M.C.Q (1 Marks)

MCQ 11 Mark

In the formula $\bar{x}=a+\frac{\sum f_i d_i}{\sum f_i}$ for finding the mean of grouped data $d_i^{\prime} s$ are deviations from $a$ of

A
upper limits of the classes
B
lower limits of the classes
✓
mid points of the classes
D
frequencies of the class marks

Answer

Correct option: C.

mid points of the classes

(c) mid points of the classes
Explanation : We know that, $d _{ i }= x _{ i }- a$
Where,
$x _{ i }$ are data or class mark and "a" is the assumed mean
i.e. $d_i$ are the deviations of observations from assumed mean.

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MCQ 21 Mark

Cards marked with numbers 1, 2, 3, ..., 25 are placed in a box and mixed thoroughly and one card is drawn at random from the box. The probability that the number on the card is a multiple of 3 or 5 is

A
$\frac{8}{25}$
✓
$\frac{12}{25}$
C
$\frac{4}{25}$
D
$\frac{1}{5}$

Answer

Correct option: B.

$\frac{12}{25}$

(b) $\frac{12}{25}$
Explanation : Number of multiples of 3 = 8 ( 3 6 9 12 15 18 21 24)
Number of multiples of 5 = 5 ( 5 10 15 20 25)
Number of possible outcomes (multiples of 3 or 5)
= 12 ( 3,5,6,9,10,12,15,18,20,21,24,25 )
Number of Total outcomes = 25
$\therefore$ Required Probability $=\frac{12}{25}$

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MCQ 31 Mark

What is the probability that a leap year has 52 Mondays?

✓
$\frac{5}{7}$
B
$\frac{6}{7}$
C
$\frac{2}{7}$
D
$\frac{4}{7}$

Answer

Correct option: A.

$\frac{5}{7}$

(a) $\frac{5}{7}$
Explanation : No. of days in a leap year $=366$
No. of Mondays $=52$
Extra days $=366-52 \times 7$
$=366-364=2$
$\therefore$ Remaining days in the week $=7-2=5$
$\therefore$ Probability of being 52 Mondays in the leap year $=\frac{5}{7}$

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MCQ 41 Mark

A chord of a circle of radius $10 \ cm$ subtends a right angle at the centre. The area of the minor segments $($given, $\pi$
$=3.14)$ is

A
$32.5 \ cm^2$
B
$34.5 \ cm^2$
C
$30.5 \ cm^2$
✓
$28.5 \ cm^2$

Answer

Correct option: D.

$28.5 \ cm^2$

$\operatorname{ar}($ minor segment A C B A $)=\operatorname{ar}($ sector O A C B O) $-\operatorname{ar}(\Delta O A B)$
$=\left(\frac{\pi r^2 \theta}{360}-\frac{1}{2} \times r \times r\right)$

$=\left(\frac{3.14 \times 10 \times 10 \times 90}{360}-\frac{1}{2} \times 10 \times 10\right) \ cm ^2$
$=(78.5-50) \ cm ^2=28.5 \ cm^2$

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MCQ 51 Mark

In the figure, $\text{ABDCA}$ represents a quadrant of a circle of radius $7 \ cm$ a with centre A. Find the area of the shaded portion.

A
$14 \ cm^2$
✓
$31.5 \ cm^2$
C
$24.5 \ cm^2$
D
$38.5 \ cm^2$

Answer

Correct option: B.

$31.5 \ cm^2$

$\text { Explanation: Area of quadrant }=\frac{1}{4} \pi r^2$
$=\frac{1}{4} \times \frac{22}{7} \times(7)^2=\frac{77}{2} \ cm^2=38.5 \ cm^2$
Area of $\triangle BAE =\frac{1}{2} \times$ base $\times$ height
$=\frac{1}{2} \times AB \times AE=\frac{1}{2} \times 7 \times 2=7 \ cm^2$
Hence, area of the shaded portion $=$ Area of the quadrant $A B D C A ~-$ Area of $\triangle B A E$
$=(38.5-7) \ cm^2=31.5 \ cm^2$

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MCQ 61 Mark

The tops of two poles of height 16 m and 10 m are connected by a wire. If the wire makes an angle of $30^{\circ}$ with the horizontal, then the length of the wire is

✓
12 m
B
$10 \sqrt{3} m$
C
16 m
D
10 m

Answer

Correct option: A.

12 m

(a) 12 m

Given: Two poles $BC =16 m$ and $AD =10 m$
And $\angle CDE =30^{\circ}$
To find: Length of wire $CD =x$
$\therefore$ In triangle CDE,
$\sin 30^{\circ}=\frac{ CE }{ CD }$
$\Rightarrow \frac{1}{2}=\frac{ BC - BE }{ CD }$
$\Rightarrow \frac{1}{2}=\frac{16-10}{x}$
$\Rightarrow \frac{1}{2}=\frac{6}{x}$
$\Rightarrow x=12 m$
Therefore, the length of the wire is 12 m .

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MCQ 71 Mark

The ratio of $\text{HCF}$ to $\text{LCM}$ of the least composite number and the least prime number is:

A
$1: 1$
B
$2: 1$
✓
$1: 2$
D
$1: 3$

Answer

Correct option: C.

$1: 2$

Least composite number is $4$ and the least prime number is $2 .$
$\operatorname{LCM}(4,2)=4$
$\operatorname{HCF}(4,2)=2$
The ratio of $\text{HCF}$ to $\text{LCM}=2: 4$ or $1: 2$.

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MCQ 81 Mark

If $\sec \theta+\tan \theta= p$, then the value of $\sin \theta$ is

A
$\frac{1-p^2}{p^2+1}$
✓
$\frac{p^2-1}{p^2+1}$
C
$\frac{1+p^2}{p^2-1}$
D
$\frac{p^2+1}{p^2-1}$

Answer

Correct option: B.

$\frac{p^2-1}{p^2+1}$

Given: $\sec \theta+\tan \theta= p$
$\Rightarrow \frac{1}{\cos \theta}+\frac{\sin \theta}{\cos \theta}=p$
$\Rightarrow \frac{1+\sin \theta}{\cos \theta}=p$
Squaring both sides, we get
$\Rightarrow \frac{(1+\sin \theta)^2}{\cos ^2 \theta}=p^2$
$\Rightarrow \frac{(1+\sin \theta)^2}{1-\sin \theta}=p^2$
$\Rightarrow \frac{(1+\sin \theta)^2}{(1+\sin \theta)(1-\sin \theta)}=p^2$
$\Rightarrow \frac{1+\sin \theta}{1-\sin \theta}=p^2$
$\Rightarrow 1+\sin \theta=p^2(1-\sin \theta)$
$\Rightarrow 1+\sin \theta=p^2-p^2 \sin \theta$
$\Rightarrow \sin \theta+p^2 \sin \theta=p^2-1$
$\Rightarrow \sin \theta\left(1+p^2\right)=p^2-1$
$\Rightarrow \sin \theta=\frac{p^2-1}{p^2+1}$

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MCQ 91 Mark

In the given figure, the perimeter of $\triangle \text{ABC}$ is:

A
$15 \ cm$
✓
$30 \ cm$
C
$60 \ cm$
D
$45 \ cm$

Answer

Correct option: B.

$30 \ cm$

$\ce{AQ = AR =}4$
Similarly,
$\ce{PC=CQ}=5$
Similarly,
$\ce{BP=BR}=6$
Perimeter $= \ce{AB + BC + CA}$
Perimeter $= \ce{AR + RB + BP + PC + CQ + QA}$
$=4+6+6+5+5+4$
$=30 \ cm$

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MCQ 101 Mark

In the given figure, $\text{BOA}$ is a diameter of a circle and the tangent at a point $P$ meets $BA$ extended at $T$ . If $\angle PBO =30^{\circ},$ then $\angle PTA$ is equals to:

A
$20^{\circ}$
✓
$30^{\circ}$
C
$40^{\circ}$
D
$60^{\circ}$

Answer

Correct option: B.

$30^{\circ}$

In triangle $\text{POB} , OP = OB \ [$Radii of the same circle$]$
$\Rightarrow \angle BPO=\angle PBO$
$=30^{\circ} \ [$Opposite angles of equal sides are equal$]$
Also $\angle APB =90^{\circ}$
[Angle in semicircle]
$\therefore \angle OPA =90^{\circ}-30^{\circ}=60^{\circ}$
In triangle $POA , OP = OA \ [$Radii of the same circle$]$
$\Rightarrow \angle OPA =\angle OAP =60^{\circ} \ [$Opposite angles of equal sides are equal$]$
Now, $OP \perp TP,$ then $\angle OPT =90^{\circ}$
$\therefore \angle APT=90^{\circ}-60^{\circ}=30^{\circ}$
Also $, \angle BAP+\angle PAT=180^{\circ}$
$\Rightarrow 60^{\circ}+\angle PAT=180^{\circ}$
$\Rightarrow \angle PAT=120^{\circ}$
Now, in triangle $\text{APT}, \angle PTA +\angle APT +\angle PAT =180^{\circ}$
$\Rightarrow \angle PTA+30^{\circ}+120^{\circ}=180^{\circ}$
$\Rightarrow \angle PTA=30^{\circ}$

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MCQ 111 Mark

In the given figure if $\ce{PS\|QR}$ and $\ce{PQ\|SR}$ and $\ce{AT = AQ = 6, AS = 3, TS = 4,}$ then

A
$x = 2, y = 3.$
B
$x = 1, y = 2.$
✓
$x = 3, y = 4.$
D
$x = 4, y = 5.$

Answer

Correct option: C.

$x = 3, y = 4.$

In triangles $\text{APQ}$ and $\text{ATS,}$ $\angle \text{PAQ} =\angle \text{TAS}$
$[$Vertically opposite angles$] \angle \text{PQA} =\angle \text{ATS}[$ Alternate angles$]$
$\therefore \Delta \text{APQ} \sim \Delta \text{AST[AA}$ similarity $]$
$\therefore \frac{AQ}{AT}=\frac{AP}{AS}$
$\Rightarrow \frac{6}{6}=\frac{x}{3}$
$\Rightarrow x=\frac{6 \times 3}{6}=3$
And $\frac{AQ}{AT}=\frac{PQ}{ST}$
$\Rightarrow \frac{6}{6}=\frac{y}{4}$
$\Rightarrow y=\frac{4 \times 6}{6}=4$
Therefore, $x=3, y=4$

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MCQ 121 Mark

The zeroes of the polynomial $p(x)=x^2+4 x+3$ are given by:

A
$-1,3$
B
$1,-3$
C
$1,3$
✓
$-1,-3$

Answer

Correct option: D.

$-1,-3$

Given, $P( x )= x ^2+4 x +3$
$=x^2+3 x+x+3$
$=x(x+3)+1(x+3)$
$=(x+1)(x+3)$
For zeroes of polynominal $(P(x)=0$
$(x+1)(x+3)=0$
$x=-1,-3$

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MCQ 131 Mark

A circle drawn with origin as the centre passes through $\left(\frac{13}{2}, 0\right)$. The point which does not lie in the interior of the circle is

A
$\frac{-3}{4}, 1$
B
$2, \frac{7}{3}$
C
$5, \frac{-1}{2}$
✓
$\left(-6, \frac{5}{2}\right)$

Answer

Correct option: D.

$\left(-6, \frac{5}{2}\right)$

Distance between $(0,0)$ and $\left(-6, \frac{5}{2}\right)$
$d=\sqrt{(-6-0)^2+\left(\frac{5}{2}-0\right)^2}$
$=\sqrt{36+\frac{25}{4}}$
$=\sqrt{\frac{144+25}{4}}$
$=\sqrt{\frac{169}{4}}$
$=\frac{13}{2}$
$=6.5$
So, the point $\left(-6, \frac{5}{2}\right)$ does not lie in the circle.

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MCQ 141 Mark

Find the sum of the progression $: (5 + 13 + 21 + ... + 181)$

✓
$2139$
B
$2337$
C
$2219$
D
$2476$

Answer

Correct option: A.

$2139$

Here, $a =5, d=(13-5)=8$ and $l =181$
Let the number of terms be $n$.
Then, $T _{ n }=181$
$\Rightarrow a+(n-1) d=181$
$\Rightarrow 5+(n-1) \times 8=181$
$\Rightarrow 8 n=184$
$\Rightarrow n=23$
$\therefore$ Required sum $=\frac{n}{2}(a+l)$
$=\frac{23}{2}(5+181)=23 \times 93=2139$
Hence, the required sum is $2139$

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MCQ 151 Mark

Let $b=a+c$. Then the equation $ax ^2+ bx + c =0$ has equal roots if

A
$a = -c$
✓
$a = c$
C
$a = -2c$
D
$a = 2c$

Answer

Correct option: B.

$a = c$

Since, If $a x^2+b x+c=0$ has equal roots, then
$b^2-4 ac=0$
$\Rightarrow(a+c)^2-4 ac=0 \ldots[\text { Given: } b=a+c]$
$\Rightarrow a^2+c^2+2 ac-4 ac=0$
$\Rightarrow a^2+c^2-2 ac=0$
$\Rightarrow(a-c)^2=0$
$\Rightarrow a-c=0$
$\Rightarrow a=c$

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MCQ 161 Mark

A sphere is placed inside a right circular cylinder so as to touch the top, base and lateral surface of the cylinder. If the radius of the sphere is r, then the volume of the cylinder is

✓
$2 \pi r ^3$
B
$8 \pi r^3$
C
$\frac{8}{3} \pi r^3$
D
$4 \pi r^3$

Answer

Correct option: A.

$2 \pi r ^3$

(a) $2 \pi r^3$
Explanation : Volume of a sphere $=(4 / 3) \pi r ^3$
Volume of a cylinder $=\pi r^2 h$
Given, sphere is placed inside a right circular cylinder so as to touch the top, base and lateral surface of the cylinder and the radius of the sphere is $r$.
Thus, height of the cylinder $=$ diameter $=2 r$ and base radius $= r$
Volume of the cylinder $=\pi \times r ^2 \times 2 r =2 \pi r ^3$

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MCQ 171 Mark

If $a x^2+b x+c=0$ has equal roots, then c is equal to

A
$\frac{b^2}{2 a}$
✓
$\frac{b^2}{4 a}$
C
$\frac{-b^2}{4 a}$
D
$-\frac{b^2}{2 a}$

Answer

Correct option: B.

$\frac{b^2}{4 a}$

If $a x^2+b x+c=0$ has equal roots, then
$b^2-4 a c=0$
$\Rightarrow 4 a c=b^2$
$\Rightarrow c=\frac{b^2}{4 a}$

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MCQ 181 Mark

A bag contains 8 red, 2 black and 5 white balls. One ball is drawn at random. What is the probability that the ball drawn is not black?

✓
$\frac{13}{15}$
B
$\frac{1}{3}$
C
$\frac{8}{15}$
D
$\frac{2}{15}$

Answer

Correct option: A.

$\frac{13}{15}$

(a) $\frac{13}{15}$
Explanation: Total number of balls in the bag $=8+2+5=15$.
Number of non-black balls $=8+5=13$.
$\therefore P ($ getting a non-black ball $)=\frac{13}{15}$

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Maths M.C.Q (1 Marks)

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