M.C.Q (1 Marks)

MCQ 11 Mark

If every term of the statistical data consisting of n terms is decreased by 2, then the mean of the data:

A
decreases by 1
B
remains unchanged
✓
decreases by 2
D
decreases by 2n

Answer

Correct option: C.

decreases by 2

View full question & answer→

MCQ 21 Mark

One card is drawn at random from a well-shuffled deck of 52 cards. What is the probability of getting a 6?

✓
$\frac{1}{13}$
B
$\frac{3}{26}$
C
$\frac{1}{52}$
D
$\frac{4}{52}$

Answer

Correct option: A.

$\frac{1}{13}$

(A) $\frac{1}{13}$
Number of 6 s = 4.
$\therefore P ($ getting a 6$)=\frac{4}{52}=\frac{1}{13}$

View full question & answer→

MCQ 31 Mark

An event is unlikely to happen. Its probability is closest to

✓
1.0E-5
B
0.0001
C
0.1
D
1

Answer

Correct option: A.

1.0E-5

(A) 0.00001
Explanation: An event is unlikely to happen. Its probability is very very close to zero but not zero, So it is equal to 0.00001

View full question & answer→

MCQ 41 Mark

A car has two wipers which do not overlap. Each wiper has a blade of length 42 cm sweeping through an angle of $120^{\circ}$. Find the total area cleaned at each sweep of the blades.

A
$5544 cm^2$
✓
$3696 cm^2$
C
$4224 cm^2$
D
$1848 cm^2$

Answer

Correct option: B.

$3696 cm^2$

(B) $3696 cm^2$
Explanation: Clearly, each wiper sweeps a sector of a circle of radius 42 cm and sector angle $120^{\circ}$
$\therefore$ Total area cleaned at each sweep $=2 \times \frac{\theta}{360^{\circ}} \times \pi r^2$
$=2 \times \frac{120^{\circ}}{360^{\circ}} \times \frac{22}{7} \times 42 \times 42 cm^2=3696 cm^2$

View full question & answer→

MCQ 51 Mark

A horse is grazing in a field. It is tied to a pole with a rope of length $6 m.$ The horse moves from point $A$ to point $B$ making an arch with an angle of $70^\circ .$ Find the area of the sector grazed by the horse.

A
$22.99 m$
B
$20.99 m$
✓
$21.99 m$
D
$21 m$

Answer

Correct option: C.

$21.99 m$

The area of the sector $=\frac{x^{\circ}}{360^{\circ}} \times \pi r^2$
$=\frac{70^{\circ}}{360^{\circ}} \times \frac{22}{7} \times 6^2$
$21.99 m$

View full question & answer→

MCQ 61 Mark

At some time of the day, the height and length of the shadow of a man are equal. The sun's altitude is:

✓
$45^{\circ}$
B
$30^{\circ}$
C
$90^{\circ}$
D
$60^{\circ}$

Answer

Correct option: A.

$45^{\circ}$

(A) $45^{\circ}$
Explanation: $45^{\circ}$

View full question & answer→

MCQ 71 Mark

If $\cos A =\frac{\sqrt{3}}{2}, 0^{\circ}< A <90^{\circ}$, then $A$ is equal to

✓
$30^{\circ}$
B
$1$
C
$\frac{\sqrt{3}}{2}$
D
$60^{\circ}$

Answer

Correct option: A.

$30^{\circ}$

$: \operatorname{Cos} A=\frac{\sqrt{3}}{2}$
$\operatorname{Cos} A =\operatorname{Cos} 30^{\circ}$
$A =30^{\circ}$

View full question & answer→

MCQ 81 Mark

If $2 x=\sec A$ and $\frac{2}{x}=\tan A$ then $2\left(x^2-\frac{1}{x^2}\right)=$ ?

✓
$\frac{1}{2}$
B
$\frac{1}{4}$
C
$\frac{1}{16}$
D
$\frac{1}{8}$

Answer

Correct option: A.

$\frac{1}{2}$

We know that $\sec ^2 A-\tan ^2 A=1$
$\therefore(2 x)^2-\left(\frac{2}{x}\right)^2=1 $
$\Rightarrow 4 x^2-\frac{4}{x^2}=1 $
$\Rightarrow 4\left(x^2-\frac{1}{x^2}\right)=1$
$\Rightarrow\left(x^2-\frac{1}{x^2}\right)=\frac{1}{4} $
$\Rightarrow 2\left(x^2-\frac{1}{x^2}\right)$
$=2 \times \frac{1}{4}$
$=\frac{1}{2}$

View full question & answer→

MCQ 91 Mark

A tangent to a circle is a line that touches the circle at:

A
three points
✓
one point only
C
infinite number of points
D
two points

Answer

Correct option: B.

one point only

(B) one point only
Explanation: one point only

View full question & answer→

MCQ 101 Mark

In the given figure, $O$ is the centre of the circle and $PA$ is a tangent to the circle. If $\angle OAB =60^{\circ}$, then $\angle OPA$ is equal to:

✓
$30^{\circ}$
B
$20^{\circ}$
C
$15^{\circ}$
D
$60^{\circ}$

Answer

Correct option: A.

$30^{\circ}$

Explanation:
$\angle OAB =60^{\circ}\ ($given$)$
$\angle OAB =\angle OBA \ (\because OA = OB = r )$
$\therefore \angle OBA =60^{\circ}$
Now, in $\triangle OAB$
$\angle AOB =180^{\circ}-60^{\circ}-60^{\circ}$
$\angle AOB =60^{\circ}$
Now, In $\triangle AOP$
$\angle OPA +\angle OAP +\angle AOP =180^{\circ}\ ($angle sum property of } \triangle$) $
$\angle OPA +90^{\circ}+60^{\circ}=180^{\circ}$
$\angle OPA =180^{\circ}-150^{\circ}$
$\angle OPA =30^{\circ}$

View full question & answer→

MCQ 111 Mark

In the given figure if $\angle ADE =\angle ABC,$ then $CE$ is equal to

✓
$4.5$
B
$3$
C
$2$
D
$5$

Answer

Correct option: A.

$4.5$

Explanation:
$\angle ADE =\angle ABC$ and $\angle DAE =\angle BAC$.
Hence $\triangle ADE \sim \triangle ABC \ ( AA$ similarity$)$
hence the corresponding sides are in proportion
$\frac{A D}{A B}=\frac{A E}{A c}$
$\Rightarrow \frac{2}{5}=\frac{3}{C E+3}$
$\Rightarrow C E=4.5$

View full question & answer→

MCQ 121 Mark

The coordinates of the point A, where AB is the diameter of the circle whose centre is (3, -2) and B(7, 4) is:

A
(1, 8)
✓
(-1, -8)
C
(-1, 8)
D
(1, -8)

Answer

Correct option: B.

(-1, -8)

(B) (-1, -8)
Explanation: (-1, -8)

View full question & answer→

MCQ 131 Mark

The distance between the points (6, 2) and (-6, 2) is:

✓
12 units
B
$6 \sqrt{2}$ units
C
$2 \sqrt{6}$ units
D
6 units

Answer

Correct option: A.

12 units

(A) 12 units
Explanation: 12 units

View full question & answer→

MCQ 141 Mark

In an $A P$, if $d=-4, n=7$ and $a_n=4$, then the value of $a$ is

A
$20$
B
$6$
C
$7$
✓
$28$

Answer

Correct option: D.

$28$

Given: $d =-4, n =7$ and $a _{ n }=4$
$\therefore a_n=a+(n-1) d$
$\Rightarrow 4=a+(7-1) \times(-4)$
$\Rightarrow 4=a+6 \times-4$
$\Rightarrow 4=a-24$
$\Rightarrow a=28$

View full question & answer→

MCQ 151 Mark

If $p$ is a root of the quadratic equation $x^2-(p+q) x+k=0, $ then the value of $k$ is

A
$p + q$
B
$p$
✓
$pq$
D
$q$

Answer

Correct option: C.

$pq$

Let the roots of given quadratic equation be $\alpha$ and $\beta$.
On comparing equation $x^2-(p-q) x+k=0$
with $a x^2+b x+c=0$,we have
$a=1, b=-(p+q), c=k$
We know that
$\Rightarrow \alpha+\beta=\frac{-b}{a}$
Put the value $a$ and $b$
$\Rightarrow \alpha+\beta=\frac{p+q}{1}$
$\Rightarrow \alpha+\beta= p + q \ldots \text { (i) }$
Given $\alpha= p$
Put the value of $\alpha$ in equation $(i),$
$\Rightarrow p +\beta= p + q$
$\Rightarrow \beta= q $
But we know that
$\alpha \cdot \beta=\frac{c}{a}$
Then, $k = pq.$

View full question & answer→

MCQ 161 Mark

The number of solutions for two linear equations representing parallel lines is/are

A
2
B
$\infty$
C
1
✓
0

Answer

Correct option: D.

(D) 0
Explanation: The number of solutions of two linear equations representing parallel lines is 0 because two linear equations representing parallel lines has no solution and they are inconsistent.

View full question & answer→

MCQ 171 Mark

If the diagram in Fig. shows the graph of the polynomial $f(x)=a x^2+b x+c$, then

A
a < 0, b < 0 and c < 0
B
a < 0, b > 0 and c > 0
✓
a < 0, b < 0 and c > 0
D
a < 0, b > 0 and c < 0

Answer

Correct option: C.

a < 0, b < 0 and c > 0

(C) a < 0, b < 0 and c > 0
Explanation: Clearly, $f(x)=a x^2+b x+c$ represent a parabola opening downwards.
Clearly a $<0$
Let, $y=a x^2+b x+c$ cuts $y$-axis at $P$ which lies on $O Y$.
Putting $x =0$ in $y=a x^2+b x+c$, we get $y = c$. So the coordinates of P are $(0, c)$.
Clearly, P lies on OY. Therefore $c >0$
The vertex $\left(\frac{-b}{2 a}, \frac{-D}{4 a}\right)$ of the parabola is in the second quadrant.
Therefore, $\frac{-b}{2 a}<0, b<0$
Therefore $a <0, b<0$ and $c >0$.

View full question & answer→

MCQ 181 Mark

The least positive integer divisible by 20 and 24 is

A
480
B
240
C
360
✓
120

Answer

Correct option: D.

120

(D) 120
Explanation: Least positive integer divisible by 20 and 24 is
LCM of (20, 24).
$\begin{aligned} 20 & =2^2 \times 5 \\ 24 & =2^3 \times 3\end{aligned}$
$\therefore \operatorname{LCM}(20,24)=2^3 \times 3 \times 5=120$
Thus 120 is divisible by 20 and 24.

View full question & answer→

Maths M.C.Q (1 Marks)

Test yourself on this topic