datavorous/entrance-exam-dataset huggingface dataset:
A husband and wife appear in an interview for two vacancies in the same post. The probability of husband's selection is $1 / 5$ and that of wife's selection is $1 / 3$. What is the probability that only one of them will be selected?

Options:
A $1 / 5$
B $2 / 5$
C $3 / 5$
D $4 / 5$

The correct answer is:
$2 / 5$

Solution:
Let $\mathrm{A}=$ Event of husband's selection and $\mathrm{B}=$ Event of wife's selection
Given : $P(A)=\frac{1}{5} \Rightarrow P(\bar{A})=1-\frac{1}{5}=\frac{4}{5}$
$P(B)=\frac{1}{3} \Rightarrow P(\bar{B})=1-\frac{1}{3}=\frac{2}{3}$
$P($ Only one of them selected $)$ $=P(A) \times P(\bar{B})+P(\bar{A}) \times P(B)$
$=\frac{1}{5} \times \frac{2}{3}+\frac{4}{5} \times \frac{1}{3}=\frac{2}{15}+\frac{4}{15}=\frac{6}{15}=\frac{2}{5}$

Topic: Mathematics, Probability, NDA, NDA 2011 (Phase 2)
a. In the young dicot stem, cambium present in patches as a single layer between xylem and phloem.
b. Vascular and cork cambium both are from the lateral meristem.
c. The medullary cells are derived from the interfascicular cambium.

In given statements, which of the following are correct?

Options:
A a and c
B a and b
C b and c
D a,b and c

The correct answer is:
a,b and c

Solution:
Intrafascicular cambium is present in between xylem and phloem of vascular bundle. It combines with interfascicular cambium to form complete vascular cambium ring. Both vascular cambium and cork cambium are example of lateral meristem and contribute to secondary growth.

Topic: Biology, Anatomy of Flowering Plants, NEET
A is a certain positive acute angle which satisfies the following equation:
Number of degrees in A + Number of radians in $\mathrm{A}=(180+\pi) / 3$
What is the angle A?

Options:
A $20^{\circ}$
B $40^{\circ}$
C $60^{\circ}$
D $80^{\circ}$

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

Solution:
Given that number of degrees in A+ Number of radians
in $\mathrm{A}=\frac{180^{\circ}+\pi}{3}=\frac{180^{\circ}}{3}+\frac{\pi}{3}=60^{\circ}+\frac{\pi}{3}$
$\Rightarrow$ Angle $\mathrm{A}=60^{\circ}$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2006 (Phase 2)
A is a square matrix of order 3 such that its determinant is 4. What is the determinant of its transpose?

Options:
A 64
B 36
C 32
D 4

The correct answer is:
4

Solution:
The determinant of transpose will not change. So, the determinant is equal to 4.

Topic: Mathematics, Determinants, NDA, NDA 2019 (Phase 1)
A is an angle in the fourth quadrant. If satisfies the trigonometric equation $3\left(3-\tan ^{2} \mathrm{~A}-\cot \mathrm{A}\right)^{2}=1$.
Which one of the following is a value of A?

Options:
A $300^{\circ}$
B $315^{\circ}$
C $330^{\circ}$
D $345^{\circ}$

The correct answer is:
$300^{\circ}$

Solution:
Checking through options $300^{\circ}=-60^{\circ}$
So, $3\left[3-\tan ^{2}\left(-60^{\circ}\right)-\cot \left(-60^{\circ}\right)\right]^{2}$
$=3\left[3-3+\frac{1}{\sqrt{3}}\right]^{2}=3 \times \frac{1}{3}=1$

Topic: Mathematics, Trigonometric Equations, NDA, NDA 2018 (Phase 2)
A is associated with

Options:
A $1,3,5,8,9,12$
B $2,4,6,8,10,11$
C $1,3,5,7,10,11$
D None of these

The correct answer is:
$1,3,5,8,9,12$

Solution:
$\sin x$ is periodic, continuous at every point on $(-\infty, \infty)$, differentiable at every point on $(-\infty, \infty)$, has a period $2 \pi, \sin x$ increases on $\left(0, \frac{\pi}{2}\right)$ and decreases on $\left(\frac{\pi}{2}, \pi\right)$.

Topic: Mathematics, Continuity and Differentiability, NDA, NDA 2010 (Phase 1)
A KMnO4 solution can be standardised by titration against As2O3(s). A 0.1156 g sample of As2O3 requires 27.08 mL of the KMnO4 (aq) for its titration. What is the molarity of the KMnO4(aq) [As = 75].

Options:
A 0.0172 M
B 1.0172 M
C 0.172 M
D 0.9172 M

The correct answer is:
0.0172 M

Solution:
Meq. of As2O3 = Meq of KMnO 4 As 2 3+ â2 As 5+ +4 e â
0.1156 198/4 ×1000 = M×5×27.08 Mn 7+ +5 e − → Mn 2+
â´ M = 0.0172
â´ MMnO4- = 0.0172 M

Topic: Chemistry, Redox Reactions, NEET
A karyotype is used to study

I) Number of chromosomes
II) Position of centromere
III) Position of chromosome
IV) Length of the arms
V) Secondary constriction

Options:
A I, II and III
B I, II, IV, V
C All of these
D I, IV, V

The correct answer is:
I, II, IV, V

Solution:
Karyotyping is the process by which photographs of chromosomes are taken to determine the chromosome complement of an individual, including the number of chromosomes and any abnormalities. A karyotype is used to analyze the number of chromosomes to determine the position of the centromere, to measure the length of the arms and to observe the secondary constriction.

Topic: Biology, Principles of Inheritance and Variation, NEET
Al3+(aq)+3e-→Al(s); E°=-1.66V
Cu2+(aq)+2e-→Cu(s); E°=+0.34V
What voltage is produced under standard conditions by combining the half reaction with these standard electrode potentials?

Options:
A 1.32 V
B 2.00 V
C 2.30 V
D 4.34 V

The correct answer is:
2.00 V

Solution:
The cell will be
Al(s)|Al3+(aq)||Cu2+(aq)|Cu(s)
Ecello=Ecathodeo-Eanodeo
=ECu2+/Cuo-EAl3+/Alo
=+0.34-(-1.66)=+2.00V .

Topic: Chemistry, Electrochemistry, NEET
A lamp post stands on a horizontal plane. From a point situated at a distance $150 \mathrm{~m}$ from its foot, the angle of elevation of the top is $30^{\circ}$. What is the height of the lamp post ?

Options:
A $50 \mathrm{~m}$
B $50 \sqrt{3} \mathrm{~m}$
C $\frac{50}{\sqrt{3}} \mathrm{~m}$
D $100 \mathrm{~m}$

The correct answer is:
$50 \sqrt{3} \mathrm{~m}$

Solution:
In $\Delta \mathrm{ABC}$, we have $\mathrm{BC}=150 \mathrm{~m}, \mathrm{AB}=\mathrm{h}$
$\angle \mathrm{C}=30^{\circ}$
$\tan 30^{\circ}=\frac{\mathrm{h}}{150}$
$\Rightarrow \frac{1}{\sqrt{3}}=\frac{\mathrm{h}}{150}$
$\therefore \mathrm{h}=\frac{150}{\sqrt{3}} \mathrm{~m}$
$=\frac{150 \times \sqrt{3}}{3}=50 \sqrt{3} \mathrm{~m}$

Topic: Mathematics, Heights and Distances, NDA, NDA 2014 (Phase 2)
A large number of fish are suddenly found floating dead on a lake. There is no evidence of toxic dumping but you find an abundance of phytoplankton. Suggest a reason for the fish kill.

Options:

None

Solution:
Excessive phytoplankton (organic pollutants such as leaves, grass, trash etc. ) present in water is biodegradable. A large population of bacteria decomposes this organic matter in water. During this process, they consume the oxygen dissolved in water. Water has already limited dissolved oxygen ( \(10 \mathrm{ppm})\). Thus, it is further depleted. When the level of dissolved oxygen falls below 6 ppm, the fish cannot survive. Hence, they die and float dead on the lake.

Topic: Chemistry, Environmental Chemistry
A large number of particles are placed around the origin, each at a distance R from the origin. The distance of the center of mass of the system from the origin is

Options:
A equal to R
B less than or equal to R
C greater than R
D greater than or equal to R

The correct answer is:
less than or equal to R

Solution:
A large number of particles are situated at a distance R from the origin. If particles are uniformly distributed and make a circular boundary around the origin, the centre of mass will lie at the origin.
Whereas, if the particles are not uniformly distributed, then distance can be any value between zero and R
This implies the distance between the centre of mass and the origin is always less than or equal to R

Topic: Physics, Center of Mass Momentum and Collision, NEET
A large number of particles are placed around the origin, each at a distance $R$ from the origin. The distance of the centro of mass of the system from the origin is

Options:
A equal to $R$
B less than equal to $R$
C greater than $R$
D greater than equal to $R$

The correct answer is:
less than equal to $R$

Solution:
As large number of particles is situated at a distance $R$ from the ongin If particles are uniformly distubuted and make a circulat boundary atound the ongm, then centre of mass will be at the ongin While if the patticles are not uniformly distnbuted, then centre of mass will lie between particle and origin. This implies distance between centre of mass and ongin is always less than equal to $R$.

Topic: Physics, Center of Mass Momentum and Collision, WBJEE, WBJEE 2015
A line L passes through the point $\mathrm{P}(5,-6,7)$ and is parallel to the planes $x+y+z=1$ and $2 x-y-2 z=3$. What are the direction ratios of the line of intersection of the given planes?

Options:
A $ < 1,4,3\rangle$
B $ < -1,-4,3>$
C $ < 1,-4,3>$
D $ < 1,-4,-3>$

The correct answer is:
$ < 1,-4,3>$

Solution:
Let $a, b, c$ be the direction ratios of the line Then its equation is
$\frac{x-5}{a}=\frac{y+6}{b}=\frac{z-7}{c}$
Since (i) is parallel to the planes $x+y+z=1$ and
$2 x-v-2 z=3$ then
$a(1)+b(1)+c(1)=0$ and $a(2)+b(-1)+c(-2)=0$
Bycross multiplication
$\frac{a}{-1}=\frac{b}{4}=\frac{c}{-3}=\lambda$
$\Rightarrow a=-\lambda, b=4 \lambda, c=-3 \lambda$
$\Rightarrow$ Direction ratios of the line are $ < -1,4,-3>= < 1,-4,3>$

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2016 (Phase 2)
A line L passes through the point $\mathrm{P}(5,-6,7)$ and is parallel to the planes $x+y+z=1$ and $2 x-y-2 z=3$. What is the equation of the line $\mathrm{L}$ ?

Options:
A $\frac{x-5}{-1}=\frac{y+6}{4}=\frac{z-7}{-3}$
B $\frac{x+5}{-1}=\frac{y-6}{4}=\frac{z+7}{-3}$
C $\frac{x-5}{-1}=\frac{y+6}{-4}=\frac{z-7}{3}$
D $\frac{x-5}{-1}=\frac{y+6}{-4}=\frac{7-7}{-3}$

The correct answer is:
$\frac{x-5}{-1}=\frac{y+6}{4}=\frac{z-7}{-3}$

Solution:
Substituting $\mathrm{a}, \mathrm{b}, \mathrm{c}$ in (i), we get
$\frac{x-5}{-1}=\frac{y+6}{4}=\frac{z-7}{-3}$
Hence, equation of the line is
$\frac{x-5}{-1}=\frac{y+6}{4}=\frac{z-7}{-3}$

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2016 (Phase 2)
A line makes $45^{\circ}$ with positive $\mathrm{x}$ -axis and makes equal angles with positive $y, z$ axes, respectively. What is the sum of the three angles which the line makes with positive $x, y$ and $z$ axes?

Options:
A $180^{\circ}$
B $165^{\circ}$
C $150^{\circ}$
D $135^{\circ}$

The correct answer is:
$165^{\circ}$

Solution:
We know that sum of square of direction cosines $=1$ i.e. $\cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma=1$
$\Rightarrow \quad \cos ^{2} 45^{\circ}+\cos ^{2} \beta+\cos ^{2} \beta=1$
(As given $\alpha=45^{\circ}$ and $\beta=\gamma$ )
$\Rightarrow \quad \frac{1}{2}+2 \cos ^{2} \beta=1$
$\Rightarrow \cos ^{2} \beta=\frac{1}{4}$
$\Rightarrow \cos \beta=\pm \frac{1}{2}$, Negative value is discarded, since the line
makes angle with positive axes.
Hence, $\cos \beta=\frac{1}{2}$
$\Rightarrow \cos \beta=\cos 60^{\circ}$
$\Rightarrow \quad \beta=60^{\circ}$
Required sum $=\alpha+\beta+\gamma=45^{\circ}+60^{\circ}+60^{\circ}=165^{\circ}$

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2006 (Phase 1)
A line makes angles $\theta, \phi$ and $\psi$ with $\mathrm{x}, \mathrm{y}, \mathrm{z}$ axes respectively.
Consider the following
1 . $\sin ^{2} \theta+\sin ^{2} \phi=\cos ^{2} \psi$
2 . $\quad \cos ^{2} \theta+\cos ^{2} \phi=\sin ^{2} \psi$
3 . $\sin ^{2} \theta+\cos ^{2} \phi=\cos ^{2} \psi$
Which of the above is/are correct?

Options:
A 1 only
B 2 only
C 3 only
D 2 and $3 \quad$

The correct answer is:
2 only

Solution:
If a line makes angle $\theta, \phi$ and $\psi$ with $\mathrm{x}, \mathrm{y}, \mathrm{z}$ axes respectively, then $\cos ^{2} \theta+\cos ^{2} \phi+\cos ^{2} \psi=1$
$\Rightarrow \cos ^{2} \theta+\cos ^{2} \phi=1-\cos ^{2} \psi \Rightarrow \cos ^{2} \theta+\cos ^{2} \phi=\sin ^{2} \psi$
Statement $(2)$ is correct

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2007 (Phase 2)
A line makes the same angle $\alpha$ with each of the $x$ and $y$ axes. If the angle $\theta$, which it makes with the $z$ -axis, is such that $\sin ^{2} \theta=2 \sin ^{2} \alpha$, then what is the value of $\alpha$ ?

Options:
A $\pi / 4$
B $\pi / 6$
C $\pi / 3$
D $\pi / 2$

The correct answer is:
$\pi / 4$

Solution:
Since $l^{2}+m^{2}+n^{2}=1$
$\therefore \cos ^{2} \alpha+\cos ^{2} \alpha+\cos ^{2} \theta=1$
$(\because A$ line makes the same angle $\alpha$ with $x$ and $y$ -axes and $\theta$ with z-axis) Also, $\sin ^{2} \theta=2 \sin ^{2} \alpha$
$\Rightarrow 1-\cos ^{2} \theta=2\left(1-\cos ^{2} \alpha\right)\left(\because \sin ^{2} A+\cos ^{2} A=1\right)$
$\Rightarrow \cos ^{2} \theta=2 \cos ^{2} \alpha-1$
From Eq. (i) and (ii) $2 \cos ^{2} \alpha+2 \cos ^{2} \alpha-1=1$
$\Rightarrow 4 \cos ^{2} \alpha=2 \Rightarrow \cos ^{2} \alpha=\frac{1}{2}$
$\Rightarrow \cos \alpha=\pm \frac{1}{\sqrt{2}}$
$\Rightarrow \alpha=\frac{\pi}{4}, \frac{3 \pi}{4}$

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2009 (Phase 2)
A line passes through $(2,2)$ and is perpendicular to the line $3 x+y=3$. Its $y$ -intercept is

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

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

Solution:
A line passes through $(2,2)$ and is perpendicular to the line $3 x+y=3$ Slope of line $3 \mathrm{x}+\mathrm{y}=3$ is $-3$
Slope of line which passes through $(2,2)$ is $\frac{1}{3}$
$\therefore$ Equation of line passes through $(2,2)$ and having
slope $\left(\frac{1}{3}\right)$ is
$(\mathrm{y}-2)=\frac{1}{3}(\mathrm{x}-2)$
$\therefore \quad 3 y-6=x-2$
$\therefore \quad x-3 y+4=0$
In order to find y-intercept of the line Put $x=0$ in $x-3 y+4=0$ $\therefore \quad-3 \mathrm{y}=-4$
$\mathrm{y}=\frac{4}{3}$
$\therefore$ Option (b) is correct.

Topic: Mathematics, Straight Lines, NDA, NDA 2015 (Phase 1)
A line passes through the points $(6,-7,-1)$ and $(2,-3,1)$. What are the direction ratios of the line?

Options:
A $ < 4,-4,2>$
B $ < 4,4,2>$
C $ < -4,4,2>$
D $ < 2,1,1>$

The correct answer is:
$ < -4,4,2>$

Solution:
Direction ratios $ < (2-6),(-3+7),(1+1)>$ $=\langle-4,4,2>$

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2014 (Phase 1)
A logic circuit provides the output $Y$ as per the following truth table :
\begin{array}{|l|l|l|}\hlineA & B & Y \\\hline 0 & 0 & 1 \\0 & 1 & 0 \\1 & 0 & 1 \\1 & 1 & 0 \\\hline\end{array}
The expression for the output $Y$ is :

Options:
A $A \cdot \bar{B}+\bar{A}$
B $\bar{B}$
C $B$
D $A \cdot B+\bar{A}$

The correct answer is:
$\bar{B}$

Solution:
\begin{array}{|l|l|l|}\hlineA & B & Y \\\hline 0 & 0 & 1 \\0 & 1 & 0 \\1 & 0 & 1 \\1 & 1 & 0 \\\hline\end{array}
According to given truth table, output is independent on value of $A$
$\therefore$ Output $Y=\bar{B}$

Topic: Physics, Semiconductors, NEET, NEET 2024
A lot of 4 white and 4 red balls is randomly divided into two halves. What is the probability that there will be 2 red and 2 white balls in each half?

Options:
A $18 / 35$
B $3 / 35$
C $1 / 2$
D None of these

The correct answer is:
$18 / 35$

Solution:
Required probability $=\frac{{ }^{4} C_{2} \times{ }^{4} C_{2}}{{ }^{8} C_{2}}=\frac{36}{70}=\frac{18}{35}$

Topic: Mathematics, Probability, NDA, NDA 2010 (Phase 2)
A low-loss transformer has 230 V applied to the primary and gives 4.6 V in the secondary. The secondary is connected to a load, which draws 5 A of current. The current (in ampere) in the primary is

Options:
A 0.1
B 1.0
C 10
D 250

The correct answer is:
0.1

Solution:
For the transformer
Power output = power input
Es is=Epip
ESEP=iPiS⇒iP=ESEP×iS=4.6230×5=0.1 A

Topic: Physics, Alternating Current, NEET
A machine has three parts, A, B and $\mathrm{C}$, whose chances of being defective are $0.02,0.10$ and $0.05$ respectively. The machine stops working if any one of the parts becomes defective. What is the probability that the machine will not stop working?

Options:
A $0.06$
B $0.16$
C $0.84$
D $0.94$

The correct answer is:
$0.84$

Solution:
Probability that machine stops working $=P(A \cup B \cup C)$
$\Rightarrow P(A \cup B \cup C)=P(A)+P(B)+P(C)$
$-P(A \cap B)-P(A \cap C)-P(B \cap C)$
$+P(A \cap B \cap C)$
$\begin{aligned} \Rightarrow P(A \cup B \cup C)=& P(A)+P(B)+P(C) \\ &-P(A) P(B)-P(A) P(C) \\ &-P(B) P(C)+P(A) P(B) P(C) \end{aligned}$
$(\because A, B \& C$ are independent events $)$ $\Rightarrow P(A \cup B \cup C)=0.02+0.1+0.05-(0.02 \times 0.1)$
$-(0.02 \times 0.05)-(0.1 \times 0.05)$
$+(0.02 \times 0.05 \times 0.1)$
$\Rightarrow P(A \cup B \cup C)=0.16$
$\therefore$ Probability that the machine will not stop working $=1-P(A \cup B \cup C)=1-0.16=0.84$

Topic: Mathematics, Probability, NDA, NDA 2016 (Phase 2)
A man observes the elevation of a balloon to be $30^{\circ} . \mathrm{He}$, then walks $1 \mathrm{~km}$ towards the balloon and finds that the elevation is $60^{\circ}$. What is the height of the balloon?

Options:
A $1 / 2 \mathrm{~km}$
B $\sqrt{3} / 2 \mathrm{~km}$
C $1 / 3 \mathrm{~km}$
D $1 \mathrm{~km}$

The correct answer is:
$\sqrt{3} / 2 \mathrm{~km}$

Solution:
Let the height of the balloon be $\mathrm{h}$ and new distance between $\mathrm{BC}$ be y as shown in figure given below.
$\tan 60^{\circ}=\frac{C D}{B C}$
$\Rightarrow \sqrt{3}=\frac{h}{y} \Rightarrow h=\sqrt{3} y$
and now in $\triangle \mathrm{ADC}, \quad \tan 30^{\circ}=\frac{C D}{A C}$
$\Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{1+y}$
$\Rightarrow 1+y=h \sqrt{3} \Rightarrow 1+y=3 y$
$\Rightarrow y=\frac{1}{2}$
$\therefore \quad h=\frac{\sqrt{3}}{2}$

Topic: Mathematics, Heights and Distances, NDA, NDA 2009 (Phase 1)
A man running round a racecourse notes that the sum of the distances of two flag-posts from him is always $10 \mathrm{~m}$ and the distance between the flag-posts is $8 \mathrm{~m}$. The area of the path he encloses is

Options:
A $18 \pi$ square metres
B $15 \pi$ square metres
C $12 \pi$ square metres
D $8 \pi$ square metres

The correct answer is:
$15 \pi$ square metres

Solution:
Given that sum of the distances of two flag-posts from him is always $10 \mathrm{~m}$. So, the race course is in the shape of ellipse. From the given figure, $\mathrm{Mf}_{1}+\mathrm{Mf}_{2}=10$
Let 'a' be the length of semi major axis and 'b' be the length of semi minor axis. $\mathrm{Mf}_{1}+\mathrm{Mf}_{2}=10 \Rightarrow 2 \mathrm{a}=10 \Rightarrow \mathrm{a}=$
Also, $\mathrm{f}_{1} \mathrm{f}_{2}=8$
Let $\mathrm{f}_{1}=(\mathrm{C}, 0)$ and $\mathrm{f}_{2}=(-\mathrm{C}, 0)$.
$\therefore \mathrm{f}_{1} \mathrm{f}_{2}=8 \Rightarrow 2 \mathrm{C}=8 \Rightarrow \mathrm{C}=4$
We knoy $=9=3^{2}$
$\therefore \mathrm{b}=3$
Area of the racecourse $=\pi \mathrm{ab}=\pi \times 5 \times 3=15 \pi$ sq. $\mathrm{m}$

Topic: Mathematics, Ellipse, NDA, NDA 2017 (Phase 2)
A man standing on the bank of a river observes that the angle of elevation of the top of a tree just on the opposite bank is $60^{\circ}$. The angle of elevation is $30^{\circ}$ from a point at a distance $y m$ from the bank of the river. What is the height of the tree?

Options:
A $\mathrm{ym}$
B $2 \mathrm{ym}$
C $\frac{\sqrt{3} y}{2} m$
D $\frac{y}{2} m$

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

Solution:
Let DC be the tree of height $h$ metre. Let a man is standing on the point $B$ (bank of a river) Let $\mathrm{BC}=x$ and angle of elevation i.e. $\angle \mathrm{DBC}=60^{\circ} .$
$$
\text { Also, let } \mathrm{AB}=y \text { and } \angle \mathrm{DAC}=30^{\circ}
$$
$\operatorname{In} \Delta \mathrm{ACD}$
$\tan 30^{\circ}=\frac{\mathrm{CD}}{\mathrm{AC}}$
$\Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{x+y}$
$\Rightarrow x+y=h \sqrt{3}$
and in $\Delta \mathrm{BCD}$,
$$
\tan 60^{\circ}=\frac{\mathrm{CD}}{\mathrm{BC}} \Rightarrow \sqrt{3}=\frac{h}{x}
$$
$\Rightarrow x=\frac{h}{\sqrt{3}}$
From eqns. (i) and(ii),
$$
\frac{h}{\sqrt{3}}+y=h \sqrt{3}
$$
$\Rightarrow y=h\left(\sqrt{3}-\frac{1}{\sqrt{3}}\right)=\frac{2 h}{\sqrt{3}} \Rightarrow h=\frac{\sqrt{3} y}{2} \mathrm{~m}$

Topic: Mathematics, Heights and Distances, NDA, NDA 2011 (Phase 1)
A man walks $10 \mathrm{~m}$ towards a lamp post and notices that the angle of elevation of the top of the post increases from $30^{\circ}$ to $45^{\circ}$. The height of the lamp posts is :

Options:
A $10 \mathrm{~m}$
B $(5 \sqrt{3}+5) \mathrm{m}$
C $(5 \sqrt{3}-5) \mathrm{m}$
D $(10 \sqrt{3}+10) \mathrm{m}$

The correct answer is:
$(5 \sqrt{3}+5) \mathrm{m}$

Solution:
Let $\mathrm{AB}$ bethe lamp post of height ${ }^{\prime} \mathrm{h}$ '. Let $\mathrm{BC}=\mathrm{x}$ meter In $\Delta \mathrm{ABC}$
$\tan 45^{\circ}=\frac{\mathrm{h}}{\mathrm{x}} \Rightarrow 1=\frac{\mathrm{h}}{\mathrm{x}} \Rightarrow \mathrm{x}=\mathrm{h}$
and in $\triangle \mathrm{ABD}$,
$\tan 30^{\circ}=\frac{\mathrm{h}}{\mathrm{x}+10}$
$\Rightarrow \frac{1}{\sqrt{3}}=\frac{\mathrm{h}}{\mathrm{h}+10} \quad(\because \mathrm{x}=\mathrm{h})$
$\Rightarrow \mathrm{h}+10=\sqrt{3} \mathrm{~h}$
$\Rightarrow \mathrm{h}=\left(\frac{10}{\sqrt{3}-1}\right) \mathrm{m}=\frac{10 \sqrt{3}+10}{2}$

Topic: Mathematics, Heights and Distances, NDA, NDA 2013 (Phase 1)
A mapping $f: R \rightarrow R$ which is defined as $[2009-I I]$
$f(x)=\cos x ; x \in R$ is

Options:
A One-one only
B Onto only
C One-one onto
D Neither one-one nor onto

The correct answer is:
Neither one-one nor onto

Solution:
Let $x_{1}, x_{2} \in \mathrm{R}$
Then, $f\left(x_{1}\right)=f\left(x_{2}\right)$
$\Rightarrow \cos x_{1}=\cos x_{2}$
$\Rightarrow x_{1}=2 n \pi \pm x_{2}$
So, $x_{1} \neq x_{2}$
Hence, $\cos \mathrm{x}$ is not one-one function. Now, let $\mathrm{y}=\cos \mathrm{x}$
We know, $-1 \leq \cos x \leq 1$
$\therefore \mathrm{y} \in[-1,1]$
$[-1,1] \subset \mathrm{R}$. So, $\cos \mathrm{x}$ is into function, not onto. Hence, $\mathrm{f}(\mathrm{x})=\cos \mathrm{x}$ is neither one-one nor onto.
#

Topic: Mathematics, Functions, NDA, NDA 2009 (Phase 2)
A matrix $X$ has $(a+b)$ rows and $(a+2)$ columns; and a matrix $Y$ has $(b+1)$ rows and $(a+3)$ columns. If both $X Y$ and $Y X$ exist, then what are the values of $a, b$ respectively?

Options:
A 3,2
B 2,3
C 2,4
D 4,3

The correct answer is:
2,3

Solution:
The order of a given matrices are
$[X]_{(a+b)\times(a+2)}$ and $[Y]_{(b+1)\times(a+3)}$
As $[X Y]$ and $[Y X]$ exist
$\therefore a+2=b+1$ and $a+3=a+b$
$\Rightarrow a+3=a+b$
$\Rightarrow b=3$
Hence, $a=3+1-2=2$

Topic: Mathematics, Matrices, NDA, NDA 2009 (Phase 2)
A medicine is known to be $75 \%$ effective to cure a patient. If the medicine is given to 5 patients, what is the probability that at least one patient is cured by this medicine?

Options:
A $\frac{1}{1024}$
B $\frac{243}{1024}$
C $\frac{1023}{1024}$
D $\frac{781}{1024}$

The correct answer is:
$\frac{1023}{1024}$

Solution:
Probabilty of medicine to cure a patient $=\frac{75}{100}=\frac{3}{4}$ Probability of curing at least one patient
$=1-\left[1-\frac{3}{4}\right]^{5}=1-\left(\frac{1}{4}\right)^{5}=1-\frac{1}{1024}=\frac{1023}{1024}$

Topic: Mathematics, Probability, NDA, NDA 2016 (Phase 2)
A meeting is to be addressed by 5 speakers A, B, C, D, E. In how many ways can the speakers be ordered, if B must not precede A(immediately or otherwise)?

Options:
A 120
B 24
C 60
D $5^{4} \times 4$
#

The correct answer is:
24

Solution:
According to given restriction:
B must not precede A(immediately or otherwise),
$\Rightarrow$ A must follow B, i.e., B should addressed the meeting at first place So, rest of the four speakers can address in $4 !$ ways.
$\therefore \quad$ Required number of ways $=4 !=24$

Topic: Mathematics, Permutation Combination, NDA, NDA 2007 (Phase 2)