datavorous/entrance-exam-dataset huggingface dataset:
For 10 observations on price $(\mathrm{x})$ and supply (y), the following data was obtained :
$\sum \mathrm{x}=130, \sum \mathrm{y}=220$
$\sum \mathrm{x}^{2}=2288, \sum \mathrm{y}^{2}=5506$ and $\sum \mathrm{xy}=3467$.
What is line of regression of $y$ on $x ?$

Options:
A $\mathrm{y}=0.91 \mathrm{x}+8.74$
B $\mathrm{y}=1.02 \mathrm{x}+8.74$
C $\mathrm{y}=1.02 \mathrm{x}-7.02$
D $\mathrm{y}=0.91 \mathrm{x}-7.02$

The correct answer is:
$\mathrm{y}=1.02 \mathrm{x}+8.74$

Solution:
Line of regression of $y$ on $x$ is:
$y-\bar{y}=b_{y x}(x-\bar{x})$
$\bar{y}=\frac{\Sigma y}{n} ; \bar{x}=\frac{\Sigma x}{n} \Rightarrow \bar{y}=\frac{220}{10}=22 ; \bar{x}=\frac{130}{10}=13$
$b_{y x}=\mathbf{r} \cdot \frac{\sigma_{y}}{\sigma_{x}}$
$\mathrm{r}=\frac{n \Sigma x y-(\Sigma x)(\Sigma y)}{\sqrt{\left[n \Sigma x^{2}-(\Sigma x)^{2}\right]\left[n \Sigma y^{2}-(\Sigma y)^{2}\right]}}$
$=\frac{10(3467)-(130)(220)}{\sqrt{\left[(10 \times 2288)-130^{2}\right]\left[(10 \times 5506)-\left(220^{2}\right)\right]}}$
$r=0.962$
$\sigma_{y}=\sqrt{\frac{\Sigma y^{2}}{n}-\left(\frac{\Sigma y}{n}\right)^{2}} \Rightarrow \sigma_{y}=8.2 ; \sigma_{x}=7.73$
$\Rightarrow b_{y x}=0.962 \times \frac{8.2}{7.73}=1.02$
$\Rightarrow$ Line of regression of $y$ on $x$ is
$y-22=1.02(x-13)$
$\Rightarrow y=1.02 x+8.74$

Topic: Mathematics, Statistics, NDA, NDA 2016 (Phase 2)
For 20 observations of variable $x$, if $\sum\left(x_{\mathrm{i}}-2\right)=20$ and $\sum\left(x_{\mathrm{i}}-2\right)^2=100$, then the standard deviation of variable $x$ is

Options:
A $2$
B $3$
C $4$
D $9$

The correct answer is:
$2$

Solution:
Note that standard derivation is independent of change of origin.
$\therefore \quad$ S.D. of $x_{\mathrm{i}}=$ S.D. of $\left(x_{\mathrm{i}}-2\right)$
$$
\begin{aligned}
\therefore \quad \text { S.D. of }\left(x_{\mathrm{i}}-2\right) & =\sqrt{\frac{1}{\mathrm{n}} \sum_{\mathrm{i}}^{20}\left(x_{\mathrm{i}}-2\right)^2-\left[\frac{\sum\left(x_{\mathrm{i}}-2\right)}{\mathrm{n}}\right]^2} \\
& =\sqrt{\frac{100}{20}-(1)^2} \\
& =2 \\
\Rightarrow \text { Required S.D } & =2
\end{aligned}
$$

Topic: Mathematics, Statistics, MHT CET, MHT CET 2023 (12 May Shift 2)
For a binomial distribution $B(n, p)$ mean $=200$, standard deviation $=10$, then $n^2+\frac{1}{p^2}+\frac{1}{q^2}$ is equal to

Options:
A $160004$
B $160006$
C $160008$
D $160002$

The correct answer is:
$160008$

Solution:
In binomial distribution mean $=n p$ and
standard deviation $=\sqrt{n p q}$
$\therefore \quad 200=n p$ and $10=\sqrt{n p q}$
$200=n p$ ...(i)
$100=n p q$ ...(ii)
From Eqs. (i) and (ii), we get
$q=\frac{1}{2}, n=400$ and $p=\frac{1}{2}$
$\therefore \quad n^2+\frac{1}{p^2}+\frac{1}{q^2}$
$(400)^2+(2)^2+(2)^2$
$=160000+4+4=160008$

Topic: Mathematics, Probability, AP EAMCET, AP EAMCET 2021 (24 Aug Shift 2)
For a binomial distribution $B(n, p), \eta p=4$ and variance $n p q$ $=4 / 3 .$ What is the probability $P(x \geq 5)$ equal to?

Options:
A $(2 / 3)^{6}$
B $(1 / 3)^{6}$
C $(1 / 3)^{6}$
D $Q^{8} / 3^{6} y$

The correct answer is:
$Q^{8} / 3^{6} y$

Solution:
Given, $n p=4$ and $n p q=\frac{4}{3}$
$\therefore \quad 4 q=\frac{4}{3} \Rightarrow q=\frac{1}{3}$
$\therefore \quad p=1-\frac{1}{3}=\frac{2}{3}$
$\Rightarrow \quad n=\frac{4 \times 3}{2}=6$
Now, $P(X \geq 5)={ }^{6} C_{5}(p)^{5}(q)^{1}+{ }^{6} C_{6} p^{6} q^{0}$
$={ }^{6} C_{5}\left(\frac{2}{3}\right)^{5}\left(\frac{1}{3}\right)+{ }^{6} C_{6}\left(\frac{2}{3}\right)^{6}$
$=\frac{6 \times 32}{3^{6}}+\frac{64}{3^{6}}=\frac{256}{3^{6}}=\frac{2^{8}}{3^{6}}$

Topic: Mathematics, Probability, NDA, NDA 2009 (Phase 1)
For a cell reaction,
$A(s)+B^{2+}(a q) \longrightarrow A^{2+}(a q)+B(s)$; the
standard emf of the cell is $0.295 \mathrm{~V}$ at $25^{\circ} \mathrm{C}$. The equilibrium constant at $25^{\circ} \mathrm{C}$ will be

Options:
A $1 \times 10^{10}$
B 10
C $2.95 \times 10^{-2}$
D $2.95 \times 10^{-10}$

The correct answer is:
$1 \times 10^{10}$

Solution:
For this given reaction, $n=2$. At $25^{\circ} \mathrm{C}$,
$\begin{aligned}
\therefore \quad K & =\text { antilog }\left[\frac{n E^{\circ}}{0.059}\right]=\operatorname{antilog}\left[\frac{2 \times(-0.295)}{0.059}\right] \\
K & =1 \times 10^{10}
\end{aligned}$

Topic: Chemistry, Electrochemistry, COMEDK, COMEDK 2023
For a cell reaction to be spontaneous, the standard free energy change of the reaction must be

Options:
A zero
B positive
C infinite
D negative

The correct answer is:
negative

Solution:
For spontaneous reaction free energy change is negative.
$\Delta \mathrm{G}=-\mathrm{nFE}$

Topic: Chemistry, Electrochemistry, VITEEE, VITEEE 2006
For a certain organ pipe, three successive resonance frequencies are observed at 425,595 and $765 \mathrm{~Hz}$, respectively. The length of the pipe is (speed of sound in air $=340 \mathrm{~ms}^{-1}$ )

Options:
A $0.5\ m$
B $1\ m$
C $1.5\ m$
D $2\ m$

The correct answer is:
$1\ m$

Solution:
For closed organ pipe only odd harmonies are possible.
$\therefore$ Fundamental frequency
Fundamental frequency for a closed organ pipe
$\begin{aligned}
& v_0=\frac{v_0}{4 l} \Rightarrow 85=\frac{340}{4 \times l} \\
& {\left[\because v_0=\frac{595-425}{2}=\frac{765-595}{2}=85 \mathrm{~Hz}\right]} \\
& l=\frac{340}{4 \times 85}=1 \mathrm{~m}
\end{aligned}$

Topic: Physics, Waves and Sound, AP EAMCET, AP EAMCET 2016
For a chemical reaction, the standard Gibbs energy change, $\Delta G^{\circ}$ is $-7.64 \times 10^4 \mathrm{~J} \mathrm{~mol}^1$. What is the value of equilibrium constant (K)?

Options:
A $K=1$
B $k>1$
C $K < 1$
D $K=0$

The correct answer is:
$k>1$

Solution:
Standard Gibbs free energy,
$$
\Delta G^{\circ}=-2.303 R T \log K
$$
Given, $\Delta G^{\circ}=-7.64 \times 10^4 \mathrm{~J} / \mathrm{mol}$
$$
\begin{aligned}
\log K & =-\frac{\Delta G^{\circ}}{2.303 R T} \\
K & =\operatorname{antilog}-\left(\frac{\Delta G^{\circ}}{2.303 R T}\right) \\
K & =\text { antilog }-\left(\frac{(-) 7.64 \times 10^4}{2.303 \times 8.314 \times 298}\right)
\end{aligned}
$$
Here, $K$ is greater than one.
Thus, option (2) is correct.

Topic: Chemistry, Thermodynamics (C), TS EAMCET, TS EAMCET 2019 (04 May Shift 1)
For a circular coil of radius $R$ and $N$ turns carrying current $I$, the magnitude of the magnetic field at a point on its axis at a distance $\mathbf{x}$ from its centre is given by,
$$
\mathbf{B}=\frac{\mu_0 \mathrm{IR}^2 \mathrm{~N}}{2\left(\mathrm{x}^2+\mathrm{R}^2\right)^{3 / 2}}
$$
(a) Show that this reduces to the familiar result for field at the centre of the coil.
(b) Consider two parallel co-axial circular coils of equal radius $R$, and number of turns $N$, carrying equal currents in the same direction, and separated by a distance R. Show that the field on the axis around the mid-point between the coils is uniform over a distance that is small as compared to $R$, and is given by, $\mathbf{B}=0.72 \frac{\mu_0 \mathrm{NI}}{\mathrm{R}}$, approximately.
(Such an arrangement to produce a nearly uniform magnetic field over a small region is known as Helmholtz coils.)

Options:

None

Solution:
(a) $\mathrm{B}=\frac{\mu_0 \mathrm{IR}^2 \mathrm{~N}}{2\left(\mathrm{x}^2+\mathrm{R}^2\right)^{3 / 2}}$
At the centre of the coil $x=0$
$B=\frac{\mu_0 I^2 \mathrm{~N}}{2 \mathrm{R}^3}=\frac{\mu_0 \mathrm{IN}}{2 \mathrm{R}}$-which is same standard result.
(b) In a small region of length $2 \mathrm{~d}$ about the mid-point between the coils,
$$
\begin{gathered}
\mathrm{B}=\frac{\mu_0 \mathrm{IR}^2 \mathrm{~N}}{2} \times\left[\left\{\left(\frac{\mathrm{R}}{2}+\mathrm{d}\right)^2+\mathrm{R}^2\right\}^{-3 / 2}\right. \\
\left.+\left\{\left(\frac{\mathrm{R}}{2}-\mathrm{d}\right)^2+\mathrm{R}^2\right\}^{-3 / 2}\right] \\
=\frac{\mu_0 \mathrm{IR}^2 \mathrm{~N}}{2} \times\left(\frac{5 \mathrm{R}^2}{4}\right)^{3 / 2} \\
\left.\approx \frac{\mu_0 \mathrm{IR}^2 \mathrm{~N}}{2 \mathrm{R}^3} \times\left(\frac{4}{5}\right)^{-3 / 2} \times\left[1-\frac{4 \mathrm{~d}}{5 \mathrm{R}}\right)^{-3 / 2}+\left(1-\frac{3 \mathrm{~d}}{5 \mathrm{R}}\right)^{-3 / 2}\right]
\end{gathered}
$$
(Neglecting term with $\mathrm{d}^2 / \mathrm{R}^2$ and higher powers of $\mathrm{d} / \mathrm{R}$ since $\mathrm{d} / \mathrm{R}< < 1$ ).
$$
\therefore \quad \mathrm{B}=\left(\frac{4}{5}\right)^{3 / 2} \frac{\mu_0 \mathrm{IN}}{\mathrm{R}} \approx 0.72 \frac{\mu_0 \mathrm{IN}}{\mathrm{R}}
$$

Topic: Physics, Magnetic Effects of Current
For a critical study of secondary growth in plants, which one of the following pairs is suitable?

Options:
A teak and pine
B deodar and fern
C wheat and maiden hair fern
D sugarcane and sunflower.

The correct answer is:
teak and pine

Solution:
Teak and pine is most suitable for the study of critical secondary growth because in secondary growth, secondary tissues are formed from lateral meristem which is well developed in these two cases andsecondary growth occurs in gymnosperms and dicots. Wheat, sugarcane are monocots, maiden hair fern and fern are pteridophytes, deodar and sunflower are dicots.

Topic: Biology, Organisms and Populations, NEET, NEET 2007
For a group of 100 students, the mean x¯1 and the standard deviation σ1 of their marks were found to be 40 and 15 respectively. Later it was observed that the scores 40 and 50 were misread as 30 and 60 respectively. If the mean and the standard deviation with the corrected observations of the scores are x¯2 and σ2 respectively, then

Options:
A x¯1=x¯2;σ1=σ2
B x¯1=x¯2;σ1<σ2
C x¯1=x¯2;σ1>σ2
D x¯1>x¯2;σ1=σ2

The correct answer is:
x¯1=x¯2;σ1>σ2

Solution:
It is given that.

x¯1=40

σ=15

Now, the mean is,

x¯1=∑xin

⇒40=∑xi100

⇒∑xi=4000

Corrected summation,

∑xi'=∑xi-30-60+40+50=4000

The corrected mean,

x¯2=∑xi'100

⇒x¯2=4000100=40

σ12=∑xi2n-∑xin2

⇒225=∑xi2100-40001002

⇒∑xi2=182500

Corrected summation,

∑xi2=182500-302-602+402+502

⇒∑xi2=182100

The corrected standard deviation,

σ22=∑xi2n-∑xin2

⇒σ22=182500100-40001002

⇒σ22=1821-1600

⇒σ2=14.86

Thus, x¯1=x¯2 and σ1>σ2.

Topic: Mathematics, Statistics, AP EAMCET, AP EAMCET 2019 (21 Apr Shift 2)
For a half-cell containing a Pt rod immersed in a solution of $1 \mathrm{M} \mathrm{H} A, \mathrm{O}_2(g)$ is bubbled at $1 \mathrm{~atm}$. The standard reduction potential for water formation is $1.23 \mathrm{~V}$. Given a dissociation constant, $K_a=1 \times 10^{-4}$ for $\mathrm{H} A$, what is $E_{\text {Half-cell }}$ at $298 \mathrm{~K}$ in V ?

Options:
A 1.289
B 1.171
C 1.348
D 1.112

The correct answer is:
1.348

Solution:
Given, Standard reduction potential for water formation $=1.23 \mathrm{~V}$
Dissociation constant $\left(K_a\right)$ for $\mathrm{H} A=1 \times 10^{-4}$ $E_{\text {half-cell }}$ at $298 \mathrm{~K}=$ ?
Number of electrons $=2$
$$
\begin{aligned}
& E=E^{\circ}-\frac{0.0592}{2} \log Q\left(K_a\right) \\
& E=1.23-\frac{0.0592}{2} \log \left(1 \times 10^{-4}\right) \\
& E=1.23-\frac{0.0592}{2} \times-4 \\
& E=1.348 \mathrm{~V}
\end{aligned}
$$

Topic: Chemistry, Electrochemistry, TS EAMCET, TS EAMCET 2019 (04 May Shift 2)
For a phase change
$\mathrm{H}_2 \mathrm{O}$
(l)
(s)

Options:
A $\Delta G=0$
B $\Delta S=0$
C $\Delta H=0$
D $\Delta U=0$.

The correct answer is:
$\Delta G=0$

Solution:
$\Delta G^{\circ}=-R T \ln K$
[ $\Delta G^{\circ}=$ standard free energy change, $K=$ equilibrium constant]
If a substance is in equilibrium between two phases at constant temperature and pressure, its chemical potential must have the same value in both the phases.
$$
\therefore \Delta G^{\circ}=0 \text {. }
$$

Topic: Chemistry, Thermodynamics (C), AIIMS, AIIMS 2006
For a point of inflection of $y=f(x)$, which one of the following is correct?

Options:
A $\frac{d y}{d x}$ must be zero
B $\frac{d^{2} y}{d x^{2}}$ must be zero
C $\frac{d y}{d x}$ must be non-zero
D $\frac{d^{2} y}{d x^{2}}$ must be non-zero

The correct answer is:
$\frac{d^{2} y}{d x^{2}}$ must be zero

Solution:
Lety $=\mathrm{f}(\mathrm{x})$ Now, for a point of inflection of $\mathrm{y}=\mathrm{f}(\mathrm{x})$,
$\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}$ must be zero.

Topic: Mathematics, Application of Derivatives, NDA, NDA 2010 (Phase 2)
For a positive integer $n$, what is the value of $i^{4 n+1}$ ?

Options:
A 1
B -1
C i
D -i

The correct answer is:
i

Solution:
$i^{4 n+1}=\left(i^{4}\right)^{n} \times i=(1)^{n} \times i=1 \times i=i$

Topic: Mathematics, Complex Number, NDA, NDA 2008 (Phase 2)
For a reaction $A+B \longrightarrow P$, the following data are provided
The rate constant for this reaction in standard unit is

Options:
A 5
B 1.2
C $2.4 \times 10^{-4}$
D 50

The correct answer is:
50

Solution:
(d) $A+B \longrightarrow P$
According to rate law expression
$$
\text { Rate }=k[A]^\alpha[B]^\beta(K \text { is constant }) \ldots (i)
$$
Taking ratio of (1) and (2) entry,
$$
\begin{aligned}
\frac{4 \times 10^{-2}}{2 \times 10^{-2}} & =\frac{[0.04]^\beta[0.02]^\alpha}{[0.02]^\beta[0.02]^\alpha} \\
2 & =[2]^\beta \\
{[2]^1 } & =[2]^\beta \\
\beta & =1
\end{aligned}
$$
Taking ratio of (2) and (3) entry,
$$
\begin{aligned}
\frac{8 \times 10^{-2}}{4 \times 10^{-2}} & =\frac{[0.04]^\alpha[0.04]^\beta}{[0.02]^\alpha[0.04]^\beta} \\
{[2] } & =[2]^\alpha \Rightarrow \alpha=1
\end{aligned}
$$
Put value of $\alpha$ and $\beta$ in (1) entry
$$
\begin{aligned}
2 \times 10^{-2} & =k[A]^1[B]^1 \\
k & =\frac{2 \times 10^{-2}}{0.02 \times 0.02} \\
& =\frac{2 \times 10^{-2} \times 10^4}{4}=\frac{100}{2}=50
\end{aligned}
$$

Topic: Chemistry, Chemical Kinetics, TS EAMCET, TS EAMCET 2021 (06 Aug Shift 1)
For a reversible reaction,
$$
X(g)+3 Y(g) \rightleftharpoons 2 Z(g)
$$
$\Delta H=-40 \mathrm{~kJ}$ the standard entropies of $X, Y$ and $Z$ are 60,40 and $50 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$ respectively.
The temperature at which the above reaction attains equilibrium is about

Options:
A $400 \mathrm{~K}$
B $500 \mathrm{~K}$
C $273 \mathrm{~K}$
D $373 \mathrm{~K}$

The correct answer is:
$500 \mathrm{~K}$

Solution:
$\Delta H=-40 \mathrm{~kJ}=-40 \times 10^{3} \mathrm{~J}$
$\begin{aligned} \Delta S &=\Delta S_{\text {Product }}-\Delta S_{\text {Reactant }} \\ &=2(50)-3(40)-60 \\ &=100-120-60=-80 \\ \Delta G &=\Delta H-T \Delta S \\ 0 &=-40 \times 10^{3}-T(-80) \\ T &=\frac{40 \times 10^{3}}{80}=500 \mathrm{~K} \end{aligned}$

Topic: Chemistry, Thermodynamics (C), COMEDK, COMEDK 2018
For a set $\mathrm{A}$, consider the following statements:
$\begin{array}{lll}\text { 1. } \mathrm{A} \cup \mathrm{P}(\mathrm{A})=\mathrm{P}(\mathrm{A}) & \text { 2. } & \{\mathrm{A}\} \cap \mathrm{P}(\mathrm{A})=\mathrm{A}\end{array}$
3.$\mathrm{P}(\mathrm{A})-\{\mathrm{A}\}=\mathrm{P}(\mathrm{A})$
where P denotes power set.
Which of the statements given above is/are correct?

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

The correct answer is:
1 only

Solution:
Since, Power set is the collection of all the subsets of the set A therefore $\mathrm{AUP}(\mathrm{A})=\mathrm{P}(\mathrm{A})$
$\therefore$ statement (1) iscorrect.

Topic: Mathematics, Sets and Relations, NDA, NDA 2010 (Phase 1)
For a set of discrete numbers, three measures of central $\begin{array}{ll}\text { tendency are given below } & {}\end{array}$
1 . Arithmetic mean 2 Median
3 . Geometric mear Which of the above measures may not have a meaningful definition?

Options:
A 1 only
B 2 only
C 3 only
D All of them are meaningfully defined

The correct answer is:
All of them are meaningfully defined

Solution:
(i) Arithmetic mean $=\frac{\text { Sum of all observations }}{\text { Total no. of obseryation }}$
(ii) Median = The midpoint of the data after being ranked (arranged in ascending order)
(iii) Geometric mean $=$ If $x_{1}, x_{2}, x_{3}, \ldots, x_{n}$ are $n$ values of $a$ variate $x$, none of them being zero, then the geometric mean $G$ is defined as $G=\left(x_{1} x_{2} x_{3} \ldots x_{n}\right)^{x}$ Thuse, all of them are meaningfully defined.

Topic: Mathematics, Statistics, NDA, NDA 2008 (Phase 2)
For a square matrix $\mathrm{A}$, which of the following properties hold?
$1.$ $\left(\mathrm{A}^{-1}\right)^{-1}=\mathrm{A}$
$2 \quad \operatorname{det}\left(\mathrm{A}^{-1}\right)=\frac{1}{\operatorname{det} \mathrm{A}}$
$3.$ $(\lambda \mathrm{A})^{-1} \lambda \mathrm{A}^{-1}$ where $\lambda$ is a scalar Select the correct answer using the code given below:

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

The correct answer is:
1 and 2 only

Solution:
Statement 1 and 2 are correct. Statement 3 is incorrect because
$(\lambda \mathrm{A})^{-1}=\frac{1}{\lambda} \mathrm{A}^{-1}, \lambda \neq 0$

Topic: Mathematics, Matrices, NDA, NDA 2018 (Phase 2)
For a stationary wave, $Y=10 \sin \left(\frac{\pi x}{15}\right) \cos (48 \pi t) \mathrm{cm}$, the distance between a node and the successive antinode is

Options:
A $7.5 \mathrm{~cm}$
B $30 \mathrm{~cm}$
C $15 \mathrm{~cm}$
D $60 \mathrm{~cm}$

The correct answer is:
$7.5 \mathrm{~cm}$

Solution:
Comparing given equation with standard equation $y=2 a \sin \left(\frac{2 \pi x}{\lambda}\right) \cos (2 \pi f t)$
where, $a$ is the amplitude, $k=\frac{2 \pi}{\lambda}$ the propagation constant and $f$ the frequency.
Therefore, we obtain: $\frac{2 \pi}{\lambda}=\frac{\pi}{15} \mathrm{~cm}^{-1}$
$\Rightarrow \lambda=30 \mathrm{~cm}$
Distance between nearest node and antinode $=\frac{\lambda}{4}=\frac{30}{4}=7.5 \mathrm{~cm}$

Topic: Physics, Waves and Sound, MHT CET, MHT CET 2022 (06 Aug Shift 2)
For a transformed, the turns ratio is 3 and its efficiency is $0.75$. The current flowing in the primary coil is $2 \mathrm{~A}$ and the voltage applied to it is $100 \mathrm{~V}$. Then the voltage and the current flowing in the secondary coil are ... respectively

Options:
A $150 \mathrm{~V}, 1.5 \mathrm{~A}$
B $300 \mathrm{~V}, 0.5 \mathrm{~A}$
C $300 \mathrm{~V}, 1.5 \mathrm{~A}$
D $150 \mathrm{~V}, 0.5 \mathrm{~A}$

The correct answer is:
$300 \mathrm{~V}, 0.5 \mathrm{~A}$

Solution:
Given, $\frac{n_{s}}{n_{P}}=3 ; \quad \eta=0.75 ; I_{P}=2 \mathrm{~A}$ and $V_{P}=100 \mathrm{~V}$
We know that,
$\begin{aligned}
&\qquad \eta=\frac{V_{S}}{V_{P}} \frac{I_{S}}{I_{P}} \\
&\text { or } \quad \eta=\left(\frac{n_{s}}{n_{P}}\right) \frac{I_{s}}{I_{P}} \quad\left(\because \frac{n_{s}}{n_{p}}=\frac{V_{s}}{V_{p}}\right) \\
&\Rightarrow \quad 0.75=\frac{3 \times I_{s}}{2} \\
&\Rightarrow \quad I_{s}=0.5 \mathrm{~A} \\
&\text { By the relation, } \frac{n_{s}}{n_{P}}=\frac{V_{s}}{V_{P}} \\
&\Rightarrow \quad 3=\frac{V_{s}}{100} \Rightarrow \quad V_{s}=300 \mathrm{~V}
\end{aligned}$

Topic: Physics, Alternating Current, KCET, KCET 2013
For all $n \in N, 2^{4 n}-15 n-1$ is divisible by

Options:
A 125
B 225
C 450
D None of the above

The correct answer is:
225

Solution:
Let $\mathrm{P}(n): 2^{4 n}-15 n-1$
Put $n=2$ $\mathrm{P}(2)=2^{8}-30-1=225$ which is divisible by $225 .$ Let us assume,
$\mathrm{P}(n)$ is true for $n=k$ is $\mathrm{P}(\mathrm{k}): 2^{4 \mathrm{k}}-15 \mathrm{k}-1$ is divisible by
$225 .$
$\Rightarrow 2^{4 k}-15 k-1=225 \lambda, \lambda \in R, k \in N$
To prove for $n=k+1$
Consider
$2^{4 k+4}-15 k-15-1=2^{4 k} \cdot 2^{4}-15 k-16$
$=2^{4}\left[\begin{array}{ll}225 \lambda+1+15 k]-15 k-16 & (\text { from }(i))\end{array}\right.$
$=2^{4} \cdot 225 \lambda \cdot 2^{4}+15 \cdot 2^{4} \cdot k-15 k-16$
$=24 \cdot 225 \lambda+225 k$
$=225\left[2^{4} \lambda+k\right]$
$=225 r$ where $r=2^{4} \lambda+k$ is a constant
Hence, $2^{4 n}-15 n-1$ is divisible by 225.

Topic: Mathematics, Binomial Theorem, NDA, NDA 2011 (Phase 2)
For an AP with first term u and common difference $v$, the $\mathrm{p}^{\text {th }}$ term is 15 uv more than the $\mathrm{q}^{\text {th }}$ term. Which one of the following is correct?

Options:
A $\mathrm{p}=\mathrm{q}+15 \mathrm{v}$
B $\mathrm{p}=\mathrm{q}+15 \mathrm{u}$
C $\mathrm{p}=\mathrm{q}+14 \mathrm{v}$
D $\mathrm{p}=\mathrm{q}+14 \mathrm{u}$

The correct answer is:
$\mathrm{p}=\mathrm{q}+15 \mathrm{u}$

Solution:
Since, first term and common difference of an $\mathrm{AP}$ are u and $\mathrm{v}$ respectively. $\mathrm{p}^{\text {th }}$ term, $\mathrm{Tp}=\mathrm{u}+(\mathrm{p}-1) \mathrm{v}$
and $q^{\text {th }}$ term, $\mathrm{T}_{\mathrm{q}}=\mathrm{u}+(\mathrm{q}-1) \mathrm{v}$ According to condition given in question, ...(ii)
$\Rightarrow \quad \mathrm{T}_{\mathrm{p}}=\mathrm{T}_{\mathrm{q}}+15 \mathrm{uv}$
$\Rightarrow \quad T_{n}^{r}-T_{0}^{4}=15 \mathrm{uv}$
$\Rightarrow \quad \mathrm{u}^{\mathrm{p}}+(\mathrm{p}-1) \mathrm{v}-\mathrm{u}-(\mathrm{q}-1) \mathrm{v}=15 \mathrm{uv}$
$\Rightarrow \mathrm{v}(\mathrm{p}-1-\mathrm{q}+1)=15 \mathrm{uv}$
$\Rightarrow \mathrm{v}(\mathrm{p}-\mathrm{q})=15 \mathrm{uv}$
$\Rightarrow \mathrm{p}-\mathrm{q}=15 \mathrm{u} \Rightarrow \mathrm{p}=\mathrm{q}+15 \mathrm{u}$

Topic: Mathematics, Sequences and Series, NDA, NDA 2006 (Phase 2)
For any \(a, b, c \in \mathbf{R}\), the determinant \(\left|\begin{array}{lll}b c & b+c & 1 \\ c a & c+a & 1 \\ a b & a+b & 1\end{array}\right|\) is equal to

Options:
A \(a\left(b^2-c^2\right)+b\left(c^2-a^2\right)+c\left(a^2-b^2\right)\)
B \(a(b-c)+b(c-a)+c(a-b)\)
C \((a-b)(b-c)(c-a)\)
D \(a b c\)

The correct answer is:
\((a-b)(b-c)(c-a)\)

Solution:
For any \(a, b, c \in \mathbf{R}\), the given determinant
\(\Delta=\left|\begin{array}{lll}
b c & b+c & 1 \\
c a & c+a & 1 \\
a b & a+b & 1
\end{array}\right|\)
On applying \(R_2 \rightarrow R_2-R_1\) and \(R_3 \rightarrow R_3-R_1\), we have
\(\begin{gathered}
\Delta=\left|\begin{array}{ccc}
b c & b+c & 1 \\
c(a-b) & a-b & 0 \\
b(a-c) & a-c & 0
\end{array}\right| \\
=(a-b)(a-c)\left|\begin{array}{ccc}
b c & b+c & 1 \\
c & 1 & 0 \\
b & 1 & 0
\end{array}\right|
\end{gathered}\)
On applying \(R_3 \rightarrow R_3-R_2\), we have
\(\begin{aligned}
\Delta & =(a-b)(a-c)\left|\begin{array}{ccc}
b c & b+c & 1 \\
c & 1 & 0 \\
b-c & 0 & 0
\end{array}\right| \\
& =(a-b)(a-c)[0-(b-c)]=(a-b)(b-c)(c-a)
\end{aligned}\)
Hence option (c) is correct.

Topic: Mathematics, Determinants, AP EAMCET, AP EAMCET 2020 (21 Sep Shift 2)
For any two events $A$ and $B$, which one of the following holds ?

Options:
A $\quad \mathrm{P}(\mathrm{A} \cap \mathrm{B}) \leq \mathrm{P}(\mathrm{A}) \leq \mathrm{P}(\mathrm{A} \cup \mathrm{B}) \leq \mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})$
B $\quad \mathrm{P}(\mathrm{A} \cup \mathrm{B}) \leq \mathrm{P}(\mathrm{A}) \leq \mathrm{P}(\mathrm{A} \cap \mathrm{B}) \leq \mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})$
C $\quad \mathrm{P}(\mathrm{A} \cup \mathrm{B}) \leq \mathrm{P}(\mathrm{B}) \leq \mathrm{P}(\mathrm{A} \cap \mathrm{B}) \leq \mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})$
D $\quad \mathrm{P}(\mathrm{A} \cap \mathrm{B}) \leq \mathrm{P}(\mathrm{B}) \leq \mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B}) \leq \mathrm{P}(\mathrm{A} \cup \mathrm{B})$

The correct answer is:
$\quad \mathrm{P}(\mathrm{A} \cap \mathrm{B}) \leq \mathrm{P}(\mathrm{A}) \leq \mathrm{P}(\mathrm{A} \cup \mathrm{B}) \leq \mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})$

Solution:
Clearly, $\mathrm{A} \cap \mathrm{B} \subseteq \mathrm{A}$
$\Rightarrow P(A \cap B) \leq \bar{P}(A)$
$\mathrm{A} \subset \mathrm{A} \cup \mathrm{B}$
$\Rightarrow \mathrm{P}(\mathrm{A}) \leq \mathrm{P}(\mathrm{A} \cup \mathrm{B})$
We know that, $\mathrm{P}(\mathrm{A} \cup \mathrm{B})=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})$
$\Rightarrow \mathrm{P}(\mathrm{A} \cup \mathrm{B}) \leq \mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})$
From eqs. (i), (ii) and (iii), we get $P(A \cap B) \leq P(A) \leq P(A \cup B) \leq P(A)+P(B)$

Topic: Mathematics, Probability, NDA, NDA 2014 (Phase 2)