datavorous/entrance-exam-dataset huggingface dataset:
Ifn! $3 \times(n !)$ and $(n+1) !$ are in GP, then the value of $n$ will be

Options:
A 3
B 4
C 8
D 10

The correct answer is:
8

Solution:
Let n!, $3(\mathrm{n} !)$ and $(\mathrm{n}+1) !$ are in G. $P$. Then, $[3(\mathrm{n} !)]^{2}=(\mathrm{n} !)(\mathrm{n}+1) !$
$\Rightarrow 9 \times n ! \times n !=(n !)(n+1) n !$
$\Rightarrow 9=(\mathrm{n}+1) \Rightarrow \mathrm{n}=8$

Topic: Mathematics, Sequences and Series, NDA, NDA 2011 (Phase 2)
IfO, Pare the points $(0,0,0),(2,3,-1)$ respectively, then what is the equation to the plane through $\mathrm{P}$ at right angles to $\mathrm{OP}$ ?

Options:
A $2 x+3 y+z=16$
B $2 x+3 y-z=14$
C $2 x+3 y+z=14$
D $2 x+3 y-z=0$

The correct answer is:
$2 x+3 y-z=14$

Solution:
Since, coordinates of points $\mathrm{O}$ and $\mathrm{P}$ are $(0,0,0)$ and $(2,3,-1)$ respectively. Direction ratios of $\mathrm{OP}$ are $ < 2,3,-1\rangle$
The plane is perpendicular to OP. So, its equation is $2 x+3 y-z+d=0$
Since, this plane passes through $(2,3,-1) ; 2 \times 2+3 \times$
$3-1 \times-1+d=0$
$\Rightarrow 4+9+1+\mathrm{d}=0$
$\Rightarrow \mathrm{d}=-14$
On putting the value of $\mathrm{d}$ in Eq. (i) $2 x+3 y-z-14=0$
$\Rightarrow 2 x+3 y-z=14$
which is required equation of plane

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2006 (Phase 2)
IfP(E) denotes the probability of an event $\mathrm{E}$, then $\mathrm{E}$ is called certain event if:

Options:
A $\quad \mathrm{P}(\mathrm{E})=0$
B $\mathrm{P}(\mathrm{E})=1$
C $\mathrm{P}(\mathrm{E})$ is either 0 or 1
D $\mathrm{P}(\mathrm{E})=1 / 2$

The correct answer is:
$\mathrm{P}(\mathrm{E})=1$

Solution:
Prob. (certain event) $=1$ $\Rightarrow \mathrm{P}(\mathrm{E})=1$

Topic: Mathematics, Probability, NDA, NDA 2012 (Phase 2)
IfP,Q are $(2,5,-7),(-3,2,1)$ respectively, then what are the direction ratios of the line $\mathrm{PQ}$ ?

Options:
A $ < 10,6,-16>$
B $ < 5,3,8>$
C $ < -5,-3,-8>$
D None of these

The correct answer is:
None of these

Solution:
Since, coordinates of points $\mathrm{P}$ and $\mathrm{Q}$ are $(2,5,-7)$ and $(-3,2,1)$ respectively. Direction ratios of $\mathrm{PQ}$ are $ < -3-2,2-5,1+7>$ i.e., $ < -5,-3,8>$

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2006 (Phase 2)
Ifthe latus rectum of an ellipse is equal to half its minor axis, then what is its eccentricity?

Options:
A $\frac{1}{2}$
B $\sqrt{3}$
C $\frac{\sqrt{3}}{2}$
D $\frac{1}{\sqrt{2}}$

The correct answer is:
$\frac{\sqrt{3}}{2}$

Solution:
Let the equation of ellipse be $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$
Length of minor axis $=2 \mathrm{~b}$ and length of latus rectum $=\frac{2 \mathrm{~b}^{2}}{\mathrm{a}}$
According to the question,
$$
\frac{2 b^{2}}{a}=b \Rightarrow 2 b=a \Rightarrow 4 b^{2}=a^{2}
$$
Now, eccentricity of ellipse $e=\frac{\sqrt{a^{2}-b^{2}}}{a}$
$$
e=\frac{\sqrt{4 b^{2}-b^{2}}}{2 b}=\frac{\sqrt{3} b}{2 b}=\frac{\sqrt{3}}{2}
$$
$\Rightarrow \quad e=\frac{\sqrt{3}}{2}$

Topic: Mathematics, Ellipse, NDA, NDA 2010 (Phase 2)
Ifthe matrix
$A=\left[\begin{array}{ll}
\alpha & \beta \\
\beta & \alpha
\end{array}\right]$
is such that $A^{2}=I$, then which one of the following is correct?

Options:
A $\alpha=0, \beta=1$ or $\alpha=1, \beta=0$
B $\alpha=0, \beta \neq 1$ or $\alpha \neq 1, \beta=1$
C $\alpha=1, \beta \neq 0$ or $\alpha \neq 1, \beta=1$
D $\alpha \neq 0, \beta \neq 0$

The correct answer is:
$\alpha=0, \beta=1$ or $\alpha=1, \beta=0$

Solution:
$\begin{array}{ll}\text { Let } & A=\left[\begin{array}{ll}\alpha & \beta \\ \beta & \alpha\end{array}\right] \\ \therefore & \mathrm{A}^{2}=\left[\begin{array}{ll}\alpha & \beta \\ \beta & \alpha\end{array}\right]\left[\begin{array}{ll}\alpha & \beta \\ \beta & \alpha\end{array}\right] \\ & =\left[\begin{array}{rc}\alpha^{2}+\beta^{2} & 2 \alpha \beta \\ 2 \alpha \beta & \alpha^{2}+\beta^{2}\end{array}\right]\end{array}$
Now, $\mathrm{A}^{2}=\mathrm{I}$
$\Rightarrow \quad\left[\begin{array}{cc}\alpha^{2}+\beta^{2} & 2 \alpha \beta \\ 2 \alpha \beta & \alpha^{2}+\beta^{2}\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
$\Rightarrow \quad \alpha^{2}+\beta^{2}=1, \quad \alpha \beta=0$
$\Rightarrow \quad \alpha=0, \beta=1$
or $\quad \beta=0, \alpha=1$

Topic: Mathematics, Matrices, NDA, NDA 2011 (Phase 1)
Ifthe roots of a quadratic equation $a x^{2}+b x+c=0$ are $\alpha$ and $\beta$, then the quadratic equation having roots $\alpha^{2}$ and $\beta^{2}$ is

Options:
A $x^{2}-\left(b^{2}-2 a c\right) x+c=0$
B $\quad a^{2} x^{2}-\left(b^{2}-2 a c\right) x+c=0$
C $a x^{2}-\left(b^{2}-2 a c\right) x+c^{2}=0$
D $a^{2} x^{2}-\left(b^{2}-2 a c\right) x+c^{2}=0$

The correct answer is:
$a^{2} x^{2}-\left(b^{2}-2 a c\right) x+c^{2}=0$

Solution:
Quadratic equation is given by $\mathrm{x}^{2}-($ sum of roots $) \mathrm{x}+($ product of roots $)=0$ Required equation is $\mathrm{x}^{2}-\mathrm{x}\left(\alpha^{2}+\beta^{2}\right)+(\alpha \beta)^{2}=0$
$\Rightarrow \mathrm{x}^{2}-\mathrm{x}\left[(\alpha+\beta)^{2}-2 \alpha \beta\right]+(\alpha \beta)^{2}=0$
$\Rightarrow \mathrm{x}^{2}-\mathrm{x}\left(\frac{\mathrm{b}^{2}}{\mathrm{a}^{2}}-\frac{2 \mathrm{c}}{\mathrm{a}}\right)+\frac{\mathrm{c}^{2}}{\mathrm{a}^{2}}=0$
$\Rightarrow \mathrm{a}^{2} \mathrm{x}^{2}-\mathrm{x}\left(\mathrm{b}^{2}-2 \mathrm{ac}\right)+\mathrm{c}^{2}=$

Topic: Mathematics, Quadratic Equation, NDA, NDA 2013 (Phase 1)
Ifthe roots of a quadratic equation are $\mathrm{m}+\mathrm{n}$ and $\mathrm{m}-\mathrm{n}$, then the quadratic equation will be

Options:
A $x^{2}+2 m x+m^{2}-m n+n^{2}=0$
B $x^{2}+2 m x+(m-n)^{2}=0$
C $x^{2}-2 m x+m^{2}-n^{2}=0$
D $x^{2}+2 m x+m^{2}-n^{2}=0$

The correct answer is:
$x^{2}-2 m x+m^{2}-n^{2}=0$

Solution:
Let $\alpha=\mathrm{m}+\mathrm{n}, \beta=\mathrm{m}-$
Now, $\alpha+\beta=\mathrm{m}+\mathrm{n}+\mathrm{m}-\mathrm{n}=2 \mathrm{~m}$
$\alpha \times \beta=(m+n)(m-n)=m^{2}-n^{2}$
Now, Required Quadratic Equation will be, $\mathrm{x}^{2}-(\alpha+\beta) \mathrm{x}+\alpha \beta=0$
$x^{2}-2 m x+m^{2}-n^{2}=0$

Topic: Mathematics, Quadratic Equation, NDA, NDA 2012 (Phase 2)
Ifthe roots of the equation $3 \mathrm{ax}^{2}+2 \mathrm{bx}+\mathrm{c}=0$ are in the ratio $2: 3$, then which one of the following is correct?

Options:
A $8 a c=25 b$
B $8 a c=9 b^{2}$
C $8 b^{2}=9 a c$
D $8 b^{2}=25 a c$

The correct answer is:
$8 b^{2}=25 a c$

Solution:
Let roots of equation be $2 \alpha$ and $3 \alpha$
$2 \alpha+3 \alpha=-\frac{2 b}{3 a}$
$\Rightarrow \alpha=\frac{-2 b}{15 a}$...(i)
$2 \alpha \cdot 3 \alpha=\frac{\mathrm{c}}{3 \mathrm{a}} \Rightarrow \alpha^{2}=\frac{\mathrm{c}}{18 \mathrm{a}}$...(ii)
Now, put value of $\alpha$ in equation (ii) $\left(\frac{-2 b}{15 a}\right)^{2}=\frac{c}{18 a} \Rightarrow \frac{4 b^{2}}{225 a^{2}}=\frac{c}{18 a}$
$\Rightarrow 8 b^{2}=25 a c$

Topic: Mathematics, Quadratic Equation, NDA, NDA 2013 (Phase 1)
Ifthe roots of the equation $x^{2}-b x+c=0$ are two consecutiv integers, then what is the value of $b^{2}-4 c$ ?

Options:
A 1
B 2
C -2
D 3

The correct answer is:
1

Solution:
Let $\alpha$ and $(\alpha+1)$ be the roots of equation $x^{2}-b x+c=0$
$\therefore \alpha+(\alpha+1)=b$
$\Rightarrow 2 \alpha+1=b$
and $\alpha(\alpha+1)=c$
Now, $b^{2}-4 c=(2 \alpha+1)^{2}-4[\alpha(\alpha+1)]$
$=4 \alpha^{2}+1+4 \alpha-4 \alpha^{2}-4 \alpha=1$

Topic: Mathematics, Quadratic Equation, NDA, NDA 2008 (Phase 2)
Ifthe roots of the quadratic equation $3 \mathrm{x}^{2}-5 \mathrm{x}+\mathrm{p}=0$ are real and unequal, then which one of the following is correct?

Options:
A $\quad \mathrm{p}=25 / 12$
B $\mathrm{p} < 25 / 12$
C $p>25 / 12$
D $p \leq 25 / 12$

The correct answer is:
$\mathrm{p} < 25 / 12$

Solution:
The given equation is. $3 x^{2}-5 x+p=0$
We have, $\mathrm{a}=3, \mathrm{~b}=-5, \mathrm{c}=$
$\mathrm{D}=\mathrm{b}^{2}-4 \mathrm{ac}=25-12 \mathrm{p}$
For Real and unequal, $\mathrm{D}>0$ $\therefore 25-12 \mathrm{p}>0$
$\Rightarrow \quad 25>12 \mathrm{p} \Rightarrow \mathrm{p} < \frac{25}{12}$

Topic: Mathematics, Quadratic Equation, NDA, NDA 2012 (Phase 2)
Ifx and y are real numberssuch that $x>y$ and $|x|>|y|$, then which one of the following is correct?

Options:
A $x>0$
B $\mathrm{y}>0$
C $y < 0$
D $\mathrm{x} < 0$

The correct answer is:
$x>0$

Solution:
As given: $\mathrm{x}$ and $\mathrm{y}$ are real numbers such that $\mathrm{x}>\mathrm{y}$ and $|\mathrm{x}|>|\mathrm{y}|$ then $\mathrm{x}$ can not be negative of zero $\Rightarrow \mathrm{x}>0 .$

Topic: Mathematics, Basic of Mathematics, NDA, NDA 2007 (Phase 2)
In a $50 \mathrm{gm}$ living tissue, what would be the amount of water?

Options:
A $15-25 \mathrm{gm}$
B $25-30 \mathrm{gm}$
C $35-45 \mathrm{gm}$
D $70-90 \mathrm{gm}$

The correct answer is:
$35-45 \mathrm{gm}$

Solution:
Water is the most abundant chemical compound in living organism. In a cell, water occupies $70-90 \%$ of the cellular mass. So, in 50 $\mathrm{gm}$ of living tissue 35-45 gm of water would be present.

Topic: Biology, Transport in Plants, AIIMS, AIIMS 2011
In a $\triangle A B C$, $a+b=3(1+\sqrt{3}) \mathrm{cm}$ and $a-b=3(1-\sqrt{3}) \mathrm{cm}$.
If angle $A$ is $30^{\circ}$, then what is the angle $B$ ?

Options:
A $120^{\circ}$
B $90^{\circ}$
C $75^{\circ}$
D $60^{\circ}$

The correct answer is:
$60^{\circ}$

Solution:
Given, $a+b=3(1+\sqrt{3}) \ldots \ldots \ldots$. (1)
and $a-b=3(1-\sqrt{3}) \ldots \ldots \ldots .(2)$
By adding (1) and (2) we get
$(a+b)+(a-b)=3+3 \sqrt{3}+3-3 \sqrt{3}$
$\Rightarrow 2 a=6 \Rightarrow a=3$
$\therefore b=3-3+3 \sqrt{3}=3 \sqrt{3}$
By using sine rule, $\Rightarrow \frac{a}{\sin A}=\frac{b}{\sin B}$
Given $\angle A=30 \quad \Rightarrow \frac{3}{\sin 30^{\circ}}=\frac{3 \sqrt{3}}{\sin B}$
$\Rightarrow \sin B=\sqrt{3} \times \frac{1}{2} \Rightarrow \sin B=\sin 60^{\circ}$
$\Rightarrow B=60^{\circ}$

Topic: Mathematics, Properties of Triangles, NDA, NDA 2009 (Phase 2)
In a $\Delta \mathrm{ABC}$, angle $\mathrm{B}$ is obtuse and $\mathrm{D}, \mathrm{E}, \mathrm{F}$ are the middle points of sides $\mathrm{BC}, \mathrm{CA}, \mathrm{AB}$ respectively. Which one of the following vectors has the greatest magnitude?

Options:
A $\overrightarrow{\mathrm{BC}}$
B $\overrightarrow{\mathrm{CA}}$
C $\overrightarrow{\mathrm{AB}}$
D $\overrightarrow{\mathrm{AD}}$

The correct answer is:
$\overrightarrow{\mathrm{CA}}$

Solution:
Since, side opposite to greatest angle is longest and $\angle \mathrm{B}$ is greatest angle in $\Delta \mathrm{ABC}$. Thus, $\overline{\mathrm{CA}}$ has greatest magnitude.

Topic: Mathematics, Vector Algebra, NDA, NDA 2006 (Phase 1)
In a $\triangle \mathrm{ABC}$, if $\mathrm{c}=2, \mathrm{~A}=120^{\circ}, \mathrm{a}=\sqrt{6}$, then what is $\mathrm{C}$ equal
to ?

Options:
A $30^{\circ}$
B $45^{\circ}$
C $60^{\circ}$
D $75^{\circ}$

The correct answer is:
$45^{\circ}$

Solution:
Let $c=2, \angle A=120^{\circ}$ and $a=\sqrt{6}$ in $\Delta A B C$,
$\therefore$ By Sine rule, we have
$\begin{aligned} & \frac{a}{\sin A}=\frac{c}{\sin C} \Rightarrow \frac{\sqrt{6}}{\sin 120^{\circ}}=\frac{2}{\sin C} \\ \Rightarrow & \sin C=\frac{2 \times \sqrt{3}}{\sqrt{6} \times 2}=\frac{1}{\sqrt{2}} \\ \Rightarrow \sin C=\sin 45^{\circ} & \Rightarrow \quad \angle C=45^{\circ} \end{aligned}$

Topic: Mathematics, Properties of Triangles, NDA, NDA 2011 (Phase 1)
In a binary number system, assume that $a=00111$ and
$b=01110$, then in a decimal system $\frac{b}{a}$, which is equal to

Options:
A 1
B 2
C 4
D 5

The correct answer is:
2

Solution:
Let $a=00111=2^{2} \times 1+2^{1} \times 1+2^{0} \times 1$
$\begin{aligned} &=4+2+1=7 \\ \text { Let } b &=01110=2^{3} \times 1+2^{2} \times 1+2^{1} \times 1+2^{0} \times 0 \\ &=8+4+2=14 \end{aligned}$
$\frac{b}{a}=\frac{14}{7}=2$

Topic: Mathematics, Sets and Relations, NDA, NDA 2011 (Phase 1)
In a binomial distribution, the mean is $\frac{2}{3}$ and the variance
is $\frac{5}{9} .$ What is the probability that $\mathrm{X}=2$ ?

Options:
A $\frac{5}{36}$
B $\frac{25}{36}$
C $\frac{25}{216}$
D $\frac{25}{54}$

The correct answer is:
$\frac{25}{216}$

Solution:
Mean $=\frac{2}{3}$, variance $=\frac{5}{9}$
$n p=\frac{2}{3}, n p q=\frac{5}{9}$
$\Rightarrow \frac{2}{3} q=\frac{5}{9} \Rightarrow q=\frac{5}{9} \times \frac{3}{2}=\frac{5}{6}$
So, $\mathrm{p}=1-\frac{5}{6}=\frac{1}{6}$
Now, $\mathrm{np}=\frac{2}{3} \Rightarrow \mathrm{n}\left(\frac{1}{6}\right)=\frac{2}{3} \Rightarrow \mathrm{n}=\frac{2}{3} \times 6=4$
$\therefore \mathrm{p}(\mathrm{x}=2)={ }^{\mathrm{n}} \mathrm{c}_{\mathrm{r}} \cdot \mathrm{p}^{\mathrm{r}} \cdot \mathrm{q}^{\mathrm{n}-\mathrm{r}}={ }^{4} \mathrm{c}_{2} \cdot\left(\frac{1}{6}\right)^{2} \cdot\left(\frac{5}{6}\right)^{4-2}$
$=6 \times \frac{1}{36} \times \frac{25}{36}$
$=\frac{25}{216}$.

Topic: Mathematics, Probability, NDA, NDA 2017 (Phase 2)
In a binomial distribution, the mean is 4 and the variance is 3 . What is the mode?

Options:
A 6
B 5
C 4
D 3

The correct answer is:
4

Solution:
As give, $n p=4$ and $n p q=3$ [where $\mathrm{p}$ is the probability of success and $\mathrm{q}$ is the probability of failure for an event to occur, and 'n' is the number of trials
$\Rightarrow \mathrm{q}=\frac{\mathrm{npq}}{\mathrm{np}}=\frac{3}{4}$
Also, $\mathrm{p}=1-\mathrm{q}=1-\frac{3}{4}=\frac{1}{4}$
$\mathrm{n}=16$
In a binomial distribution, the value of $\mathrm{r}$ for which $\mathrm{P}(\mathrm{X}=\mathrm{r})$ is maximum is the mode of binomial distribution.
hence, $(\mathrm{n}+1) \mathrm{p}-1 \leq \mathrm{r} \leq(\mathrm{n}+1) \mathrm{p}$
$\Rightarrow \frac{17}{4}-1 \leq \mathrm{r} \leq \frac{17}{4}$
$\Rightarrow \frac{13}{4} \leq \mathrm{r} \leq \frac{17}{4}$
$\Rightarrow 3.25 \leq \mathrm{r} \leq 4.25$
$\Rightarrow \mathrm{r}=4$

Topic: Mathematics, Statistics, NDA, NDA 2007 (Phase 1)
In a binomial distribution, the occurrence and the nonoccurrence of an event are equally likely and the mean is 6 . The number of trials required is

Options:
A 15
B 12
C 10
D 6

The correct answer is:
12

Solution:
$\mathrm{P}=\mathrm{q}=\frac{1}{2}, \mathrm{nP}=6$
$\frac{\mathrm{n}}{2}=6 \Rightarrow \mathrm{n}=12$

Topic: Mathematics, Probability, NDA, NDA 2013 (Phase 2)
In a binomial distribution the probability of getting success is $\frac{1}{4}$ and the standard deviation is 3 . Then, its mean is

Options:
A $6$
B $8$
C $10$
D $12$

The correct answer is:
$12$

Solution:
Given that
$$
\begin{aligned}
p=\frac{1}{4} \text { and } q=1-\frac{1}{4} & =\frac{3}{4} \\
\mathrm{SD}=3 \Rightarrow \sqrt{n p q} & =3 \\
\Rightarrow \quad n p q & =9 \\
\Rightarrow \quad n \cdot \frac{1}{4} \cdot \frac{3}{4} & =9 \\
n & =48 \\
\text { Mean }=n p=48 \times \frac{1}{4} & =12
\end{aligned}
$$

Topic: Mathematics, Probability, AP EAMCET, AP EAMCET 2002
In a binomial distribution the probability of getting success is $\frac{1}{4}$ and the standard deviation is 3 . Then, its mean is

Options:
A $6$
B $8$
C $10$
D $12$

The correct answer is:
$12$

Solution:
Given that
$$
\begin{aligned}
p=\frac{1}{4} \text { and } q=1-\frac{1}{4} & =\frac{3}{4} \\
\mathrm{SD}=3 \Rightarrow \sqrt{n p q} & =3 \\
\Rightarrow \quad n p q & =9 \\
\Rightarrow \quad n \cdot \frac{1}{4} \cdot \frac{3}{4} & =9 \\
n & =48 \\
\text { Mean }=n p=48 \times \frac{1}{4} & =12
\end{aligned}
$$

Topic: Mathematics, Probability, TS EAMCET, TS EAMCET 2002
In a Binominal distribution, the mean is three times its variance. What is the probability of exactly 3 successes out of 5 trials? $\quad$

Options:
A $\frac{80}{243}$
B $\frac{40}{243}$
C $\frac{20}{243}$
D $\frac{10}{243}$

The correct answer is:
$\frac{80}{243}$

Solution:
Mean $=\mathrm{np}$
Variance $=\mathrm{npq}$
Given, $n p=3 n p q$ $\Rightarrow q=\frac{1}{3}, p=\frac{2}{3}$.
Also, Given $n=5$ trials.
$r=3$
we know, $\mathrm{p}(\mathrm{x}=\mathrm{r})={ }^{\mathrm{n}} \mathrm{c}_{\mathrm{r}} \cdot \mathrm{p}^{\mathrm{r}} \cdot \mathrm{q}^{\mathrm{n}-\mathrm{r}}$
$\mathrm{p}(\mathrm{x}=3)={ }^{5} \mathrm{c}_{3}\left(\frac{2}{3}\right)^{3}\left(\frac{1}{3}\right)^{2}$
$=\frac{80}{243} .$

Topic: Mathematics, Statistics, NDA, NDA 2018 (Phase 1)
In a bolt factory, machines X, Y, Z manufacture bolts that are respectively $25 \%, 35 \%$ and $40 \%$ of the factory's total output. The machines $\mathrm{X}, \mathrm{Y}, \mathrm{Z}$ respectively produce $2 \%, 4 \%$ and $5 \%$ defective bolts. A bolt is drawn at random from the product and is found to be defective. What is the probability that it was manufactured by machine X?

Options:
A $\frac{5}{39}$
B $\frac{11}{39}$
C $\frac{20}{39}$
D $\frac{34}{39}$

The correct answer is:
$\frac{5}{39}$

Solution:
Required probability
$=\frac{25 \times 2}{25 \times 2+35 \times 4+40 \times 5}=\frac{5}{39}$

Topic: Mathematics, Probability, NDA, NDA 2018 (Phase 2)
In a circle of diameter $44 \mathrm{~cm}$, the length of a chord is $22 \mathrm{~cm}$. What is the length of minor arc of the chord?

Options:
A $\frac{484}{21} \mathrm{~cm}$
B $\frac{242}{21} \mathrm{~cm}$
C $\frac{121}{21} \mathrm{~cm}$
D $\frac{44}{7} \mathrm{~cm}$

The correct answer is:
$\frac{484}{21} \mathrm{~cm}$

Solution:
Given,
diameter of circle $=44 \mathrm{~cm}$.
radius of circle $=22 \mathrm{~cm}$
Chord of circle $=22 \mathrm{~cm}$
In figure $\Delta \mathrm{OAB}$ is equilateral triangle. Angle is $60^{\circ} . \mathrm{So}$.
arc is $\frac{1}{6}$ times circumstance.
Length of arc $=\frac{1}{6} \times 2 \pi r=\frac{1}{6} \times 2 \times \frac{22}{7} \times 22$
$=\frac{484}{21} \mathrm{~cm}$

Topic: Mathematics, Circle, NDA, NDA 2019 (Phase 1)
In a city, three dailynewspaper $B, C$ are published, $42 \%$ read $A ; 51 \%$ read $B ; 68 \%$ read $C$; $30 \%$ read A and B; $28 \%$ read $B$ and $C$; $36 \%$ read A and $C$; $8 \%$ do read any of the three newspaper
What is the percentage of persons who read only two papers?

Options:
A $19 \%$
B $31 \%$
C $44 \%$
D None of the above

The correct answer is:
None of the above

Solution:
$\quad(30-x)+(36-x)+(28-x)=94-3 x$
$=98-3 \times 25=2$
Hence percentage of people who read only two newspapers $=23 \%$

Topic: Mathematics, Sets and Relations, NDA, NDA 2012 (Phase 1)
In a city three dailynewspaper $B, C$ are published, $42 \%$ read $A ; 51 \%$ read $B ; 68 \%$ read $C$; $30 \%$ read A and B; $28 \%$ read $B$ and $C$; $36 \%$ read A and $C$; $8 \%$ do read any of the three newspaper
What is the percentage of who read all the three papers?

Options:
A $20 \%$
B $25 \%$
C $30 \%$
D $40 \%$

The correct answer is:
$25 \%$

Solution:
Hence people who read all the three newspapers $=25 \%$

Topic: Mathematics, Sets and Relations, NDA, NDA 2012 (Phase 1)
In a city, three dailynewspaper $B, C$ are published, $42 \%$ read $A ; 51 \%$ read $B ; 68 \%$ read $C$; $30 \%$ read A and B; $28 \%$ read $B$ and $C$; $36 \%$ read A and $C$; $8 \%$ do read any of the three newspaper
What is thepercentageof persons who read only one paper?

Options:
A $38 \%$
B $48 \%$
C $51 \%$
D None of the above

The correct answer is:
$48 \%$

Solution:
$(x-24)+(x-7)+(x+4)=3 x-27$
$3 \times 25-27=48$
Hence percentage of people who read only one newspaper $=48 \%$

Topic: Mathematics, Sets and Relations, NDA, NDA 2012 (Phase 1)
In a city, three dailynewspaper $B, C$ are published, $42 \%$ read $A ; 51 \%$ read $B ; 68 \%$ read $C$; $30 \%$ read A and B; $28 \%$ read $B$ and $C$; $36 \%$ read A and $C$; $8 \%$ do read any of the three newspaper
What is the percentage of persons who read only A but neither B nor C?

Options:
A $4 \%$
B $3 \%$
C $1 \%$
D None of the above

The correct answer is:
$1 \%$

Solution:
$x-24=25-24=1$
Hence percentage of people who read only Newspaper
A but neither $\mathrm{B}$ nor $\mathrm{C}=1 \%$

Topic: Mathematics, Sets and Relations, NDA, NDA 2012 (Phase 1)
In a class, 54 students are good in Hindi only, 63 students are good in Mathematics only and 41 students are good in English only. There are 18 students who are good in both Hindi and Mathematics. 10 students are good in all threesubjects.
What is the number of students who are good in either Hindi or Mathematics but not in English?

Options:
A 99
B 107
C 125
D 130

The correct answer is:
125

Solution:
Let us represent the given data in Venn diagram as shown
Number of students who are good in either Hindi or Maths but not in English $=54+18+63=125$

Topic: Mathematics, Sets and Relations, NDA, NDA 2018 (Phase 1)
In a class, 54 students are good in Hindi only, 63 students are good in Mathematics only and 41 students are good in English only. There are 18 students who are good in both Hindi and Mathematics. 10 students are good in all threesubjects.
What is the number of students who are good in Hindi and Mathematics but not in English?

Options:
A 18
B 12
C 10
D 8

The correct answer is:
8

Solution:
From the same Venn diagram, Number of students whoare good in Hindi and Maths but not English $=8$

Topic: Mathematics, Sets and Relations, NDA, NDA 2018 (Phase 1)
In a class of 125 students 70 passed in Mathematics, 55 passed in Statistics and 30 passed in both. What is the probability that a student selected at random from the class has passed in only one subject?

Options:
A $13 / 25$
B $3 / 25$
C $17 / 25$
D $8 / 25$

The correct answer is:
$13 / 25$

Solution:
Required probability
$=\frac{40}{125}+\frac{25}{125}=\frac{65}{125}=\frac{13}{25}$

Topic: Mathematics, Probability, NDA, NDA 2011 (Phase 2)
In a complete study of grassland ecosystem and pond ecosystem, it may be observed that

Options:
A The abiotic components are almost similar.
B The biotic components are almost similar
C Both biotic and abiotic component are different.
D Primary and secondary consumers are similar.

The correct answer is:
Both biotic and abiotic component are different.

Solution:
Biotic components, or biotic factors, can be described as any living component that affects another organism or shapes the ecosystem. This includes both animals that consume other organisms within their ecosystem. Abiotic factors include all the nonliving things find in nature. They play an important part in the lives of all organisms. The combination of abiotic and biotic factors is different for two different ecosystems. Some examples of abiotic factors include wind, sunlight, soil, temperature, climate, and water.

Topic: Biology, Ecosystem, NEET
In a coordination entity, what is the Lewis acid?

Options:
A The central atom/ion
B The coordination number
C The counter ions
D The secondary valence

The correct answer is:
The central atom/ion

Solution:
In a coordination entity, the Lewis acid is the central atom/ion.

Topic: Chemistry, Coordination Compounds, NEET
In a dicot root, vascular cambium is

Options:
A completely primary
B partially primary and partially secondary
C completely secondary
D sometimes primary sometimes secondary

The correct answer is:
completely secondary

Solution:
Dicot plants are sub group of the division Angiosperms which have two cotyledons or embryonic leaves in their seeds. Meristem are a type of tissue found in such plants which can divide indefinitely. These are generally considered undifferentiated as they can give develop into all other tissues and organs of the plant. The meristematic tissues that appear early in plant life and help in formation of the plant body is called primary meristem. The apical and intercalary meristem are included in it. Meristematic tissue occurring between the mature regions of root and shoots of angiosperms are called secondary meristem. These generally form by dedifferentiation of the permanent tissues. The vascular cambium tissues found within the vascular tissues like xylem and phloem are secondary meristem which help in lateral growth of the plant helping in increasing the girth of the root or shoot.

Topic: Biology, Anatomy of Flowering Plants, NEET
In a diffraction pattern due to a single slit of width a the first minimum is observed at an angle 30o when light of wavelength 5000 Å is incident on the slit. The first secondary maximum is observed at an angle of:

Options:
A sin-1⁡14
B sin-1⁡23
C sin-1⁡12
D sin-1⁡34

The correct answer is:
sin-1⁡34

Solution:
asin⁡30=λ
asin⁡θ=3λ2
sin⁡θsin⁡30=32
sin⁡θ=32×12
sin⁡θ=34
θ=sin⁡34

Topic: Physics, Wave Optics, NEET, NEET 2016 (Phase 1)