M.C.Q

MCQ 11 Mark

The curved surface area of a cylinder and a cone is equal. If their base radius is same, then the ratio of the slant height of the cone to the height of the cylinder is

A
$1 : 1$
B
$2 : 3$
C
$1 : 2$
✓
$2 : 1$

Answer

Correct option: D.

$2 : 1$

$\text{CSA}$ of cone $= \text{CSA}$ of cylinder
$\pi rl =2 \pi rh$
$l =2 h$
$l : h =2: 1$

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MCQ 21 Mark

In a histogram, which of the following is proportional to the frequency of the corresponding class?

A
Width of the rectangle
B
Length of the rectangle
C
Perimeter of the rectangle
D
Area of the rectangle

Answer

(b) Length of the rectangle
Explanation: In, Histogram each rectangle is drawn, where width equivalent to class interval and height equivalent to the frequency of the class.

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MCQ 31 Mark

The congruence rule, by which the two triangles in the given figure are congruent is$............$

A
$\text{ASA}$
✓
$\text{SAS}$
C
$\text{SSS}$
D
$\text{RHS}$

Answer

Correct option: B.

$\text{SAS}$

In $\triangle \text{PQR}$ and $\triangle \text{PQS}$
$\text{PR = PS} = 8 \ cm$
$\angle \text{RPQ} =\angle \text{SPQ} ($Given$)$
$\text{PQ = PQ}($Common$)$
$\therefore \triangle \text{PQR} \cong \triangle \text{PQS} ($ By $\text{SAS}$ congruency $)$

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MCQ 41 Mark

Which of the following point does not lie on the line y = 2x + 3?

A
$(-5,-7)$
B
$(-1,1)$
C
$(3,9)$
D
$(3,7)$

Answer

(d) (3, 7)
Explanation: Let us put x = 3 in the give equation,
Then, y = 2(3) + 3
y = 6 + 3 = 9
So, the point will be (3, 9)
For x = 3, y = 9. But in the given option, y = 7
So, the given point (3, 7) will not lie on the line y = 2x + 3.

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MCQ 51 Mark

The simplest form of $0.5 \overline{7}$ is

A
$\frac{26}{45}$
B
$\frac{57}{99}$
C
$\frac{57}{100}$
D
$\frac{57}{90}$

Answer

(a) $\frac{26}{45}$
Explanation : $0.5 \overline{7}=\frac{57-5}{90}$
$=\frac{52}{90}=\frac{26}{45}$

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MCQ 61 Mark

In the figure, if $\ce{\angle DAB=60^{\circ}, \angle ABD}=50^{\circ}$, then $\ce{\angle ACB}$ is equal to :

A
$80^{\circ}$
B
$60^{\circ}$
C
$50^{\circ}$
✓
$70^{\circ}$

Answer

Correct option: D.

$70^{\circ}$

In, $\ce{\triangle ABD}$
$\angle D=180^{\circ}-\angle A-\angle B$
$=180^{\circ}-110^{\circ}$
$=70^{\circ}$
Since angles made by same chord at any point of circumference are equal.
so, $\ce{\angle ACB=\angle ADB}=70^{\circ}$

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MCQ 71 Mark

Diagonals of a quadrilateral $\text{ABCD}$ bisect each other. If $\angle A=45^{\circ}$, then $\angle B=$

A
$125^{\circ}$
B
$115^{\circ}$
C
$120^{\circ}$
✓
$135^{\circ}$

Answer

Correct option: D.

$135^{\circ}$

Given,

$\text{ABCD}$ is a quadrilateral
$\angle A =45^{\circ}$,
$\because$ diagonals of quadrilateral bisects each other hence $\text{ABCD}$ is a parallelogram,
$\Rightarrow \angle A +\angle B =180^{\circ}$
$\Rightarrow 45^{\circ}+\angle B =180^{\circ}$
$\Rightarrow \angle B =180^{\circ}-45^{\circ}=135^{\circ}$

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MCQ 81 Mark

ABCD is a Rhombus such that $\angle A C B=40^{\circ}$, then $\angle A D B$ is

A
$100^{\circ}$
B
$40^{\circ}$
C
$60^{\circ}$
D
$50^{\circ}$

Answer

(d) $50^{\circ}$
Explanation : In Rhombus, digonals bisect each other right angle. By using angle sum property in any of the four triangles formed by intersection of diagonals, we get $\angle C B D=50$ and $\angle C B D=\angle A D C$ ( alternate angles).
So, $\angle ADC =50$

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MCQ 91 Mark

Express $y$ in terms of $x$ in the equation $5x - 2y = 7.$

✓
$y=\frac{5 x-7}{2}$
B
$y=\frac{7-5 x}{2}$
C
$y=\frac{7 x+5}{2}$
D
$y=\frac{5 x+7}{2}$

Answer

Correct option: A.

$y=\frac{5 x-7}{2}$

$5 x-2 y=7$
$-2 y=7-5 x$
$2 y=5 x-7$
$y=\frac{5 x-7}{2}$

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MCQ 101 Mark

Zero of the zero polynomial is -

A
every real number
B
1
C
not defined
D
$0$

Answer

(a) every real number
Explanation : Zero of the zero polynomial is any real number.
e.g., Let us consider zero polynomial be 0(x - k), where k is a real number.
For determining the zero, put $x - k =0 \Rightarrow x = k$ Hence, zero of the zero polynomial be any real number.

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MCQ 111 Mark

The diagonals $\text{AC}$ and $\text{BD}$ of a rectangle $\text{ABCD}$ intersect each other at $P. $If $\ce{\angle ABD} =50^{\circ}$, then $\ce{\angle DPC =}$

A
$70^{\circ}$
✓
$80^{\circ}$
C
$90^{\circ}$
D
$100^{\circ}$

Answer

Correct option: B.

$80^{\circ}$

Given, $\text{ABCD}$ is a rectangle

Diagonals $\text{ACBD}$ intersect each other at $P$
$\ce{\angle ABD}=50^{\circ}$
$\because$ diagonals of rectangle bisect each other and are equal in length
$\ce{\Rightarrow \angle ABD =\angle PDC}[$ alternate angles $]$
$\ce{\Rightarrow \angle PDC =\angle PCD} =50^{\circ}$
In $\ce{\triangle DPC}$
$\ce{\Rightarrow \angle DPC + \angle PCD + \angle PDC = 180^{\circ}}$
$\ce{\Rightarrow \angle DPC} + 50^{\circ}+50^{\circ}=180^{\circ}$
$\ce{\Rightarrow \angle DPC} =180^{\circ}-100^{\circ}=80^{\circ}$

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MCQ 121 Mark

In the figure $\ce{AB\ \&\ CD}$ are two straight lines intersecting at $\text{O , OP}$ is a ray. What is the measure of $\ce{\angle AOD}$.

A
$128^{\circ}$
B
$40^{\circ}$
✓
$140^{\circ}$
D
$100^{\circ}$

Answer

Correct option: C.

$140^{\circ}$

From the figure it follows that
$(3 x+7)+(x+5)+40=180$
$\Rightarrow 4 x+52=180$
$\Rightarrow 4 x=180-52=128$
$\Rightarrow x=32$
Now,
$\ce{\angle AOD=\angle COP + \angle POB}$
$\Rightarrow \ce{\angle AOD}=(3 x+7)+(x+5)$
$\Rightarrow \ce{\angle AOD}=4 x+12$
$\Rightarrow \ce{\angle AOD}=4 \times 32+12$
$\Rightarrow \ce{\angle AOD}=128+12$
$\Rightarrow \ce{\angle AOD}=140$

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MCQ 131 Mark

Euclid stated that all right angles are equal to each other in the form of

A
A postulate
B
A proof
C
An axiom
D
A definition

Answer

(a) A postulate
Explanation: Eucid's fourth postulate states that all right angles are equal to one another.

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MCQ 141 Mark

The graph of the linear equation 2x + 3y = 6 is a line which meets the x-axis at the point

A
(0,3)
B
(3,0)
C
(2, 0)
D
(0 ,2)

Answer

(b) (3,0)
Explanation : 2x + 3y = 6 meets the X-axis.
Put y = 0,
2x + 3(0) = 6
x = 3
Therefore, graph of the given line meets X-axis at (3, 0).

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MCQ 151 Mark

To draw a histogram to represent the following frequency distribution :

Class interval	5-10	10-15	15-25	25-45	45-75
Frequency	6	12	10	8	15

The adjusted frequency for the class 25-45 is

Answer

(c) 2
Explanation: Adjusted frequency $=\left(\frac{\text { frequency of the class }}{\text { width of the class }}\right) \times 5$
Therefore, Adjusted frequency of $25-45=\frac{8}{20} \times 5=2$

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MCQ 161 Mark

Two points having same abscissa but different ordinates lie on

A
y-axis
B
x-axis
C
a line parallel to y-axis
D
a line parallel to x-axis

Answer

(c) a line parallel to y-axis
Explanation: Two points having same abscissa but different ordinate always make a line which is parallel to the y-axis as abscissa is fixed and the only ordinate keeps changing.

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MCQ 171 Mark

The linear equation 3x - 5y = 15 has

A
no solution
B
infinitely many solutions
C
a unique solution
D
two solutions

Answer

(b) infinitely many solutions
Explanation:
Given linear equation: $3 x-5 y=15 Or , x=\frac{5 y+15}{3}$
When $y=0, x=\frac{15}{3}=5$
When $y=3, x=\frac{30}{3}=10$
When $y=-3, x=\frac{0}{3}=0$

xx	5	10	0
yy	0	3	-3

Plot the points A(5,0) , B(10,3) and 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.

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MCQ 181 Mark

$\pi$ is

A
a rational number
B
an integer
C
an irrational number
D
a whole number

Answer

(c) an irrational number
Explanation: $\pi=3.14159265359\ldots\ldots$ which is non-terminating non-recurring.
Hence, it is an irrational number.

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