MCQ 11 Mark
If the arithmetic mean of $x, x+3, x+6, x+9$ and $x+12$ is $10 ,$ then $x=$
AnswerMean of $x, x+3, x+6, x+9, x+12=10$
$\Rightarrow \frac{x+x+3+x+6+x+9+x+12}{5}=10$
$\Rightarrow \frac{5 x+30}{5}=10$
$\Rightarrow x+6=10$
$\Rightarrow x=10-6=4$
View full question & answer→MCQ 21 Mark
If a digit is chosen at random from the digits $1,2,3,4,5,6,7,8,9$, then the probability that it is odd and is a multiple of 3 is
- A
$\frac{1}{9}$
- ✓
$\frac{2}{9}$
- C
$\frac{2}{3}$
- D
$\frac{1}{3}$
AnswerCorrect option: B. $\frac{2}{9}$
(B) $\frac{2}{9}$
Explanation: Total numbers of digits for 1 to $9(n)=9$
Number divisible by $3(m)=3,6,9$
Odd numbers out of $3,6,9=3,9$
$\therefore$ Probability $=\frac{m}{n}=\frac{2}{9}$
View full question & answer→MCQ 31 Mark
The probability of guessing the correct answer to a certain test questions is $\frac{x}{12}$. If the probability of not guessing the correct answer to this question is $\frac{2}{3}$, then $x =$
AnswerProbability of guessing the correct answer
$=\frac{x}{12}$
and probability of not guessing the correct
answer $=\frac{2}{3}$
$\frac{x}{12}+\frac{2}{3}=1$
$\because(A+\bar{A}=1)$
$\Rightarrow \frac{x}{12}=1-\frac{2}{3}=\frac{1}{3}$
$\Rightarrow x=\frac{12}{3}=4$
$\therefore x=4$
View full question & answer→MCQ 41 Mark
If the perimeter of a sector of a circle of radius $5.2 \ cm$ is $16.4 \ cm ,$ then the area of the sector is.
- A
$15.5 \ cm^2$
- ✓
$15.6 \ cm^2$
- C
$15.9 \ cm^2$
- D
$15.1 \ cm^2$
AnswerCorrect option: B. $15.6 \ cm^2$
Perimeter of a sector of circle $=\frac{\theta}{360^{\circ}} \times 2 \pi r+2 r$
$\Rightarrow\left(\frac{\theta}{360^{\circ}} \times 2 \pi \times 5.2\right)+(2 \times 5.2)=16.4$
$\Rightarrow \frac{\theta}{360^{\circ}} \pi$
$=\frac{16.4-10.4}{10.4}=\frac{6}{10.4}$
Area of sector of circle $=\frac{\theta}{360^{\circ}} \times \pi r^2$
$=\frac{6}{10.4} \times(5.2)^2$
$=15.6 \ cm^2$
View full question & answer→MCQ 51 Mark
Three horses are tethered with 7-meter-long ropes at the three corners of a triangular field having sides $20 m, 34$ m , and 42 m . The area of the plot which can be grazed by the horses is
- ✓
$77 m^2$
- B
$80 m^2$
- C
$100 m^2$
- D
$30 m^2$
AnswerCorrect option: A. $77 m^2$
(A) $77 m^2$
Explanation: $77 m^2$
View full question & answer→MCQ 61 Mark
If the height of a tower and the distance of the point of observation from its foot, both, are increased by $10 \%$, then the angle of elevation of its top
Answer(C) remains unchanged
Let height of the tower be $h$ meters and distance of the point of observation from its foot be $x$ meters and angle of elevation be $\theta \therefore \tan \theta=\frac{h}{x}$ $\qquad$ .(i)
Now, new height $= h +10 \%$ of $h =h+\frac{10}{100} h=\frac{11 h}{10}$ And new distance $x x +10 \%$ of $x =x+\frac{10}{100} x=\frac{11 x}{10} \therefore$
$
\tan \theta=\frac{\frac{11 h}{10}}{\frac{11 x}{10}}=\frac{h}{x}
$
From eq. (i) and (ii), it is clear that the angle of elevation is same i.e., angle of elevation remains unchanged. View full question & answer→MCQ 71 Mark
The prime factorisation of the number 5488 is
- A
$2^3 \times 7^3$
- ✓
$2^4 \times 7^3$
- C
- D
AnswerCorrect option: B. $2^4 \times 7^3$
(B) $2^4 \times 7^3$
Explanation: $2^4 \times 7^3$
View full question & answer→MCQ 81 Mark
If $\sec \theta+\tan \theta= x$, then $\tan \theta=$
- A
$\frac{x^2+1}{x}$
- B
$\frac{x^2+1}{2 x}$
- ✓
$\frac{x^2-1}{x}$
- D
$\frac{x^2-1}{2 x}$
AnswerCorrect option: C. $\frac{x^2-1}{x}$
Explanation:
Given, $\sec \theta+\tan \theta= x \ldots (i)$
We know that
$\sec ^2 \theta-\tan ^2 \theta=1$
$\Rightarrow(\sec \theta+\tan \theta)(\sec \theta-\tan \theta)=1$
$\Rightarrow x(\sec \theta-\tan \theta)=1$
$\Rightarrow \sec \theta-\tan \theta=\frac{1}{x} \ldots(ii)$
Subtracting $(ii)$ from $(i)$
$2 \tan \theta=x-\frac{1}{x}=\frac{x^2-1}{x}$
$\tan \theta=\frac{x^2-1}{2 x}$
View full question & answer→MCQ 91 Mark
How many tangents can be drawn to a circle from a point on it?
MCQ 101 Mark
In the given figure, PA and PB are tangents to a circle from an external point P . If $\angle A P B=50^{\circ}$ and $AC \| PB$, then the measures of angles of triangle $A B C$ are
- ✓
$65^{\circ}, 50^{\circ}, 65^{\circ}$
- B
$50^{\circ}, 55^{\circ}, 75^{\circ}$
- C
$80^{\circ}, 60^{\circ}, 40$
- D
$50^{\circ}, 50^{\circ}, 80^{\circ}$
AnswerCorrect option: A. $65^{\circ}, 50^{\circ}, 65^{\circ}$
(A) $65^{\circ}, 50^{\circ}, 65^{\circ}$
Explanation: Since $PA = PB$ {Tangents from an external point to a circle $\}$ $
\therefore \angle PAB=\angle PBA
$
Let $\angle PAB =\angle PBA =x$
Now, in triangle APB, $x+x+50^{\circ}=180^{\circ}$
$
\Rightarrow x=65^{\circ}
$
Since $AC \| PB$ and AB is intersecting.
$
\therefore \angle PBA=\angle BAC=65^{\circ} \text { [Alternate angles] }
$
And $\angle PAB =\angle CBA =65^{\circ}$ [Alternate angles]
$\therefore \angle ACB =180^{\circ}-\left(65^{\circ}+65^{\circ}\right)=50^{\circ}$ [Angle sum property of a triangle]
$
\therefore \angle BAC=65^{\circ}, \angle ABC=65^{\circ}, \angle ACB=50^{\circ}
$
View full question & answer→MCQ 111 Mark
In a $\triangle \text{ABC}$, it is given that $\text{AD}$ is the internal bisector of $\angle A$. If $\text{AB}=10 \ cm, \text{AC}=14 \ cm$ and $\text{BC}=6 \ cm$, the $\text{CD} = ?$
- ✓
$3.5 \ cm$
- B
$7 \ cm$
- C
$4.8 \ cm$
- D
$10.5 \ cm$
AnswerCorrect option: A. $3.5 \ cm$
By using angle bisector theore in $\triangle \text{ABC}$, we have
$\frac{AB}{AC}=\frac{BD}{DC}$
$\Rightarrow \frac{60}{14}=\frac{6-x}{x}$
$\Rightarrow 10 x=84-14 x$
$\Rightarrow 24 x=84$
$\Rightarrow x=3.5$
Hence, the correct answer is $3.5.$
View full question & answer→MCQ 121 Mark
If $\alpha$ and $\beta$ are zeros of $x^2+5 x+8$, then the value of $(\alpha+\beta)$ is
Answer$\text { Explanation: } x^2+5 x+8$
$\alpha+\beta=\frac{- \text { Coefficient of x }}{\text { Coefficient of } x^2}$
$=\frac{-5}{1}$
$=-5$
View full question & answer→MCQ 131 Mark
The distance between the points $(a, a)$ and $(-\sqrt{3} a, \sqrt{3} a)$ is
- A
$2 \sqrt{2}$ units
- B
$3 \sqrt{2} a$ units
- C
- ✓
$2 a \sqrt{2}$ units
AnswerCorrect option: D. $2 a \sqrt{2}$ units
Let the points be $A (a, a)$ and $B (-\sqrt{3} a, \sqrt{3} a)$
$\therefore AB=\sqrt{(-\sqrt{3} a-a)^2+(\sqrt{3} a-a)^2}$
$=\sqrt{3 a^2+a^2+2 \sqrt{3} aa+3 a^2+a^2-2 \sqrt{3} aa}$
$=\sqrt{6 a^2+2 a^2}$
$=\sqrt{8 a^2}$
$=2 a \sqrt{2} \text { units }$
View full question & answer→MCQ 141 Mark
The 2 nd term of an AP is 13 and its 5 th term is 25. What is its 17 th term?
Answer(D) 73
Explanation: $a + d =13 \ldots$... (i)
and $a +4 d=25 \ldots$...ii)
From (i) and (ii), we get $a =9$ and $d =4$
$
\Rightarrow T_{17}=(a+16 d)=(9+16 \times 4)=73
$
View full question & answer→MCQ 151 Mark
The sum of two numbers is 17 and the sum of their reciprocals is $\frac{17}{62}$. The quadratic representation of the above situation is
- A
$\frac{1}{x}+\frac{1}{x+17}=\frac{17}{62}$
- B
$\frac{1}{x(17-x)}=\frac{17}{62}$
- ✓
$\frac{1}{x}+\frac{1}{17-x}=\frac{17}{62}$
- D
$\frac{1}{x}-\frac{1}{17-x}=\frac{17}{62}$
AnswerCorrect option: C. $\frac{1}{x}+\frac{1}{17-x}=\frac{17}{62}$
(C) $\frac{1}{x}+\frac{1}{17-x}=\frac{17}{62}$
Explanation: Let one number be $x$, As the sum of the numbers is 17, then the other number will be ( 17 - x). Their reciprocals will be $\frac{1}{x}$ and $\frac{1}{17-x}$.
$\therefore$ According to question, $\frac{1}{x}+\frac{1}{17-x}=\frac{17}{62}$
View full question & answer→MCQ 161 Mark
An icecream cone has hemispherical top. If the height of the cone is $9 \ cm$ and base radius is $2.5 \ cm ,$ then the volume of icecream is
- ✓
$91.67 \ cm^3$
- B
$96.67 \ cm^3$
- C
$90.67 \ cm^3$
- D
$91.76 \ cm^3$
AnswerCorrect option: A. $91.67 \ cm^3$
Explanation :
Height of ice $-$ cream cone is $9 \ cm$ and radius of the hemispherical top is $2.5 \ cm$ .
Now, Volume of ice $-$ cream cone $=$ Volume of cone $+$ volume of Hemispherical top
$=\frac{1}{3} \pi r^2 h+\frac{2}{3} \pi r^3$
$=\frac{1}{3} \pi r^2(h+2 r)$
$=\frac{1}{3} \times \frac{22}{7} \times 2.5 \times 2.5(9+5)$
$=\frac{1}{3} \times \frac{22}{7} \times 2.5 \times 2.5 \times 14$
$=91.67 \ cm^3$
View full question & answer→MCQ 171 Mark
The roots of a quadratic equation $x^2-4 p x+4 p^2-q^2=0$ are
- ✓
$2 p+q, 2 p-q$
- B
$p+2 q, p-2 q$
- C
$2 p + q , 2 p + q$
- D
$2 p-q, 2 p-q$
AnswerCorrect option: A. $2 p+q, 2 p-q$
Given: $x^2-4 p x+4 p^2-q^2=0$
$\Rightarrow(x-2 p)^2-q^2=0$
Using $a^2-b^2=(a+b)(a-b)$
$\Rightarrow(x-2 p+q)(x-2 p-q)=0$
$\Rightarrow x-2 p+q=0$ and $x-2 p-q=0$
$\Rightarrow x=2 p-q$ and $ x=2 p+q$
View full question & answer→MCQ 181 Mark
Cards marked with numbers $1,2,3, \ldots, 25$ 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 and 5 is
- A
$\frac{12}{25}$
- B
$\frac{4}{25}$
- ✓
$\frac{1}{25}$
- D
$\frac{8}{25}$
AnswerCorrect option: C. $\frac{1}{25}$
(C) $\frac{1}{25}$
Explanation: Multiples of $3=3,6,9,12,15,18,21,24$
Multiples of $5=5,10,15,20,25$
Number of possible outcomes (multiple of 3 and 5$)=\{15\}=1$
Number of Total outcomes $=25$
$\therefore$ Required Probability $=\frac{1}{25}$
View full question & answer→