myhelper: RS AGGARWAL CLASS 9 CHAPTER 4 LINEAR EQUATIONS IN TWO VARIABLE MCQ

RS AGGARWAL CLASS 9 CHAPTER 4 LINEAR EQUATIONS IN TWO VARIABLE MCQ

MULTIPLE CHOICE QUESTIONS

Question 1:

The equation of the x-axis is
(a) x = 0
(b) y = 0
(c) x = y
(d) x + y = 0

Answer 1:

Since, the y-coordinate of any point on x-axis is always 0.

So, the equation of the x-axis is y = 0.

Hence, the correct option is (b).

Question 2:

The equation of the y-axis is
(a) x = 0
(b) y = 0
(c) x = y
(d) x + y = 0

Answer 2:

Since, the x-coordinate of any point on y-axis is always 0.

So, the equation of the y-axis is x = 0.

Hence, the correct option is (a).

PAGE NO-163

Question 3:

The point of the form (a, a), where a ≠ 0, lies on
(a) the x-axis
(b) the y-axis
(c) the line y = x
(d) the line x + y = 0

Answer 3:

(c) the line y = x
Given, a point of the form $(a, a)$ $(a, a)$ , where $a$ $a$ ≠ 0 .
When $a = 1$ $a = 1$ , the point is (1,1)
When $a = 2$ $a = 2$ , the point is (2,2).....and so on.
Plot the points (1,1) and (2,2).......and so on. Join the points and extend them in both the direction. You will get the equation of the line $y = x$ $y = x$ .

Question 4:

The point of the form (a, −a), where a ≠ 0, lies on
(a) the x-axis
(b) the y-axis
(c) the line y = x
(d) the line x + y = 0

Answer 4:

(d) the line x + y = 0
Given, a point of the form $(a, - a)$ $(a, - a)$ , where $a$ $a$ ≠ 0.
When $a = 1$ $a = 1$ , the point is (1,-1).
When $a = 2$ $a = 2$ , the point is (2,-2).
When $a = 3$ $a = 3$ , the point is (3,-3).......and so on.
Plot these points on a graph paper. Join these points and extend them in both the directions.
You will get the equation of the line $x + y = 0$ $x + y = 0$ .

Question 5:

The linear equation 3x − 5y = has
(a) a unique solution
(b) two solutions
(c) infinitely many solutions
(d) no solution

Answer 5:

(c) infinitely many solutions
Given linear equation: $3 x - 5 y = 15$ $3 x - 5 y = 15$
Or, $x = \frac{5 y + 15}{3}$ $x = \frac{5 y + 15}{3}$

When $y = 0$ $y = 0$ , $x = \frac{15}{3} = 5$ $x = \frac{15}{3} = 5$ .

When $y = 3$ $y = 3$ , $x = \frac{30}{3} = 10$ $x = \frac{30}{3} = 10$ .

When $y = - 3$ $y = - 3$ , $x = \frac{0}{3} = 0$ $x = \frac{0}{3} = 0$ .

Thus, we have the following table:

$x$ $x$	5	10	0
$y$ $y$	0	3	-3

Plot the points

A (5, 0), B (10, 3)

$A (5, 0), B (10, 3)$ and

C (0, - 3)

$C (0, - 3)$ . Join the points and extend them in both the directions.
We get infinite points that satisfy the given equation.
Hence, the linear equation has infinitely many solutions.

Question 6:

The equation 2x + 5y = 7 has a unique solution, if x and y are
(a) natural numbers
(b) rational numbers
(c) positive real numbers
(d) real numbers

Answer 6:

Since, every point on the line represented by the equation 2x + 5y = 7 is its solution.

Therefore, there are infinite solutions of the equation the equation 2x + 5y = 7 in which the values of x and y are rational numbers, positive real numbers or real numbers.

But, as 2 + 5 = 7, i.e. x = 1 and y = 1 are the only pair of natural numbers that are the solution of the equation the equation 2x + 5y = 7.

So, the equation 2x + 5y = 7 has a unique solution, if x and y are both are natural numbers.

Hence, the correct option is (a).

Question 7:

The graph of y = 5 is a line
(a) making an intercept 5 on the x-axis
(b) making an intercept 5 on the y-axis
(c) parallel to the x-axis at a distance of 5 units from the origin
(d) parallel to the y-axis at a distance of 5 units from the origin

Answer 7:

As, the graph of y = 5 is a line parallel to x-axis i.e. y = 0.

$\Rightarrow$ $\Rightarrow$ The line represented by the equation y = 5 is parallel to x-axis and intersects y-axis at y = 5.

So, the graph of y = 5 is a line parallel to the x-axis at a distance of 5 units from the origin making an intercept 5 on the y-axis.

Hence, the correct answers are options (b) and (c).
Disclaimer: In this question, there are two correct answers.

Question 8:

The graph of x = 4 is a line
(a) making an intercept 4 on the x-axis
(b) making an intercept 4 on the y-axis
(c) parallel to the x-axis at a distance of 4 units from the origin
(d) parallel to the y-axis at a distance of 4 units from the origin

Answer 8:

As, the graph of x = 4 is a line parallel to y-axis i.e. x = 0.

$\Rightarrow$ $\Rightarrow$ The line represented by the equation x = 4 is parallel to y-axis and intersects x-axis at x = 4.

So, the graph of x = 4 is parallel to y-axis at a distance of 4 units from the origin making an intercept 4 on the x-axis.

Hence, the correct options are (a) and (d).

Disclaimer: In this question, there are two correct answers.

Question 9:

The graph of x + 3 = 0 is a line
(a) making an intercept –3 on the x-axis
(b) making an intercept –3 on the y-axis
(c) parallel to the y-axis at a distance of 3 units to the left of y-axis
(d) parallel to the x-axis at a distance of 3 units below the x-axis

Answer 9:

As, the graph of x + 3 = 0 or x = $-$ $-$ 3 is a line parallel to y-axis i.e. x = 0.

$\Rightarrow$ $\Rightarrow$ The line represented by the equation x = $-$ $-$ 3 is parallel to y-axis and intersects x-axis at x = $-$ $-$ 3.

So, the graph of x + 3 = 0 is a line parallel to the y-axis at a distance of 3 units to the left of y-axis making an intercept $-$ $-$ 3 on the x-axis.

Hence, the correct options are (a) and (c).

Disclaimer: In this question, there are two correct answers.

Question 10:

The graph of y + 2 = 0 is a line
(a) making an intercept –2 on the x-axis
(b) making an intercept –2 on the y-axis
(c) parallel to the x-axis at a distance of 2 units below the x-axis
(d) parallel to the y-axis at a distance of 2 units to the left of y-axis

Answer 10:

As, the graph of y + 2 = 0 or y = −2 is a line parallel to x-axis i.e. y = 0.

⇒ The line represented by the equation y = −2 is parallel to x-axis and intersects y-axis at y = −2.

So, the graph of y + 2 = 0 is a line parallel to the x-axis at a distance of 2 units below the x-axis making an intercept −2 on the y-axis.

Hence, the correct options are (b) and (c).

Disclaimer: In this question, there are two correct answers.

Question 11:

The graph of the linear equation 2x + 3y = 6 meets the y-axis at the point
(a) (2, 0)
(b) (3, 0)
(c) (0, 2)
(d) (0, 3)

Answer 11:

If he graph of the linear equation 2x + 3y = 6 meets the y-axis, then x = 0.

Substituting the value of x = 0 in equation 2x + 3y = 6, we get

$2 (0) + 3 y = 6 \Rightarrow 3 y = 6 \Rightarrow y = \frac{6}{3} \Rightarrow y = 2$ $2 (0) + 3 y = 6 \Rightarrow 3 y = 6 \Rightarrow y = \frac{6}{3} \Rightarrow y = 2$

So, the point of meeting is (0, 2).

Hence, the correct answer is option (c).

PAGE NO- 164

Question 12:

The graph of the linear equation 2x + 5y = 10 meets the x-axis at the point
(a) (0, 2)
(b) (2, 0)
(c) (5, 0)
(d) (0, 5)

Answer 12:

If he graph of the linear equation 2x + 5y = 10 meets the x-axis, then y = 0.

Substituting the value of y = 0 in equation 2x + 5y = 10, we get

$2 x + 5 (0) = 10 \Rightarrow 2 x = 10 \Rightarrow x = \frac{10}{2} \Rightarrow x = 5$ $2 x + 5 (0) = 10 \Rightarrow 2 x = 10 \Rightarrow x = \frac{10}{2} \Rightarrow x = 5$

So, the point of meeting is (5, 0).

Hence, the correct option is (c).

Question 13:

The graph of the line x = 3 passes through the point
(a) (0, 3)
(b) (2, 3)
(c) (3, 2)
(d) None of these

Answer 13:

Question 14:

The graph of the line y = 3 passes through the point
(a) (3, 0)
(b) (3, 2)
(c) (2, 3)
(d) none of these

Answer 14:

Since, the graph of the line y = 3 is parallel to x-axis at a distance of 3 units from the x-axis.

Or, the y-coordinate of every point on the line is always equal to 3.

So, the graph of the line y = 3 passes through the point (2, 3).

Hence, the correct option is (c).

Question 15:

The graph of the line y = −3 does not pass through the point
(a) (2, −3)
(b) (3, −3)
(c) (0, −3)
(d) (−3, 2)

Answer 15:

(d) (−3, 2)

The graph of the line y = -3 does not pass through (-3,2) since (-3,2) does not satisfy y = -3.

Question 16:

The graph of the linear equation x − y = 0 passes through the point
(a) $(\frac{- 1}{2}, \frac{1}{2})$
(b) $(\frac{3}{2}, \frac{- 3}{2})$
(c) (0, −1)
(d) (1, 1)

Answer 16:

(d) (1, 1)
Given equation: $x - y = 0$ or, $x = y$
When $x = 1$ , $y = 1$
When $x = 2$ , $y = 2$ ... and so on
Thus, we get the following table:

$x$	1	2
$y$	1	2

Plot the points on the graph paper. Join the points and extend them in both the directions.
We can see the linear equation

x - y = 0

passes through the point (1,1).

Question 17:

Each of the points (−2, 2), (0, 0), (2, −2) satisfies the linear equation
(a) x − y = 0
(b) x + y = 0
(c) −x + 2y = 0
(d) x – 2y = 0

Answer 17:

(b) x + y = 0
Given points: $(- 2, 2), (0, 0)$ and $(2, - 2)$
We have to check which equation satisfies the given points.
Let us check for (a) $x - y = 0$
Substituting $(- 2, 2)$ in the equation, we get: $x - y = - 2 - 2 = - 4$
Substituting $(0, 0)$ in the equation, we get: $x - y = 0 - 0 = 0$
Substituting $(2, - 2)$ in the equation, we get: $x - y = 2 + 2 = 4$
So, the given points do not satisfy the equation.
Now, let us check (b) $x + y = 0$
Substituting $(- 2, 2)$ in the equation, we get: $x + y = - 2 + 2 = 0$
Substituting $(0, 0)$ in the equation, we get: $x + y = 0 + 0 = 0$
Substituting $(2, - 2)$ in the equation, we get: $x + y = 2 - 2 = 0$
So, the given points satisfy the equation.
Similarly, we can check other equations and find that the given points will not satisfy the equations.
Hence, each of $(- 2, 2), (0, 0)$ and $(2, - 2)$ is a solution of the linear equation $x + y = 0$ .

Question 18:

How many linear equations in x and y can be satisfied by x = 2, y = 3?
(a) Only one
(b) Only two
(c) Infinitely many
(d) None of these

Answer 18:

Question 19:

A linear equation in two variables x and y is of the form ax + by + c = 0, where
(a) a ≠ 0, b ≠ 0
(b) a ≠ 0, b = 0
(c) a = 0, b ≠ 0
(d) a = 0, c = 0

Answer 19:

(a) a ≠ 0, b ≠ 0
A linear equation in two variables $x$ and $y$ is of the form $a x + b y + c = 0$ , where $a$ ≠ 0 and $b$ ≠ 0.

Question 20:

If (2, 0) is a solution of the linear equation 2x + 3y = k then the value of k is
(a) 6
(b) 5
(c) 2
(d) 4

Answer 20:

Since, (2, 0) is a solution of the linear equation 2x + 3y = k.

Substituting the values x = 2 and y = 0 in the equation 2x + 3y = k, we get

$2 (2) + 3 (0) = k \Rightarrow 4 = k or, k = 4$

Hence, the correct option is (d).

Question 21:

Any point on the x-axis is of the form
(a) (x, y)
(b) (0, y)
(c) (x, 0)
(d) (x, x)

Answer 21:

Question 22:

Any point on the y-axis is of the form
(a) (x, y)
(b) (0, y)
(c) (x, 0)
(d) (y, y)

Answer 22:

(b) (0, y),
Any point on the $y$ -axis is of the form $(0, y)$ , where $y$ ≠ 0.

Question 23:

x = 5, y = 2 is a solution of the linear equation
(a) x + 2y = 7
(b) 5x + 2y = 7
(c) x + y = 7
(d) 5x + y = 7

Answer 23:

Substituting the values x = 5, y = 2 in

(a) x + 2y = 7, we get

$LHS = 5 + 2 (2) = 5 + 4 = 9 \neq 7 = RHS i . e . LHS \neq RHS$

(b) 5x + 2y = 7, we get

$LHS = 5 (5) + 2 (2) = 25 + 4 = 29 \neq 7 = RHS i . e . LHS \neq RHS$

(c) x + y = 7, we get

$LHS = 5 + 2 = 7 = RHS i . e . LHS = RHS$

(d) 5x + y = 7, we get

$LHS = 5 (5) + 2 = 25 + 2 = 27 \neq 7 = RHS i . e . LHS \neq RHS$

Hence, the correct option is (c).

Question 24:

If the point (3, 4) lies on the graph of 3y = ax + 7 then the value of a is

(a) $\frac{2}{5}$

(b) $\frac{5}{3}$

(c) $\frac{3}{5}$

(d) $\frac{2}{7}$

Answer 24:

Given equation: $3 y = a x + 7$ .
Also, $(3, 4)$ lies on the graph of the equation.
Putting $x = 3, y = 4$ in the equation, we get:
$3 \times 4 = 3 a + 7$
⇒ $12 = 3 a + 7$
⇒ $3 a = 12 - 7 = 5$
⇒ $a = \frac{5}{3}$
Hence, the correct answer is option (b).