CBSE Class 10 Mathematics Pair of Linear Equations in Two Variables MCQs Set 28

Practice MCQs for Class 10 Mathematics Chapter 03 Pair of Linear Equations in Two Variables

Explore reliable objective questions for Chapter 03 Pair of Linear Equations in Two Variables tailored for Class 10 learners. Utilizing these Mathematics multiple-choice formats ensures thorough preparation and strengthens problem-solving speed for upcoming school assessments.

Access Chapter 03 Pair of Linear Equations in Two Variables Questions and Solutions

View or download the dedicated Chapter 03 Pair of Linear Equations in Two Variables MCQ resource below. Practicing these 50 objective questions regularly builds familiarity with standard exam patterns and helps secure higher marks in final Mathematics evaluations.

Multiple Choice Questions

Question 1: Solving \(\dfrac x2+\dfrac{2y}5+2=10\) and \(\dfrac{2x}7-\dfrac y2+1=9\), we get
(a) \(x=\dfrac{80}{17},\ y=-\dfrac{336}{17}\)
(b) \(x=\dfrac{336}{17},\ y=\dfrac{80}{17}\)
(c) \(x=\dfrac{336}{17},\ y=-\dfrac{80}{17}\)
(d) \(x=16,\ y=0\)
Show Answer & Explanation

Answer: (c) \(x=\dfrac{336}{17},\ y=-\dfrac{80}{17}\)

Explanation:
1. Simplify: \(5x+4y=80\) and \(4x-7y=112\)
2. Multiply the first by 7 and the second by 4 and add: \(51x=1008\), so \(x=\dfrac{336}{17}\)
3. \(4y=80-\dfrac{1680}{17}=-\dfrac{320}{17}\), so \(y=-\dfrac{80}{17}\)

Teacher's Note:
1. Remove the denominators by multiplying each equation by its LCM.
2. Check the answer by substituting it in both original equations.

Question 2: Solving \(\dfrac{ax}b-\dfrac{by}a=a+b\) and \(ax-by=2ab\), we get
(a) \(x=a,\ y=-b\)
(b) \(x=b,\ y=-a\)
(c) \(x=b,\ y=a\)
(d) \(x=-b,\ y=a\)
Show Answer & Explanation

Answer: (b) \(x=b,\ y=-a\)

Explanation:
1. Divide the second by a: \(x-\dfrac{by}a=2b\). Subtract from the first: \(x\left(\dfrac ab-1\right)=a-b\)
2. \(x\cdot\dfrac{a-b}b=a-b\), so \(x=b\)
3. Then \(b-\dfrac{by}a=2b\) gives \(y=-a\)

Teacher's Note:
1. Compare the two equations to see that they share the term by/a.
2. Check both equations with x = b, y = -a.

Question 3: Solving \(2(ax-by)+(a+4b)=0\) and \(2(bx+ay)+(b-4a)=0\), we get
(a) \(x=-\dfrac12,\ y=2\)
(b) \(x=\dfrac12,\ y=-2\)
(c) \(x=-2,\ y=\dfrac12\)
(d) \(x=-\dfrac12,\ y=-2\)
Show Answer & Explanation

Answer: (a) \(x=-\dfrac12,\ y=2\)

Explanation:
1. Multiply the first by b, the second by a and subtract: \(2(a^2+b^2)y=4(a^2+b^2)\), so \(y=2\)
2. Multiply the first by a, the second by b and add: \(2(a^2+b^2)x=-(a^2+b^2)\), so \(x=-\dfrac12\)

Teacher's Note:
1. Cross-multiplying by a and b makes a^2 + b^2 appear as a common factor.
2. Check by substituting x = -1/2, y = 2.

Question 4: Solving \(\dfrac1{x+y}+\dfrac2{x-y}=2\) and \(\dfrac2{x+y}-\dfrac1{x-y}=3\) (with \(x\pm y\ne0\)), we get
(a) \(x=\dfrac{35}{16},\ y=-\dfrac{45}{16}\)
(b) \(x=\dfrac{45}{16},\ y=\dfrac{35}{16}\)
(c) \(x=\dfrac{45}{16},\ y=-\dfrac{35}{16}\)
(d) \(x=\dfrac85,\ y=\dfrac15\)
Show Answer & Explanation

Answer: (c) \(x=\dfrac{45}{16},\ y=-\dfrac{35}{16}\)

Explanation:
1. Put \(u=\dfrac1{x+y},\ v=\dfrac1{x-y}\): \(u+2v=2\), \(2u-v=3\), so \(v=\dfrac15,\ u=\dfrac85\)
2. \(x+y=\dfrac58\) and \(x-y=5\)
3. \(x=\dfrac{45}{16},\ y=-\dfrac{35}{16}\)

Teacher's Note:
1. Remember to invert u and v to get x + y and x - y.
2. Then add and subtract.

Question 5: The solution of \(x-z=5\), \(y+z=3\), \(x-y=2\) is
(a) \(x=3,\ y=5,\ z=0\)
(b) \(x=5,\ y=3,\ z=0\)
(c) \(x=5,\ y=3,\ z=1\)
(d) \(x=6,\ y=4,\ z=1\)
Show Answer & Explanation

Answer: (b) \(x=5,\ y=3,\ z=0\)

Explanation:
1. Adding all three: \(2x=10\), so \(x=5\)
2. \(z=x-5=0\), \(y=3-z=3\)
3. Check \(x-y=5-3=2\)

Teacher's Note:
1. Adding all three equations eliminates y and z at once.
2. Check the answer by substituting it in both original equations.

Question 6: The solution of \(x+2y+z=12\), \(2x-z=4\), \(x-2y=4\) is
(a) \(x=5,\ y=\dfrac12,\ z=6\)
(b) \(x=6,\ y=1,\ z=8\)
(c) \(x=5,\ y=\dfrac12,\ z=-6\)
(d) \(x=4,\ y=0,\ z=4\)
Show Answer & Explanation

Answer: (a) \(x=5,\ y=\dfrac12,\ z=6\)

Explanation:
1. Add the first and second: \(3x+2y=16\); with \(x-2y=4\) adding gives \(4x=20\), so \(x=5\)
2. \(z=2x-4=6\)
3. \(2y=x-4=1\), so \(y=\dfrac12\)

Teacher's Note:
1. Eliminate z by adding the first two equations.
2. Check with all three equations.

Question 7: Solving \(2x+3y+8=0\) and \(4x+5y+14=0\) by cross-multiplication gives
(a) \(x=1,\ y=2\)
(b) \(x=-2,\ y=-1\)
(c) \(x=-1,\ y=2\)
(d) \(x=-1,\ y=-2\)
Show Answer & Explanation

Answer: (d) \(x=-1,\ y=-2\)

Explanation:
1. \(\dfrac x{3\cdot14-5\cdot8}=\dfrac{-y}{2\cdot14-4\cdot8}=\dfrac1{2\cdot5-4\cdot3}\)
2. \(\dfrac x2=\dfrac{-y}{-4}=\dfrac1{-2}\)
3. \(x=-1\) and \(y=-2\)

Teacher's Note:
1. Cross-multiplication needs the equations in the form ax + by + c = 0.
2. Watch the signs of the denominators.

Question 8: Solving \(2x-6y+10=0\) and \(3x-7y+13=0\) by cross-multiplication gives
(a) \(x=2,\ y=-1\)
(b) \(x=-2,\ y=-1\)
(c) \(x=-2,\ y=1\)
(d) \(x=1,\ y=-2\)
Show Answer & Explanation

Answer: (c) \(x=-2,\ y=1\)

Explanation:
1. \(\dfrac x{(-6)(13)-(-7)(10)}=\dfrac{-y}{2\cdot13-3\cdot10}=\dfrac1{2(-7)-3(-6)}\)
2. \(\dfrac x{-8}=\dfrac{-y}{-4}=\dfrac14\)
3. \(x=-2\) and \(y=1\)

Teacher's Note:
1. Compute each determinant-like difference carefully.
2. Check: 2(-2) - 6(1) + 10 = 0.

Question 9: Solving \(ax+by=a-b\) and \(bx-ay=a+b\) gives
(a) \(x=1,\ y=-1\)
(b) \(x=-1,\ y=1\)
(c) \(x=1,\ y=1\)
(d) \(x=a,\ y=-b\)
Show Answer & Explanation

Answer: (a) \(x=1,\ y=-1\)

Explanation:
1. Multiply the first by a and the second by b and add: \(a^2x+b^2x=a^2-ab+ab+b^2\)
2. \((a^2+b^2)x=a^2+b^2\), so \(x=1\)
3. Then \(by=a-b-a=-b\), so \(y=-1\)

Teacher's Note:
1. Adding the cross-multiplied equations removes y.
2. Check by substitution.

Question 10: Solving \(x+y=a-b\) and \(ax-by=a^2+b^2\) gives
(a) \(x=b,\ y=-a\)
(b) \(x=a,\ y=b\)
(c) \(x=-a,\ y=b\)
(d) \(x=a,\ y=-b\)
Show Answer & Explanation

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

Explanation:
1. Multiply the first by b and add to the second: \((a+b)x=ab-b^2+a^2+b^2=a(a+b)\)
2. So \(x=a\), and then \(y=a-b-a=-b\)

Teacher's Note:
1. Add the second equation to b times the first so y cancels.
2. Check the answer by substituting it in both original equations.

Question 11: Solving \((a-b)x+(a+b)y=a^2-2ab-b^2\) and \((a+b)(x+y)=a^2+b^2\) gives
(a) \(x=a-b,\ y=\dfrac{2ab}{a+b}\)
(b) \(x=a+b,\ y=-\dfrac{2ab}{a+b}\)
(c) \(x=a+b,\ y=\dfrac{2ab}{a+b}\)
(d) \(x=\dfrac{2ab}{a+b},\ y=a+b\)
Show Answer & Explanation

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

Explanation:
1. The second equation is \((a+b)x+(a+b)y=a^2+b^2\). Subtract the first: \(2bx=2b(a+b)\)
2. So \(x=a+b\)
3. \((a+b)y=a^2+b^2-(a+b)^2=-2ab\), so \(y=-\dfrac{2ab}{a+b}\)

Teacher's Note:
1. Subtract the equations to remove y.
2. Expand (a + b)^2 carefully.

Question 12: Solving \(a(x+y)+b(x-y)=a^2-ab+b^2\) and \(a(x+y)-b(x-y)=a^2+ab+b^2\) gives
(a) \(x=\dfrac{b^2}{2a},\ y=\dfrac{2a^2+b^2}{2a}\)
(b) \(x=\dfrac{2a^2+b^2}{2a},\ y=\dfrac{b^2}{2a}\)
(c) \(x=\dfrac{b^2}{2a},\ y=\dfrac{b^2}{2a}\)
(d) \(x=\dfrac{a^2}{2b},\ y=\dfrac{2b^2+a^2}{2b}\)
Show Answer & Explanation

Answer: (a) \(x=\dfrac{b^2}{2a},\ y=\dfrac{2a^2+b^2}{2a}\)

Explanation:
1. Adding: \(2a(x+y)=2(a^2+b^2)\), so \(x+y=\dfrac{a^2+b^2}a\)
2. Subtracting: \(2b(x-y)=-2ab\), so \(x-y=-a\)
3. \(x=\dfrac{b^2}{2a}\), \(y=\dfrac{2a^2+b^2}{2a}\)

Teacher's Note:
1. Group the terms a(x + y) and b(x - y) as new unknowns.
2. Add and subtract the equations.

Question 13: Solving \(\dfrac ax-\dfrac by=0\) and \(\dfrac{ab^2}x+\dfrac{a^2b}y=a^2+b^2\) (\(x\ne0,\ y\ne0\)) gives
(a) \(x=b,\ y=a\)
(b) \(x=a^2,\ y=b^2\)
(c) \(x=1,\ y=1\)
(d) \(x=a,\ y=b\)
Show Answer & Explanation

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

Explanation:
1. Put \(u=\dfrac ax,\ v=\dfrac by\): \(u-v=0\) and \(b^2u+a^2v=a^2+b^2\)
2. \(u=v\), so \((a^2+b^2)u=a^2+b^2\) and \(u=v=1\)
3. Hence \(x=a\) and \(y=b\)

Teacher's Note:
1. The first equation gives u = v directly.
2. Invert u and v to find x and y.

Question 14: On comparing the ratios, the pair \(6x-3y+10=0\) and \(2x-y+9=0\) represents lines that
(a) coincide
(b) are parallel
(c) intersect at exactly one point
(d) intersect on the y-axis
Show Answer & Explanation

Answer: (b) are parallel

Explanation:
1. \(\dfrac62=3,\ \dfrac{-3}{-1}=3,\ \dfrac{10}9\)
2. \(\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}\ne\dfrac{c_1}{c_2}\), so the lines are parallel

Teacher's Note:
1. The first two ratios equal but the third different means parallel lines.
2. Parallel lines have no common point.

Question 15: The coach of a cricket team buys 7 bats and 6 balls for Rs. 3800. Later he buys 3 bats and 5 balls for Rs. 1750. The cost of one bat and one ball are
(a) Rs. 500 and Rs. 50
(b) Rs. 50 and Rs. 500
(c) Rs. 400 and Rs. 150
(d) Rs. 450 and Rs. 108
Show Answer & Explanation

Answer: (a) Rs. 500 and Rs. 50

Explanation:
1. \(7x+6y=3800\) and \(3x+5y=1750\)
2. Multiply by 3 and 7 and subtract: \(17y=850\), so \(y=50\)
3. \(7x=3800-300\), so \(x=500\)

Teacher's Note:
1. x is the price of a bat and y of a ball.
2. Check the answer by substituting it in both original equations.

Question 16: Meena withdrew Rs. 2000 from a bank in notes of Rs. 50 and Rs. 100 only and got 25 notes in all. The number of notes of Rs. 50 and Rs. 100 are
(a) 15 and 10
(b) 5 and 20
(c) 12 and 13
(d) 10 and 15
Show Answer & Explanation

Answer: (d) 10 and 15

Explanation:
1. \(x+y=25\) and \(50x+100y=2000\), i.e. \(x+2y=40\)
2. Subtracting: \(y=15\), so \(x=10\)

Teacher's Note:
1. Divide the money equation by 50 to simplify.
2. Check: 500 + 1500 = 2000.

Question 17: Yash scored 40 marks in a test, getting 3 marks for each right answer and losing 1 mark for each wrong answer. Had 4 marks been awarded for each correct answer and 2 marks deducted for each wrong answer, he would have scored 50. The number of questions in the test is
(a) 15
(b) 25
(c) 20
(d) 18
Show Answer & Explanation

Answer: (c) 20

Explanation:
1. Let correct \(=x\), wrong \(=y\): \(3x-y=40\) and \(4x-2y=50\)
2. Multiply the first by 2: \(6x-2y=80\). Subtract \(4x-2y=50\): \(2x=30\), so \(x=15\)
3. \(y=3(15)-40=5\), so the total is \(15+5=20\)

Teacher's Note:
1. The question asks for the total, not for the correct answers alone.
2. Check the answer by substituting it in both original equations.

Question 18: The number that must be added to each of 5, 9, 17, 27 so that they are in proportion is
(a) 2
(b) 3
(c) 4
(d) 1
Show Answer & Explanation

Answer: (b) 3

Explanation:
1. Four numbers are in proportion if first x fourth = second x third
2. \((5+x)(27+x)=(9+x)(17+x)\): \(135+32x=153+26x\)
3. \(6x=18\), so \(x=3\) (giving 8, 12, 20, 30)

Teacher's Note:
1. The x squared terms cancel on both sides.
2. Check: 8 : 12 = 20 : 30.

Question 19: The average score of boys in an examination is 71 and that of girls is 73. The average score of the school is 71.8. The ratio of the number of boys to the number of girls is
(a) 3 : 2
(b) 2 : 3
(c) 4 : 1
(d) 1 : 1
Show Answer & Explanation

Answer: (a) 3 : 2

Explanation:
1. Let boys \(=x\), girls \(=y\): \(71x+73y=71.8(x+y)\)
2. \(1.2y=0.8x\), so \(\dfrac xy=\dfrac{1.2}{0.8}=\dfrac32\)

Teacher's Note:
1. Total marks of the school = sum of the total marks of boys and girls.
2. A mean closer to 71 means more boys.

Question 20: The difference between two numbers is 26 and one number is three times the other. The numbers are
(a) 36 and 12
(b) 42 and 16
(c) 52 and 26
(d) 39 and 13
Show Answer & Explanation

Answer: (d) 39 and 13

Explanation:
1. \(x-y=26\) and \(x=3y\)
2. \(3y-y=26\), so \(y=13\)
3. \(x=39\)

Teacher's Note:
1. Check the answer by substituting it in both original equations.
2. Three times 13 is 39.

Question 21: The sum of a two-digit number and the number obtained by reversing its digits is 99. If the units digit is 3 more than the tens digit, the number is
(a) 63
(b) 36
(c) 45
(d) 27
Show Answer & Explanation

Answer: (b) 36

Explanation:
1. Let tens digit \(=t\), units digit \(=u\): \(11(t+u)=99\), so \(t+u=9\)
2. \(u-t=3\), so \(u=6\) and \(t=3\)
3. The number is 36 (check: \(36+63=99\))

Teacher's Note:
1. The sum of a number and its reverse is 11 times the sum of the digits.
2. Read the condition carefully: the units digit is the larger one.

Question 22: The sum of the digits of a two-digit number is 9. Also, nine times the number is twice the number obtained by reversing the digits. The number is
(a) 81
(b) 27
(c) 36
(d) 18
Show Answer & Explanation

Answer: (d) 18

Explanation:
1. Let the digits be x (tens) and y (units): \(x+y=9\)
2. \(9(10x+y)=2(10y+x)\), so \(88x=11y\) and \(y=8x\)
3. \(9x=9\), so \(x=1,\ y=8\) and the number is 18

Teacher's Note:
1. Write each number as 10 x tens digit + units digit.
2. Check: 9 x 18 = 162 = 2 x 81.

Question 23: The sum of the numerator and denominator of a fraction is 4 more than twice the numerator. If the numerator and denominator are each increased by 3, the ratio becomes \(2:3\). The fraction is
(a) \(\dfrac95\)
(b) \(\dfrac47\)
(c) \(\dfrac59\)
(d) \(\dfrac38\)
Show Answer & Explanation

Answer: (c) \(\dfrac59\)

Explanation:
1. \(x+y=2x+4\), so \(y=x+4\)
2. \(\dfrac{x+3}{y+3}=\dfrac23\), so \(3x+9=2y+6\)
3. Substituting: \(3x+9=2x+14\), so \(x=5\) and \(y=9\)

Teacher's Note:
1. Translate each sentence into one equation.
2. Check: 8/12 = 2/3.

Question 24: The sum of the numerator and denominator of a fraction is 3 less than twice the denominator. If the numerator and denominator are each decreased by 1, the numerator becomes half the denominator. The fraction is
(a) \(\dfrac74\)
(b) \(\dfrac47\)
(c) \(\dfrac35\)
(d) \(\dfrac59\)
Show Answer & Explanation

Answer: (b) \(\dfrac47\)

Explanation:
1. \(x+y=2y-3\), so \(x=y-3\)
2. \(x-1=\dfrac12(y-1)\), so \(2x-2=y-1\)
3. Substituting: \(2y-6-2=y-1\), so \(y=7\) and \(x=4\)

Teacher's Note:
1. Write the fraction as x over y and form two equations.
2. Check: 3/6 = 1/2.

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Practice MCQs for Class 10 Mathematics Chapter 03 Pair of Linear Equations in Two Variables

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