M.C.Q - Maths STD 9 Questions - Vidyadip

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18 questions · auto-graded multiple-choice test.

MCQ 11 Mark

The number of spherical bullets each $5\ dm$ in diameter which can be cast from a rectangular block of lead $11 m$ long, $10 m$ broad and $5$ high is

✓
$8400$
B
$5600$
C
$6300$
D
$4200$

Answer

Correct option: A.

$8400$

Here, radius of spherical bullets $=2.5\ dm$ or $0.25 m(1 dm =0.1 m)$
Let the number of bullets be $n$
Now, volume of $n$ number of bullets $=$ volume of rectangular block
$ n \times \frac{4}{3} \pi r ^3= l \times b \times h$
$n \times \frac{4}{3} \times \frac{22}{7} \times 0.25 \times 0.25 \times 0.25=11 \times 10 \times 5$
$n =\frac{11 \times 10 \times 5 \times 21}{88 \times 0.25 \times 0.25 \times 0.25}$
$n =8400$

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MCQ 21 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

A
6
B
5
✓
2
D
3

Answer

Correct option: C.

2

(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 31 Mark

Line sgements $AB$ and $CD$ intersect at $O$ such that $AC \| DB$. If $\angle CAB =45^{\circ}$ and $\angle CDB =55^{\circ}$, then $\angle BOD =$

A
$135^{\circ}$
✓
$80^{\circ}$
C
$100^{\circ}$
D
$90^{\circ}$

Answer

Correct option: B.

$80^{\circ}$

$AC\|BD$
And, $AB$ is transverse to these parallel lines
$\text { So } \angle CAB =\angle ABD \text ($Alternate angles$)$
$\Rightarrow \angle ABD =45^{\circ}$
Now In $\triangle BOD$
$\angle BOD +\angle ODB +\angle DBA =180^{\circ}$
$\angle DBA =\angle ABD =45^{\circ}, \angle ODB =55^{\circ}$
$\text { So } \angle B O D=180^{\circ}-45^{\circ}-55^{\circ}$
$=80^{\circ}$

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MCQ 41 Mark

If a linear equation has solutions $(1, 2), (-1, -16)$ and $(0, -7)$, then it is of the form

✓
$y=9 x-7$
B
$9 x-y+7=0$
C
$x-9 y=7$
D
$x=9 y-7$

Answer

Correct option: A.

$y=9 x-7$

(a) $y=9 x-7$
Explanation:Since all the given co- ordinate $(1,2),(-1,-16)$ and $(0,-7)$ satisfy the given line $y=9 x-7$
For point (1, 2)
y = 9x - 7
2 = 9(1) - 7
2 = 9 - 7
2 = 2
Hence (2, 1) is a solution.
For point (-1, -16)
y = 9x - 7
-16 = 9(-1) - 7
-16 = -9 - 7
-16 = -16
Hence (-1, -16) is a solution.
For point (0,-7)
y = 9x - 7
-7 = 9(0) -7
-7 = -7
Hence (0, -7) is a solution.

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MCQ 51 Mark

The simplest rationalising factor of $\sqrt[3]{500}$, is

A
$\sqrt{3}$
✓
$\sqrt[3]{2}$
C
$2 \sqrt{3}$
D
$\sqrt[3]{5}$

Answer

Correct option: B.

$\sqrt[3]{2}$

$\sqrt[3]{500}$
$=\sqrt[3]{5 \times 2 \times 5 \times 2 \times 5}$
$=\sqrt[3]{5 \times 5 \times 5 \times 2 \times 2}$
$=5 \sqrt[3]{2 \times 2}$
$=5 \sqrt[3]{4}$
The simplest rationalising factor of $\sqrt[3]{500}$, is $\sqrt[3]{2}$

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MCQ 61 Mark

In the given figure, $A B$ and $C D$ are two intersecting chords of a circle. If $\angle C A B=40^{\circ}$ and $\angle B C D=80^{\circ}$, then $\angle C B D=?$

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

Answer

Correct option: A.

$60^{\circ}$

(a) $60^{\circ}$
Explanation:We have:
$\angle CDB =\angle CAB =40^{\circ}$ (Angles in the same segment of a circle)
In $\triangle CBD$, we have:
$\angle CDB +\angle BCD +\angle CBD =180^{\circ}$ (Angle sum property of a triangle $)$
$\Rightarrow 40^{\circ}+80^{\circ}+\angle CBD =180^{\circ}$
$\Rightarrow \angle CBD =\left(180^{\circ}-120^{\circ}\right)=60^{\circ}$
$\Rightarrow \angle C B D=60^{\circ}$

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MCQ 71 Mark

If bisector of $\angle A$ and $\angle B$ of a quadrilateral $\text{ABCD}$ intersect each other at $p , \angle B$ and $\angle C$ at $Q , \angle C$ and $\angle D$ at R and, $\angle D$ and $\angle A$ at S then $\text{PQRS}$ is a

✓
Rectangle
B
Parallelogram
C
Rhombus
D
Quadrilateral whose opposite angles are supplementary

Answer

Correct option: A.

Rectangle

Let's assume our quadrilateral $\text{ABCD}$ as a parallelogram :

we know
$\angle DCB +\angle ABC =180^{\circ} \ ($Co $-$ interior angles of parallelogram are supplementary$)$
$\Rightarrow \frac{1}{2} \angle DCB +\frac{1}{2} \angle ABC =90^{\circ} \ ($ Both sides divide by $2 )$
$\Rightarrow \angle 1+\angle 2=90^{\circ}.........(1)$
In, $\triangle CQB$ we know
$\Rightarrow \angle 1+\angle 2+\angle CQB =180^{\circ} \ldots(2)$
From eq $(1)$ and eq $(2),$ We get
$\Rightarrow \angle CQB =180^{\circ}-90^{\circ}$
$\Rightarrow \angle CQB =90^{\circ}$
$\Rightarrow \angle PQR =90^{\circ} (\because \angle CQB =\angle PQR ,$ vertically opposite angles $)$
Similarly, it can be shown
$\angle QPS =\angle PSR =\angle SRQ =90^{\circ}$
So, quadrilateral $\text{PQRS}$ is a rectangle.

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MCQ 81 Mark

$\text{ABCD}$ is a parallelogram in which diagonal $AC$ bisects $\angle BAD$. If $\angle BAC =35^{\circ}$, then $\angle ABC =$

A
$70^{\circ}$
B
$120^{\circ}$
✓
$110^{\circ}$
D
$90^{\circ}$

Answer

Correct option: C.

$110^{\circ}$

Given,
$\text{ABCD}$ is a parallelogram
Diagonal $\text{AC}$ bisects $\angle BAD$

$\angle B A C=35^{\circ}$
$\because \angle A +\angle B =180^{\circ} \ldots (i) [$ angle sum property of quadrilateral$]$
$\angle A =2 \angle BAC =2 \times 35^{\circ}=70^{\circ}$
Putting value of $\angle A$ in equation $(i)$
$70^{\circ}+\angle B=180^{\circ}$
$\angle B=180^{\circ}-70^{\circ}=110^{\circ}$
$\angle A B C=110^{\circ}$

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MCQ 91 Mark

Any solution of the linear equation $2x + 0y + 9 = 0$ in two variables is of the form

✓
$\left(-\frac{9}{2}, m\right)$
B
$(-9,0)$
C
$\left(0,-\frac{9}{2}\right)$
D
$\left(n,-\frac{9}{2}\right)$

Answer

Correct option: A.

$\left(-\frac{9}{2}, m\right)$

(a) $\left(-\frac{9}{2}, m\right)$
Explanation: $2 x +9=0$
$\Rightarrow x=\frac{-9}{2}$ and $y = m$, where m is any real number
Hence, $\left(-\frac{9}{2}, m\right)$ is the solution of the given equation.

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MCQ 101 Mark

$\left(x^2-4 x-21\right)=?$

✓
$(x - 7) (x + 3)$
B
$( x - 7) ( x - 3)$
C
$(x + 7) (x + 3)$
D
$(x + 7) (x - 3)$

Answer

Correct option: A.

$(x - 7) (x + 3)$

$\left(x^2-4 x-21\right)=x^2-7 x+3 x-21$
$=x(x-7)+3(x-7)$
$=(x-7)(x+3)$

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MCQ 111 Mark

In Triangle ABC which is right angled at B. Given that AB = 9cm, AC = 15cm and D, E are the mid-points of the sides AB and AC respectively. Find the length of BC?

A
13cm
B
$13.5cm$
✓
12cm
D
15cm

Answer

Correct option: C.

12cm

(c) 12 cm
Explanation:

Applying pythagoras theorem in $\triangle ABC$
$AC ^2= AB ^2+ BC ^2$
$15^2=9^2+B C^2$
$225=81+ BC ^2$
$225-81= BC ^2$
$B C^2=144$
$BC =12$

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MCQ 121 Mark

In the adjoining figure, $A O B$ is a straight line. If $x: y: z=4: 5: 6$, then $y= ?$

A
$72^{\circ}$
B
$48^{\circ}$
✓
$60^{\circ}$
D
$80^{\circ}$

Answer

Correct option: C.

$60^{\circ}$

Let
$\angle A O C=x^{\circ}=(4 a)^{\circ}, \angle C O D=y^{\circ}=(5 a)^{\circ}$ and $\angle B O D=z^{\circ}=(6 a)^{\circ}$
Then, we have
$\angle A O C+\angle C O D+\angle B O D=180^{\circ} [$Since $AOB$ is a straight line$]$
$\Rightarrow 4 a +5 a +6 a =180^{\circ}$
$\Rightarrow 15 a =180^{\circ}$
$\Rightarrow a =12^{\circ}$
$\therefore y=5 \times a=5 \times 12^{\circ}$
$=60^{\circ}$

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MCQ 131 Mark

The basic facts which are taken for granted, without proof, are called

A
theorems
✓
axioms
C
propositions
D
lemmas

Answer

Correct option: B.

axioms

(b) axioms
Explanation: An axiom is a proposition regarded as self-evidently true without proof

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MCQ 141 Mark

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

A
infinitely many lines
B
two lines
✓
a point
D
a line

Answer

Correct option: C.

a point

(c) a point
Explanation:$x -2=0$
$x =2$ is a point on the number line

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MCQ 151 Mark

A histogram is a pictorial representation of the grouped data in which class intervals and frequency are respectively taken along

A
horizontal axis only
✓
horizontal axis and vertical axis
C
vertical axis and horizontal axis
D
vertical axis only

Answer

Correct option: B.

horizontal axis and vertical axis

(b) horizontal axis and vertical axis
Explanation:In a histogram the class limits are marked on the horizontal axis and the frequency is marked on the vertical axis. Thus, a rectangle is constructed on each class interval.

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MCQ 161 Mark

In which quadrant does the point $(-7, -4)$ lie?

A
IV
B
I
C
II
✓
III

Answer

Correct option: D.

III

(d) III
Explanation: Recall that (+ ,+ ) lies in I quadrant, (- ,+) lies in II quadrant, (-, -) lies in III quadrant, (+,-) lies in IV quadrant. Since, both the coordinates of given point are negative, it lies in Quadrant III.

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MCQ 171 Mark

If the line represented by the equation $3 x+k y=9$ passes through the points $(2,3),$ then the value of $k$ is

A
$2$
✓
$1$
C
$3$
D
$4$

Answer

Correct option: B.

$1$

If the line represented by the equation $3 x+k y=9$ passes through the points $(2,3)$ then $(2,3)$ will satisy the equation $3 x+k y=9$
$3(2)+3 k=9$
$\Rightarrow 6+3 k=9$
$\Rightarrow 3 k=9-6$
$\Rightarrow 3 k=3$
$\Rightarrow k=1$

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MCQ 181 Mark

$(125)^{-1 / 3}=?$

A
$-\frac{1}{5}$
B
$-5$
✓
$\frac{1}{5}$
D
$5$

Answer

Correct option: C.

$\frac{1}{5}$

$(125)^{-1 / 3}$
$=\left(5^3\right)^{-1 / 3}$
$=5^{-1}$
$\frac{1}{5}$

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M.C.Q - Maths STD 9 Questions - Vidyadip