher's Note:
1. Opposite angles of a cyclic quadrilateral are supplementary.
2. Check that the four angles add to 360°.

Question 15: If \( 2^x+3^y=17 \) and \( 2^{x+2}-3^{y+1}=5 \), then (x, y) is:
(a) (3, 2)
(b) (3, 1)
(c) (2, 3)
(d) (4, 1)
Show Answer & Explanation

Answer: (a) (3, 2)

Explanation:
1. Put a = 2^x and b = 3^y: a + b = 17 and 4a - 3b = 5.
2. Solve: 7a = 56, so a = 8 and b = 9, i.e. x = 3, y = 2.

Teacher's Note:
1. Reduce the exponential equations to linear equations.
2. Check: 8 + 9 = 17 and 32 - 27 = 5.

Question 16: Solve \( \frac{2}{\sqrt{x}}+\frac{3}{\sqrt{y}}=2 \) and \( \frac{4}{\sqrt{x}}-\frac{9}{\sqrt{y}}=-1 \). The values of x and y are:
(a) x = 2, y = 3
(b) x = 9, y = 4
(c) x = 4, y = 9
(d) x = 16, y = 81
Show Answer & Explanation

Answer: (c) x = 4, y = 9

Explanation:
1. Put u = 1/√x, v = 1/√y: 2u + 3v = 2 and 4u - 9v = -1.
2. Multiply the first by 3 and add: 10u = 5, so u = 1/2 and v = 1/3.
3. √x = 2 and √y = 3, so x = 4 and y = 9.

Teacher's Note:
1. Substitute u and v to turn the system into a linear one.
2. Check: 2/2 + 3/3 = 2 and 4/2 - 9/3 = -1.

Question 17: The sum of two numbers is 15 and the sum of their reciprocals is 3/10. The larger number is:
(a) 5
(b) 10
(c) 12
(d) 9
Show Answer & Explanation

Answer: (b) 10

Explanation:
1. Let the numbers be a and b: a + b = 15 and (a + b)/(ab) = 3/10, so ab = 50.
2. The numbers are the roots of t² - 15t + 50 = 0, i.e. 5 and 10.
3. The larger is 10.

Teacher's Note:
1. Reciprocals add as (a + b)/(ab).
2. Check: 1/5 + 1/10 = 3/10.

Question 18: The denominator of a fraction is 4 more than twice the numerator. When both numerator and denominator are decreased by 6, the denominator becomes 12 times the numerator. The fraction is:
(a) \( \frac{7}{18} \)
(b) \( \frac{6}{16} \)
(c) \( \frac{5}{14} \)
(d) \( \frac{9}{22} \)
Show Answer & Explanation

Answer: (a) \( \frac{7}{18} \)

Explanation:
1. Let numerator n and denominator d = 2n + 4. After decreasing: d - 6 = 12(n - 6).
2. 2n - 2 = 12n - 72, so n = 7 and d = 18.
3. The fraction is 7/18.

Teacher's Note:
1. Form two equations from the two conditions.
2. Check: (18 - 6)/(7 - 6) = 12.

Question 19: A tea set sold at 5% loss and a lemon set at 15% gain give a gain of Rs. 7. Selling the tea set at 5% gain and the lemon set at 10% gain gives a gain of Rs. 13. The actual prices of the tea set and the lemon set are:
(a) Rs. 100 and Rs. 80
(b) Rs. 150 and Rs. 100
(c) Rs. 100 and Rs. 60
(d) Rs. 80 and Rs. 100
Show Answer & Explanation

Answer: (a) Rs. 100 and Rs. 80

Explanation:
1. Let the prices be T and L: -0.05T + 0.15L = 7 and 0.05T + 0.10L = 13.
2. Adding: 0.25L = 20, so L = 80; then 0.05T = 13 - 8 = 5, so T = 100.
3. Check: -5 + 12 = 7.

Teacher's Note:
1. Convert percentages to decimals and form two equations.
2. Add the equations to eliminate T.

Question 20: The sum of two numbers is 80 and the larger number is 5 more than 4 times the smaller. The numbers are:
(a) 45 and 35
(b) 50 and 30
(c) 65 and 15
(d) 60 and 20
Show Answer & Explanation

Answer: (c) 65 and 15

Explanation:
1. Let the smaller be s: s + (4s + 5) = 80, so 5s = 75 and s = 15.
2. The larger is 4(15) + 5 = 65.

Teacher's Note:
1. Express the larger number in terms of the smaller.
2. Check: 65 + 15 = 80.

Question 21: A two-digit number is 6 times the sum of its digits. If 2 is subtracted from the units digit and 1 from the tens digit, the new number is 7 times the sum of its digits. The number is:
(a) 68
(b) 64
(c) 70
(d) 54
Show Answer & Explanation

Answer: (d) 54

Explanation:
1. Let the number be 10a + b: 10a + b = 6(a + b), so 4a = 5b.Second condition: 10(a - 1) + (b - 2) = 7(a + b - 3), so 3a - 6b = -9, i.e. a - 2b = -3.
2. Solving 4a = 5b and a - 2b = -3 gives b = 4 and a = 5.
3. The number is 54.

Teacher's Note:
1. Write the number as 10a + b with digits a and b.
2. Check: 54 = 6 × 9 and 42 = 7 × 6.

Question 22: If each of the numerator and the denominator of a fraction is increased by 1 the fraction becomes 7/8, and if each is decreased by 1 it becomes 6/7. The fraction is:
(a) \( \frac{13}{15} \)
(b) \( \frac{13}{14} \)
(c) \( \frac{21}{16} \)
(d) \( \frac{26}{15} \)
Show Answer & Explanation

Answer: (a) \( \frac{13}{15} \)

Explanation:
1. Let the fraction be p/q: 8(p + 1) = 7(q + 1) and 7(p - 1) = 6(q - 1).
2. These give 8p - 7q = -1 and 7p - 6q = 1; solving, p = 13 and q = 15.
3. Check: 14/16 = 7/8 and 12/14 = 6/7.

Teacher's Note:
1. Form two equations from the two conditions.
2. Cross multiply each fraction condition.

Question 23: If 4 chairs and 7 tables cost Rs. 360 and 6 chairs and 10 tables cost Rs. 520, the cost of 1 chair and 1 table are respectively:
(a) Rs. 18 and Rs. 36
(b) Rs. 15 and Rs. 30
(c) Rs. 20 and Rs. 45
(d) Rs. 20 and Rs. 40
Show Answer & Explanation

Answer: (d) Rs. 20 and Rs. 40

Explanation:
1. 4c + 7t = 360 and 6c + 10t = 520. Multiply by 3 and 2: 12c + 21t = 1080 and 12c + 20t = 1040.
2. Subtract: t = 40, then 4c = 360 - 280 = 80, so c = 20.

Teacher's Note:
1. Make the coefficients of c equal and subtract.
2. Check: 6(20) + 10(40) = 520.

Question 24: A vendor sells a tea set at 5% loss and a lemon set at 25% profit and makes Rs. 15 profit overall. Had he sold the tea set at 10% profit and the lemon set also at 10% profit he would have made Rs. 18. The cost prices of the tea set and the lemon set are:
(a) Rs. 80 and Rs. 100
(b) Rs. 100 and Rs. 60
(c) Rs. 150 and Rs. 100
(d) Rs. 100 and Rs. 80
Show Answer & Explanation

Answer: (d) Rs. 100 and Rs. 80

Explanation:
1. -0.05T + 0.25L = 15 and 0.10T + 0.10L = 18.
2. The second gives T + L = 180; substituting into the first: -0.05T + 0.25(180 - T) = 15, so T = 100 and L = 80.

Teacher's Note:
1. Convert the percentages to decimals.
2. Check: -5 + 20 = 15.

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