Pair of Linear Equations in Two Variables MCQs

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A system of equations in two variables having a unique solution, no solutions, or an infinite number of solutions is known as a linear equation in two variables. There may be as many as 'n' variables in a linear system of equations. A straight line is the collection of answers found by solving these linear equations. The algebraic equations of the form y = mx + b, where m is the slope and b is the y-intercept, are known as linear equations in two variables. They're called first-order equations. Two-variable linear equations such as y = 2x+3 and 2y = 4x + 9.

Multiple Choice Questions

Ques. 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

Ans. 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.

Ques. 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

Ans. 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

As a result, the pairs of equations are parallel, and the lines never meet at any point, indicating that there is no solution.

Ques. 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. 4/5
d. 5/4

Ans. 15/4

Explanation: The condition for parallel lines is:

a1/a2 = b1/b2 ≠ c1/c2

Hence, 3/2 = 2k/5

k=15/4

Ques. 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

Ans. -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

Ques. 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

Ans. 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.

Ques. 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

Ans. 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.

Ques. 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
strong>d. 10km/hr and 6km/hr

Ans. 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

Ques. 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°

Ans. x=20° and y = 30°

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

Ques. 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

Ans. 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.

Ques. 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

Ans. 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.

Ques. 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

Ans. 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 = â??

Condition for having infinitely many solutions is

a1/a2 = b1/b2 = c1/c2

c/6 = ½ = â??

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.

Ques. 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

Ans. 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.

Ques. 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

Ans. 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

Read More-

Pair of Linear Equations In Two Variables Important Questions	Linear Equation in Two Variables: Definition and Graph
Graph of a Linear Equation in Two Variables	Substitution Method of Solving a Pair of Linear Equations
Cross-multiplication method of solving linear equations	Elimination method of solving linear equations
Linear Equation in standard form	Consistent system of linear equations
Differential equation	Inconsistent system

Pair of Linear Equations in Two Variables MCQs

Multiple Choice Questions

CBSE X Related Questions

1.
There are two sections A and B of Grade X. There are 28 students in Section A and 30 students in Section B. What is the minimum number of books you will acquire for the class library so that they can be distributed equally among students of Section A or Section B ?

2.
Find the H.C.F. and L.C.M. of 1530 and 2040.

3.
In the figure given above, \(\triangle ABC \sim \triangle XYZ\), then find the values of \(x\) and \(y\).

4.
The HCF of 960 and 432 is :

5.
In the given figure, PA and PB are tangents to a circle centred at O. If \( \angle AOB = 130^\circ \), then \( \angle APB \) is equal to :

6.
Prove that : \( \frac{\sec^3 \theta}{\sec^2 \theta - 1} + \frac{\text{cosec}^3 \theta}{\text{cosec}^2 \theta - 1} = \sec \theta \cdot \text{cosec} \theta (\sec \theta + \text{cosec} \theta) \)

Similar Mathematics Concepts

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Pair of Linear Equations in Two Variables MCQs

Multiple Choice Questions

CBSE X Related Questions

1.There are two sections A and B of Grade X. There are 28 students in Section A and 30 students in Section B. What is the minimum number of books you will acquire for the class library so that they can be distributed equally among students of Section A or Section B ?

2.Find the H.C.F. and L.C.M. of 1530 and 2040.

3.In the figure given above, \(\triangle ABC \sim \triangle XYZ\), then find the values of \(x\) and \(y\).

4.The HCF of 960 and 432 is :

5.In the given figure, PA and PB are tangents to a circle centred at O. If \( \angle AOB = 130^\circ \), then \( \angle APB \) is equal to :

6.Prove that : \( \frac{\sec^3 \theta}{\sec^2 \theta - 1} + \frac{\text{cosec}^3 \theta}{\text{cosec}^2 \theta - 1} = \sec \theta \cdot \text{cosec} \theta (\sec \theta + \text{cosec} \theta) \)

Similar Mathematics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
There are two sections A and B of Grade X. There are 28 students in Section A and 30 students in Section B. What is the minimum number of books you will acquire for the class library so that they can be distributed equally among students of Section A or Section B ?

2.
Find the H.C.F. and L.C.M. of 1530 and 2040.

3.
In the figure given above, \(\triangle ABC \sim \triangle XYZ\), then find the values of \(x\) and \(y\).

4.
The HCF of 960 and 432 is :

5.
In the given figure, PA and PB are tangents to a circle centred at O. If \( \angle AOB = 130^\circ \), then \( \angle APB \) is equal to :

6.
Prove that : \( \frac{\sec^3 \theta}{\sec^2 \theta - 1} + \frac{\text{cosec}^3 \theta}{\text{cosec}^2 \theta - 1} = \sec \theta \cdot \text{cosec} \theta (\sec \theta + \text{cosec} \theta) \)