datavorous/entrance-exam-dataset huggingface dataset:
If the mean of 100 observations is 50 and their standard deviation is 5 , then the sum of squares of all observations is

Options:
A 50000
B 250000
C 252500
D 255000

The correct answer is:
252500

Solution:
Given; \(\bar{x}=50\)
\(n=100\) and \(S . D(\sigma)=5\)
As, \(\bar{x}=\frac{\Sigma x_i}{n} \Rightarrow \Sigma x_i=n . \bar{x}=5000\).
Also, S.D. is given by,
\(\begin{aligned}
\sigma & =\sqrt{\frac{\Sigma x_i^2}{n}-\left(\frac{\Sigma x_i}{n}\right)^2} \\
\Rightarrow \quad \sigma^2 & =\frac{\Sigma x_i^2}{n}-(\bar{x})^2 \Rightarrow 25=\frac{\Sigma x_i^2}{100}-(50)^2 \\
\Rightarrow \quad \Sigma x_i^2 & =252500
\end{aligned}\)

Topic: Mathematics, Statistics, AP EAMCET, AP EAMCET 2020 (17 Sep Shift 2)
If the mean of a binomial distribution is 25 , then its standard deviation lies in the interval

Options:
A $[0,5)$
B $(0,5]$
C $[0,25)$
D $(0,25]$

The correct answer is:
$[0,5)$

Solution:
Standard deviation $\sigma=\sqrt{\mathrm{npq}} \geq 0$
Now mean $=n p=25$ and $q < 1$
So $\sigma=\sqrt{\mathrm{npq}} < \sqrt{\mathrm{np}}=5$
$\therefore 0 \leq \sigma < 5$

Topic: Mathematics, Probability, VITEEE, VITEEE 2016
If the mean of few observations is 40 and standard deviation is 8 , then what is the coefficient of variation? $\quad$

Options:
A $1 \%$
B $10 \%$
C $20 \%$
D $30 \%$

The correct answer is:
$20 \%$

Solution:
Mean $=40, \mathrm{S.D}=8$
$\therefore \quad$ Coeft of variation $=\frac{8}{40} \times 100$
$=\frac{1}{5} \times 100-20 \%$

Topic: Mathematics, Statistics, NDA, NDA 2012 (Phase 2)
If the mean of the discrete distribution 8,9 , $6,5, x, 4,6,5$ is 6 , then its standard deviation (nearest to two decimal places) is

Options:
A 2.50
B 1.58
C 0.51
D 0.41

The correct answer is:
1.58

Solution:
We have,
$\begin{aligned}
& \text { Mean }=6 \\
& \therefore \quad \frac{8+9+6+5+x+4+6+5}{8}=6 \\
& \Rightarrow 43+x=48 \Rightarrow x=5
\end{aligned}$
So, data are $8,9,6,5,5,4,6,5$
$\begin{array}{llrl}\therefore & & x_i=8,9,6,5,5,4,6,5 \\ \Rightarrow & x_i{ }^2 & =64,81,36,25,25,16,36,25\end{array}$
$\begin{aligned} \therefore \quad \Sigma x_i^2 & =64+81+36+25+25+16+36+25 \\ & =308\end{aligned}$
$\begin{aligned} \therefore \quad \operatorname{SD}(x) & =\sqrt{\frac{1}{N} \Sigma x_i^2-(\bar{x})^2} \\ & =\sqrt{\frac{1}{8} \times 308-(6)^2}=\sqrt{38.5-36} \\ & =\sqrt{2.5}=1.58\end{aligned}$

Topic: Mathematics, Statistics, TS EAMCET, TS EAMCET 2020 (10 Sep Shift 1)
If the mid point between the points $(a+b, a-b)$ and $(-a, b)$ lies on the line $a x+b y=k$, what is $k$ equal to?

Options:
A $\mathrm{a} / \mathrm{b}$
B $\mathrm{a}+\mathrm{b}$
C ab
D $a-b$

The correct answer is:
ab

Solution:
Given points are $(a+b, a-b)$ and $(-a, b)$ Mid point is $\left(\frac{a+b-a}{2}, \frac{a-b+b}{2}\right)=\left(\frac{b}{2}, \frac{a}{2}\right)$
Since, it lies on $a x+b y=k$ $a\left(\frac{b}{2}\right)+b\left(\frac{a}{2}\right)=k \Rightarrow k=a b$

Topic: Mathematics, Straight Lines, NDA, NDA 2012 (Phase 1)
If the monthly expenditure pattern of a person who carns monthlysalary of Rs 15000 is represented in a pie diagram then the sector angle of an item on transport expense:
measures $15^{\circ}$. What is his monthly expenditure on transport

Options:
A $\mathrm{Rs} 450$
B $\mathrm{Rs} 625$
C $\operatorname{Rs} 675$
D Cannot be computed from the given data

The correct answer is:
$\mathrm{Rs} 625$

Solution:
Since, monthly salary $=$ Rs 15000 and sector angle of expenses $=15^{\circ}$
$\therefore \quad$ Amount $=\frac{15^{\circ}}{360^{\circ}} \times 15000=\mathrm{Rs} 625$

Topic: Mathematics, Statistics, NDA, NDA 2009 (Phase 1)
If the $n$ th term of an arithmetic progression is $2 n-1$, then what is the sum upto $n$ terms?

Options:
A $n^{2}$
B $n^{2}-1$
C $n^{2}+1$
D $\frac{1}{2} n(n+1)$

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

Solution:
Given $a_{n}=2 n-1$
$\therefore S_{n}=\sum_{k=1}^{n} a_{k}=\sum_{k=1}^{n}(2 n-1)$
$=2 \sum_{k=1}^{n} n-n=2 \cdot \frac{n(n+1)}{2}-n=n^{2}+n-n=n^{2}$

Topic: Mathematics, Sequences and Series, NDA, NDA 2008 (Phase 2)
If the nth term of an arithmetic progression is $3 \mathrm{n}+7$, then what is the sum of its first 50 terms?

Options:
A 3925
B 4100
C 4175
D 8200

The correct answer is:
4175

Solution:
As given, $\mathrm{n}_{\text {th }}$ term $\mathrm{T}_{\mathrm{n}}=3 \mathrm{n}+7$
Sum of n term, $\mathrm{S}_{\mathrm{n}}=\sum \mathrm{T}_{\mathrm{n}}$
$=\sum(3 n+7)=3 \sum n+7 \sum 1$
$=\frac{3 n(n+1)}{2}+7 n=n\left[\frac{3 n+3+14}{2}\right]=n\left[\frac{3 n+17}{2}\right]$
Sum of 50 terms $S_{50}=50\left[\frac{3 \times 50+17}{2}\right]$ $=50\left[\frac{167}{2}\right]=25 \times 167=4175$

Topic: Mathematics, Sequences and Series, NDA, NDA 2007 (Phase 2)
If the number 235 in decimal system is converted into binary system, then what is the resulting number?

Options:
A $(11110011)$
B $(11101011)$
C $(11110101)$
D $(11011011)$

The correct answer is:
$(11101011)$

Solution:
\begin{array}{l|l|l}
2 & 235 & 1 \\
\hline 2 & 117 & 1 \\
\hline 2 & 58 & 0 \\
\hline 2 & 29 & 1 \\
\hline 2 & 14 & 0 \\
\hline 2 & 7 & 1 \\
\hline 2 & 3 & 1 \\
\hline & 1 &
\end{array}
So, $(235)_{10}=(11101011)_{2}$

Topic: Mathematics, Sets and Relations, NDA, NDA 2016 (Phase 2)
If the number of terms of an A.P. is $(2 n+1)$, then what is the ratio of the sum of the odd terms to the sum of even terms?

Options:
A $\frac{n}{n+1}$
B $\frac{n^{2}}{n+1}$
C $\frac{n+1}{n}$
D $\frac{n+1}{2 n}$

The correct answer is:
$\frac{n+1}{n}$

Solution:
Let the $\mathrm{AP}$ is $\mathrm{a}, \mathrm{a}+\mathrm{d}, \mathrm{a}+2 \mathrm{~d}, \ldots \ldots, \mathrm{a}+(2 \mathrm{n}-1) \mathrm{d}, \mathrm{a}+2 \mathrm{nd}$
Series of even terms. $\mathrm{a}+\mathrm{d}, \mathrm{a}+3 \mathrm{~d}, \ldots \ldots \ldots, \mathrm{a}+(2 \mathrm{n}-1) \mathrm{d}$, has $\mathrm{n}$ terms
$\begin{aligned} \text { Sum of even number } &=\frac{n}{2}[(a+d)+\{a+(2 n-1) d\}] \\ &=\frac{n}{2}[2 a+2 n d]=n[a+n d] \end{aligned}$
Series of odd terms $\mathrm{a}, \mathrm{a}+2 \mathrm{~d}, \mathrm{a}+4 \mathrm{~d}, \ldots \ldots \ldots \ldots, \mathrm{a}+2 \mathrm{nd}$, has $(\mathrm{n}+1)$ terms.
$\begin{aligned} \text { Sum of odd numbers } &=\frac{n+1}{2}[a+(a+2 n d)] \\ &=\frac{n+1}{2}(2 a+2 n d) \\ &=(n+1)(a+n d) \end{aligned}$
So, the required ratio $=\frac{n+1}{n}$

Topic: Mathematics, Sequences and Series, NDA, NDA 2008 (Phase 2)
If the numbers $\mathrm{n}-3,4 \mathrm{n}-2,5 \mathrm{n}+1$ are in AP, what is the value of $\mathrm{n} ?$

Options:
A 1
B 2
C 3
D 4

The correct answer is:
1

Solution:
$\begin{aligned} & \because(\mathrm{n}-3), 4 \mathrm{n}-2,5 \mathrm{n}+1 \text { are in A.P. } \\ & \Rightarrow(4 \mathrm{n}-2)-(\mathrm{n}-3)=(5 \mathrm{n}+1)-(4 n-2) \\ & \Rightarrow 3 \mathrm{n}+1=\mathrm{n}+3 \\ & \Rightarrow \mathrm{n}=1 \end{aligned}$

Topic: Mathematics, Sequences and Series, NDA, NDA 2013 (Phase 1)
If the p. m. f. of a random variable $\mathrm{X}$ is
\begin{array}{|c|c|c|c|c|c|}
\hline \mathrm{X} & 1 & 2 & 3 & 4 & 5 \\
\hline \mathrm{P}(\mathrm{X}=x) & k & \frac{k}{3} & \frac{k}{4} & \frac{k}{2} & \frac{k}{2} \\
\hline
\end{array}
then $k=$

Options:
A $\frac{15}{31}$
B $\frac{1}{12}$
C $\frac{11}{12}$
D $\frac{12}{31}$

The correct answer is:
$\frac{12}{31}$

Solution:
Here $\mathrm{k}+\frac{\mathrm{k}}{3}+\frac{\mathrm{k}}{4}+\frac{\mathrm{k}}{2}+\frac{\mathrm{k}}{2}=1$
$\therefore \mathrm{k}\left(\frac{12+4+3+6+6}{12}\right)=1 \Rightarrow \mathrm{k}\left(\frac{31}{12}\right)=1 \Rightarrow \mathrm{k}=\frac{12}{13}$

Topic: Mathematics, Probability, MHT CET, MHT CET 2020 (13 Oct Shift 2)
If the perimeter of a triangle $\mathrm{ABC}$ is $30 \mathrm{~cm}$, then what is the value of $a \cos ^{2}(\mathrm{C} / 2)+c \cos ^{2}(\mathrm{~A} / 2) ?$

Options:
A $15 \mathrm{~cm}$
B $10 \mathrm{~cm}$
C $\frac{15}{2} \mathrm{~cm}$
D $13 \mathrm{~cm}$

The correct answer is:
$15 \mathrm{~cm}$

Solution:
We know from properties of triangle that
$\cos \frac{A}{2}=\sqrt{\frac{s(s-a)}{b c}} \Rightarrow \cos ^{2} \frac{A}{2}=\frac{s(s-a)}{b c}$
and $\cos \frac{\mathrm{C}}{2}=\sqrt{\frac{\mathrm{s}(\mathrm{s}-\mathrm{c})}{\mathrm{ab}}} \Rightarrow \cos ^{2} \frac{\mathrm{C}}{2}=\frac{\mathrm{s}(\mathrm{s}-\mathrm{c})}{\mathrm{ab}}$
So, $a \cos ^{2} \frac{C}{2}+c \cos ^{2} \frac{A}{2}$
$=\mathrm{a}\left(\frac{\mathrm{s}(\mathrm{s}-\mathrm{c})}{\mathrm{ab}}\right)+\mathrm{c}\left(\frac{\mathrm{s}(\mathrm{s}-\mathrm{a})}{\mathrm{bc}}\right)=\frac{\mathrm{s}(\mathrm{s}-\mathrm{c}+\mathrm{s}-\mathrm{a})}{\mathrm{b}}$
$=\frac{s(2 s-a-c)}{b}=\frac{s(a+b+c-a-c)}{b}=\frac{s \cdot b}{b}=s$
$=\frac{30}{2}=15 \mathrm{~cm}$ [given that $2 \mathrm{~s}=30$ ]

Topic: Mathematics, Inverse Trigonometric Functions, NDA, NDA 2006 (Phase 2)
If the p.m.f. of a r.v. $X$ is
then, the standard deviation of $\mathrm{X}$ is (given $p+q=1$ )

Options:
A $2 \sqrt{q}$
B $\sqrt{2 p q}$
C $2 \sqrt{p}$
D $\sqrt{p q}$

The correct answer is:
$\sqrt{2 p q}$

Solution:
(B)
\begin{array}{|c|c|c|c|}
\hline \mathrm{x}_{\mathrm{i}} & \mathrm{p}_{\mathrm{i}} & \mathrm{p}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}} & \mathrm{p}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}^{2} \\
\hline 0 & \mathrm{q}^{2} & 0 & 0 \\
\hline 1 & 2 \mathrm{pq} & 2 \mathrm{pq} & 2 \mathrm{pq} \\
\hline 2 & \mathrm{p}^{2} & 2 \mathrm{p}^{2} & 4 \mathrm{p}^{2} \\
\hline Total & & 2 \mathrm{pq}+2 \mathrm{p}^{2} & 2 \mathrm{pq}+4 \mathrm{p}^{2} \\
\hline
\end{array}
$\operatorname{Mean}(\mu)=\mathrm{E}(\mathrm{x})=\Sigma \mathrm{p}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}} \quad=2 \mathrm{p}(\mathrm{p}+\mathrm{q})=2 \mathrm{p} \quad \ldots[\because \mathrm{p}+\mathrm{q}=1$, given $]$
Variance $\left(\sigma_{\mathrm{x}}^{2}\right)=\sum \mathrm{p}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}{ }^{2}-\mu^{2}$
$=2 \mathrm{pq}+4 \mathrm{p}^{2}-4 \mathrm{p}^{2}=2 \mathrm{pq}$
Standard deviation $\left(\sigma_{x}\right)=\sqrt{2 \mathrm{pq}}$

Topic: Mathematics, Probability, MHT CET, MHT CET 2020 (19 Oct Shift 1)
If the point $(a, a)$ lies between the lines $|x+y|=2$, then which one of the following is correct?

Options:
A $|\mathrm{a}| < 2$
B $|\mathrm{a}| < \sqrt{2}$
C $|\mathrm{a}| < 1$
D $|\mathrm{a}| < \frac{1}{\sqrt{2}}$

The correct answer is:
$|\mathrm{a}| < 1$

Solution:
$|\mathrm{x}+\mathrm{y}|=2 \Rightarrow \mathrm{x}+\mathrm{y}=\pm 2$
$\Rightarrow \quad x+y+2=0$ and $x+y-2=0$
$\Rightarrow-2 < 2 a < 0$ and $-1 < a < 0$
$\Rightarrow \quad|a| < 1$

Topic: Mathematics, Straight Lines, NDA, NDA 2016 (Phase 2)
If the point $z_{1}=1+i$ where $i=\sqrt{-1}$ is the reflection of a point $\mathrm{z}_{2}=\mathrm{x}+$ iy in the line $\mathrm{i} \overline{\mathrm{z}}-\mathrm{i} \mathrm{z}=5$, then the point $\mathrm{z}_{2}$ is

Options:
A $\quad 1+4 \mathrm{i}$
B $4+\mathrm{i}$
C $\quad 1-i$
D $-1-i$

The correct answer is:
$\quad 1+4 \mathrm{i}$

Solution:
Let $z=a+b i$ 74
$\Rightarrow \bar{z}=\mathrm{a}-\mathrm{bi}$
$\therefore \mathrm{i} \overline{\mathrm{z}}-\mathrm{iz}=\mathrm{i}[(\mathrm{a}-\mathrm{bi})-(\mathrm{a}+\mathrm{bi})]=5$
$\Rightarrow \mathrm{i}[-2 \mathrm{bi}]=5$
$\Rightarrow \mathrm{b}=\frac{5}{2}$
So from figure it is clear that
$x=1, y=\frac{5}{2}+\frac{3}{2}=4$
$\mathrm{z}_{2}=1+4 \mathrm{i}$

Topic: Mathematics, Complex Number, NDA, NDA 2015 (Phase 2)
If the points $(2,4),(2,6)$ and $(2+\sqrt{3}, \mathrm{k})$ are the vertices of an equilateral triangle, then what is the value of $\mathrm{k}$ ?

Options:
A 6
B 5
C $-3$
D 1

The correct answer is:
5

Solution:
Since, $\mathrm{ABC}$ is an equilateral $\Delta$. $\mathrm{AB}=\mathrm{BC}=\mathrm{CA}$
Consider $\mathrm{BC}=\mathrm{AC}$
$\Rightarrow \sqrt{3+(\mathrm{k}-6)^{2}}=\sqrt{3+(\mathrm{k}-4)^{2}}$
$\Rightarrow \sqrt{39+\mathrm{k}^{2}-12 \mathrm{k}}=\sqrt{\mathrm{k}^{2}-8 \mathrm{k}+19}$
$\Rightarrow \quad \mathrm{k}^{2}-12 \mathrm{k}+39=\mathrm{k}^{2}-8 \mathrm{k}+19$
$\Rightarrow \quad-4 \mathrm{k}=-20$
$\mathrm{k}=5$

Topic: Mathematics, Straight Lines, NDA, NDA 2012 (Phase 2)
If the points $\mathrm{A}(1,2), \mathrm{B}(2,4)$ and $C(3, \mathrm{a})$ are collinear, what is the length $\mathrm{BC}$ ?

Options:
A $\sqrt{2}$ unit
B $\sqrt{3}$ unit
C $\sqrt{5}$ unit
D 5 unit

The correct answer is:
$\sqrt{5}$ unit

Solution:
Since the points are collinear.
$\left|\begin{array}{lll}1 & 2 & 1 \\ 2 & 4 & 1 \\ 3 & a & 1\end{array}\right|=0$
Expanding the determinant
$\Rightarrow 1\left|\begin{array}{ll}4 & 1 \\ \mathrm{a} & 1\end{array}\right|-2\left|\begin{array}{ll}2 & 1 \\ 3 & 1\end{array}\right|+1\left|\begin{array}{ll}2 & 4 \\ 3 & \mathrm{a}\end{array}\right|=0$
$\Rightarrow(4-a)-2(2-3)+1(2 a-12)=0$
$\Rightarrow \quad 4-a+2+2 a-12=0$
$\Rightarrow \quad a-6=0$
$\Rightarrow \quad a=6$
Thus, Coordinates of $\mathrm{C}$ are $(3,6)$. Thus, $\mathrm{BC}=\sqrt{(3-2)^{2}+(6-4)^{2}}$
$=\sqrt{1+4}=\sqrt{5}$ unit

Topic: Mathematics, Straight Lines, NDA, NDA 2006 (Phase 2)
If the points with the coordinates $(\mathrm{a}, \mathrm{ma}),\{\mathrm{b},(\mathrm{m}+1) \mathrm{b}\}$, $\{\mathrm{c},(\mathrm{m}+2) \mathrm{c}\}$ are collinear, then which one of the following is correct?

Options:
A $\mathrm{a}, \mathrm{b}, \mathrm{c}$ are in arithmetic progression for all $\mathrm{m}$
B $\mathrm{a}, \mathrm{b}, \mathrm{c}$ are in geometric progression for all $\mathrm{m}$
C $\mathrm{a}, \mathrm{b}, \mathrm{c}$ are in harmonic progression for all $\mathrm{m}$
D $\mathrm{a}, \mathrm{b}, \mathrm{c}$ are in arithmetic progression only for $\mathrm{m}=1$

The correct answer is:
$\mathrm{a}, \mathrm{b}, \mathrm{c}$ are in harmonic progression for all $\mathrm{m}$

Solution:
As given :
Points $\mathrm{A}(\mathrm{a}, \mathrm{ma}), \mathrm{B}[\mathrm{b},(\mathrm{m}+1) \mathrm{b}]$ and $\mathrm{C}[\mathrm{c},(\mathrm{m}+2) \mathrm{c}]$ are
collinear. $\Rightarrow \quad \mathrm{a}\{(\mathrm{m}+1) \mathrm{b}-(\mathrm{m}+2) \mathrm{c}\}+\mathrm{b}\{(\mathrm{m}+2) \mathrm{c}-\mathrm{ma}\}$
$+\mathrm{c}\{\mathrm{ma}-(\mathrm{m}+1) \mathrm{b}\}=0$
$\Rightarrow \quad \mathrm{mab}+\mathrm{ab}-\mathrm{mac}-2 \mathrm{ac}+\mathrm{mbc}+2 \mathrm{bc}-\mathrm{mab}+\mathrm{mac}$
$\quad-\mathrm{mbc}-\mathrm{bc}=0$
$\Rightarrow \mathrm{ab}-2 \mathrm{ac}+2 \mathrm{bc}-\mathrm{bc}=0$
$\Rightarrow \mathrm{ab}+\mathrm{bc}=2 \mathrm{ac}$
Dividing both the sides by abc, we get
$\frac{1}{c}+\frac{1}{a}=\frac{2}{b}$
$\Rightarrow \frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in A.P
$\Rightarrow \mathrm{a}, \mathrm{b}, \mathrm{c}$, are in Harmonic progression for all $\mathrm{m}$.

Topic: Mathematics, Sequences and Series, NDA, NDA 2007 (Phase 2)
If the position vector $\vec{a}$ of the point $(5, \mathrm{n})$ is such that $|\overrightarrow{\mathrm{a}}|=13$, then the value/values of $n$ an be

Options:
A $\pm 8$
B $\pm 12$
C 8 only
D 12 only

The correct answer is:
$\pm 12$

Solution:
$\sqrt{(5-0)^{2}+(n-0)^{2}}=13$
$25+n^{2}=169$
$n^{2}=144$
$n=\pm 12$

Topic: Mathematics, Vector Algebra, NDA, NDA 2015 (Phase 2)
If the position vector of a point $P$ with respect to origin $O$ is
$\hat{\mathbf{i}}+3 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}$ and that of a point $Q$ is $3 \hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}}$, then
what is the position vector of a point on the bisector of the angle $P O Q$ ?

Options:
A $\hat{\mathbf{i}}-\hat{\mathbf{j}}-\hat{\mathbf{k}}$
B $\hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}$
C $\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}$
D None of these

The correct answer is:
$\hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}$

Solution:
Let $\overrightarrow{\mathrm{OP}}=\hat{i}+3 \hat{j}-2 \hat{k}$ and $\overrightarrow{\mathrm{OQ}}=3 \hat{i}+\hat{j}-2 \hat{k}$
Let $\hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}}$ be required position vector of the bisector of the angle $P O Q$ since, it is the bisector of
$\angle P O Q$ therefore. It will make equal angles with $\overline{O P}$
and $\overrightarrow{\mathrm{OQ}}$.
Let Angle between $\hat{i}+3 \hat{j}-2 \hat{k}$ and $\hat{i}+\hat{j}-\hat{k}$ is
$\theta=\cos ^{-1}\left(\frac{1+3+2}{\sqrt{1+9+4} \sqrt{1+1+1}}\right)$
$=\cos ^{-1}\left(\frac{6}{\sqrt{14} \sqrt{3}}\right)$
and angle between $3 \hat{i}+\hat{j}-2 \hat{k}$ and $\hat{i}+\hat{j}-\hat{k}$, is
$\phi=\cos ^{-1}\left(\frac{1+3+2}{\sqrt{9+1+4} \sqrt{1+1+1}}\right)=\cos ^{-1}\left(\frac{6}{\sqrt{14} \sqrt{3}}\right)$
Hence, $\theta=\phi$

Topic: Mathematics, Vector Algebra, NDA, NDA 2010 (Phase 2)
If the position vectors of points $\mathrm{A}$ and $\mathrm{B}$ are $3 \hat{i}-2 \hat{j}+\hat{k}$
and $2 \hat{i}+4 \hat{j}-3 \hat{k}$ respectively, then what is the length of
$\overrightarrow{A B} ?$

Options:
A $\sqrt{14}$
B $\sqrt{29}$
C $\sqrt{43}$
D $\sqrt{53}$

The correct answer is:
$\sqrt{53}$

Solution:
$\begin{aligned} \text { Given, } \overrightarrow{O A}=3 \hat{i}-2 \hat{j}+\hat{k} \\ & \overrightarrow{O B}=2 \hat{i}+4 \hat{j}-3 \hat{k} \\ & \overrightarrow{A B}=\overrightarrow{O B}-\overrightarrow{O A} \\ &=(2 \hat{i}+4 \hat{j}-3 \hat{k})-(3 \hat{i}-2 \hat{j}+\hat{k}) \\ &=\hat{i}(2-3)+\hat{j}(4+2)+\hat{k}(-3-1) \\ &=-\hat{i}+6 \hat{j}-4 \hat{k} \\ & \text { Length of } \mathrm{AB}=\sqrt{(-1)^{2}+(6)^{2}+(-4)^{2}} \\ &=\sqrt{1+36+16}=\sqrt{53} \end{aligned}$

Topic: Mathematics, Vector Algebra, NDA, NDA 2019 (Phase 1)
If the positive integers $a, b, c, d$ are in $A P$, then the numbers $a b c, a b d, a c d, b c d$ are in

Options:
A $H P$
B $A P$
C $G P$
D None of the above

The correct answer is:
$H P$

Solution:
Given, $\mathrm{a}, \mathrm{b}, \mathrm{c}, \mathrm{d}$ are in A.P.
$\Rightarrow \frac{1}{a}, \frac{1}{b}, \frac{1}{c}, \frac{1}{d}$ are in H.P.
$\Rightarrow \frac{1}{d}, \frac{1}{c}, \frac{1}{b}, \frac{1}{a}$ are also in H.P.
Now, multiply each term by abcd.
$\frac{\text { abcd }}{\mathrm{d}}, \frac{\text { abcd }}{\mathrm{c}}, \frac{\text { abcd }}{\mathrm{b}}, \frac{\text { abcd }}{\mathrm{a}}$
abc, abd, acd, bed, are in H.P.

Topic: Mathematics, Sequences and Series, NDA, NDA 2013 (Phase 2)
If the probability of simultaneous occurrence of two events A and B is p and the probability that exactly one of $\mathrm{A}$. B occurs is q, then which of the following is/are correct?
$1.$ $\quad \mathrm{P}(\overline{\mathrm{A}})+\mathrm{P}(\overline{\mathrm{B}})=2-2 \mathrm{p}-\mathrm{q}$
$2.$ $\quad \mathrm{P}(\overline{\mathrm{A}} \cap \overline{\mathrm{B}})=1-\mathrm{p}-\mathrm{q}$
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:
Both 1 and 2

Solution:
$\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-2 \mathrm{P}(\mathrm{A} \cap \mathrm{B})=q$
$\mathrm{P}(\mathrm{A} \cap \mathrm{B})=p$
$\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})=2 p+q$
$1-\mathrm{P}(\overline{\mathrm{A}})+1-\mathrm{P}(\overline{\mathrm{B}})=p+q$
$\mathrm{P}(\overline{\mathrm{A}})+\mathrm{P}(\overline{\mathrm{B}})=2-2 p-q$
$\mathrm{P}(\overline{\mathrm{A}} \cap \overline{\mathrm{B}})=1-\mathrm{P}(\mathrm{A} \cup \mathrm{B})$
$=1-(q+p)=1-p-q$

Topic: Mathematics, Probability, NDA, NDA 2018 (Phase 2)
If the product of the roots of the equation $x^{2}-5 x+k=15$ is
$-3$, then what is the value of $k$ ?

Options:
A 12
B 15
C 16
D 18

The correct answer is:
12

Solution:
The given quadratic equation is
$x^{2}-5 x+k=15 \Rightarrow x^{2}-5 x+k-15=0$
Let $\alpha$ and $\beta$ be the roots of the equation $x^{2}-5 x+k-15=0$
Now, product of roots $=\alpha \beta=k-15$ But $\alpha \beta=-3 \Rightarrow-3=k-15$
$\Rightarrow k=15-3=12$

Topic: Mathematics, Quadratic Equation, NDA, NDA 2010 (Phase 1)
If the projections of a straight line segment on the coordinate axes are $2,3,6$, then the length of the segment is

Options:
A 5 units
B 7 units
C 11 units
D 49 units

The correct answer is:
7 units

Solution:
Position vector of line segment $=2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+6 \hat{\mathrm{k}}$
length $=\sqrt{2^{2}+3^{2}+6^{2}}=\sqrt{49}=7$ units

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2013 (Phase 2)
If the radius of the sphere $x^{2}+y^{2}+z^{2}-6 x-8 y+10 z+\lambda=0$ is unity, what is the value of $\lambda ?$

Options:
A 49
B 7
C $-49$
D $-7$

The correct answer is:
49

Solution:
Given sphere : $x^{2}+y^{2}+z^{2}-6 x-8 y+10 z+\lambda=0$ Its radius $=1$
$\Rightarrow \sqrt{(-3)^{2}+\left(-4^{2}\right)+(5)^{2}-\lambda}=1$
$\Rightarrow 9+16+25-\lambda=1$
$\therefore \lambda=49$

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2008 (Phase 2)
If the rate of change in volume of spherical soap bubble is uniform, then the rate of change of surface area varies as

Options:
A $\begin{array}{ll}\text { } \text { square of radius } & {}\end{array}$
B square root of radius
C inversely proportional to radius
D cube of the radius

The correct answer is:
inversely proportional to radius

Solution:
Let volume $=\mathrm{V}=\frac{4}{3} \pi r^{3}$ and surface area $=\mathrm{S}=4 \pi r^{2}$
Now, $(1) \Rightarrow \frac{d v}{d t}=\frac{4}{3} \times 3 \pi r^{2} \times \frac{d r}{d t}$
$\quad=4 \pi r^{2} \frac{d r}{d t}$
(2) $\Rightarrow \frac{d s}{d t}=4 \pi \times 2 \times r \frac{d r}{d t}=\frac{8 \pi r^{2}}{r} \frac{d r}{d t}$
$=\frac{2}{r}\left[4 \pi r^{2} \frac{d r}{d t}\right]=\frac{2}{r} \frac{d v}{d t} \quad($ from 3$)$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2011 (Phase 2)
If the ratio of AM to GM of two positive numbers $a$ and $b$ is $5: 3$, then $\mathrm{a}: \mathrm{b}$ is equal to

Options:
A $3: 5$
B $2: 9$
C $9: 1$
D $5: 3$

The correct answer is:
$9: 1$

Solution:
A.M of two numbers a, $\mathrm{b}=\frac{\mathrm{a}+\mathrm{b}}{2}$
G. $\mathrm{M}$ of two numbers $\mathrm{a}, \mathrm{b}=\sqrt{\mathrm{ab}}$.
Given, $\frac{\mathrm{A} \cdot \mathrm{M}}{\mathrm{G} \cdot \mathrm{M}}=\frac{5}{3}$
$\Rightarrow \frac{\left(\frac{a+b}{2}\right)}{\sqrt{a b}}=\frac{5}{3}$
$\Rightarrow \frac{a+b}{\sqrt{a b}}=\frac{10}{3}$
$\Rightarrow \frac{\mathrm{a}^{2}+\mathrm{b}^{2}+2 \mathrm{ab}}{\mathrm{ab}}=\frac{100}{9} \quad$ (Squaring on both sides)
$\Rightarrow \quad 9 a^{2}+9 b^{2}+18 a b=100 a b$
$\Rightarrow \quad 9 a^{2}-82 a b+9 b^{2}=0$
$\Rightarrow \quad 9 a^{2}-81 a b-a b+9 b^{2}=0$
$\Rightarrow \quad 9 a(a-9 b)-b(a-9 b)=0$
$\Rightarrow \quad(9 a-b)(a-9 b)=0$
$\Rightarrow \quad 9 a-b=0 ; a-9 b=0$
$\Rightarrow \mathrm{b}=9 \mathrm{a} ; \mathrm{a}=9 \mathrm{~b}$
$\Rightarrow \frac{\mathrm{a}}{\mathrm{b}}=\frac{1}{9} ; \frac{\mathrm{a}}{\mathrm{b}}=\frac{9}{1}$

Topic: Mathematics, Sequences and Series, NDA, NDA 2018 (Phase 1)
If the regression coefficient of $x$ on $\mathrm{vand}$ $x$ are $-\frac{1}{2}$
and $-\frac{1}{2}$ respectively, then what is the correlation
coefficient between $\mathrm{x}$ and $\mathrm{y} ?$

Options:
A $-\frac{1}{4}$
B $-\frac{1}{16}$
C $\frac{1}{16}$
D $\frac{1}{4}$

The correct answer is:
$-\frac{1}{4}$

Solution:
Correlation coefficient $=\sqrt{\frac{-1}{2} \times \frac{-1}{8}}$
$\sqrt{\frac{1}{16}}=-\frac{1}{4}$
Since, both the regression coefficients are negative, correlation coefficient is negative.

Topic: Mathematics, Statistics, NDA, NDA 2017 (Phase 1)
If the regression coefficient of Yon $\mathrm{X}$ is $-6$, and the correlation coefficient between $\mathrm{X}$ and $\mathrm{Y}-\frac{1}{2}$, then the regression coefficient of $X$ on Y would be $\quad$

Options:
A $\frac{1}{24}$
B $-\frac{1}{24}$
C $-\frac{1}{6}$
D $\frac{1}{6}$

The correct answer is:
$-\frac{1}{24}$

Solution:
$b_{y x}=-6, r=-\frac{1}{2}$
$\left(-\frac{1}{2}\right)^{2}=-6 \times b_{x y}$
$\Rightarrow b_{x y}=-\frac{1}{24}$

Topic: Mathematics, Statistics, NDA, NDA 2018 (Phase 2)
If the roots of $a x^{2}+b x+c=0$ are $\sin \alpha$ and $\cos \alpha$ for some $\alpha$, then which one of the following is correct?

Options:
A $a^{2}+b^{2}=2 a c$
B $b^{2}-c^{2}=2 a b$
C $b^{2}-a^{2}=2 a c$
D $b^{2}+c^{2}=2 a b$

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

Solution:
Let $\sin \alpha$ and $\cos \alpha$ be the roots of $a x^{2}+b x+c=0$ Now, $\sin \alpha+\cos \alpha=\frac{-b}{a}$ and $\sin \alpha \cos \alpha=\frac{c}{a}$
Consider $\sin \alpha+\cos \alpha=\frac{-b}{a}$
Squaring both side, $(\sin \alpha+\cos \alpha)^{2}=\frac{b^{2}}{a^{2}}$
$\Rightarrow \sin ^{2} \alpha+\cos ^{2} \alpha+2 \sin \alpha \cos \alpha=\frac{\mathrm{b}^{2}}{\mathrm{a}^{2}}$
$\Rightarrow 1+\frac{2 c}{a}=\frac{b^{2}}{a^{2}}$
$\Rightarrow \frac{a+2 c}{a}=\frac{b^{2}}{a^{2}} \Rightarrow a+2 c=\frac{\mathrm{b}^{2}}{\mathrm{a}}$
$\Rightarrow a^{2}+2 a c=b^{2}
\Rightarrow b^{2}-a^{2}=2 a c$

Topic: Mathematics, Quadratic Equation, NDA, NDA 2009 (Phase 2)