M.C.Q (1 Marks) - Maths STD 10 Questions - Vidyadip

Questions

M.C.Q (1 Marks)

🎯

Test yourself on this topic

18 questions · timed · auto-graded

MCQ 11 Mark

Consider the following table:

Class interval	$10-14$	$14-18$	$18-22$	$22-26$	$26-30$
Frequency	$5$	$11$	$16$	$25$	$19$

The mode of the above data is

A
$25$
B
$23.5$
✓
$24.4$
D
$24$

Answer

Correct option: C.

$24.4$

Maximum frequency $= 25$
Hence, modal class is $22 - 26$
Now, Mode $=x_k+h\left\{\frac{\left(f_k-f_{k-1}\right)}{\left(2 f_k-f_{k-1}-f_{k+1}\right)}\right\}$
$=22+4\left\{\frac{(25-16)}{(2(25)-16-19)}\right\}$
$=22+4 \times \frac{9}{15}$
$= 22 + 2.4$
$= 24.4$

View full question & answer→

MCQ 21 Mark

A cylindrical vessel of radius $4 \ cm$ contains water. A solid sphere of radius $3 \ cm$ is lowered into the water until it is completely immersed. The water level in the vessel will rise by

✓
$\frac{9}{4} \ cm$
B
$\frac{9}{2} \ cm$
C
$\frac{2}{9} \ cm$
D
$\frac{4}{9} \ cm$

Answer

Correct option: A.

$\frac{9}{4} \ cm$

Radius of sphere $\left(r_1\right)=3 \ cm$
$\therefore$ Volume $=\frac{4}{3} \pi r^3$
$=\frac{4}{3} \pi(3)^3 \ cm^3$
$=36 \pi \ cm^3$
$\therefore$ Volume of water in the cylinder $=36 \pi \ cm^3$
Radius of cylindrical vessel $\left(r_2\right)=4 \ cm$
Let $h$ be its height, then
$\pi r_2^2 h=36 \pi$
$\Rightarrow \pi(4)^2 h=36 \pi$
$\Rightarrow 16 \pi h=36 \pi$
$\Rightarrow h=\frac{36 \pi}{16 \pi}=\frac{9}{4} \ cm$

View full question & answer→

MCQ 31 Mark

Consider the following frequency distribution:

Class	0-5	6-11	12-17	18-23	24-29
Frequency	13	10	15	8	11

The upper limit of the median class is

A
18.5
✓
17.5
C
18
D
17

Answer

Correct option: B.

17.5

(B) 17.5
Explanation: Given, classes are not continuous, so we make continuous by subtracting 0.5 from lower limit and adding 0.5 to upper limit of each class.

Class	Frequency	Cumulative frequency
-0.5-5.5	13	13
5.5-11.5	10	23
11.5-17.5	15	38
17.5-23.5	8	46
23.5-29.5	11	57

Here, $\frac{N}{2}=\frac{57}{2}=28.5$, which lies in the interval 11.5 - 17.5.
Hence, the upper limit is 17.5 .

View full question & answer→

MCQ 41 Mark

In a single throw of a die, the probability of getting a multiple of $3$ is

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

Answer

Correct option: D.

$\frac{1}{3}$

A die is thrown, the possible number of events $(n) = 6$
Now multiple of $3$ are $3, 6$ which are $2$
$\therefore m =2$
$\therefore \text { Probability }=\frac{m}{n}=\frac{2}{6}=\frac{1}{3}$

View full question & answer→

MCQ 51 Mark

Pankaj has a motorcycle with wheels of diameter 91 cm. There are 22 spokes in the wheel. Find the length of arc between two adjoining spokes.

A
13 cm
B
26 cm
C
15 cm
D
18 cm

Answer

13 cm
Explanation: Radius of wheel $=\frac{91}{2} cm$
Angle between two adjoining spokes, $\theta=\frac{360^{\circ}}{22}$
$\therefore$ Length of arc $=\frac{\theta}{360^{\circ}} \times 2 \pi r$
$=\frac{360^{\circ}}{360^{\circ} \times 22} \times 2 \times \frac{22}{7} \times \frac{91}{2}=13 cm$

View full question & answer→

MCQ 61 Mark

A circular disc of radius $6 \ cm$ is divided into three sectors with central angles $90^{\circ}, 120^{\circ}$ and $150^{\circ}$. The ratio of the areas of the three sectors is

A
$4 : 5 : 6$
✓
$3 : 4 : 5$
C
$1 : 5 : 6$
D
$2 : 3 : 4$

Answer

Correct option: B.

$3 : 4 : 5$

Area of sector Having central angle $90$ degree $:$ Area of sector Having central angle $120$ degree $:$ Area of sector Having central angle $150$ degree
$=\frac{90}{360} \times \pi \times 6^2: \frac{120}{360} \times \pi \times 6^2: \frac{150}{360} \times \pi \times 6^2$
$=\frac{1}{4}: \frac{1}{3}: \frac{5}{12}$
$=3: 4: 5$

View full question & answer→

MCQ 71 Mark

$9 \sec ^2 A-9 \tan ^2 A=$

✓
$9$
B
$1$
C
$0$
D
$99$

Answer

Correct option: A.

$9$

$\text { Explanation: Given: } 9 \sec ^2 A-9 \tan ^2 A$
$=9\left(\sec ^2 A-\tan ^2 A\right)$
$=9 \times 1=9 \ldots\left[\because \sec ^2 \theta-\tan ^2 \theta=1\right]$

View full question & answer→

MCQ 81 Mark

The angle subtended by a vertical pole of height $100\ m$ at a point on the ground $100 \sqrt{3}$ m from the base is, has measure of

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

Answer

Correct option: D.

$30^{\circ}$

$30^{\circ}$
$\text { Explanation: } \tan \theta=\frac{100}{100 \sqrt{3}}$
$\tan \theta=\frac{1}{\sqrt{3}}$
$\theta=30^{\circ}$

View full question & answer→

MCQ 91 Mark

If $X \sin ^3 \theta+ Y \cos ^3 \theta=\sin \theta \cos \theta$ and $X \sin \theta= Y \cos \theta$, then $........$

A
$X^4+Y^4=1$
✓
$X ^2+ Y ^2=1$
C
$X ^2- Y ^2=1$
D
$X^3+Y^3=1$

Answer

Correct option: B.

$X ^2+ Y ^2=1$

We have, $X \sin ^3 \theta+ Y \cos ^3 \theta=\sin \theta \cos \theta \ldots (i)$
$X \sin \theta= Y \cos \theta \ldots (ii)$
Using $(ii)$ in $(i),$ we get
$\Rightarrow Y \cos \theta \sin ^2 \theta+y \cos ^3 \theta=\sin \theta \cos \theta$
$\Rightarrow Y \sin ^2 \theta+Y \cos ^2 \theta=\sin \theta$
$\Rightarrow Y=\sin \theta$
Now, $X \sin \theta=\sin \theta \times \cos \theta$
$\Rightarrow X =\cos \theta$
$\therefore X ^2+ Y ^2=1$

View full question & answer→

MCQ 101 Mark

In Figure, if $\text{PQR}$ is the tangent to a circle at Qwhose centre is $O, AB$ is a chord parallel to $PR$ and $\angle BQR =70^{\circ},$ then $\angle AQB$ is equal to

✓
$40^{\circ}$
B
$20^{\circ}$
C
$35^{\circ}$
D
$45^{\circ}$

Answer

Correct option: A.

$40^{\circ}$

Given $, AB \| PR$
$\angle ABQ =\angle BQR =70^{\circ} \ [$alternate angles$]$
Also $QD$ is perpendicular to $AB$ and $QD$ bisects $AB$.
In $\triangle ODA$ and $\triangle ODB$
$AD = BD$
$QD = QD\ [$common side$]$
$\therefore \triangle ADQ \cong \triangle BDQ \ldots[$ by $\text{SAS}$ similarity criterion$]$
Then, $\angle Q A D=\angle Q B D \ldots (i) \ [ c , p , c , t ]$
Also , $\angle ABQ =\angle BQR \ [$alternate interior angle$]$
$\angle A B Q=70^{\circ} \ldots\left[B Q R=70^{\circ}\right]$
Hence, $\angle QAB =70^{\circ} \ [$from Eq. $(i)]$
Now, in $\triangle ABQ$
$\angle A+\angle B+\angle Q=180^{\circ}$
$\Rightarrow \angle Q=180^{\circ}-\left(70^{\circ}+70^{\circ}\right)=40^{\circ}$

View full question & answer→

MCQ 111 Mark

In $\triangle ABC , PQ \| BC$. If $PB =6 cm, AP =4 cm, AQ =8 cm$, find the length of AC .

A
12 cm
B
14 cm
✓
20 cm
D
6 cm

Answer

Correct option: C.

20 cm

(C) 20 cm
Explanation: In $\triangle ABC , PQ \| BC$
$\therefore \frac{ AP }{ PB }=\frac{ AQ }{ QC }$ (By proportionality theorem)
$\Rightarrow \frac{4}{6}=\frac{8}{Q C}$
$\Rightarrow QC =\frac{8 \times 6}{4}=12 cm$
Now, AC = AQ + QC
= 8 + 12
= 20 cm

View full question & answer→

MCQ 121 Mark

$\triangle PQR \sim \triangle XYZ$ and the perimeters of $\triangle PQR$ and $\triangle XYZ$ are 30 cm and 18 cm respectively. If $QR =9 cm$, then, YZ is equal to

A
4.5 cm.
✓
5.4 cm.
C
12.5 cm.
D
9.5 cm.

Answer

Correct option: B.

5.4 cm.

(B) 5.4 cm.
Explanation: Given: $\triangle PQR \sim \triangle XYZ$
$\therefore \frac{\text { Perimeter of } \triangle PQR }{\text { Perimeter of } \triangle XYZ }=\frac{ QR }{ YZ }$
$\Rightarrow \frac{30}{18}=\frac{9}{ YZ }$
$\Rightarrow YZ =5.4 cm$

View full question & answer→

MCQ 131 Mark

The distance of the point (4, 7) from the y-axis is

A
11
✓
4
C
$\sqrt{65}$
D
7

Answer

Correct option: B.

4

(B) 4
Explanation: The distance of the point (4, 7) from y-a x is = 4

View full question & answer→

MCQ 141 Mark

One of the two students, while solving a quadratic equation in x, copied the constant term incorrectly and got the roots 3 and 2. The other copied the constant term and coefficient of $x^2$ correctly as -6 and 1 respectively. The correct roots are ________.

✓
6, -1
B
3, -2
C
(-3, 2)
D
(-6, -1)

Answer

Correct option: A.

6, -1

(A) 6, -1
Explanation: Let the equation be $x^2+a x+b=0$
Its roots are 3 and 2
Sum of roots, 5 = -a
and product of roots, 6 = b
$\therefore$ Equation is $x^2-5 x+6=0$
Now constant term is wrong and it is given that correct constant term is -6.
$\therefore x ^2-5 x -6=0$ is the correct equation.
Its roots are -1 and 6.

View full question & answer→

MCQ 151 Mark

If am = bl and $bn \neq cm$, then the system of equations
$a x+b y=c$
$Ix + my = n$

A
Has a unique solution.
B
Has infinitely many solutions.
✓
Has no solution.
D
May or may not have a solution.

Answer

Correct option: C.

Has no solution.

(C) Has no solution.
Explanation: We have, ax + by - c and lx + my = n
Now, $\frac{a}{l}=\frac{b}{m} \neq \frac{c}{n}$ (given)
$\therefore$ The given system of equations has no solution.

View full question & answer→

MCQ 161 Mark

Two numbers whose sum is 12 and the absolute value of whose difference is 4 are the roots of the equation ________.

A
$2 x^2-24 x+43=0$
B
$2 x^2-6 x+7=0$
✓
$x^2-12 x+32=0$
D
$x^2-12 x+30=0$

Answer

Correct option: C.

$x^2-12 x+32=0$

(C) $x^2-12 x+32=0$
Explanation: Let the two roots be a and b, then
a + b = 12 ...(i)
and a - b = 4 ...(ii)
$\Rightarrow a =8$ and $b =4$ (from (i) and (ii))
$\therefore$ Required equation is $x^2-12 x+32=0$

View full question & answer→

MCQ 171 Mark

If $a=2^3 \times 3, b=2 \times 3 \times 5, c=3^n \times 5$ and LCM $( a , b , c )=2^3 \times 3^2 \times 5$, then $n =$

A
$1$
B
$4$
C
$3$
✓
$2$

Answer

Correct option: D.

$2$

$\text{LCM} (a, b, c) =2^3 \times 3^2 \times 5 \ldots (I)$
we have to find the value of $n$
Also we have
$a=2^3 \times 3$
$b=2 \times 3 \times 5$
$c=3^n \times 5$
We know that the while evaluating $\text{LCM,}$ we take greater exponent of the prime numbers in the factorisation of the number.
Therefore, by applying this rule and taking $n \geq 1$ we get the $\text{LCM}$ as
$\operatorname{LCM}( a , b , c )=2^3 \times 3^n \times 5 \ldots ..(II)$
On comparing $(I)$ and $(II)$ sides, we get:
$2^3 \times 3^2 \times 5=2^3 \times 3^n \times 5$
$n=2$

View full question & answer→

MCQ 181 Mark

The LCM and HCF of two rational numbers are equal, then the numbers must be

✓
equal
B
prime
C
co-prime
D
composite

Answer

Correct option: A.

equal

(A) equal
Explanation: If we assume that a and b are equal and consider a = b = k
Then,
HCF (a, b)= k
LCM (a, b) = k

View full question & answer→

Generate Paper Free All MODEL PAPER 3 (BASIC) topics

M.C.Q (1 Marks) - Maths STD 10 Questions - Vidyadip