Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables MCQs

Class 10 Maths MCQs for Chapter 3 (Pair of Linear Equations in Two Variables) are provided here online with answers. These multiple choice questions will help students to score good marks in the board exam. These objective questions are prepared as per the latest CBSE syllabus and NCERT textbook. Practising these questions will help them to develop problem-solving skills. To get chapter-wise MCQs, click here.

Class 10 Maths MCQs for Pair of Linear Equations in Two Variables

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Below are the MCQs for Chapter 3

1. The pairs of equations x+2y-5 = 0 and -4x-8y+20=0 have:

(a) Unique solution

(b) Exactly two solutions

(c) Infinitely many solutions

(d) No solution

Answer: (c) Infinitely many solutions

Explanation:

a1/a2 = 1/-4

b1/b2 = 2/-8 = 1/-4

c1/c2 = -5/20 = -¼

This shows:

a1/a2 = b1/b2 = c1/c2

Therefore, the pair of equations has infinitely many solutions.

2. If a pair of linear equations is consistent, then the lines are:

(a) Parallel

(b) Always coincident

(c) Always intersecting

(d) Intersecting or coincident

Answer: (d) Intersecting or coincident

Explanation: Because the two lines definitely have a solution.

3. The pairs of equations 9x + 3y + 12 = 0 and 18x + 6y + 26 = 0 have

(a) Unique solution

(b) Exactly two solutions

(c) Infinitely many solutions

(d) No solution

Answer: (d) No solution

Explanation: Given, 9x + 3y + 12 = 0 and 18x + 6y + 26 = 0

a1/a2 = 9/18 = 1/2

b1/b2 = 3/6 = 1/2

c1/c2 = 12/26 = 6/13

Since, a1/a2 = b1/b2 ≠ c1/c2

So, the pairs of equations are parallel and the lines never intersect each other at any point, therefore there is no possible solution.

4. If the lines 3x+2ky – 2 = 0 and 2x+5y+1 = 0 are parallel, then what is the value of k?

(a) 4/15

(b) 15/4

(c) ⅘

(d) 5/4

Answer: (b) 15/4

Explanation: The condition for parallel lines is:

a1/a2 = b1/b2 ≠ c1/c2

Hence, 3/2 = 2k/5

k=15/4

5. If one equation of a pair of dependent linear equations is -3x+5y-2=0. The second equation will be:

(a) -6x+10y-4=0

(b) 6x-10y-4=0

(c) 6x+10y-4=0

(d) -6x+10y+4=0

Answer: (a) -6x+10y-4=0

Explanation: The condition for dependent linear equations is:

a1/a2 = b1/b2 = c1/c2

For option a,

a1/a2 = b1/b2 = c1/c2= ½

6.The solution of the equations x-y=2 and x+y=4 is:

(a) 3 and 1

(b) 4 and 3

(c) 5 and 1

(d) -1 and -3

Answer: (a) 3 and 1

Explanation: x-y =2

x=2+y

Substituting the value of x in the second equation we get;

2+y+y=4

2+2y=4

2y = 2

y=1

Now putting the value of y, we get;

x=2+1 = 3

Hence, the solutions are x=3 and y=1.

7. A fraction becomes 1/3 when 1 is subtracted from the numerator and it becomes 1/4 when 8 is added to its denominator. The fraction obtained is:

(a) 3/12

(b) 4/12

(c) 5/12

(d) 7/12

Answer: (c) 5/12

Explanation: Let the fraction be x/y

So, as per the question given,

(x -1)/y = 1/3 => 3x – y = 3…………………(1)

x/(y + 8) = 1/4 => 4x –y =8 ………………..(2)

Subtracting equation (1) from (2), we get

x = 5 ………………………………………….(3)

Using this value in equation (2), we get,

4×5 – y = 8

y= 12

Therefore, the fraction is 5/12.

8. The solution of 4/x+3y=14 and 3/x-4y=23 is:

(a) ⅕ and -2

(b) ⅓ and ½

(c) 3 and ½

(d) 2 and ⅓

Answer: (a) ⅕ and -2

Explanation: Let 1/x = m

4m + 3y = 14

3m – 4y = 23

By cross multiplication we get;

m/(-69-56) = y/(-42-(-92)) = 1/(-16-9)

m/-125=y/50=-1/25

m/-125 = -1/25 and y/50=-1/25

m=5 and y=-2

m=1/x or x=1/m = ⅕

9. Ritu can row downstream 20 km in 2 hours, and upstream 4 km in 2 hours. Her speed of rowing in still water and the speed of the current is:

(a) 6km/hr and 3km/hr

(b) 7km/hr and 4km/hr

(c) 6km/hr and 4km/hr

(d) 10km/hr and 6km/hr

Answer: (c) 6km/hr and 4km/hr

Explanation: Let, Speed of Ritu in still water = x km/hr

Speed of Stream = y km/hr

Now, speed of Ritu, during,

Downstream = x + y km/h

Upstream = x – y km/h

As per the question given,

2(x+y) = 20

Or x + y = 10……………………….(1)

And, 2(x-y) = 4

Or x – y = 2………………………(2)

Adding both the equations, we get,

2x=12

x = 6

Putting the value of x in eq.1, we get,

y = 4

Therefore,

Speed of Ritu is still water = 6 km/hr

Speed of Stream = 4 km/hr

10. The angles of cyclic quadrilaterals ABCD are: A = (6x+10), B=(5x)°, C = (x+y)° and D=(3y-10)°. The value of x and y is:

(a) x=20° and y = 10°

(b) x=20° and y = 30°

(c) x=44° and y=15°

(d) x=15° and y=15°

Answer: (b) x=20° and y = 30°

Explanation: We know, in cyclic quadrilaterals, the sum of the opposite angles are 180°.

Hence,

A + C = 180°

6x+10+x+y=180 =>7x+y=170°

And B+D=180°

5x+3y-10=180 =>5x+3y=190°

By solving the above two equations we get;

x=20° and y = 30°.

11. The pair of equations x = a and y = b graphically represents lines which are

(a) parallel 

(b) intersecting at (b, a)

(c) coincident 

(d) intersecting at (a, b)

Answer: (d) intersecting at (a, b)

The pair of equations x = a and y = b graphically represents lines which are intersecting at (a, b).

12. The pair of equations 5x – 15y = 8 and 3x – 9y = 24/5 has

(a) one solution 

(b) two solutions 

(c) infinitely many solutions

(d) no solution

Answer: (c) infinitely many solutions

Explanation:

The given pair of equations are 5x – 15y = 8 and 3x – 9y = 24/5.

Comparing with the standard form,

a1 = 5, b1 = -15, c1 = -8

a2 = 3, b2 = -9, c2 = -24/5

a1/a2 = 5/3

b1/b2 = -15/-9 = 5/3

c1/c2 = -8/(-24/5) = 5/3

Thus, a1/a2 = b1/b2 = c1/c2

Hence, the given pair of equations has infinitely many solutions.

13. The pair of equations x + 2y + 5 = 0 and –3x – 6y + 1 = 0 have

(a) a unique solution 

(b) exactly two solutions

(c) infinitely many solutions 

(d) no solution

Answer: (d) no solution

Explanation:

Given pair of equations are x + 2y + 5 = 0 and –3x – 6y + 1 = 0.

Comparing with the standard form,

a1 = 1, b1 = 2, c1 = 5

a2 = -3, b2 = -6, c2 = 1

a1/a2 = -1/3

b1/b2 = 2/-6 = -1/3

c1/c2 = 5/1

Thus, a1/a2 = b1/b2 ≠ c1/c2

Hence, the given pair of equations has no solution.

14. The value of c for which the pair of equations cx – y = 2 and 6x – 2y = 3 will have infinitely many solutions is

(a) 3 

(b) -3 

(c) -12 

(d) no value

Answer: (d) no value

Explanation:

Given pair of equations are cx – y = 2 and 6x – 2y = 3.

Comparing with the standard form,

a1 = c, b1 = -1, c1 = -2

a2 = 6, b2 = -2, c2 = -3

a1/a2 = c/6

b1/b2 = -1/-2 = 1/2

c1/c2 = -2/-3 = ⅔

Condition for having infinitely many solutions is

a1/a2 = b1/b2 = c1/c2

c/6 = ½ = ⅔

Therefore, c = 3 and c = 4

Here, c has different values.

Hence, for no value of c the pair of equations will have infinitely many solutions.

15. If the lines representing the pair of linear equations a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 are coincident, then

(a) a1/a2 = b1/b2

(b) a1/a2 = b1/b2 = c1/c2

(c) a1/a2 ≠ b1/b2

(d) a1/a2 = b1/b2 ≠ c1/c2

Answer: (b) a1/a2 = b1/b2 = c1/c2

If the lines representing the pair of linear equations a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 are coincident, then a1/a2 = b1/b2 = c1/c2.

16. A pair of linear equations which has a unique solution x = 2, y = -3 is

(a) x + y = -1; 2x – 3y = -5

(b) 2x + 5y = -11; 4x + 10y = -22

(c) 2x – y = 1; 3x + 2y = 0

(d) x – 4y – 14 = 0; 5x – y – 13 = 0

Answer: (b) 2x + 5y = -11; 4x + 10y = -22

Explanation:

If x = 2, y = -3 is a unique solution of any pair of equations, then these values must satisfy that pair of equations.

By verifying the options, option (b) satisfies the given values.

LHS = 2x + 5y = 2(2) + 5(- 3) = 4 – 15 = -11 = RHS

LHS = 4x + 10y = 4(2) + 10(- 3)= 8 – 30 = -22 = RHS

17. The father’s age is six times his son’s age. Four years hence, the age of the father will be four times his son’s age. The present ages, in years, of the son and the father are, respectively

(a) 4 and 24 

(b) 5 and 30

(c) 6 and 36 

(d) 3 and 24

Answer: (c) 6 and 36

Explanation:

Let x years be the present age of father and y years be the present age of son. 

According to the given,

(x + 4) = 4(y + 4) 

x + 4 = 4y + 16 

x- 4y + 4 – 16 = 0

x – 4y – 12 = 0….(i) 

Also, x = 6y….(ii) 

From (i) and (ii),

6y – 4y – 12 = 0 

2y = 12 

y = 6

Substituting y = 6 in (ii),

x = 6(6) = 36

18. If the pair of linear equations has a unique solution, then the lines representing these equations will 

(a) coincide

(b) intersect at one point

(c) parallel to each other

(d) parallel to x-axis

Answer: (b) intersect at one point

If the pair of linear equations has a unique solution, then the lines representing these equations will intersect at one point.

19. Which of the following method(s) is/are used to find the solution of a pair of linear equations algebraically?

(a) Substitution Method

(b) Elimination Method

(c) Cross- multiplication Method

(d) All the above

Answer: (d) All the above

The methods used to find the solution of a pair of linear equations algebraically are:

Substitution Method

Elimination Method

Cross- multiplication Method

20. The graphical representation of a pair of equations 4x + 3y – 1 = 5 and 12x + 9y = 15 will be

(a) parallel lines

(b) coincident lines

(c) intersecting lines

(d) perpendicular lines

Answer: (a) parallel lines

Explanation:

Given pair of equations are 4x + 3y – 1 = 5 and 12x + 9y = 15.

Comparing with the standard form,

a1 = 4, b1 = 3, c1 = -6

a2 = 12, b2 = 9, c2 = -15

a1/a2 = 4/12 = 1/3

b1/b2 = 3/9 = 1/3

c1/c2 = -6/-15 = 2/5

Thus, a1/a2 = b1/b2 ≠ c1/c2

Hence, the given pair of equations has no solution.

That means, the lines representing the given pair of equations are parallel to each other.

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