MCQ(1M)

MCQ 11 Mark

$x = 2, y = -1$ is a solution of the linear equation:

✓
$x + 2y = 0$
B
$x + 2y = 4$
C
$2x + y = 0$
D
$2x + y = 5$

Answer

Correct option: A.

$x + 2y = 0$

Substituting $x = 2$ and $y = -1$ in the following equations:
$\text{L.H.S}. = x + 2y = 2 + 2(-1) = 2 - 2 = 0 = \text{R.H.S.}$
$\text{L.H.S}. = x + 2y = 2 + 2(-1) = 2 - 2 = 0 \neq 4 \neq \text{R.H.S}.$
$\text{L.H.S}. = 2x + y = 2(2) + (-1) = 4 - 1 = 3 \neq 0 \neq \text{R.H.S}.$
$\text{L.H.S}. = 2x + y = 2(2) + (-1) = 4 - 1 = 3 \neq 5 \neq \text{R.H.S.}$
Hence, correct option is $(a)$.

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MCQ 21 Mark

If $(a, 4)$ lies on the graph of $3x + y = 10,$ then the value of a is:

A
$3$
B
$1$
✓
$2$
D
$4$

Answer

Correct option: C.

$2$

$3x + y = 10$
If $(a, 4)$ lies on its graph, then it must satisfy the equation.
Thus, we have
$3(a) + 4 = 10$
i.e. $3a = 6$
i.e. $a = 2$
Hence, correct option is $(c).$

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MCQ 31 Mark

If $(4, 19)$ is a solution of the equation $y = ax + 3$, then $a =$

A
$3$
✓
$4$
C
$5$
D
$6$

Answer

Correct option: B.

$4$

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MCQ 41 Mark

If $(2k - 1, k)$ is a solution of the equation $10x - 9y = 12,$ then $k =$

A
$1$
✓
$2$
C
$3$
D
$4$

Answer

Correct option: B.

$2$

If $(2k - 1, k)$ is solution of equation $10x - 9y = 12,$ then it must satisfy this equation.
Thus, we have
$10(2k - 1) - 9k = 12$
$20k - 10 - 9k = 12$
$11k = 22$
$k = 2$
Hence, correct option is $(b).$

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MCQ 51 Mark

The distance between the graph of the equations $x = -3$ and $x = 2$ is:

A
$1$
B
$2$
C
$3$
✓
$5$

Answer

Correct option: D.

$5$

The distance between the graph of the equations $x = -3$ and $x = 2$
$= 2 - (-3)$
$= 2 + 3$
$= 5$
Hence, correct option is $(d)$.

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MCQ 61 Mark

lf the graph of the equation $4x + 3y = 12$ cuts the coordinate axes at $A$ and $B,$ then hypotenuse of right triangle $\text{AOB}$ is of length:

A
$4$ units.
B
$3$ units.
✓
$5$ units.
D
None of these.

Answer

Correct option: C.

$5$ units.

$4x + 3y = 12$
At $x = 0, 3y = 12 $
$\Rightarrow y = 4$ units
At $y = 0, 4x = 12 $
$\Rightarrow x = 3$ units
The triangle formed is $\triangle\text{AOB},$ where
$OB = 4 $ units
$OA = 3$ units
Hypotenuse $=\text{AB}=\sqrt{\text{OB}^2+\text{OA}^2}$
$=\sqrt{16+9}=5\text{ units}$
Hence, correct option is $(c)$.

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MCQ 71 Mark

The graph of the linear equation $2x - y = 4$ cuts $x-$ axis at:

✓
$(2, 0)$
B
$(-2, 0)$
C
$(0, -4)$
D
$(0, 4)$

Answer

Correct option: A.

$(2, 0)$

On $x-$ axis, the $y-$ co $-$ ordinate is always $0$.
So $,2x - y = 4$ will cut the $x-$ axis where $y = 0$
i.e. $2x = 4$
i.e. $x = 2$
Thus $, 2x - y = 4 $ will cut the $x-$ axis at $(2, 0).$
Hence, correct option is $(a).$

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MCQ 81 Mark

The distance between the graphs of the equations $y = -1$ and $y = 3$ is:

A
$2$
✓
$4$
C
$3$
D
$1$

Answer

Correct option: B.

$4$

The distance between given two graphs
$= 3 - (-1)$
$= 3 + 1$
$= 4$
Hence, correct option is $(b)$.

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MCQ 91 Mark

The equation $x - 2 = 0$ on number line is represented by :

A
A line.
✓
A point.
C
Infinitely many lines.
D
Two lines.

Answer

Correct option: B.

A point.

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

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MCQ 101 Mark

How many linear equations are satisfied by $x = 2$ and $y = -3$ ?

A
Only one.
B
Two.
C
Three.
✓
Infinitely many.

Answer

Correct option: D.

Infinitely many.

From Point $(2, -3)$ there are infinitely many lines passing in every $-$ direction.
So $(2, -3)$ is satisfied with infinite linear equations.
Hence, correct option is $(d).$

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