datavorous/entrance-exam-dataset huggingface dataset:
The sum of the first $(2 \mathrm{p}+1)$ terms of an APis $\{(\mathrm{p}+1)$. (2p $+1$ )\}. Which one of the following inferences can be
drawn?

Options:
A The $(\mathrm{p}+1)^{\text {th }}$ term of the AP is $(2 \mathrm{p}+1)$
B The $(2 \mathrm{p}+1)^{\text {th }}$ term of the APis $(2 \mathrm{p}+1)$
C The $(2 \mathrm{p}+1)^{\text {th }}$ term of the $\mathrm{AP}$ is $(\mathrm{p}+1)$
D The $(p+1)^{\text {th }}$ term of the $\mathrm{AP}$ is $(\mathrm{p}+1)$

The correct answer is:
The $(p+1)^{\text {th }}$ term of the $\mathrm{AP}$ is $(\mathrm{p}+1)$

Solution:
Let the first term and common difference of an AP be a and d respectively. Then, as given $(p+1)(2 p+1)=\left(\frac{2 p+1}{2}\right)\{2 a+(2 p+1-1) d\}$
$\Rightarrow \quad(p+1)=\frac{1}{2}\{2 a+2 p d\}$
$\Rightarrow(\mathrm{p}+1)=\mathrm{a}+\mathrm{pd}$
$\Rightarrow \quad p+1=a+[(p+1)-1] d=t_{p+1}$
Hence, the inference is: the $(p+1)^{\text {th }}$ term of the $A P$ is $(p+1)$

Topic: Mathematics, Sequences and Series, NDA, NDA 2006 (Phase 2)
The sum of the first five terms and the sum of the first ten terms of an $A P$ are same. Which one of the following is the correct statement?

Options:
A The first term must be negative
B The common difference must be negative
C Either the first term or the common difference is negative but not both
D Both the first term and the common difference are negative

The correct answer is:
Either the first term or the common difference is negative but not both

Solution:
$\begin{aligned} & \mathrm{S}_{5}=\mathrm{S}_{10} \\ & \Rightarrow \frac{5}{2}[2 \mathrm{a}+4 \mathrm{~d}]=\frac{10}{2}[2 \mathrm{a}+9 \mathrm{~d}] \\ & \Rightarrow 5 \mathrm{a}+10 \mathrm{~d}=10 \mathrm{a}+45 \mathrm{~d} \\ & \Rightarrow \mathrm{a}=-7 \mathrm{~d} \text { or } \mathrm{d}=\frac{-1}{7} \mathrm{a} \end{aligned}$
We see, If $\mathrm{d}$ is positive, then first term should be negative and common difference should be positive. If $\mathrm{d}$ is negative, then first term should be position and common difference should be negative.

Topic: Mathematics, Sequences and Series, NDA, NDA 2013 (Phase 2)
The sum of the first n terms of the series $\frac{1}{2}+\frac{3}{4}+\frac{7}{8}+\frac{15}{16}+\ldots .$ is equal to

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

The correct answer is:
$2^{-\mathrm{n}}+\mathrm{n}-1$

Solution:
$\begin{aligned} & \frac{1}{2}+\frac{3}{4}+\frac{7}{8}+\frac{15}{16}+\ldots . . \\ &=\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{4}\right)+\left(1-\frac{1}{8}\right)+\left(1-\frac{1}{16}\right)+\ldots \ldots \\ &=(1+1+1+\ldots . \mathrm{n})-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\ldots . .\right) \end{aligned}$
$=n-\left(\frac{1}{2}+\frac{1}{2^{2}}+\frac{1}{2^{3}}+\frac{1}{2^{4}}+\ldots .\right)$
$=n-\frac{\frac{1}{2}\left(1-\frac{1}{2^{n}}\right)}{1-\frac{1}{2}} \quad\left(\because\right.$ G.P. $\left.a=\frac{1}{2}, r=\frac{1}{2}\right)$
$=n-\frac{\frac{1}{2}\left(1-\frac{1}{2^{n}}\right)}{\frac{1}{2}}$
$=n-1+2^{-n}$
$=2^{-n}+n-1$

Topic: Mathematics, Sequences and Series, NDA, NDA 2017 (Phase 1)
The sum of the focal distances of a point on an ellipse is constant and equal to the

Options:
A length of minor axis
B length of major axis
C length of latus rectum
D sum of the lengths of semi-major and semi-minor axes

The correct answer is:
length of major axis

Solution:
No Solution

Topic: Mathematics, Ellipse, NDA, NDA 2019 (Phase 1)
The sum of the focal distances of a point on the ellipse
$\frac{\mathrm{x}^{2}}{4}+\frac{\mathrm{y}^{2}}{9}=1$ is:

Options:
A 4 units
B 6 units
C 8 units
D 10 units

The correct answer is:
4 units

Solution:
Given equation of ellipse is
$\frac{x^{2}}{4}+\frac{y^{2}}{9}=1$
$\Rightarrow \quad \frac{x^{2}}{(2)^{2}}+\frac{y^{2}}{(3)^{2}}=1$
$\Rightarrow \quad a=2$ and $b=3$
Length of major axis $=2 \mathrm{a}=4$
Since, we have Sum of the focal distances of a point on ellipse $=$ length of major axis. Required Ans $=4$ units.

Topic: Mathematics, Ellipse, NDA, NDA 2012 (Phase 2)
The sum of the roots of the equation $a x^{2}+x+c=0$ (where a and $\mathrm{c}$ are non-zero) is equal to the sum of the reciprocals of their squares. Then $\mathrm{a}, \mathrm{ca}^{2}, \mathrm{c}^{2}$ are in

Options:
A $\mathrm{AP}$
B GP
C $\mathrm{HP}$
D None of the above

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

Solution:
Let $\alpha, \beta$ be roots of $a x^{2}+x+c=0$. Given, $\alpha+\beta$
$=\frac{1}{\alpha^{2}}+\frac{1}{\beta^{2}}=\frac{\alpha^{2}+\beta^{2}}{\alpha^{2} \beta^{2}}=\frac{(\alpha+\beta)^{2}-2 \alpha \beta}{(\alpha \beta)^{2}}$
$\Rightarrow \frac{-1}{a}=\frac{\left(\frac{-1}{a}\right)^{2}-2\left(\frac{c}{a}\right)}{\left(\frac{c}{a}\right)^{2}}$
$\Rightarrow \frac{-1}{a}=\frac{\frac{1}{a^{2}}-\frac{2 c}{a}}{\frac{c^{2}}{a^{2}}}$
$\Rightarrow \frac{-1}{\mathrm{a}}=\frac{1-2 \mathrm{ac}}{\mathrm{c}^{2}} \Rightarrow-\mathrm{c}^{2}=\mathrm{a}-2 \mathrm{a}^{2} \mathrm{c}$
$\Rightarrow 2 \mathrm{a}^{2} \mathrm{c}=\mathrm{a}+\mathrm{c}^{2}$
So, $a, a^{2} c, c^{2}$ are in A.P.

Topic: Mathematics, Sequences and Series, NDA, NDA 2017 (Phase 1)
The sum of the roots of the equation $x^{2}+b x+c=0$ (where $\mathrm{b}$ and $\mathrm{c}$ are non-zero) is equal to the sum of the reciprocals
of their squares. Then $\frac{1}{\mathrm{c}}, \mathrm{b}, \frac{\mathrm{c}}{\mathrm{b}}$ are in

Options:
A AP
B GP
C HP
D None of the above

The correct answer is:
HP

Solution:
Let $\alpha, \beta$ be the roots of $x^{2}+b x+c=0$ Given, $\alpha+\beta=\frac{1}{\alpha^{2}}+\frac{1}{\beta^{2}}$
$\alpha+\beta=-b ; \alpha, \beta=c$
$\alpha+\beta=\frac{1}{\alpha^{2}}+\frac{1}{\beta^{2}}$
$\Rightarrow \alpha+\beta=\frac{\alpha^{2}}{\alpha^{2}}+\frac{\beta^{2}}{\beta^{2}}=\frac{(\alpha+\beta)^{2}-2 \alpha \beta}{\alpha^{2} \beta^{2}}$
$\Rightarrow-b=\frac{b^{2}-2 c}{c^{2}}$
$\Rightarrow-b c^{2}=b^{2}-2 c \Rightarrow 2 c=b^{2}+b c^{2}$
$\Rightarrow \frac{2 \mathrm{c}}{\mathrm{b}}=\mathrm{b}+\mathrm{c}^{2} \Rightarrow \frac{2}{\mathrm{~b}}=\frac{\mathrm{b}}{\mathrm{c}}+\mathrm{c}$
We know, if $a, b, c$ are in A.P. $2 b=a+c$.
Similarly, we got $\frac{2}{b}=\frac{b}{c}+c$, which means $\frac{b}{c}, \frac{1}{b}, c$ are in A.P.
So, $\frac{\mathrm{c}}{\mathrm{b}}, \mathrm{b}, \frac{1}{\mathrm{c}}$ are in H.P.

Topic: Mathematics, Sequences and Series, NDA, NDA 2017 (Phase 1)
The sum of the series $3-1+\frac{1}{3}-\frac{1}{9}+\ldots .$ is equal to

Options:
A $\frac{20}{9}$
B $\frac{9}{20}$
C $\frac{9}{4}$
D $\frac{4}{9}$

The correct answer is:
$\frac{9}{4}$

Solution:
The series is $\frac{3}{3^{0}}-\frac{3}{3^{1}}+\frac{3}{3^{2}}-\frac{3}{3^{3}}+\ldots . \infty$
This is a G.P. with first term $a=3$
and common ratio $r=-\frac{1}{3}$
$\therefore \operatorname{Sum} \mathrm{S}=\frac{3}{1-\left(-\frac{1}{3}\right)}=\frac{3 \times 3}{3+1}=\frac{9}{4}$

Topic: Mathematics, Sequences and Series, NDA, NDA 2018 (Phase 2)
The sum of the series formed by the sequence $3, \sqrt{3}, 1 \ldots \ldots$ upto infinity is :

Options:
A $\frac{3 \sqrt{3}(\sqrt{3}+1)}{2}$
B $\frac{3 \sqrt{3}(\sqrt{3}-1)}{2}$
C $\frac{3(\sqrt{3}+1)}{2}$
D $\frac{3(\sqrt{3}-1)}{2}$

The correct answer is:
$\frac{3 \sqrt{3}(\sqrt{3}+1)}{2}$

Solution:
$3, \sqrt{3}, 1, \frac{1}{\sqrt{3}}, \ldots \ldots, \infty$
This is a Geometric Progression with $\mathrm{a}=3, \mathrm{r}=\frac{1}{\sqrt{3}}$
$\begin{aligned} \mathrm{S}_{\infty} &=\frac{\mathrm{a}}{1-\mathrm{r}}=\frac{3}{1-\frac{1}{\sqrt{3}}} \\ &=\frac{3 \sqrt{3}}{\sqrt{3}-1} \times \frac{\sqrt{3}+1}{\sqrt{3}+1}=\frac{3 \sqrt{3}(\sqrt{3}+1)}{2} \end{aligned}$

Topic: Mathematics, Sequences and Series, NDA, NDA 2014 (Phase 1)
The sum of twonumbers is 20 . What are the numbers if the product of the square of one and the cube of the other is maxinmm?

Options:
A 6,14
B 15,5
C 12,8
D 10,10

The correct answer is:
12,8

Solution:
Let the numbers are $\mathrm{x}$ and $\mathrm{y}$. So, $x+y=20$; Let $P=x^{2} y^{3} \quad$ (As given) $=x^{2}(20-x)^{3}$
Differentiating w. r. t. $\mathrm{x}$ $\begin{aligned} \frac{d P}{d x} &=x^{2} \cdot 3(20-x)^{2}(-1)+(20-x)^{3} \cdot 2 x \\ &=(20-x)^{2}\left[-3 x^{2}+40 x-2 x^{2}\right] \\ &=(20-x)^{2}\left[40 x-5 x^{2}\right] \\ \frac{d^{2} P}{d x^{2}} &=(20-x)^{2}[40-10 x]+\left(40 x-5 x^{2}\right) 2(20-x)(-1) \end{aligned}$
$\frac{d p}{d x}=0$ for maxima or minima. So, $(20-x)^{2}\left[40 x-5 x^{2}\right]=0$
$\Rightarrow(20-\mathrm{x})^{2} \times(\mathrm{x})(40-5 \mathrm{x})=0 \Rightarrow \mathrm{x}=20,0,8$
We get, $\left(\frac{\mathrm{d}^{2} \mathrm{P}}{\mathrm{dx}^{2}}\right)_{\mathrm{x}=8} < 0 ;\left(\frac{\mathrm{d}^{2} \mathrm{p}}{\mathrm{dx}^{2}}\right)_{\mathrm{x}=0}>0$ and
$\left(\frac{\mathrm{d}^{2} \mathrm{p}}{\mathrm{dx}^{2}}\right)_{\mathrm{x}=20}=0$
Hence. $\mathrm{P}$ is maximumat $\mathrm{x}=8$
and, Numbers are 12 and 8 .

Topic: Mathematics, Application of Derivatives, NDA, NDA 2006 (Phase 1)
The supernatant left after sedimentation of sewage water during primary treatment is known as

Options:
A affluent
B flocs
C sludge
D effluent

The correct answer is:
effluent

Solution:
Primary treatment is a physical process which involves filteration and sedimentation. After sedimentation the part which settles down is called as primary sludge and the water which comes above is called effluent, it contains dissolved organic substance and is transferred to secondary treatment plant.

Topic: Biology, Microbes in Human Welfare, NEET
The suspended solids in sewage can be removed at which step in the process of sewage treatment?

Options:
A Secondary treatment
B Primary treatment
C Sludge treatment
D Tertiary treatment

The correct answer is:
Primary treatment

Solution:
Before disposal, sewage is treated in sewage treatment plants (STPs) to make it less polluting. The initial steps basically involve the physical removal of particles â large and small â from the sewage through filtration and sedimentation, which is known as the Primary Treatment. These are removed in stages: initially, floating debris is removed by sequential filtration. Then the grit (soil and small pebbles) are removed by sedimentation. All solids that settle form the primary sludge, and the supernatant forms the effluent. The effluent from the primary settling tank is taken for secondary treatment.

Topic: Biology, Microbes in Human Welfare, NEET
The sweet potato family is included in the order

Options:
A Polymoniales
B Sapindales
C Poales
D Asparagales

The correct answer is:
Polymoniales

Solution:
Ipomoea batatas or sweet potato is a plant which belongs to the family Convolvulaceae. It is also called the Morning glory family. Along with the potato family, Solanaceae is included in the order Polymoniales.

Topic: Biology, The Living World, NEET
The symmetric part of the matrix \( A=\left(\begin{array}{ccc}1 & 2 & 4 \\ 6 & 8 & 2 \\ 2 & -2 & 7\end{array}\right) \) is

Options:
A \( \left(\begin{array}{lll}1 & 4 & 3 \\ 2 & 8 & 0 \\ 3 & 0 & 7\end{array}\right) \)
B \( \left(\begin{array}{lll}1 & 4 & 3 \\ 4 & 8 & 0 \\ 3 & 0 & 7\end{array}\right) \)
C \( \left(\begin{array}{ccc}0 & -2 & -1 \\ -2 & 0 & -2 \\ -1 & -2 & 0\end{array}\right) \)
D \( \left(\begin{array}{ccc}0 & -2 & 1 \\ 2 & 0 & 2 \\ -1 & 2 & 0\end{array}\right) \)

The correct answer is:
\( \left(\begin{array}{lll}1 & 4 & 3 \\ 4 & 8 & 0 \\ 3 & 0 & 7\end{array}\right) \)

Solution:
Given that \( A=\left[\begin{array}{ccc}1 & 2 & 4 \\ 6 & 8 & 2 \\ 2 & -2 & 7\end{array}\right] \)
Symmetric part of A matrix is \( \frac{1}{2}\left(A+A^{\prime}\right) \)
\( =\frac{1}{2}\left[\begin{array}{ccc}1 & 2 & 4 \\ 6 & 8 & 2 \\ 2 & -2 & 7\end{array}\right]+\left[\begin{array}{ccc}1 & 6 & 2 \\ 2 & 8 & -2 \\ 4 & 2 & 7\end{array}\right] \)
\( =\left[\begin{array}{lll}1 & 4 & 3 \\ 4 & 8 & 0 \\ 3 & 0 & 7\end{array}\right] \)

Topic: Mathematics, Indefinite Integration, KCET, KCET 2014
The system of equation
$2 x+y-3 z=5$
$3 x-2 y+2 z=5$ and
$5 x-3 y-z=16$

Options:
A is inconsistent
B is consistent, with a unique solution
C is consistent, with infinitely many solutions
D has its solution lying along $\mathrm{x}$ -axis in three - dimensional space

The correct answer is:
is consistent, with a unique solution

Solution:
Since $\left|\begin{array}{ccc}2 & 1 & -3 \\ 3 & -2 & 2 \\ 5 & -3 & -1\end{array}\right|=26 \neq 0$
$\Rightarrow$ System is consistent with unique solution.

Topic: Mathematics, Matrices, NDA, NDA 2018 (Phase 2)
The system of equations $k x+y+z=1, x+k y+z=k$ and $x+y+k z=k^{2}$ has no solution if $k$ equals

Options:
A 0
B 1
C $-1$
D $-2$

The correct answer is:
$-2$

Solution:
$\begin{array}{l}
\mathrm{kx}+\mathrm{y}+\mathrm{z}=1 \\
\mathrm{x}+\mathrm{ky}+\mathrm{z}=\mathrm{k} \\
\mathrm{x}+\mathrm{y}+\mathrm{k} z=\mathrm{k}^{2}
\end{array}$
These equations will have no solution of $\left|\begin{array}{ccc}\mathrm{k} & 1 & 1 \\ 1 & \mathrm{k} & 1 \\ 1 & 1 & \mathrm{k}\end{array}\right|=0$.
$\Rightarrow \mathrm{k}\left(\mathrm{k}^{2}-1\right)-1(\mathrm{k}-1)+(1-\mathrm{k})=0$
$\Rightarrow \mathrm{k}(\mathrm{k}+1)(\mathrm{k}-1)-1(\mathrm{k}-1)-(\mathrm{k}-1)=0$
$\Rightarrow(\mathrm{k}-1)[\mathrm{k}(\mathrm{k}+1)-1-1]=0$
$\Rightarrow(\mathrm{k}-1)\left(\mathrm{k}^{2}+\mathrm{k}-2\right)=0$
$\Rightarrow \mathrm{k}=1$ or $-2$
For $\mathrm{k}=1$, all equations are same and have infinite solution. So, for $\mathrm{k}=-2$, equations have no solution.

Topic: Mathematics, Determinants, NDA, NDA 2017 (Phase 2)
The system of linear equations $\mathrm{kx}+\mathrm{y}+\mathrm{z}=1, \mathrm{x}+\mathrm{ky}+\mathrm{z}=1$ and
$\mathrm{x}+\mathrm{y}+\mathrm{kz}=1$ has a unique solution under which one of the following conditions?

Options:
A $k \neq 1$ and $k \neq-2$
B $\mathrm{k} \neq 1$ and $\mathrm{k} \neq 2$
C $\mathrm{k} \neq-1$ and $\mathrm{k} \neq-2$
D $\mathrm{k} \neq-1$ and $\mathrm{k} \neq 2$

The correct answer is:
$k \neq 1$ and $k \neq-2$

Solution:
Linear equations $K x+y+z=1$
$x+K y+z=1$
$x+y+K z=1$
$\left[\begin{array}{ccc}K & 1 & 1 \\ 1 & K & 1 \\ 1 & 1 & K\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]$
$A X=B$
$X=A^{-1} B$
Linear equantion will have unique solution when $A^{-1}$ exist:
$|A| \neq 0$
$\left[\begin{array}{lll}K & 1 & 1 \\ 1 & K & 1 \\ 1 & 1 & K\end{array}\right] \neq 0$
$K\left(K^{2}-1\right)-1(K-1)+1(1-K) \neq 0$
$\Rightarrow K^{3}-K-K+1+1-K \neq 0$
$\Rightarrow K^{3}-3 K+2 \neq 0$
$(K-1)\left(K^{2}+K-2\right) \neq 0$
$(K-1)(K-1)(K+2) \neq 0$
$K \neq 1, K \neq 1$ and $K \neq-2$
$\Rightarrow K \neq 1$ and $K \neq-2$

Topic: Mathematics, Determinants, NDA, NDA 2016 (Phase 1)
The table below gives an incomplete frequency distribution with two missing frequencies $f_{1}$ and $f_{2}$ \begin{array}{|c|c|}
\hline Value of x & Frequency \\
\hline 0 & f_{1} \\
1 & f_{2} \\
2 & 4 \\
3 & 4 \\
4 & 3 \\
\hline
\end{array} The total frequency is 18 and the arithmetic mean of $x$ is 2 .
What is the value of $f_{2}$ ?

Options:
A 4
B 3
C 2
D 1

The correct answer is:
4

Solution:
\begin{array}{|c|c|c|}
\hlinex & f & x f \\
\hline 0 & f_{1} & 0 \\
1 & f_{2} & f_{2} \\
2 & 4 & 8 \\
3 & 4 & 12 \\
4 & 3 & 12 \\
\hline Total & f_{1}+f_{2}+11 & 32+f_{2} \\
\hline
\end{array} Since, total frequency is 18
$f_{1}+f_{2}+11=18$
$\Rightarrow f_{1}+f_{2}=7$
As we have, Mean $=\frac{\Sigma x f}{\Sigma f}=2$
(i)
$\frac{32+f_{2}}{18}=2 \Rightarrow f_{2}=36-32=4$
On putting the value of $f_{2}$ in Eq. (i), we get $f_{1}=7-4=3$
$f_{2}=4$

Topic: Mathematics, Statistics, NDA, NDA 2010 (Phase 1)
The table below gives an incomplete frequency distribution with two missing frequencies $f_{1}$ and $f_{2}$ \begin{array}{|c|c|}
\hline Value of x & Frequency \\
\hline 0 & f_{1} \\
1 & f_{2} \\
2 & 4 \\
3 & 4 \\
4 & 3 \\
\hline
\end{array} The total frequency is 18 and the arithmetic mean of $x$ is 2 .
What is the standard deviation? $\quad$

Options:
A $\frac{\sqrt{5}}{2}$
B $\frac{\sqrt{5}}{3}$
C $\frac{4}{3}$
D $\frac{16}{9}$

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

Solution:
\begin{array}{|c|c|c|}
\hlinex & f & x f \\
\hline 0 & f_{1} & 0 \\
1 & f_{2} & f_{2} \\
2 & 4 & 8 \\
3 & 4 & 12 \\
4 & 3 & 12 \\
\hline Total & f_{1}+f_{2}+11 & 32+f_{2} \\
\hline
\end{array} Since, total frequency is 18
$f_{1}+f_{2}+11=18$
$\Rightarrow f_{1}+f_{2}=7$
As we have, Mean $=\frac{\Sigma x f}{\Sigma f}=2$
(i)
$\frac{32+f_{2}}{18}=2 \Rightarrow f_{2}=36-32=4$
On putting the value of $f_{2}$ in Eq. (i), we get $f_{1}=7-4=3$
Since, mean $=\bar{x}=2$ ( given)
\begin{array}{|c|c|c|c|c|}
\hlinex & x-x & (x-x)^{2} & f & f(x-x)^{2} \\
\hline 0 & -2 & 4 & 3 & 12 \\
1 & -1 & 1 & 4 & 4 \\
2 & 0 & 0 & 4 & 0 \\
3 & 1 & 1 & 4 & 4 \\
4 & 2 & 4 & 3 & 12 \\
\hline Total & & & 18 & 32 \\
\hline
\end{array}
Now, $S D=\sqrt{\frac{\sum f(x-\bar{x})^{2}}{N}}$
where $N=$ sum of all frequencies
$=\sqrt{\frac{32}{18}}=\sqrt{\frac{16}{9}}=\frac{4}{3}$

Topic: Mathematics, Statistics, NDA, NDA 2010 (Phase 1)
The table below gives an incomplete frequency distribution with two missing frequencies $f_{1}$ and $f_{2}$ \begin{array}{|c|c|}
\hline Value of x & Frequency \\
\hline 0 & f_{1} \\
1 & f_{2} \\
2 & 4 \\
3 & 4 \\
4 & 3 \\
\hline
\end{array} The total frequency is 18 and the arithmetic mean of $x$ is 2 .
What is the coefficient of variance? $\quad$

Options:
A $\frac{200}{3}$
B $\frac{50 \sqrt{5}}{9}$
C $\frac{600}{\sqrt{5}}$
D 150

The correct answer is:
$\frac{200}{3}$

Solution:
\begin{array}{|c|c|c|}
\hlinex & f & x f \\
\hline 0 & f_{1} & 0 \\
1 & f_{2} & f_{2} \\
2 & 4 & 8 \\
3 & 4 & 12 \\
4 & 3 & 12 \\
\hline Total & f_{1}+f_{2}+11 & 32+f_{2} \\
\hline
\end{array} Since, total frequency is 18
$f_{1}+f_{2}+11=18$
$\Rightarrow f_{1}+f_{2}=7$
As we have, Mean $=\frac{\Sigma x f}{\Sigma f}=2$
(i)
$\frac{32+f_{2}}{18}=2 \Rightarrow f_{2}=36-32=4$
On putting the value of $f_{2}$ in Eq. (i), we get $f_{1}=7-4=3$
Coefficient of variance $=\frac{\sigma}{\bar{x}} \times 100$ where $\sigma=S . D$
$=\frac{4}{3} \times \frac{1}{2} \times 100=\frac{200}{3}$

Topic: Mathematics, Statistics, NDA, NDA 2010 (Phase 1)
The temperature of a body is measured both in \({ }^{\circ} \mathrm{C}\) and \({ }^{\circ} \mathrm{F}\). A graph is plotted with \({ }^{\circ} \mathrm{F}\) on \(X\)-axis and \({ }^{\circ} \mathrm{C}\) on \(Y\)-axis. Then, the cosine of angle between the graph and the \(X\)-axis is

Options:
A 0
B \(\frac{9}{5}\)
C \(\frac{5}{\sqrt{106}}\)
D \(\frac{9}{\sqrt{106}}\)

The correct answer is:
\(\frac{9}{\sqrt{106}}\)

Solution:
The relation between celsius scale temperature and fahrenheit scale temperature is given as
\(\begin{aligned}
& & \frac{C}{5} & =\frac{F-32}{9} \\
\Rightarrow & & 9 C & =5 F-160 \\
\Rightarrow & & C & =\frac{5}{9} F-\frac{160}{9} \quad \ldots (i)
\end{aligned}\)
The graph between \(C\) and \(F\) is shown in the figure
Comparing Eq. (i) with line \(y=m x+C\), we get Slope of graph, \(m=\tan \theta=\frac{5}{9}\)
\(\therefore \quad \cos \theta=\frac{9}{\sqrt{106}}\)

Topic: Physics, Thermal Properties of Matter, AP EAMCET, AP EAMCET 2020 (18 Sep Shift 2)
The term 'bark' refers to

Options:
A phellem, phelloderm and vascular cambium
B periderm and secondary xylem
C cork cambium and cork
D phellogen, phellem, phelloderm and secondary phloem

The correct answer is:
phellogen, phellem, phelloderm and secondary phloem

Solution:
All the dead cells lying outside the phellogen constitute the bark of plant. It may include epidermis, cork, hypodermis and even part of cortex depending upon the position of origin of cork cambium or phellogen.

Topic: Biology, Anatomy of Flowering Plants, JIPMER, JIPMER 2011
The term 'keel' is used for special type of

Options:
A sepals
B petals
C stamens
D carpels

The correct answer is:
petals

Solution:
Keel is a special type of petal found in papilionaceous type of corolla. Papilionaceous type is butterfly like, the corolla has one large posterior standard, two lateral wings, two innermost and smaller keels.

Topic: Biology, Morphology of Flowering Plants, JIPMER, JIPMER 2011
The term used for permanent cells that have reobtain meristematic property to perform cell division is

Options:
A Differentation
B Dedifferentiation
C Redifferentiation
D determination

The correct answer is:
Dedifferentiation

Solution:
Dedifferentiation is the phenomenon by which a permanent mature cell that has lost the power of division becomes meristematic and starts division again. For example during secondary growth in dicot stem fully differentiated parenchyma becomes meristematic to form interfascicular cambium and cork cambium.

Topic: Biology, Plant - Growth and Development, NEET
The test that distinguishes primary amines from other amines is

Options:
A Iodoform test
B Victor Meyer test
C Lucas test
D Carbylamine test

The correct answer is:
Carbylamine test

Solution:
The carbylamine reaction is also known as Hoffman isocyanide synthesis. It is the reaction of a primary amine, chloroform and a base to synthesize isocyanides or carbylamines.

The dichlorocarbene intermediate is very important for this conversion.

The carbylamine reaction is not given by secondary and tertiary amines, so primary amines can be distinguished from tertiary and secondary amines.

R-NH2+CHCl3+3KOH→R-NC+3KCl+3H2O

Topic: Chemistry, Amines, TS EAMCET, TS EAMCET 2021 (05 Aug Shift 1)
The test that is done for the differentiation of primary amines from secondary and tertiary amine is:

Options:
A Hell-Volhard Zelinsky reaction
B Tollen’s reagent
C Azo dye test
D Carbylamine test

The correct answer is:
Carbylamine test

Solution:
The Carbylamine reaction is a chemical test for detection of primary amines. In this reaction, the analyte is heated with alcoholic potassium hydroxide and chloroform. If a primary amine is present, the isocyanide is formed.

For example, the reaction with ethylamine:

C 2 H 5 NH 2 + CHCl 3 +3KOH → C 2 H 5 NC+3KCl+3 H 2 O Ethylamine   Chloroform             Ethyl isocyanide

Topic: Chemistry, Amines, NEET
The third term of a GP is 3 . What is the product of the first five terms?

Options:
A 216
B 226
C 243
D Cannot be determined due to insufficient data

The correct answer is:
243

Solution:
Given, $3^{\text {rd }}$ term of G.P. =3 Let ' $a$ ' be the first term and ' $\mathrm{r}$ ' be the common ratio. $\therefore \mathrm{ar}^{2}=3$
We know, $\mathrm{T}_{1}=\mathrm{a}, \mathrm{T}_{2}=\mathrm{ar}, \mathrm{T}_{3}=\mathrm{ar}^{2}, \mathrm{~T}_{4}=\mathrm{ar}^{3}, \mathrm{~T}_{5}=\mathrm{ar}^{4}$
$\mathrm{T}_{1} \cdot \mathrm{T}_{2} \cdot \mathrm{T}_{3} \cdot \mathrm{T}_{4} \cdot \mathrm{T}_{5}=(\mathrm{a})(\operatorname{ar})\left(\mathrm{ar}^{2}\right)\left(\mathrm{ar}^{3}\right)\left(\mathrm{ar}^{4}\right)$
$=\left(\mathrm{ar}^{2}\right)^{5}$
$=3^{5}$
$=243$

Topic: Mathematics, Sequences and Series, NDA, NDA 2018 (Phase 1)
The three lines $4 \mathrm{x}+4 \mathrm{y}=1,8 \mathrm{x}-3 \mathrm{y}=2, \mathrm{y}=0$ are

Options:
A the sides of an isosceles triangle
B concurrent
C mutually perpendicular
D the sides of an equilateral triangle

The correct answer is:
concurrent

Solution:
So from the figure it is clear that all the three lines are concurrent at point $\mathrm{P}$.

Topic: Mathematics, Straight Lines, NDA, NDA 2015 (Phase 2)
The three stable isotopes of neon: ${ }_{10}^{20} \mathrm{Ne},{ }_{10}^{21} \mathrm{Ne}$ and ${ }_{10}^{22}$ Ne have respective abundances of $\mathbf{9 0 . 5 1 \%} \% \mathbf{0 . 2 7 \%}$ and $9.22 \%$. The atomic masses of the three isotopes are $19.99 \mathrm{u}, 20.99 \mathrm{u}$ and $21.99 \mathrm{u}$, respectively. Obtain the average atomic mass of neon.

Options:

None

Solution:
The masses of the three isotopes are $19.99 \mathrm{u}, 20.99 \mathrm{u}$ and 21.99u.
Their abundance are $90.51 \%, 0.27 \%$ and $9.22 \%$
$\therefore \quad$ Average atomic mass of neon is
$$
\begin{aligned}
&\frac{90.51 \times 19.99+0.27 \times 20.99+9.22 \times 21.99}{90.51+0.27+9.22} \\
&=\frac{2017.7}{100}=20.17 \mathrm{u} .
\end{aligned}
$$

Topic: Physics, Nuclear Physics
The tissue that will be present in the maximum amount in the stem of a many years old woody tree is

Options:
A Secondary Xylem
B Secondary cortex
C Primary xylem
D Primary phloem

The correct answer is:
Secondary Xylem

Solution:
In an old trunk, increase in thickness occurs due to secondary growth, secondary growth in stellar region occurs due to activity of vascular cambium which forms more amount of secondary xylem and this form heart of the trunk.

Topic: Biology, Anatomy of Flowering Plants, NEET