M.C.Q (1 Marks)

MCQ 11 Mark

The median class for the data given below is:

Class	20- 40	40 - 60	60 - 80	80 - 100	100 - 120
Frequency	10	12	14	13	17

A
80 - 100
✓
60 - 80
C
20 - 40
D
40 - 60

Answer

Correct option: B.

60 - 80

(B) 60 - 80
Explanation: Total frequencies (N) = 10 + 12 + 14 + 13 + 17= 66
So, $\frac{N}{2}=\frac{66}{2}=33$
c.f. Just greater than 33 is 36 and the corresponding class is 60 - 80
hence, median class = 60 - 80

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MCQ 21 Mark

A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to $1\ cm$ and the height of the cone is equal to its radius. The volume of the solid is

✓
$\pi \ cm ^3$
B
$4 \pi \ cm^3$
C
$2 \pi \ cm^3$
D
$3 \pi \ cm^3$

Answer

Correct option: A.

$\pi \ cm ^3$

Radii of cone $= r =1 \ cm$
Radius of hemisphere $= r =1 \ cm( h )=1 \ cm$
Height of cone $(h)=1 h=1 \ cm$
Volume of solid $=$ Volume of cone $+$ Volume of a hemisphere
$=\frac{1}{3} \pi r^2 h+\frac{2}{3} \pi r^3$
$=\frac{1}{3} \pi r^2(h+2 r)$
$=\frac{1}{3} \times \pi \times(1)^2(1+2 \times 1)$
$=\frac{1}{3} \times \pi \times 3$
$=\pi \ cm ^3$

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MCQ 31 Mark

In a data, if $l =60, h=15, f _1=16, f _0=6, f _2=6$, then the mode is

✓
$67.5$
B
$72$
C
$60$
D
$62$

Answer

Correct option: A.

$67.5$

$=l+\left(\frac{f_1-f_0}{2 f_1-f_0-f_2}\right) \times h$
$=60+\frac{16-6}{2 \times 16-6-6} \times 15$
$=60+\frac{10}{32-12} \times 15$
$=60+\frac{10}{20} \times 15$
$=60+7.5$
$=67.5$

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MCQ 41 Mark

One card is drawn at random from a well-shuffled deck of 52 cards. What is the probability of getting a black face card?

✓
$\frac{3}{13}$
B
$\frac{3}{14}$
C
$\frac{3}{26}$
D
$\frac{1}{26}$

Answer

Correct option: A.

$\frac{3}{13}$

(A) $\frac{3}{26}$
Explanation: Total number of cards = 52.
Number of black face cards = 6
( 2 kings +2 queens +2 jacks).
$\therefore P ($ getting a face card $)=\frac{6}{52}=\frac{3}{26}$

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MCQ 51 Mark

Find the area of the sector if the radius is $5 \ cm$ and with an angle of $50^{\circ}$.

✓
$10.90 \ cm$
B
$12.90 \ cm$
C
$13.90 \ cm$
D
$11.90 \ cm$

Answer

Correct option: A.

$10.90 \ cm$

The area of the sector $=\frac{x^{\circ}}{360^{\circ}} \times \pi r^2$
$=\frac{50^{\circ}}{360^{\circ}} \times \frac{22}{7} \times 5^2$
$=10.90 \ cm$

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MCQ 61 Mark

The area of a quadrant of a circle whose circumference is $616 \ cm$ will be

✓
$7546 \ cm^2$
B
$7500 \ cm^2$
C
$7564 \ cm^2$
D
$7456 \ cm^2$

Answer

Correct option: A.

$7546 \ cm^2$

$2 \pi R =616$
$ R =\frac{(616 \times 7)}{(2 \times 22)}$
$R =98 \ cm$
Area of quadrant $=\frac{\pi r^2}{4}$
$=\frac{(22 \times 98 \times 98)}{(7 \times 4)}$
$=7546 \ cm^2$

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MCQ 71 Mark

If $\cos \theta=\frac{2}{3}$, then $2 \sec ^2 \theta+2 \tan ^2 \theta-7$ is equal to

A
$1$
B
$4$
C
$0$
✓
$3$

Answer

Correct option: D.

$3$

Given,
$\cos \theta=\frac{2}{3}=\frac{b}{h}= k$
$2 \sec ^2 \theta+2 \tan ^2 \theta-7$
$b=2 k , h =3 k $

In $\triangle \text{ABC}$
$h ^2= p ^2+ b ^2$
$\Rightarrow(3 k )^2= p ^2+(2 k )^2$
$\Rightarrow 9 k ^2= p ^2+4 k ^2$
$\Rightarrow p ^2=9 k ^2-4 k ^2$
$\Rightarrow p ^2=5 k ^2$
$\Rightarrow p =\sqrt{ } 5 k $
Then,
$\sec \theta=\frac{3 k}{2 k}=\frac{3}{2}$ and $\tan \theta=\frac{\sqrt{5} k}{2 k}=\frac{\sqrt{5}}{2}$
$\Rightarrow 2 \sec ^2 \theta+2 \tan ^2 \theta-7$
$\Rightarrow 2\left(\frac{3}{2}\right)^2+2\left(\frac{\sqrt{5}}{2}\right)^2-7$
$\Rightarrow 2 \times \frac{9}{4}+2 \times \frac{5}{4}-7$
$\Rightarrow \frac{9}{2}+\frac{5}{2}-7$
$\Rightarrow \frac{9+5-14}{2}=0$

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MCQ 81 Mark

There is a small island in the middle of a 50 m wide river. A tall tree stands on the island. P and Q are points directly opposite to each other on the two banks, and in line with the tree. If the angles of elevation of the top of the tree from P and Q are respectively $60^{\circ}$ and $30^{\circ}$, then find the height of the tree.

A
22.65 m
B
23.56 m
C
24.69 m
D
21.65 m

Answer

(D) 21.65 m
Explanation: Let the height of the tree be h.
In $\triangle PAT , \tan 60^{\circ}=\frac{h}{x} \Rightarrow \sqrt{3}=\frac{h}{x} \Rightarrow h=\sqrt{3} x$
In $\triangle QAT , \tan 30^{\circ}=\frac{h}{50-x} \Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{50-x}$

$\Rightarrow \sqrt{3} h=50-\frac{h}{\sqrt{3}} \Rightarrow h=\frac{50 \sqrt{3}}{4}=21.65 m\left[\because x =\frac{h}{\sqrt{3}}\right]$
$\Rightarrow$ The height of the tree is 21.65 m

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MCQ 91 Mark

If $\sqrt{3} \tan 2 \theta-3=0$ then $\theta=?$

✓
$30^{\circ}$
B
$60^{\circ}$
C
$15^{\circ}$
D
$45^{\circ}$

Answer

Correct option: A.

$30^{\circ}$

$\sqrt{3} \tan 2 \theta-3=0$
$\Rightarrow \sqrt{3} \tan 2 \theta=3$
$\Rightarrow \tan 2 \theta=\frac{3}{\sqrt{3}}$
$\Rightarrow \tan 2 \theta=\sqrt{3}$
$\Rightarrow \tan 2 \theta=\tan 60^{\circ}$
$\Rightarrow 2 \theta=60^{\circ}$
$\Rightarrow \theta=30^{\circ}$

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MCQ 101 Mark

A tangent $PQ$ at point of contact $P$ to a circle of radius $12 \ cm$ meets the line through centre $O$ to a point $Q$ such that $OQ = 20 \ cm,$ length of tangent $PQ$ is:

A
$15 \ cm$
B
$12 \ cm$
C
$13 \ cm$
✓
$16 \ cm$

Answer

Correct option: D.

$16 \ cm$

Since op is perpendicular to $PQ,$ the $\angle \text{OPQ} =90^{\circ}$
Now, in right angled triangle $\text{OPQ},$
$ OQ ^2= OP ^2+ PQ ^2$
$\Rightarrow(20)^2=(12)^2+ PQ ^2$
$\Rightarrow PQ ^2=400-144$
$\Rightarrow PQ ^2=256$
$\Rightarrow PQ =16 \ cm$

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MCQ 111 Mark

In the given figure, DE || BC and all measurements are given in centimetres. The length of AE is:

A
2.75 cm
B
2.5 cm
C
2 cm
✓
2.25 cm

Answer

Correct option: D.

2.25 cm

(D) 2.25 cm
Explanation: By BPT
$\frac{A D}{D B}=\frac{A E}{E C}$
$\frac{3}{4}=\frac{A E}{3}$
$AE =\frac{9}{4}$
AE = 2.25 cm

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MCQ 121 Mark

$\triangle ABC$ is such that AB = 3 cm, BC = 2 cm and CA = 2.5 cm. If $\triangle D E F \sim \triangle A B C$ and EF = 4 cm, then perimeter of $\triangle DEF$ is

A
30 cm
B
15 cm
C
22.5 cm
D
7.5 cm

Answer

(B) 15 cm
Explanation:

$\triangle DEF \sim \triangle ABC$
AB = 3CM, BC = 2 CM, CA = 2.5 CM, EF = 4CM
Since $\triangle^{\prime}$ s are similar, we have
$\frac{ DE }{ AB }=\frac{ EF }{ BC }=\frac{ FD }{ CA }$
$\Rightarrow \frac{ DE }{3}=\frac{4}{2}=\frac{ FD }{2.5}$
Now $\frac{ DE }{3}=\frac{4}{2}$
$\Rightarrow DE =\frac{3 \times 4}{2}=6 cm$
and $FD =\frac{4}{2} \Rightarrow FD =\frac{4 \times 2.5}{2}=5 cm$
perimeter of $\triangle D E F$
= 6 + 4 + 5 = 15cm

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MCQ 131 Mark

If (a, 0), (0, b) and (x, y) are collinear, then

A
ay - bx = 1
B
ax + by = 1
C
ay + bx = ab
✓
ax - by = ab

Answer

Correct option: D.

ax - by = ab

(D) ay + bx = ab
Explanation: If given points are collinear, then the area of the triangle formed by these three points is 0.
$\therefore$ Area $=\frac{1}{2}|a(b-y)+0(y-0)+x(0-b)|=0$
$\Rightarrow \frac{1}{2}|a b-a y-b x|=0$
$\Rightarrow a b-a y-b x=0$
$\Rightarrow a y+b x=a b$

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MCQ 141 Mark

A quadratic equation whose one root is 3 is

A
$x^2-5 x+6=0$
B
$x^2-6 x-6=0$
C
$x^2-5 x-6=0$
✓
$x^2-5 x+6=0$

Answer

Correct option: D.

$x^2-5 x+6=0$

(D) $x^2-5 x+6=0$
Explanation: since 3 is the root of the equation, x = 3 must satisfy the equation.
Applying x = 3 in the equation $x^2-5 x+6=0$
gives, $(3)^2-5(3)+6=0$
$\Rightarrow 9-15+6=0$
$\Rightarrow 15-15=0$
$\Rightarrow 0=0$
$\Rightarrow$ L.H.S. = R.H.S.
Hence, $x^2-5 x+6=0$ is a required equation which has 3 as root.

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MCQ 151 Mark

The value of k for which the system of linear equations x + 2y = 3, 5x + ky + 7 = 0 is inconsistent is:

✓
$-\frac{14}{3}$
B
5
C
$\frac{2}{5}$
D
10

Answer

Correct option: A.

$-\frac{14}{3}$

(A) 10
Explanation:For a system of equations $a_1 x+b_1 y+c_1=0 ; a_2 x+b_2 y+c_2=0$ to have no solution, the condition to be satisfied is
$\frac{a_1}{a_2}=\frac{b_1}{b_2} \neq \frac{c_1}{c_2}$
$\Rightarrow \frac{1}{5}=\frac{2}{k} \neq \frac{-3}{7}$
$\therefore$ For k = 10, the given system of equation is inconsistent.

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MCQ 161 Mark

If the equation $9 x^2+6 k x+4=0$ has equal roots then k = ?

A
-2 or 0
B
0 only
C
2 or 0
✓
2 or -2

Answer

Correct option: D.

2 or -2

(D) 2 or -2
Explanation: Since the roots are equal, we have D = 0.
$\therefore 36 k ^2-4 \times 9 \times 4=0 \Rightarrow 36 k ^2=144 \Rightarrow k ^2=4 \Rightarrow k =2$ or -2

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MCQ 171 Mark

120 can be expressed as a product of its prime factors as

A
$15 \times 2^3$
✓
$5 \times 2^3 \times 3$
C
$5 \times 8 \times 3$
D
$10 \times 2^2 \times 3$

Answer

Correct option: B.

$5 \times 2^3 \times 3$

(B) $5 \times 2^3 \times 3$
Explanation:We have,
$120=5 \times 2^3 \times 3$

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MCQ 181 Mark

If a is rational and $\sqrt{b}$ is irrational, then $a+\sqrt{b}$ is:

✓
an irrational number
B
an integer
C
a natural number
D
a rational number

Answer

Correct option: A.

an irrational number

(A) an irrational number
Explanation: Let a be rational and $\sqrt{b}$ is irrational.
If possible let $a+\sqrt{b}$ be rational.
Then $a+\sqrt{b}$ is rational and a is rational.
$\Rightarrow[(a+\sqrt{b})-a]$ is rational [Difference of two rationals is rational]
$\Rightarrow \sqrt{b}$ is rational.
The contradiction arises by assuming that $a+\sqrt{b}$ is rational.
Therefore, $a+\sqrt{b}$ is irrational.

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Maths M.C.Q (1 Marks)

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