datavorous/entrance-exam-dataset huggingface dataset:
The following question consist of two statements, one labelled as the 'Assertion (A)' and the other as 'Reason (R)'. You are to examine these two statements carefully and select the answer.
Assertion (A) : $\int_{1}^{e} \ln ^{2} x d x=e-2$
Reason $(\mathrm{R}): \mathrm{I}_{\mathrm{n}}=\int_{1}^{\mathrm{e}} \ln ^{\mathrm{n}} \mathrm{x} \mathrm{dx}=\mathrm{e}-\mathrm{nI}_{\mathrm{n}-1}$

Options:
A Roth $A$ and $R$ are individually true and $R$ is the correct explanation of $\mathrm{A}$
B Both $\mathrm{A}$ and $\mathrm{R}$ are individually true but $\mathrm{R}$ is not the correct explanation A
C $\mathrm{A}$ is true but $\mathrm{R}$ is false
D A is false but R is true

The correct answer is:
Roth $A$ and $R$ are individually true and $R$ is the correct explanation of $\mathrm{A}$

Solution:
Assertion (A): $\int_{1}^{e} \ln ^{2} x \cdot d x=\int_{1}^{e} \ln ^{2} x \cdot 1 d x$
$\int \ln ^{2} x \cdot 1 d x=x \cdot \ln ^{2}(x)-\int 2 \cdot \ln x \cdot d x$
$=x \cdot \ln ^{2} x-2 x \cdot \ln x+2 x$
$\therefore \int_{1}^{e} \ln ^{2} x \cdot d x=\left[x \cdot \ln ^{2} x-2 x \ln x+2 x\right]_{1}^{e}$
$=\left(e \cdot \ln ^{2} e-2 e \cdot \ln e+2 e\right)-\left(\ln ^{2} 1-2 \ln 1+2\right)$
$=e-2 e+2 e-0+0-2 \quad(\because \ln e=1$ and $\ln 1=0)$
$=e-2$
So, $\mathrm{A}$ is true.
Reason (R): $\mathrm{I}_{n}=\int_{1}^{e} \ln ^{n} x \cdot d x$
$\therefore \int \ln ^{n} x \cdot d x=x \times \ln ^{n} x-n \int \ln ^{n-1} x \cdot d x$
$\quad=x \cdot \ln ^{n} x-n \times \mathrm{I}_{n-1}$
So, $\int_{1}^{e} \ln ^{n} x \cdot d x=\left[x \cdot \ln ^{n} x-n \times \mathrm{I}_{n-1}\right]_{1}^{e}$
$=e \cdot \ln ^{n} e-\ln ^{n} 1-n \times \mathrm{I}_{n-1}$
$=e-n \times \mathrm{I}_{n-1} \quad(\because \ln e=1$ and $\ln 1=0)$
So, $\mathrm{R}$ is true and $\mathrm{R}$ is correct explanation of $\mathrm{A}$.

Topic: Mathematics, Definite Integration, NDA, NDA 2006 (Phase 2)
The following question consist of two statements, one labelled as the 'Assertion (A)' and the other as 'Reason $(R)^{\prime}$. You are to examine these two statements carefully and select the answer. Assertion(A): If $ < 1, \mathrm{~m}, \mathrm{n}>$ are direction cosines of a line, there can be a line whose direction cosines are
$\left\langle\sqrt{\frac{1^{2}+m^{2}}{2}}, \sqrt{\frac{m^{2}+n^{2}}{2}}, \sqrt{\frac{n^{2}+1^{2}}{2},}\right\rangle .$
Reason(R): The sum of direction cosines of a line is unity.

Options:
A Both A and R individually true, and $\mathrm{R}$ is the correct explanation of A.
B Both A and $\mathrm{R}$ are individually true but $\mathrm{R}$ is not the correct explanation of $\mathrm{A}$.
C $\mathrm{A}$ is true but $\mathrm{R}$ is false.
D $\quad A$ is false but $R$ is true. $\quad$

The correct answer is:
$\mathrm{A}$ is true but $\mathrm{R}$ is false.

Solution:
Sum of directions cosines of a line i.e. $\ell+m+n \neq 1$. So, $\mathrm{R}$ is false Since sum of squares of direction cosines is unity
$=\left(\sqrt{\frac{\ell^{2}+\mathrm{m}^{2}}{2}}\right)^{2}+\left(\sqrt{\frac{\mathrm{m}^{2}+\mathrm{n}^{2}}{2}}\right)^{2}+\left(\sqrt{\frac{\mathrm{n}^{2}+\ell^{2}}{2}}\right)^{2}$
$=\frac{\ell^{2}+\mathrm{m}^{2}}{2}+\frac{\mathrm{m}^{2}+\mathrm{n}^{2}}{2}+\frac{\mathrm{n}^{2}+\ell^{2}}{2}$
$\Rightarrow \frac{2\left(\ell^{2}+\mathrm{m}^{2}+\mathrm{n}^{2}\right)}{2}=\ell^{2}+\mathrm{m}^{2}+\mathrm{n}^{2}=1$
Hence, assertion A is true.
So, $A$ is true but $R$ is false.

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2007 (Phase 1)
The following question consist of two statements, one labelled as the 'Assertion (A)' and the other as 'Reason (R)'. Youare to examine these twostatements carefully and select the answer. Assertion (A): Ifevents, $A, B, C, D$ are mutually exhaustive,
then $(\mathrm{A} \cup \mathrm{B} \cup \mathrm{C})^{\mathrm{C}}=\mathrm{D}$.
Reason $(\mathbf{R}):(\mathrm{A} \cup \mathrm{B} \cup \mathrm{C})^{\mathrm{C}}=\mathrm{D}$.implies if any element is
excluded from the sets $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$, then it is included in $\mathrm{D}$.

Options:
A Both Aand $\mathbf{R}$ are individually true, and $\mathbf{R}$ is the correct explanation of $\mathbf{A}$
B Both $\mathbf{A}$ and $\mathbf{R}$ are individually true but $\mathbf{R}$ is not the correct explanation of $\mathbf{A}$.
C $\mathbf{A}$ is true but $\mathbf{R}$ is false.
D A is false but $\mathbf{R}$ is true.

The correct answer is:
Both Aand $\mathbf{R}$ are individually true, and $\mathbf{R}$ is the correct explanation of $\mathbf{A}$

Solution:
Both (A) and (R) are true and $R$ is the correct explanation of A.

Topic: Mathematics, Sets and Relations, NDA, NDA 2007 (Phase 2)
The following question consist of two statements, one labelled as the 'Assertion $(A)^{r}$ and the other as 'Reason $(R)^{\prime} .$ You are to examine these two statements carefully and select the answer. While constructing the cumulative frequency column of a frequency distribution, it is noticed that these cumulative frequencies are in arithmetic progression Assertion (A): All the class frequencies are equal. Reason (R): When all the class frequencies are equal, the cumulative frequencies are in arithmetic progression

Options:
A Both $\mathbf{A}$ and $\mathbf{R}$ are individually true, and $\mathbf{R}$ is the correct explanation of $\mathbf{A}$
B Both $\mathbf{A}$ and $\mathbf{R}$ are individually true but $\mathbf{R}$ is not the correct explanation of $\mathbf{A}$
C $\mathbf{A}$ is true but $\mathbf{R}$ is false.
D A is false but R is true. $\quad$

The correct answer is:
Both $\mathbf{A}$ and $\mathbf{R}$ are individually true, and $\mathbf{R}$ is the correct explanation of $\mathbf{A}$

Solution:
From the given statement
$\Rightarrow$ Both $(\mathrm{A}) \operatorname{and}(\mathrm{R})$ are true and $\mathrm{R}$ is the corred explanation of $\mathrm{A}$

Topic: Mathematics, Statistics, NDA, NDA 2007 (Phase 2)
The following questions consist of two statements, one labelled as the "Assertion $(A)^{\prime}$ and the other as 'Reason
(R). You are to examine these two statements canefilly and select the answe? Assertion (A): The probability of drawing either an ace or a king from a deck of card in a single draw is $\frac{2}{13}$ Reason $(\mathrm{R}):$ For two events $E_{1}$ and $E_{2}$, which are not mutually exclusive probability is given by $P\left(E_{1}+E_{2}\right)=P\left(E_{1}\right)+P\left(E_{2}\right)-P\left(E_{1} \cap E_{2}\right)$

Options:
A Both $\mathrm{A}$ and $\mathrm{R}$ are true and $\mathrm{R}$ is the correct explanation of $A$
B Both Aand Rare true but Risnot the correct explanation of $\mathrm{A}$
C A is true but $\mathrm{R}$ is false
D A is false but $\mathrm{R}$ is true

The correct answer is:
Both Aand Rare true but Risnot the correct explanation of $\mathrm{A}$

Solution:
(A) Total no. of cards $=52$ Total no. of ace cards $=4$ Total no. of king cards $=4$
$\mathrm{P}($ drawn an ace $)=\frac{4}{52}, \mathrm{P}($ drawn an king $)=\frac{4}{52}$
Thus, $\mathrm{P}$ (drawing either an ace or a king) $=\mathrm{P}($ an ace $)+\mathrm{P}($ a king $)$
$=\frac{4}{52}+\frac{4}{52}=\frac{8}{52}=\frac{4}{26}=\frac{2}{13}$
Both A and $\mathrm{R}$ are true but $\mathrm{R}$ is not the correct explanation of $\mathrm{A}$.

Topic: Mathematics, Probability, NDA, NDA 2009 (Phase 1)
The following questions consist of two statements, one labelled as the 'Assertion $(A)^{\prime}$ and the other as 'Reason $(R)^{\prime}$. You are to examine these two statements canefully and select the answer.
$Assertion(A)$: If $A=\left(\begin{array}{ll}\cos \alpha & \sin \alpha \\ \cos \alpha & \sin \alpha\end{array}\right)$ and
$\mathrm{B}=\left(\begin{array}{ll}\cos \alpha & \cos \alpha \\ \sin \alpha & \sin \alpha\end{array}\right)$, then $\mathrm{AB} \neq 1 .$
$Reason(R)$: The product of two matrices can never be equal to an identity matrix.

Options:
A Both A and $\mathrm{R}$ are individually true and $\mathrm{R}$ is the correct explanation of A
B Both A and $\mathrm{R}$ are individually true but $\mathrm{R}$ is not the correct explanation of A
C A is true but $R$ is false
D $\mathrm{A}$ is false but $\mathrm{R}$ is true

The correct answer is:
A is true but $R$ is false

Solution:
$A=\left(\begin{array}{ll}\cos \alpha & \sin \alpha \\ \cos \alpha & \sin \alpha\end{array}\right)$
$\mathrm{B}=\left(\begin{array}{ll}\cos \alpha & \cos \alpha \\ \sin \alpha & \sin \alpha\end{array}\right)$
$\mathrm{AB}=\left(\begin{array}{cc}\cos \alpha & \sin \alpha \\ \cos \alpha & \sin \alpha\end{array}\right)\left(\begin{array}{ll}\cos \alpha & \cos \alpha \\ \sin \alpha & \sin \alpha\end{array}\right)$
$\left(\begin{array}{cc}\cos ^{2} \alpha+\sin ^{2} \alpha & \cos ^{2} \alpha+\sin ^{2} \alpha \\ \cos ^{2} \alpha+\sin ^{2} \alpha & \cos ^{2} \alpha+\sin ^{2} \alpha\end{array}\right)$
$=\left(\begin{array}{ll}1 & 1 \\ 1 & 1\end{array}\right) \neq \mathrm{I}$
So, A is true. Since product of two matrix may be equal to identity matrix. so, $R$ is false and $A$ is true.

Topic: Mathematics, Matrices, NDA, NDA 2006 (Phase 2)
The following questions consist of two statements, one labelled as the 'Assertion $(A)^{\prime}$ and the other as 'Reason $(R)^{\prime}$. You are to examine these two statements canefully and select the answer.
$Assertion$ $(\mathbf{A}):$ If $\mathrm{A}=\left(\begin{array}{ll}2 & 3 \\ 1 & 4\end{array}\right), \mathrm{B}=\left(\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right)$, then $(\mathrm{A}+\mathrm{B})^{2}$
$=\mathrm{A}^{2}+\mathrm{B}^{2}+2 \mathrm{AB}$
$Reason(R)$: In the above $A B=B A$

Options:
A Both A and $\mathrm{R}$ are individually true and $\mathrm{R}$ is the correct explanation of A
B Both A and $\mathrm{R}$ are individually true but $\mathrm{R}$ is not the correct explanation of A
C A is true but $R$ is false
D $\mathrm{A}$ is false but $\mathrm{R}$ is true

The correct answer is:
Both A and $\mathrm{R}$ are individually true and $\mathrm{R}$ is the correct explanation of A

Solution:
$A=\left(\begin{array}{ll}2 & 3 \\ 1 & 4\end{array}\right), B=\left(\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right)$
$\mathrm{AB}=\left(\begin{array}{ll}2 & 3 \\ 1 & 4\end{array}\right)\left(\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right)$
$=\left(\begin{array}{ll}2+0 & 0+3 \\ 1+0 & 0+4\end{array}\right)=\left(\begin{array}{ll}2 & 3 \\ 1 & 4\end{array}\right)$
$\mathrm{BA}=\left(\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right)\left(\begin{array}{ll}2 & 3 \\ 1 & 4\end{array}\right)$
$=\left(\begin{array}{ll}2+0 & 3+0 \\ 0+1 & 0+4\end{array}\right)=\left(\begin{array}{ll}2 & 3 \\ 1 & 4\end{array}\right)$
Also, $A+B=\left(\begin{array}{ll}3 & 3 \\ 1 & 5\end{array}\right)$
$(\mathrm{A}+\mathrm{B})^{2}=\left(\begin{array}{cc}9+3 & 9+15 \\ 3+5 & 3+25\end{array}\right)=\left(\begin{array}{cc}12 & 24 \\ 8 & 28\end{array}\right)$
$\mathrm{A}^{2}=\left(\begin{array}{ll}2 & 3 \\ 1 & 4\end{array}\right)\left(\begin{array}{ll}2 & 3 \\ 1 & 4\end{array}\right)=\left(\begin{array}{ll}4+3 & 6+12 \\ 2+4 & 3+16\end{array}\right)=\left(\begin{array}{ll}7 & 18 \\ 6 & 19\end{array}\right)$
$\mathrm{B}^{2}=\left(\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right)\left(\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right)=\left(\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right)$
$\mathrm{A}^{2}+\mathrm{B}^{2}+2 \mathrm{AB}$
$=\left(\begin{array}{ll}7 & 18 \\ 6 & 19\end{array}\right)+\left(\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right)+2\left(\begin{array}{ll}2 & 3 \\ 1 & 4\end{array}\right)$
$=\left(\begin{array}{cc}12 & 24 \\ 8 & 28\end{array}\right)=(\mathrm{A}+\mathrm{B})^{2}$
So, Assertion $\mathrm{A}$ is correct $\mathrm{R}$ is $\mathrm{AB}=\mathrm{BA}$
Hence, $\mathrm{R}$ is correct. Since this leads from Assersion A, then both A and $R$ are individually true and $\mathrm{R}$ is the correct explanation of $\mathrm{A}$.

Topic: Mathematics, Matrices, NDA, NDA 2006 (Phase 2)
The following questions consist of two statements, one labelled as the 'Assertion $(\mathrm{A})$ ' and the other as 'Reason $(\mathrm{R})$ ' You are to examine these two statements carefully and select the answer. Assertion (A) : For a binomial distribution $\mathrm{B}(\mathrm{n}, \mathrm{p})$, Mean $>$ Variance Reason $(\mathrm{R}):$ Probability is less than or equal to 1

Options:
A Both $\mathrm{A}$ and $\mathrm{R}$ are individually tue and $\mathrm{R}$ is the correct explanation of $A$
B Both $\mathrm{A}$ and $\mathrm{R}$ are individually true but $\mathrm{R}$ is not the correct explanation of $A$
C A is true but $\mathrm{R}$ is false
D A is false but $\mathrm{R}$ is true

The correct answer is:
Both $\mathrm{A}$ and $\mathrm{R}$ are individually true but $\mathrm{R}$ is not the correct explanation of $A$

Solution:
Mean $=\mathrm{np}$ and Variance $=\mathrm{npq} < \mathrm{np}(\because \mathrm{q} < 1)$

Topic: Mathematics, Probability, NDA, NDA 2008 (Phase 2)
The following reaction

Is known by the name:

Options:
A Friedel-Craft's reaction
B Perkin's reaction
C Acetylation reaction
D Schotten-Baumen reaction

The correct answer is:
Schotten-Baumen reaction

Solution:
The reaction is known by the name Schotten-Baumen reaction. It is a benzoylation reaction. When aniline is treated with benzoyl chloride in the presence of NaOH, benzanilide (N-benzoylaniline) is obtained.This reaction is given by primary and secondary amines.

Benzoylation of aniline is an example of Schotten-Baumen reaction.

Topic: Chemistry, Amines, NEET, NEET 2015 (Phase 2)
The following table gives the monthly expenditure of two familie:
\begin{array}{|c|c|c|}
\hline \multicolumn{3}{|c|} { Expenditure (in \boldsymbol{y}} \\
\hline Items & Family A & Family B \\
\hline Food & 3,500 & 2,700 \\
\hline Clothing & 500 & 800 \\
\hline Rent & 1,500 & 1,000 \\
\hline Education & 2,000 & 1,800 \\
\hline Miscellaneous & 2,500 & 1,800 \\
\hline
\end{array} In constructing a pie diagram to the above data, the radii of the circles are to be chosen by which one of the followin:
ratios?

Options:
A $1: 1$
B $10: 9$
C $100: 91$
D $5: 4$

The correct answer is:
$10: 9$

Solution:
Total expenditure of $\mathrm{A}=3,500+500+1,500+2,000$
$=10,000$
Total expenditure of $\mathrm{B}=2,700+800+1,000+1,800$
$+1,800$
$=8,100$
Area of A: Area of $\mathrm{B}=10,000: 8,100$
$=100: 8]$
$\Rightarrow$ radius of A: radius of $\mathrm{B}=\sqrt{100}: \sqrt{81}$
$=10: 9$

Topic: Mathematics, Statistics, NDA, NDA 2017 (Phase 2)
The foot of a tower of height $h \mathrm{~m}$ is in a direct line between two observers $A$ and $B$. If the angles of elevation of the top of the tower as seen from $A$ and $B$ are $\alpha$ and $\beta$ respectively and if $A B=d \mathrm{~m}$, then what is $h / d$ equal to?

Options:
A $\frac{\tan (\alpha+\beta)}{(\cot \alpha \cot \beta-1)}$
B $\frac{\cot (\alpha+\beta)}{(\cot \alpha \cot \beta-1)}$
C $\frac{\tan (\alpha+\beta)}{(\cot \alpha \cot \beta+1)}$
D $\frac{\cot (\alpha+\beta)}{(\cot \alpha \cot \beta+1)}$

The correct answer is:
$\frac{\cot (\alpha+\beta)}{(\cot \alpha \cot \beta-1)}$

Solution:
Let $\begin{aligned} & A D=x \\ & \therefore D B=d-x \\ & \operatorname{In} \Delta A D C \end{aligned}$
$\tan \alpha=\frac{h}{x}$
$\Rightarrow x=h \cot \alpha$
$\operatorname{In} \Delta C D B$
$\tan \beta=\frac{h}{d-x}$
$\Rightarrow d-x=h \cot \beta$
From equations (i) and (ii), we get
$d=h(\cot \alpha+\cot \beta)$
We know,
$\cot (\alpha+\beta)=\frac{\cot \alpha \cot \beta-1}{\cot \beta+\cot \alpha}$
$\Rightarrow \cot \beta+\cot \alpha=\frac{\cot \alpha \cot \beta-1}{\cot (\alpha+\beta)}$
$\therefore \quad d=h\left[\frac{\cot \alpha \cot \beta-1}{\cot (\alpha+\beta)}\right]$
$\Rightarrow \frac{h}{d}=\frac{\cot (\alpha+\beta)}{\cot \alpha \cot \beta-1}$

Topic: Mathematics, Heights and Distances, NDA, NDA 2008 (Phase 2)
The foot of the perpendicular drawn from the origin to a plane is the point $(1,-3,1)$. What is the intercept cut on the $x$ -axis by the plane? $\quad$

Options:
A $\sqrt{3}$
B 3
C $\sqrt{11}$
D 11

The correct answer is:
11

Solution:
Equation of plane passing through $(1,-3,1)$ and whose normal $(1,-3,1)$ is $1(x-1)-3(y+3)+1(z-1)=0$
$\Rightarrow x-3 y+z-11=0$
$\Rightarrow \frac{x}{11}-\frac{y}{11 / 3}+\frac{z}{11}=0$
The above plane intercept the $x$ -axis at 11 .

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2009 (Phase 2)
The formal potential of $Fe^{3+}/Fe^{2+}$ in a sulphuric acid and phosphoric acid mixture $\left(E^{\circ}=+0.61 \mathrm{V}\right)$ is much lower than the standard potential $\left(E^{\circ}=+0.77 \mathrm{V}\right)$. This is due to

Options:
A formation of the species $\left[\mathrm{FeHPO}_{4}\right]^{+}$
B lowering of potential upon complexation
C formation of the species $\left[\mathrm{FeSO}_{4}\right]^{+}$
D high acidity of the medium

The correct answers are:
formation of the species $\left[\mathrm{FeHPO}_{4}\right]^{+}$, lowering of potential upon complexation, high acidity of the medium

Solution:
Since, $\mathrm{H}_{2} \mathrm{SO}_{4}$ and $\mathrm{H}_{3} \mathrm{PO}_{4}$, both are strong acids so their presence makes the medium more acidic. The $\mathrm{Fe}^{3+}$ ions form complex, IFeHPO $\left._{4}\right]^{+}$ with these acid and hence, the concentration of $\mathrm{Fe}^{3+}$ ion reduces which results in lowering in the formal reduction potential.

Topic: Chemistry, Electrochemistry, WBJEE, WBJEE 2014
The formula $\sin ^{-1}\left\{2 x\left(1-x^{2}\right)\right\}=2 \sin ^{-1} x$ is true for all
values of $x$ lying in the interval

Options:
A $[-1,1]$
B $[0,1]$
C $[-1,0]$
D $[-1 / \sqrt{2}, 1 / \sqrt{2}]$

The correct answer is:
$[-1 / \sqrt{2}, 1 / \sqrt{2}]$

Solution:
$\sin ^{-1}\left\{2 x\left(1-x^{2}\right)\right\}=2 \sin ^{-1} x$ is true
$\forall x \in\left[-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right]$

Topic: Mathematics, Inverse Trigonometric Functions, NDA, NDA 2009 (Phase 1)
The four points $(0,4,1),(2,3,-1),(4,5,0)(2,6,2)$ are the vertices of which one of the following figures?

Options:
A Rhombus
B Rectangle
C Square
D Parallelogram

The correct answer is:
Square

Solution:
Let the coordinates of $\mathrm{A}, \mathrm{B}, \mathrm{C}$ and $\mathrm{D}$ are $(0,4,1)$, $(2,3,-1)(4,5,0)$ and $(2,6,2)$ respectively.
We find its sides and diagonals as below
$\begin{aligned} \mathrm{AB} &=\sqrt{(2-0)^{2}+(3-4)^{2}+(-1-1)^{2}} \\ &=\sqrt{4+1+4}=\sqrt{9}=3 \\ \mathrm{BC} &=\sqrt{(4-2)^{2}+(5-3)^{2}+(0+1)^{2}} \\ &=\sqrt{4+4+1}=\sqrt{9}=3 \\ \mathrm{CD} &=\sqrt{(2-4)^{2}+(6-5)^{2}+(2-0)^{2}} \\ &=\sqrt{4+1+4}=\sqrt{9}=3 \\ \mathrm{DA} &=\sqrt{(0-2)^{2}+(4-6)^{2}+(1-2)^{2}} \\ &=\sqrt{4+4+1}=\sqrt{9}=3 \\ \mathrm{AC} &=\sqrt{(4-0)^{2}+(5-4)^{2}+(0-1)^{2}} \\ &=\sqrt{16+1+1}=\sqrt{18} \\ \text { and } \mathrm{BD} &=\sqrt{(2-2)^{2}+(6-3)^{2}+(2+1)^{2}} \\ &=\sqrt{9+9}=\sqrt{18} \end{aligned}$
Since, $\mathrm{AB}=\mathrm{BC}=\mathrm{CD}=\mathrm{DA}$, sides are equal and $\mathrm{AC}=\mathrm{BD}$, diagonals are also equal. Hence, $\mathrm{A}, \mathrm{B}, \mathrm{C}$ and $\mathrm{D}$ are the vertices of a square.

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2006 (Phase 2)
The four sketches (A, B, C, and D) given below represent four different types of animal tissues.

Which one of these is correctly identified in the options given along with its correct location and function?

Tissue	Location	Function

(A)	Simple squamous epithelium	Trachea	Diffusion boundary

(B)	Unicellular gland	Alimentary canal	Secretion

(C)	Bone	Larynx	Secretion

(D)	Compound epithelium	Skin	Protection

Options:
A A
B B
C C
D D

The correct answer is:
D

Solution:
Compound squamous epithelium is a type of tissue that makes up skin layers and lines some body cavities and organs. Its structure makes it an excellent protective layer.

Topic: Biology, Structural Organisation in Animals, NEET
The frequency curve for the distribution of income in a region is positively skewed as shown in the figure above. Then, for this distribution

Options:
A Mean $ < $ Mode $ < $ Median
B Mode $ < $ Median $ < $ Mean
C Mode $ < $ Mean $ < $ Median
D $\quad$ Median $ < $ Mean $ < $ Mode $\quad$

The correct answer is:
$\quad$ Median $ < $ Mean $ < $ Mode $\quad$

Solution:
For the given distribution which is positively seowed, Median $ < $ Mean $ < $ Mode

Topic: Mathematics, Statistics, NDA, NDA 2008 (Phase 1)
The frequency distribution of life
$90 \mathrm{TV}$ tubes whose median life is $17 \mathrm{months}$ is as follows
\begin{array}{|c|c|}
\hline Life of TV tubes (in months) & No. of TV tubes \\
\hline 0-5 & 3 \\
\hline 5-10 & 12 \\
\hline 10-15 & x \\
\hline 15-20 & 35 \\
\hline 20-25 & y \\
\hline 25-30 & 4 \\
\hline
\end{array} $\mathrm{n}=90$
$\frac{\mathrm{n}}{2}=45$
\begin{array}{|c|c|c|}
\hline Calss & Frequency & c f \\
\hline 0-5 & 3 & 3 \\
\hline 5-10 & 12 & 15 \\
\hline 10-15 & x & 15+x \\
\hline 15-20 & 35 & 50+x \\
\hline 20-25 & y & 50+x+y \\
\hline 25-30 & 4 & 54+x+y \\
\hline
\end{array}
What is the missing frequency $y$ ?

Options:
A 20
B 16
C 15
D 12

The correct answer is:
20

Solution:
\begin{array}{|c|c|c|}
\hline Class & Frequency & C. \mathbf{F} \\
\hline 0-5 & 3 & 3 \\
\hline 5-10 & 12 & 15 \\
\hline 10-15 & x & 15+x \\
\hline 15-20 & 35 & 50+x \\
\hline 20-25 & y & 50+x+y \\
\hline 25-30 & 4 & 54+x+y \\
\hline
\end{array}
Since, $n=90 \therefore \frac{n}{2}=45$
Missing frequency $y$ is 20 .

Topic: Mathematics, Statistics, NDA, NDA 2010 (Phase 2)
The frequency distribution of life
$90 \mathrm{TV}$ tubes whose median life is $17 \mathrm{months}$ is as follows
\begin{array}{|c|c|}
\hline Life of TV tubes (in months) & No. of TV tubes \\
\hline 0-5 & 3 \\
\hline 5-10 & 12 \\
\hline 10-15 & x \\
\hline 15-20 & 35 \\
\hline 20-25 & y \\
\hline 25-30 & 4 \\
\hline
\end{array} $\mathrm{n}=90$
$\frac{\mathrm{n}}{2}=45$
\begin{array}{|c|c|c|}
\hline Calss & Frequency & c f \\
\hline 0-5 & 3 & 3 \\
\hline 5-10 & 12 & 15 \\
\hline 10-15 & x & 15+x \\
\hline 15-20 & 35 & 50+x \\
\hline 20-25 & y & 50+x+y \\
\hline 25-30 & 4 & 54+x+y \\
\hline
\end{array}
What is the lower limit of the median class? $\quad$

Options:
A 10
B 15
C 20
D 25

The correct answer is:
15

Solution:
\begin{array}{|c|c|c|}
\hline Class & Frequency & C. \mathbf{F} \\
\hline 0-5 & 3 & 3 \\
\hline 5-10 & 12 & 15 \\
\hline 10-15 & x & 15+x \\
\hline 15-20 & 35 & 50+x \\
\hline 20-25 & y & 50+x+y \\
\hline 25-30 & 4 & 54+x+y \\
\hline
\end{array}
Since, $n=90 \therefore \frac{n}{2}=45$
$\because$ Lower limit of median class is 15 .

Topic: Mathematics, Statistics, NDA, NDA 2010 (Phase 2)
The frequency distribution of life
$90 \mathrm{TV}$ tubes whose median life is $17 \mathrm{months}$ is as follows
\begin{array}{|c|c|}
\hline Life of TV tubes (in months) & No. of TV tubes \\
\hline 0-5 & 3 \\
\hline 5-10 & 12 \\
\hline 10-15 & x \\
\hline 15-20 & 35 \\
\hline 20-25 & y \\
\hline 25-30 & 4 \\
\hline
\end{array} $\mathrm{n}=90$
$\frac{\mathrm{n}}{2}=45$
\begin{array}{|c|c|c|}
\hline Calss & Frequency & c f \\
\hline 0-5 & 3 & 3 \\
\hline 5-10 & 12 & 15 \\
\hline 10-15 & x & 15+x \\
\hline 15-20 & 35 & 50+x \\
\hline 20-25 & y & 50+x+y \\
\hline 25-30 & 4 & 54+x+y \\
\hline
\end{array}
What is the cumulative frequency of the modal class ? $\quad$

Options:
A 31
B 35
C 66
D Cannot be determined as the given data is insufficient.

The correct answer is:
66

Solution:
\begin{array}{|c|c|c|}
\hline Class & Frequency & C. \mathbf{F} \\
\hline 0-5 & 3 & 3 \\
\hline 5-10 & 12 & 15 \\
\hline 10-15 & x & 15+x \\
\hline 15-20 & 35 & 50+x \\
\hline 20-25 & y & 50+x+y \\
\hline 25-30 & 4 & 54+x+y \\
\hline
\end{array}
Since, $n=90 \therefore \frac{n}{2}=45$
Cumulative frequency of model class is 66

Topic: Mathematics, Statistics, NDA, NDA 2010 (Phase 2)
The frequency of a tuning fork A is 2% more than the frequency of a standard tuning fork. The frequency of another fork B is 3% less than the frequency of a standard tuning fork. If 6 beats s-1 are heard when the two tuning forks A and B are excited, the frequency of A is

Options:
A 120 Hz
B 122.4 Hz
C 116.4 Hz
D 130 Hz

The correct answer is:
122.4 Hz

Solution:
Let the frequency of standard fork =x
∴nA=102100x, nB=97100x
Number of beats per second =nA-nB=6
102100x-97100x=6
x=6×1005=120 Hz

∴nA=102100x=122.4 Hz

Topic: Physics, Waves and Sound, NEET