R.D. Sharma Solutions Class 9th: Ch 13 Linear Equations in Two Variables MCQ's

Chapter 13 Linear Equations in Two Variables R.D. Sharma Solutions for Class 9th MCQ's

Exercise 13.3

Mark the correct alternative in each of the following:

1. If (4, 19) is a solution of the equation y = ax + 3, then a=
(a) 3
(b) 4
(c) 5
(d) 6 

Solution

We are given (4,19)as the solution of equation.
 y = ax+3
Substituting x= 4 and y= 19, we get
19 = 4a+3
⇒ 4a=19-3
⇒ 4a=16
⇒ a=4
Therefore, the correct answer is (b). 

2. If (a, 4) lies on the graph of 3x + y = 10, then the value of a is
(a) 3
(b) 1
(c) 3
(d) 4

Solution

We are given (a, 4) lies on the graph of linear equation 3x+ y= 10.
So, the given co-ordinates are the solution of the equation 3x+ y= 10.
Therefore, we can calculate the value of a by substituting the value of given co-ordinates in equation 3x+ y= 10.
Substituting x= a and y= 4 in equation 3x+ y= 10, we get
3×a + 4 = 10
⇒ 3a = 10-4
⇒ 3a=6
⇒ a=6/3
⇒ a=2
No, option is correct. 

3. The graph of the linear equation 2x − y = 4 cuts x- axis at
(a) (2, 0)
(b) (−2, 0)
(c) (0, −4)
(d) (0, 4)

Solution

Given,
2x- y = 4
we get, 
y = 2x-4
We will substitute y =0 in y = 2x-4 to get the co-ordinates for the graph of 2x- y = 4 at x axis
0 = 2x-4
⇒ 2x=0+4
⇒ 2x=4 x=2
Co-ordinates for the graph of 2x- y = 4 are (2,0). 
Therefore, the correct answer is (a). 

4. How many linear equations are satisfied by x = 2  and y = −3?
(a) Only one
(b) Two
(c) Three
(d) Infinitely many

Solution
There are infinite numbers of linear equations that are satisfied by x= 2 and y = -3 as
(i) Every solution of the linear equation represent a point on the line.
(ii) Every point on the line is the solution of the linear equation.
Therefore, the correct answer is (d).

5. The equation x − 2 = 0 on number line is represented by
(a) a line
(b) a point
(c) infinitely many lines
(d) two lines

Solution

The equation x - 2 = 0  is represented by a point on the number line.
Therefore, the correct answer is (b).

6. x = 2, y = −1 is a solution of the linear equation
(a) x + 2y = 0
(b) x + 2y = 4
(c) 2x + y = 0
(d) 2x + y = 5

Solution

We are given x=2; y = -1 as the solution of linear equation, which we have to find.
The equation is x+2y = 0 which can he proved by
Substituting x=2 and y= -1 in the equation x+2y= 0, we get
2 + 2×(-1) = 0
2-2 = 0
0 = 0
RHS = LHS
Therefore, the correct answer is (a).

7. If (2k − 1, k) is a solution of the equation 10x − 9y = 12, then k =
(a) 1
(b) 2
(c) 3
(d) 4

Solution

We are given (2k-1, k)as the solution of equation
10x - 9y = 12
Substituting x=2k-1 and y = k , we get
10 ×(2k-1)- 9×k =12
⇒ 20k - 10-9k =12
⇒ 11k = 12+10
⇒ 11k = 22
⇒ k = 22/11
⇒ k = 2
Therefore, the correct answer is (b). 

8. The distance between the graph of the equations x = −3 and x = 2 is
(a) 1
(b) 2
(c) 3
(d) 5

Solution

Distance between the graph of equations x= -3 and x = 2.
D = Distance of co-ordinate on negative side of x axis + Distance of co-ordinate on positive side of x axis
Distance of co-ordinate on negative side of x axis = x = 3 units
Distance of co-ordinate on positive side of x axis = x = 2 units 
D = 3+2
D = 5 units
Therefore, the correct answer is d 

9. The distance between the graphs of the equations y = −1 and y = 3 is
(a) 
(b) 4
(c) 3
(d) 1

Solution

Here, you can see, the distance is 4 units. The distance can be calculated by subtracting y coordinates. 3-(-1) =4. So, (b) is the correct answer.

10. If the graph of the equation 4x + 3y = 12 cuts the coordinate axes at A and B, then hypotenuse of right triangle AOB is of length
(a) 4 units
(b) 3 units
(c) 5 units
(d) none of these

Solution



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