datavorous/entrance-exam-dataset huggingface dataset:
Let the correlation coefficient between $\mathrm{X}$ and $\mathrm{Y}$ be $0.6$. Random variables $\mathrm{Z}$ and $\mathrm{W}$ are defined as $\mathrm{Z}=\mathrm{X}+5$ and $\mathrm{W}$
$=\frac{Y}{3}$. What is the correlation coefficient between $\mathrm{Z}$ and $\mathrm{W}$ ?

Options:
A $\underline{0.1}$
B 2
C $0.36$
D $0.6$

The correct answer is:
$0.6$

Solution:
Given, $\mathrm{r}_{x y}=0.6$
$\mathrm{z}=\mathrm{x}+5 ; w=\frac{y}{3}$
$\Rightarrow \mathrm{b}_{\mathrm{z}}=1 \Rightarrow b_{w y}=\frac{1}{3}$
$b_{x x} b_{v y}=(1)\left(\frac{1}{3}\right)=\frac{1}{3}$
$\Rightarrow \frac{r_{z w}}{r_{x y}}=\frac{1}{3} \Rightarrow r_{z w}=\frac{r_{x y}}{3}=\frac{0.6}{3}=0.2$

Topic: Mathematics, Statistics, NDA, NDA 2019 (Phase 1)
Let the random variable $\mathrm{X}$ follow $\mathrm{B}(6, \mathrm{p})$. If $16 \mathrm{P}(\mathrm{X}=4)=$ $\mathrm{P}(\mathrm{X}=2)$, then what is the value of $\mathrm{p}$ ?

Options:
A $\frac{1}{3}$
B $\frac{1}{4}$
C $\frac{1}{5}$
D $\frac{1}{6}$

The correct answer is:
$\frac{1}{5}$

Solution:
$\begin{aligned} & \text { X follows } B(6, p)=16 P(x=4)=P(x=2) \\ & \Rightarrow 16{ }^{6} C_{4} \cdot p^{4}(1-p)^{6-4}={ }^{6} C_{2} \cdot p^{2}(1-p)^{6-2} \\ & \Rightarrow 16 \times \frac{6 !}{4 ! 2 !} p^{4}(1-p)^{2}=\frac{6 !}{2 ! 4 !} p^{2}(1-p)^{4} \\ & \Rightarrow 16 p^{2}=(1-p)^{2} \Rightarrow 16 p^{2}=1+p^{2}-2 p \\ & \Rightarrow 15 p^{2}+2 p-1=0 \Rightarrow 15 p^{2}+5 p-3 p-1=0 \\ & \Rightarrow p=\frac{1}{5},-\frac{1}{3} \\ & \text { As } P>0 \Rightarrow P=\frac{1}{5} \end{aligned}$

Topic: Mathematics, Probability, NDA, NDA 2016 (Phase 2)
Let the sample space consist of non-negative integers up to $50, \mathrm{X}$ denote the numbers which are multiples of 3 and $\mathrm{Y}$ denote the odd numbers. Which of the following is/are correct?
$1.$ $\mathrm{P}(\mathrm{X})=\frac{8}{25}$
$2.$ $\mathrm{P}(\mathrm{Y})=\frac{1}{2}$
Select the correct answer using the code given below.

Options:
A 1 only
B 2 only
C Both 1 and 2
D Neither 1 nor 2

The correct answer is:
Neither 1 nor 2

Solution:
Given $\mathrm{S}=51$ (includes 0 ) $x$ denotes the multiples of 3 upto 51 . $\mathrm{n}(\mathrm{x})=16$
y denotes the odd numbers upto 51 . $\mathrm{n}(\mathrm{y})=25$
$\therefore \mathrm{P}(\mathrm{x})=\frac{16}{51}, \mathrm{P}(\mathrm{y})=\frac{25}{51}$

Topic: Mathematics, Probability, NDA, NDA 2017 (Phase 2)
let $\mu=$ the set of all triangles, $\mathrm{P}=$ the set of all isosceles triangles, $\mathrm{Q}=$ the set of all equilateral triangles, $\mathrm{R}=$ the set of all right-angled triangles. What do the sets $\mathrm{P} \cap \mathrm{Q}$ and $\mathrm{R}$ - P represents respectively?

Options:
A The set of isosceles triangles; the set of non- isosceles right angled triangles
B The set of isosceles triangles; the set of right angled triangles
C The set of equilateral triangles; the set ofright angled triangles
D The set of isosceles triangles; the set of equilateral triangles

The correct answer is:
The set of isosceles triangles; the set of non- isosceles right angled triangles

Solution:
As given : $\mu=$ the set of all triangles $\mathrm{P}=$ the set of all isosceles triangles $\mathrm{Q}=$ the set of all equilateral triangles $\mathrm{R}=$ the set of all right angled triangles
$\therefore \quad \mathrm{P} \cap \mathrm{Qrepresents}$ the set of isosceles triangles and $\mathrm{R}-\mathrm{P}$ represents the set of non-isosceles right angled triangles.

Topic: Mathematics, Sets and Relations, NDA, NDA 2007 (Phase 1)
Let the slope of the curve $y=\cos ^{-1}(\sin x)$ be $\tan \theta$ : Then the value of $\theta$ in the interval $(0, \pi)$ is

Options:
A $\frac{\pi}{6}$
B $\frac{3 \pi}{4}$
C $\frac{\pi}{4}$
D $\frac{\pi}{2}$

The correct answer is:
$\frac{3 \pi}{4}$

Solution:
$\begin{aligned} y &=\cos ^{-1}(\sin x) \\ &=\cos ^{-1} \cos \left(\frac{\pi}{2}-x\right) \\ &=\frac{\pi}{2}-x \end{aligned}$
From question, slope of the curve $m=\tan \theta$
$\tan \theta=-1 \Rightarrow \theta=\frac{3 \pi}{4}$

Topic: Mathematics, Inverse Trigonometric Functions, NDA, NDA 2018 (Phase 2)
Let $\overrightarrow{\mathbf{u}}=\hat{\mathbf{i}}-\hat{\mathbf{j}}, \overrightarrow{\mathbf{v}}=2 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}, \overrightarrow{\mathbf{w}}=4 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}$ and $\overrightarrow{\mathbf{p}}=\overrightarrow{\mathbf{u}}+\overrightarrow{\mathbf{v}}+\overrightarrow{\mathbf{w}}$.
Which one of the following is correct?

Options:
A $-3 \overrightarrow{\mathbf{u}}+2 \overrightarrow{\mathbf{v}}=\overrightarrow{\mathbf{p}}$
B $3 \overrightarrow{\mathbf{u}}-2 \overrightarrow{\mathbf{v}}=\overrightarrow{\mathbf{p}}$
C $3 \overrightarrow{\mathbf{u}}+2 \overrightarrow{\mathbf{v}}=\overrightarrow{\mathbf{p}}$
D $-3 \overrightarrow{\mathbf{u}}-2 \overrightarrow{\mathbf{v}}=\overrightarrow{\mathbf{p}}$

The correct answer is:
$3 \overrightarrow{\mathbf{u}}+2 \overrightarrow{\mathbf{v}}=\overrightarrow{\mathbf{p}}$

Solution:
$$
\begin{aligned}
\text { Since, } \overrightarrow{\mathrm{p}}=\overrightarrow{\mathrm{u}}+\overrightarrow{\mathrm{v}}+\overrightarrow{\mathrm{w}} \\
\Rightarrow & \overrightarrow{\mathrm{p}}=(\hat{\mathrm{i}}-\hat{\mathrm{j}})+(2 \hat{\mathrm{i}}+5 \hat{\mathrm{j}})+(4 \hat{\mathrm{i}}+3 \hat{\mathrm{j}})=7 \hat{\mathrm{i}}+7 \hat{\mathrm{j}} \\
& \text { Now, } 3 \overrightarrow{\mathrm{u}}+2 \overrightarrow{\mathrm{v}}=3(\hat{\mathrm{i}}-\hat{\mathrm{j}})+2(2 \hat{\mathrm{i}}+5 \hat{\mathrm{j}})=\\
& \hat{3}-3 \hat{\mathrm{j}}+4 \hat{\mathrm{i}}+10 \hat{\mathrm{j}}=7 \hat{\mathrm{i}}+7 \hat{\mathrm{j}} \\
\Rightarrow & 3 \overrightarrow{\mathrm{u}}+2 \overrightarrow{\mathrm{v}}=\overrightarrow{\mathrm{p}}
\end{aligned}
$$

Topic: Mathematics, Vector Algebra, NDA, NDA 2007 (Phase 1)
Let $\mathrm{U}=\{\mathrm{x} \in \mathrm{N}: 1 \leq \mathrm{x} \leq 10\}$ be the universal set, $\mathrm{N}$ being the set of natural numbers. If $\mathrm{A}=\{1,2,3,4\}$ and $\mathrm{B}=\{2,3,6$. $10\}$ then what is the complement of $(\mathrm{A}-\mathrm{B})$?

Options:
A $\{6,10$
B $\{1,4\}$
C $\{2,3,5,6,7,8,9,10\}$
D $\{5,6,7,8,9,10\}$

The correct answer is:
$\{2,3,5,6,7,8,9,10\}$

Solution:
$\begin{array}{l}\text { A-B }=\{1,4\} \\ & (\mathrm{A}-\mathrm{B})^{\mathrm{c}}=\mathrm{U}-(\mathrm{A}-\mathrm{B})=\{2,3,5,6,7,8,9,10\}\end{array}$

Topic: Mathematics, Sets and Relations, NDA, NDA 2012 (Phase 1)
Let $\mathrm{x}_1, \mathrm{x}_2, \mathrm{x}_3, \mathrm{x}_4$ and $\mathrm{x}_5$ be the observations with mean $\mathrm{m}$ and standard deviations. Then, standard deviation of the observations $\mathrm{kx}_1, \mathrm{kx}_2, \mathrm{kx}_3, \mathrm{kx}_4$ and $\mathrm{kx}_5$ is

Options:
A
$\mathrm{k}+\mathrm{s}$

B
$\mathrm{s} / \mathrm{k}$

C
$\mathrm{ks}$

D
$\mathrm{s}$

The correct answer is:
$\mathrm{ks}$

Solution:
Standard deviation is dependent on change of scale. Therefore, the standard deviation of $\mathrm{kx}_1, \mathrm{kx}_2, \mathrm{kx}_3, \mathrm{kx}_4, \mathrm{kx}_5$ is $\mathrm{ks}$.

Topic: Mathematics, Statistics
Let $x_1, x_2, \ldots x_n$ be $n$ observations. Let $w_i=l x_i+k$ for $i=1$, $2, \ldots, n$, where $l$ and $k$ are constants. If the mean of $x_i^{\prime} \operatorname{sis} 48$ and their standard deviation is 12 , the mean of $w_i$ 's is 55 and standard deviation of $w_i^{\prime}$ 's is 15 , then the value of $l$ and $k$ should be

Options:
A
$l=1.25, k=-5 & \text$

B
$l=-1.25, k=5\end{array}$

C
$l=2.5, k=-5$

D
$l=2.5, k=5$

The correct answer is:
$l=1.25, k=-5 & \text$

Solution:
Since, $w_i=l x_l+k, \bar{x}_i=48, \sigma x_i=12, w_i=55$ and $\sigma w_i=15$
So, $\bar{w}_i=l \bar{x}_i+k$
$\Rightarrow 55=48 l+k$
Now, $\sigma \omega_i=l \sigma x_i \Rightarrow l=\frac{15}{2}=1.25$
After putting the value of 1 in equation (i), we get $k=55-1.25 \times 48=-5$

Topic: Mathematics, Statistics
Let $X$ and $Y$ be two non-empty sets and let $R_{1}$ and $R_{2}$ be two relationsfrom $\mathrm{X}$ into $\mathrm{Y}$. Then, which one of the following is correct?

Options:
A $\left(\mathrm{R}_{1} \cap \mathrm{R}_{2}\right)^{-1} \subset \mathrm{R}_{1}^{-1} \cap \mathrm{R}_{2}^{-1}$
B $\left(\mathrm{R}_{1} \cap \mathrm{R}_{2}\right)^{-1} \supset \mathrm{R}_{1}^{-1} \cap \mathrm{R}_{2}^{-1}$
C $\left(\mathrm{R}_{1} \cap \mathrm{R}_{2}\right)^{-1}=\mathrm{R}_{1}{ }^{-1} \cap \mathrm{R}_{2}^{-1}$
D $\left(\mathrm{R}_{1} \cap \mathrm{R}_{2}\right)^{-1}=\mathrm{R}_{1}^{-1} \cup \mathrm{R}_{2}^{-1}$

The correct answer is:
$\left(\mathrm{R}_{1} \cap \mathrm{R}_{2}\right)^{-1}=\mathrm{R}_{1}^{-1} \cup \mathrm{R}_{2}^{-1}$

Solution:
The correct relations as per De Morgan's theorem is $\left(\mathrm{R}_{1} \cap \mathrm{R}_{2}\right)^{-1}=\mathrm{R}_{1}^{-1} \cup \mathrm{R}_{2}^{-1}$

Topic: Mathematics, Sets and Relations, NDA, NDA 2006 (Phase 1)
Let $\mathrm{X}$ and $\mathrm{Y}$ be two related variables. The two regression lines are given by $\mathrm{x}-\mathrm{y}+1=0$ and $2 \mathrm{x}-\mathrm{y}+4=0$. The two regression lines pass through the point : $\quad$

Options:
A $(-4,-3)$
B $(-6,-5)$
C $(3,-2)$
D $(-3,-2)$

The correct answer is:
$(-3,-2)$

Solution:
Given regression lines are $x-y+1=0$ and $2 x-y+4$ $=0$ only point $(-3,-2)$ given in option (d) satisfies the both equations. Hence, The two regression lines pass through the point $(-3,-2)$

Topic: Mathematics, Statistics, NDA, NDA 2012 (Phase 2)
Let $\mathrm{X}$ be a discrete random variable whose probability distribution is defined as follows.
$$
\mathrm{P}(\mathrm{X}=\mathrm{x})= \begin{cases}\mathrm{k}(\mathrm{x}+1), & \text { for } \mathrm{x}=1,2,3,4 \\ 2 \mathrm{kx}, & \text { for } \mathrm{x}=5,6,7 \\ 0, & \text { otherwise }\end{cases}
$$
where, $\mathrm{k}$ is a constant, calculate
(i) the value of $k$. (ii) $\mathrm{E}(\mathrm{X})$
(iii) standard deviation of X.

Options:

None

Solution:
We have following table
\begin{array}{|l|c|c|c|c|c|c|c|}
\hline \mathbf{X} & \mathbf{1} & \mathbf{2} & \mathbf{3} & \mathbf{4} & \mathbf{5} & \mathbf{6} & \mathbf{7} \\
\hline \mathbf{P}(\mathbf{X}) & 2 \mathrm{k} & 3 \mathrm{k} & 4 \mathrm{k} & 5 \mathrm{k} & 10 \mathrm{k} & 12 \mathrm{k} & 14 \mathrm{k} \\
\mathbf{X P}(\mathbf{X}) & 2 \mathrm{k} & 6 \mathrm{k} & 12 \mathrm{k} & 20 \mathrm{k} & 50 \mathrm{k} & 72 \mathrm{k} & 98 \mathrm{k} \\
\mathbf{X} \mathbf{P}(\mathbf{X}) & 2 \mathrm{k} & 12 \mathrm{k} & 36 \mathrm{k} & 80 \mathrm{k} & 250 \mathrm{k} & 432 \mathrm{k} & 686 \mathrm{k} \\
\hline
\end{array}
(i) We know that
$\Rightarrow \mathrm{k}(2+3+4+5+10+12+14)=1$
$\Rightarrow \mathrm{k}=\frac{1}{50} \quad\left[\because \Sigma \mathrm{P}_{\mathrm{i}}=1\right]$
(ii) $\because \mathrm{E}(\mathrm{X})=\Sigma \mathrm{XP}(\mathrm{X})$
$\begin{aligned} \therefore \mathrm{E}(\mathrm{X})=2 \mathrm{k}+6 \mathrm{k}+12 \mathrm{k}+20 \mathrm{k}+& 50 \mathrm{k} \\ &+72 \mathrm{k}+98 \mathrm{k}+0=260 \mathrm{k} \end{aligned}$
$=260 \times \frac{1}{50}=\frac{26}{5}=5.2$
(iii) We know that,
$$
\begin{aligned}
& \operatorname{Var}(\mathrm{X})=\Sigma \mathrm{X}^2 \mathrm{P}(\mathrm{X})-\left[\sum\{\mathrm{XP}(\mathrm{X})\}\right]^2 \\
=& {[2 \mathrm{k}+12 \mathrm{k}+36 \mathrm{k}+80 \mathrm{k}+250 \mathrm{k}+432 \mathrm{k}} \\
&+686 \mathrm{k}+0]-[5.2]^2[\mathrm{Using} \text { Eq. (i) }] \\
=& {\left[1498 \times \frac{1}{50}\right]-27.04\left[\because \mathrm{k}=\frac{1}{50}\right] } \\
&=29.96-24.04=2.92
\end{aligned}
$$
And standard deviation of $X=\sqrt{\operatorname{Var}(X)}$ $=\sqrt{2.92}=1.7088=1.7$ (approx)

Topic: Mathematics, Probability
Let $\mathrm{X}$ be a non-empty set and let $\mathrm{A}, \mathrm{B}, \mathrm{C}$ be subsets of $\mathrm{X}$. Consider the following statments:
(1) $\mathrm{A} \subset \mathrm{C} \Rightarrow(\mathrm{A} \cap \mathrm{B}) \subset(\mathrm{C} \cap \mathrm{B})(\mathrm{A} \cup \mathrm{B}) \subset(\mathrm{C} \cap \mathrm{B})$
(2) $(\mathrm{A} \cup \mathrm{B}) \subset(\mathrm{C} \cap \mathrm{B})$ for all sets $\mathrm{B} \Rightarrow \mathrm{A} \subset \mathrm{C}$
(3) $(\mathrm{A} \cup \mathrm{B}) \subset(\mathrm{C} \cup \mathrm{B})$ for all sets $\mathrm{B} \Rightarrow \mathrm{A} \subset \mathrm{C}$
Which of the above statements are correct?

Options:
A 1 and 2 only
B 2 and 3 only
C 1 and 3 only
D 1,2 and 3

The correct answer is:
2 and 3 only

Solution:
Statements $(2)$ and $(3)$ are correct.

Topic: Mathematics, Sets and Relations, NDA, NDA 2019 (Phase 1)
Let $\mathrm{X}$ be any non-empty set containing $\mathrm{n}$ elements. Then what is the number of relations on $\mathrm{X}$ ?

Options:
A $2^{\mathrm{n}^{2}}$
B $2^{\text {n }}$
C $2^{2 n}$
D $\mathrm{n}^{2}$

The correct answer is:
$2^{\mathrm{n}^{2}}$

Solution:
Number of elements in $\mathrm{X}$, is $\mathrm{n}$, then the number of relations on $\mathrm{X}$ means, number of elements of cartesian product $\mathrm{X} \times \mathrm{X}$
Since, $n(X)=n$ So, $\mathrm{n}(\mathrm{X} \times \mathrm{X})=\mathrm{n}, \mathrm{n}$
then the total number of relations is $2^{
\mathrm{n} \mathrm{n}}=2^{\mathrm{n}^{2}}$

Topic: Mathematics, Sets and Relations, NDA, NDA 2006 (Phase 1)
Let $\bar{x}$ be the mean of n observations $x_{1}, x_{2}, \ldots . ., x_{n}$. If $(a-b)$ is added to each observation, then what is the mean of new set of observations?

Options:
A 0
B $\overline{\mathrm{x}}$
C $\bar{x}-(a-b)$
D $\overline{\mathrm{x}}+(\mathrm{a}-\mathrm{b}) \quad\mathrm{}$

The correct answer is:
$\overline{\mathrm{x}}+(\mathrm{a}-\mathrm{b}) \quad\mathrm{}$

Solution:
Let $\bar{x}$ is the mean of $n$ observation $x_{1}, x_{2}, \ldots ., x_{n}$
$\Rightarrow \bar{x}=\frac{x_{1}+x_{2}+x_{3}+\ldots .+x_{n}}{n}$
Now, $(a-b)$ is added to each term. New mean $=\frac{\mathrm{x}_{1}+(\mathrm{a}-\mathrm{b})+\mathrm{x}_{2}+(\mathrm{a}-\mathrm{b})+\ldots . .+\mathrm{x}_{\mathrm{n}}+(\mathrm{a}-\mathrm{b})}{\mathrm{n}}$
$=\frac{\mathrm{x}_{1}+\mathrm{x}_{2}+\ldots .+\mathrm{x}_{\mathrm{n}}}{\mathrm{n}}+\frac{\mathrm{n}(\mathrm{a}-\mathrm{b})}{\mathrm{n}}$
$=\overline{\mathrm{x}}+(\mathrm{a}-\mathrm{b})$

Topic: Mathematics, Statistics, NDA, NDA 2008 (Phase 1)
Let $\bar{x}$ be the mean of $x_{1}, x_{2}, x_{3}, \ldots ., x_{n} .$ If $x_{i}=a+c y_{i}$ for some constants a and $\mathrm{c}$, then what will be the mear $\mathrm{y}_{2}, \mathrm{y}_{3}, \ldots, \mathrm{y}_{\mathrm{n}}$

Options:
A $\quad a+c \bar{x}$
B $\mathrm{a}-\frac{1}{\mathrm{c}} \overline{\mathrm{x}}$
C $\quad \frac{1}{x} \bar{x}-a$
D $\underline{\mathrm{X}-}$

The correct answer is:
$\underline{\mathrm{X}-}$

Solution:
Given, Mean of $x_{i}=\bar{x}$ Also, Given $x_{i}=a+c y_{i}$ Mean of $\mathrm{a}+\mathrm{cy}_{\mathrm{i}}=\overline{\mathrm{x}}$
$\Rightarrow$ Mean of $\mathrm{cy}_{\mathrm{i}}=\overline{\mathrm{x}}-\mathrm{a}$
$\Rightarrow$ Mean of $y_{i}=\frac{\bar{x}-a}{c}$

Topic: Mathematics, Statistics, NDA, NDA 2018 (Phase 1)
Let $x$ be the number of integers lying between 2999 and 8001 which have at least two digits equal. Then $\mathrm{x}$ is equal to

Options:
A 2480
B 2481
C 2482
D 2483

The correct answer is:
2481

Solution:
Number of numbers between 2999 and 8001 $=8001-2999-1=5001$
Number ofnumbers with all digitdistinct and having
as starting digit
Number ofnumbers with all digitdistinct and having 4 as starting digi $4=50$
Similarly number of numbers with starting digit 5,6 and 7 respectively are 504,504 and 504 . Total numbers $=5 \times 504=2520$
Hence, required number $=5001-2520=2481$

Topic: Mathematics, Sets and Relations, NDA, NDA 2018 (Phase 2)
Let $\mathrm{X}$ be the set of all citizensof India. Elements $\mathrm{x}, \mathrm{y}$ in $\mathrm{X}$ are said to be related if the difference of their age is 5 years Which one of the following is correct?

Options:
A The relation is an equivalence relation on.
B The relation is symmetric but neither reflexive nor transitive.
C The relation is reflexive but neither, symmetric nor transitive
D None of the above

The correct answer is:
The relation is symmetric but neither reflexive nor transitive.

Solution:
$\mathrm{X}=$ Set of all citizens of India $\mathrm{R}=\{(\mathrm{x}, \mathrm{y}): \mathrm{x}, \mathrm{y} \in \mathrm{X},|\mathrm{x}-\mathrm{y}|=5\}$
$|\mathrm{x}-\mathrm{x}|=0 \neq 5 \quad(\mathrm{R}$ is not reflexive)
$x R y \Rightarrow|x-y|=5$
$\Rightarrow|y-x|=5 \quad$ (R is symmetric) $\mathrm{xRy} \Rightarrow|\mathrm{x}-\mathrm{y}|=5$
$y R z \Rightarrow|y-z|=5$
But $|x-z| \neq 5$
(R is not transitive)

Topic: Mathematics, Sets and Relations, NDA, NDA 2014 (Phase 1)
Let $X$ be the set of all graduates in India. Elements $x$ and $y$ in $X$ are said to berelated if they are graduates of the sameuniversity. Which one of the following statements is correct?

Options:
A Relation is symmetric and transitive only
B Relation is reflexive and transivtive only
C Relation is reflexive andsymmetric only
D Relation is reflexive symmetric and transitive

The correct answer is:
Relation is reflexive symmetric and transitive

Solution:
$x R y \Leftrightarrow x$ and $y$ are graduates of the same university.
Reflexive $x R x \Leftrightarrow x$ and $x$ are graduates of the same
university. Relation is reflexive.
Symmetric $x R y \Leftrightarrow x$ and $y$ are graduates of the
same university
$\Rightarrow y R x \Leftrightarrow y$ and $x$ are graduates of the same
university. $\therefore$ Relation issymmetric. Transitive $x R y, y R z \Leftrightarrow x R z$
It means $x$ and $y, y$ and $z$ are graduates of the same university, then $x$ and $z$ are also graduates of the same university.
- Relation is transitive Hence, relation is reflexive, symmetric and transitive

Topic: Mathematics, Sets and Relations, NDA, NDA 2010 (Phase 1)
Let $X$ be the set of all persons living in a city. Persons $x$, in $X$ are said to be related as $x < y$ if $y$ is at least 5 years older than $x$. Which one of the following is correct?

Options:
A The relation is an equivalence relation on
B The relation is transitive but neither reflexive nor symmetric
C The relation is reflexive but neither transitive nor symmetric
D The relation is symmetric but neither transitive nor reflexive

The correct answer is:
The relation is transitive but neither reflexive nor symmetric

Solution:
Given that $x < y$ if $y \geq x+5$ For Reflexive:
$x < x$
Hence, relation is not reflexive For Symmetry:
if $x < y$
then $y < x$
Hence, relation is not symmetry.
ForTransitive:
if $x < y$ and $y < z$
then $\mathrm{x} < 2$
Hence, relation is transitive

Topic: Mathematics, Sets and Relations, NDA, NDA 2015 (Phase 1)
Let $\mathrm{X}$ be theset of all persons living in Delhi. Thepersons $a$ and $b$ in $X$ are said to be related if the difference in their ages
is at most 5 years. The relation is

Options:
A an equivalence relation
B reflexiveand transitive but not symmetric
C symmetricand transitive but not reflexive
D retlexive and symmetric but not transitive

The correct answer is:
retlexive and symmetric but not transitive

Solution:
$a-b \leq 5$ Let $b-c \leq 5$
so $a-c \leq 10$
as $(a-b) \in R$ and $(b-c) \in R$
but $(a-c) \notin R$
So it is not a transitive relation. By considering all the options, we come to the conclusion that only option (d) is correct.

Topic: Mathematics, Sets and Relations, NDA, NDA 2015 (Phase 2)
Let $[\mathrm{x}]$ denote the greatest integer function. What is the number of solutions of the equation $x^{2}-4 x+[x]=0$ in the interval [0,2]?

Options:
A Zero(Nosolution)
B One
C Two
D Three

The correct answer is:
One

Solution:
$\mathrm{x}^{2}-4 \mathrm{x}+[\mathrm{x}]=0$
Given internal, $[0,2]$
Case $1:$ Let $0 \leq \mathrm{x} < 1$
$[\mathrm{x}]=0$
$\therefore \mathrm{x}^{2}-4 \mathrm{x}+0=0 \Rightarrow \mathrm{x}(\mathrm{x}-4)=0 \Rightarrow \mathrm{x}=0, \mathrm{x}=4$
$\mathrm{x}=4$ can't be taken in $0 \leq \mathrm{x} < 1$
$x=0$
Case $2:$ Let $1 \leq x < 2$ $[\mathrm{x}]=1$
$\therefore \mathrm{x}^{2}-4 \mathrm{x}+1=0$
roots are $\frac{4 \pm \sqrt{16-4}}{2}=\frac{4 \pm \sqrt{12}}{2}=2 \pm \sqrt{3}$
In internal $1 \leq \mathrm{x} < 2,2 \pm \sqrt{3}$ are not the roots. Case $3:$ Let $x=2$ $[\mathrm{x}]=2$
$\therefore x^{2}-4 x+2=0$
roots are $\frac{4 \pm \sqrt{16-8}}{2}=\frac{4 \pm \sqrt{8}}{2}=2 \pm \sqrt{2}$
Since, $x=2$, roots can't be $2 \pm \sqrt{2}$
$\therefore$ There is only one solution, $x=0$

Topic: Mathematics, Quadratic Equation, NDA, NDA 2018 (Phase 1)
Let $X$ denote the number of scores which exceed 4 in 18 , tosses of a symmetrical die. Consider the following statements : $\quad[2014-I]$
1 . The arithmetic mean of $X$ is 6
2 . The standard deviation of $X$ is 2 . Which of the above statements is/are correct?

Options:
A 1 only
B 2 only
C Both 1 and 2
D Neither 1 nor 2

The correct answer is:
Both 1 and 2

Solution:
Statement 1 $n(X)=2$
$\mathrm{p}=\frac{\mathrm{n}(\mathrm{X})}{\mathrm{n}(\mathrm{S})}=\frac{2}{6}=\frac{1}{3}$
$q=1-p=1-\frac{1}{3}=\frac{2}{3}$
arithmetric mean of $\mathrm{X}=\mathrm{n} \mathrm{p}=18 \times \frac{1}{3}=6$
Statement $2:$ Standard deviation of
$\mathrm{x}-\sqrt{\text { variance of } \mathrm{X}}-\sqrt{18 \times \frac{1}{3} \times \frac{2}{3}}-\sqrt{4}-2$
Hence, statements 1 and 2 both are correct.
Sol. $(128-130):$ \begin{array}{|c|cc|c|}
\hline Numbers (\mathrm{x}) & Frequency (\mathrm{f}) & cf. & \Sigma \mathrm{fx} \\
\hline 0 & 14 & 14 & 0 \\
1 & 21 & 35 & 21 \\
2 & 25 & 60 & 50 \\
3 & 43 & 103 & 129 \\
4 & 51 & 154 & 204 \\
5 & 40 & 194 & 200 \\
6 & 39 & 233 & 234 \\
7 & 12 & 245 & 84 \\
\hline & \mathrm{N}=245 & & \Sigma \mathrm{fx} =922 \\
\hline
\end{array}

Topic: Mathematics, Statistics, NDA, NDA 2014 (Phase 1)
Let $\mathbf{X}$ denote the sum of the numbers obtained when two fair dice are rolled. Find the variance and standard deviation of $\mathbf{X}$.

Options:

None

Solution:
$S=\left[\begin{array}{llllll}11 & 21 & 31 & 41 & 51 & 61 \\ 12 & 22 & 32 & 42 & 52 & 62 \\ 13 & 23 & 33 & 43 & 53 & 63 \\ 14 & 24 & 34 & 44 & 54 & 64 \\ 15 & 25 & 35 & 45 & 55 & 65 \\ 16 & 26 & 36 & 46 & 56 & 66\end{array}\right]$
$$
n(S)=36
$$
Let A denotes the sum of the numbers $=2$
$B$ denotes the sum of the numbers $=3$
$\mathrm{C}$ denotes the sum of the numbers $=4$
D denotes the sum of the numbers $=5$
E denotes the sum of the numbers $=6$
$F$ denotes the sum of the numbers $=7$
$\mathrm{G}$ denotes the sum of the numbers $=8$
$\mathrm{H}$ denotes the sum of the numbers $=9$
I denotes the sum of the numbers $=10$
$J$ denotes the sum of the numbers $=11$
$\mathrm{K}$ denotes the sum of the numbers $=12$
$A=\{11\}, P(A)=\frac{1}{36}$
$$
\mathrm{B}=\{12,21\}, \mathrm{P}(\mathrm{B})=\frac{2}{36}
$$
$$
\mathrm{C}=\{13,22,31\}, \mathrm{P}(\mathrm{C})=\frac{3}{36}
$$
$$
\begin{aligned}
&\mathrm{D}=\{14,23,32,41\}, P(D)=\frac{4}{36} \\
&E=\{15,24,33,42,51\}, P(E)=\frac{5}{36} \\
&F=\{16,25,34,43,52,61\}, P(F)=\frac{6}{36}
\end{aligned}
$$
$$
\mathrm{G}=\{26,35,44,53,62\}, \mathrm{P}(\mathrm{G})=\frac{5}{36}
$$
$$
\mathrm{H}=\{36,45,54,63\}, \mathrm{P}(\mathrm{H})=\frac{4}{36}
$$
$$
\mathrm{I}=\{46,55,64\}, \mathrm{P}(\mathrm{I})=\frac{3}{36}
$$
$$
\begin{aligned}
&J=\{56,65\}, P(J)=\frac{2}{36} \\
&K=\{6,6\}, P(K)=\frac{1}{36}
\end{aligned}
$$
Probability distribution:
Mean $=\mu=\sum p_i x_i$
$=\frac{1}{36}(2+6+12+20+30+42+40+36+30+22+12)=\frac{252}{36}=7$
Now, $\sum p_i x_i^2$
$=\frac{1}{36}[4+18+48+100+180+294+320+$
$=\frac{1}{36} \times 1974=\frac{329}{6}$
Variance $=\sum p_i x_i^2-\left(\sum p_i x_i\right)^2$
$=54.83-49=5.83$
Standard deviation $=\sqrt{8.83}=2.4$ (nearly).

Topic: Mathematics, Probability