M.C.Q (1 Marks)

MCQ 11 Mark

The distribution below gives the marks obtained by 80 students on a test:

Marks	Less than 10	Less than 20	Less than 30	Less than 40	Less than 50	Less than 60
Number of Students	3	12	27	57	75	80

The modal class of this distribution is:

✓
30 - 40
B
20 - 30
C
50 - 60
D
$10 - 20$

Answer

Correct option: A.

30 - 40

(a) 30-40
Explanation: 30 - 40

View full question & answer→

MCQ 21 Mark

3 rotten eggs are mixed with 12 good ones. One egg is chosen at random. The probability of choosing a rotten egg is

A
$\frac{1}{15}$
B
$\frac{4}{5}$
✓
$\frac{1}{5}$
D
$\frac{2}{5}$

Answer

Correct option: C.

$\frac{1}{5}$

(c) $\frac{1}{5}$
Explanation: Number of possible outcomes $=3$
Number of Total outcomes $=15$
$\therefore$ Required Probability $=\frac{3}{15}=\frac{1}{5}$

View full question & answer→

MCQ 31 Mark

A die is thrown once. The probability of getting an even number is

A
$\frac{1}{3}$
B
$\frac{5}{6}$
C
$\frac{1}{6}$
✓
$\frac{1}{2}$

Answer

Correct option: D.

$\frac{1}{2}$

(d) $\frac{1}{2}$
Explanation: Number of all possible outcomes $=6$.
Even numbers are 2,4, 6. Their number is 3 .
$\therefore P$ (getting an even number) $=\frac{3}{6}=\frac{1}{2}$

View full question & answer→

MCQ 41 Mark

A pendulum swings through an angle of $30^{\circ}$ and describes an arc $8.8 \ cm$ in length. Find the length of the pendulum.

A
$8.8 \ cm$
B
$17 \ cm$
C
$15.8 \ cm$
✓
$16.8 \ cm$

Answer

Correct option: D.

$16.8 \ cm$

Length of the pendulum $=$ Radius of a sector of the circle
Arc length $=8.8$
$\frac{\theta}{360}(2 \pi r)=8.8$
$\frac{30}{360} \times 2 \times \frac{22}{7} \times r=8.8$
$r=16.8 \ cm$

View full question & answer→

MCQ 51 Mark

In a circle of radius $14 \ cm$ , an arc subtends an angle of $120^{\circ}$ at the centre. If $\sqrt{3}=1.73$ then the area of the segment of the circle is

A
$124.63 \ cm^2$
B
$130.57 \ cm^2$
✓
$120.56 \ cm^2$
D
$118.24 \ cm^2$

Answer

Correct option: C.

$120.56 \ cm^2$

$ar($ segment $)=\left(\frac{\pi r^2 \theta}{360}-r^2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}\right)$
$=\left(\frac{22}{7} \times 14 \times 14 \times \frac{120}{360}\right)-\left(14 \times 14 \times \sin 60^{\circ} \cos 60^{\circ}\right)$
$=\left(\frac{616}{3}-\frac{\sqrt{3}}{2} \times \frac{1}{2} \times 14 \times 14\right) \ cm^2$
$=(205.33-49 \times 1.73) \ cm^2$
$=(205.33-84.77) \ cm^2$
$=120.56 \ cm^2$

View full question & answer→

MCQ 61 Mark

A ladder makes an angle of $60^{\circ}$ with the ground when placed against a wall. If the foot of the ladder is 2 m away from the wall, then the length of the ladder (in metres) is

A
$2 \sqrt{2}$
B
$4 \sqrt{3}$
C
4
D
$\frac{4}{\sqrt{3}}$

Answer

In right $\triangle AOB , \sec 60^{\circ}=\frac{x}{2}$
$\Rightarrow 2=\frac{x}{2} \Rightarrow=4 m$

View full question & answer→

MCQ 71 Mark

If $\cos \theta=\frac{4}{5}$ then $\tan \theta=?$

✓
$\frac{3}{4}$
B
$\frac{5}{3}$
C
$\frac{4}{3}$
D
$\frac{3}{5}$

Answer

Correct option: A.

$\frac{3}{4}$

$\cos \theta=\frac{4}{5}=\frac{A B}{A C}$
$\therefore B C^2=A C^2-A B^2$
$=25-16$
$=9$
$\Rightarrow B C=3$
$\therefore \tan \theta=\frac{B C}{A B}=\frac{3}{4}$

View full question & answer→

MCQ 81 Mark

$1+\frac{\cot ^2 \alpha}{1+\operatorname{cosec} \alpha}=$

A
$\sin \alpha$
B
$\sec \alpha$
✓
$\operatorname{cosec} \alpha$
D
$\tan \alpha$

Answer

Correct option: C.

$\operatorname{cosec} \alpha$

$=1+\frac{\cot ^2 \alpha}{1+\operatorname{cosec} \alpha}$
$=1+\frac{\operatorname{cosec}^2 \alpha-1}{1+\operatorname{cosec\alpha }}$
$=1+\frac{(\operatorname{cosec} \alpha-1)(\operatorname{cosec} \alpha+1)}{1+\operatorname{cosec} \alpha}$
$=1+\operatorname{cosec} \alpha-1$
$=\operatorname{cosec} \alpha$

View full question & answer→

MCQ 91 Mark

A circle inscribed in $\triangle \text{ABC}$ having $AB =10 \ cm, BC =12 \ cm, CA =28 \ cm$ touching sides at $\text{D, E, F} ($respectively$)$. Then $\text{AD + BE + CF} = ........$

A
$22 \ cm$
✓
$25 \ cm$
C
$18 \ cm$
D
$20 \ cm$

Answer

Correct option: B.

$25 \ cm$

$x+y=10 \ cm \ldots \text { (i) }$
$y+z=12 \ cm \ldots \text { (ii) }$
$x+z=28 \ cm \ldots \text { (iii) }$
Adding $(i), (ii)$ and $(iii),$ we get
$2(x+y+z)=50$
$\Rightarrow x+y+z=25 \ cm$

View full question & answer→

MCQ 101 Mark

In the given figure, PQ is tangent to the circle centred at O . If $\angle AOB =95^{\circ}$, then the measure of $\angle ABQ$ will be

A
$85^{\circ}$
✓
$47.5^{\circ}$
C
$95^{\circ}$
D
$42.5^{\circ}$

Answer

Correct option: B.

$47.5^{\circ}$

(b) $47.5^{\circ}$
Explanation: $47.5^{\circ}$

View full question & answer→

MCQ 111 Mark

In the figures find the measures of $\angle P$ and $\angle R$

✓
$20^{\circ}, 30^{\circ}$
B
$50^{\circ}, 40^{\circ}$.
C
$30^{\circ}, 20^{\circ}$.
D
$40^{\circ}, 50^{\circ}$.

Answer

Correct option: A.

$20^{\circ}, 30^{\circ}$

Explanation :
In triangle $\text{ABC} , \angle A +\angle B +\angle C =180^{\circ}$
$\Rightarrow \angle A+30^{\circ}+20^{\circ}=180^{\circ}$
$\Rightarrow \angle A=130^{\circ}$
In triangle $\text{ABC}$ and $\text{QRP}, \frac{A B}{Q R}=\frac{A C}{P Q}$
$\Rightarrow \frac{45}{5}=\frac{63}{7} \Rightarrow \frac{9}{1}=\frac{9}{1}$
Since sides of triangles $\text{ABC}$ and $\text{QRP}$ are proportional, and included angles are equal,
therefore by $\text{SAS}$ similarity criteria,
$\triangle ABC \sim \Delta QRP$
$\angle A=\angle Q, \angle B=\angle R, \angle C=\angle P$
$\Rightarrow \angle P=20^{\circ}, \angle R=30^{\circ}$

View full question & answer→

MCQ 121 Mark

In what ratio, does x-axis divide the line segment joining the points A(3, 6) and B(-12, -3)?

✓
$ 2 : 1$
B
$1 : 2$
C
$4 : 1$
D
$1 : 4$

Answer

Correct option: A.

$ 2 : 1$

(a) $2: 1$
Explanation: $2: 1$

View full question & answer→

MCQ 131 Mark

$\text{ABCD}$ is a rectangle whose three vertices are $B (4,0), C (4,3)$ and $D (0,3).$ The length of one of its diagonals is

✓
$5$
B
$3$
C
$4$
D
$25$

Answer

Correct option: A.

$5$

Three vertices of a rectangle $\text{ABCD}$ are $B (4,0), C (4,3)$ and $D (0,3)$ length of one of its diagonals
$BD=\sqrt{(4-0)^2+(0-3)^2}$
$=\sqrt{4^2+3^2}$
$=\sqrt{16+9}$
$=\sqrt{25}$
$=5$

View full question & answer→

MCQ 141 Mark

What is the common difference of an $AP$ in which $a _{18}- a _{14}=32$ ?

A
$-8$
B
$4$
C
$-4$
✓
$8$

Answer

Correct option: D.

$8$

$a _{18}- a _{14}=32$
$(18-14) d =32$
$\Rightarrow 4 d=32$
$\Rightarrow d=8$

View full question & answer→

MCQ 151 Mark

In the Maths Olympiad of 2020 at Animal Planet, two representatives from the donkey’s side, while solving a
quadratic equation, committed the following mistakes.
i. One of them made a mistake in the constant term and got the roots as 5 and 9.
ii. Another one committed an error in the coefficient of x and he got the roots as 12 and 4.
But in the meantime, they realised that they are wrong and they managed to get it right jointly. Find the quadratic equation.

A
$2 x^2+7 x-24=0$
B
$x^2+4 x+14=0$
C
$3 x^2-17 x+52=0$
✓
$x^2-14 x+48=0$

Answer

Correct option: D.

$x^2-14 x+48=0$

(d) $x^2-14 x+48=0$
Explanation: For $1^{\text {st }}$ one,
Let the equation be $x ^2+ ax + b =0$
Since roots are 5 and 9
$\therefore a=-14 \text { and } b=45$
For $2^{\text {nd }}$ one,
Let the equation be $x ^2+ px + q =0$
Since roots are 12 and 4.
$\therefore p=-16 \text { and } q=48$
Now, according to the question, b and p both are wrong.
Therefore, the correct equation would be
$x^2-14 x+48=0$

View full question & answer→

MCQ 161 Mark

The pair of equations $x+2 y+5=0$ and $-3 x-6 y+1=0$ have

A
a unique solution
B
infinitely many solutions
✓
no solution
D
exactly two solutions

Answer

Correct option: C.

no solution

Given, equations are
$x+2 y+5=0,$ and
$-3 x-6 y+1=0$
Comparing the equations with general form:
$a_1 x+b_1 y+c_1=0$
$a_2 x+b_2 y+c_2=0$
Here, $a_1=1, b_1=2, c_1=5$
And $a_2=-3, b_2=-6, c_2=1$
Taking the ratio of coefficients to compare
$\frac{a_1}{a_2}=\frac{-1}{3}, \frac{b_1}{b_2}=\frac{-1}{3}, \frac{c_1}{c_2}=\frac{5}{1}$
So $\frac{a_1}{a_2}=\frac{b_1}{b_2} \neq \frac{c_1}{c_2}$
This represents a pair of parallel lines.
Hence, the pair of equations has no solution.

View full question & answer→

MCQ 171 Mark

Figure show the graph of the polynomial $f(x)=a x^2+b x+c$ for which

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

Answer

Correct option: A.

a > 0, b < 0 and c > 0

(a) a $>0$, b $<0$ and c $>0$
Explanation: Clearly, $f(x)=a x^2+b x+c$ represent a parabola opening upwards.
Therefore, $a>0$
The vertex of the parabola is in the fourth quadrant, therefore $b <0$
$y=a x^2+b x+c$ cuts Y axis at P which lies on OY .
Putting $x =0$ in $y=a x^2+b x+c$, we get $y = c$.
So the coordinates of P is $(0, c )$.
Clearly, P lies on OY.$\Rightarrow c >0$
Hence, $a>0, b<0$ and $c>0$

View full question & answer→

MCQ 181 Mark

(HCF × LCM) for the numbers 30 and 70 is:

A
21
B
70
✓
2100
D
210

Answer

Correct option: C.

2100

(c) 2100
Explanation: As we know HCF $\times$ LCM $=$ Product of the Numbers
Hence HCF $\times \operatorname{LCM}(30,70)=30 \times 70=2100$

View full question & answer→

Maths M.C.Q (1 Marks)

Test yourself on this topic