datavorous/entrance-exam-dataset huggingface dataset:
The binomial distribution whose mean is 9 and whose standard deviation is \(\frac{3}{2}\) is equal to

Options:
A \(\left(\frac{1}{4}+\frac{3}{4}\right)^{12}\)
B \(\left(\frac{3}{4}+\frac{1}{4}\right)^{12}\)
C \(\left(\frac{1}{2}+\frac{3}{2}\right)^{12}\)
D \(\left(\frac{3}{2}+\frac{1}{2}\right)^{12}\)

The correct answer is:
\(\left(\frac{3}{4}+\frac{1}{4}\right)^{12}\)

Solution:
We know that mean \(=n p=9\) (given)
and standard deviation \(=\sqrt{n p q}=\frac{3}{2}\)
\(\therefore \quad q=\frac{1}{4}\) and \(p=\frac{3}{4}\) and \(n=12\)
So, required binomial distribution \((p+q)^n\)
\(=\left(\frac{3}{4}+\frac{1}{4}\right)^{12}\)

Topic: Mathematics, Binomial Theorem, AP EAMCET, AP EAMCET 2020 (18 Sep Shift 1)
The biogeographic region in India having maximum endemism for amphibians is

Options:
A The Western Ghats
B The Indo-Burma region
C The Himalayan range
D The Gangetic plains

The correct answer is:
The Western Ghats

Solution:
Ecologically endemism is considered as a distribution of a species in a particular region of an ecosystem. The Western Ghats are the hot spots of our country and preserves many species in them. The tropical rainforest and isolated hill ranges of the Western Ghats act more like an island preserving the 179 lineages of frogs (87%) that are found nowhere else in the world. Rare species are those whose population is originally small and scattered. These can any time enter into the categories of vulnerable and endangered species.

Topic: Biology, Biodiversity and its Conservation, NEET
The biomass available for consumption by the herbivores and the decomposers is called

Options:
A net primary productivity
B secondary productivity
C standing crop
D gross primary productivity

The correct answer is:
net primary productivity

Solution:
Net Primary Productivity (NPP) is the weight of organic matter stored by producers in a unit area/volume per unit time. NPP is equal to the rate of organic matter created by photosynthesis minus the rate of respiration and other loses. Stored biomass is transferred from one trophic level to another trophic level. Secondary Productivity is the rate of energy storage at consumers levels-herbivores, carnivorous and decomposers.
Gross Primary Productivity (GPP) is referred as total rate of photosynthesis including the organic matter used up in respiration during the measurement period.

Topic: Biology, Ecosystem, NEET, NEET 2010 (Screening)
The bisector of the acute angle between the straight lines $3 \mathrm{x}$ $-4 \mathrm{y}-3=0$ and $12 \mathrm{x}+5 \mathrm{y}+6=0$ passes through which one of the following points?

Options:
A $(5,3)$
B $(-3,6)$
C $(2,7)$
D $(-1,4)$

The correct answer is:
$(2,7)$

Solution:
The equations of given straight lines are
$3 x-4 y-3=0$...(1)
and $12 \mathrm{x}+5 \mathrm{y}+6=0$...(2)
Re-writing equations, so that constant term in both have same sign. We write second equation so that its constant term is negative. Then, equation of bisector of the acute angle between the given straight lines is
$$
\begin{aligned}
& \frac{3 x-4 y-3}{\sqrt{3^{2}+4^{2}}}=-\frac{(12 x+5 y+6)}{\sqrt{12^{2}+5^{2}}} \\
\Rightarrow & \frac{3 x-4 y-3}{5}=-\frac{-12 x-5 y-6}{13}
\end{aligned}
$$
$\Rightarrow 39 \mathrm{x}-52 \mathrm{y}-39=-60 \mathrm{x}-25 \mathrm{y}-30$
$\Rightarrow-60 x-25 y-30-39 x+52 y+39=0$
$\Rightarrow \quad-99 \mathrm{x}+27 \mathrm{y}+9=0$
$\Rightarrow-11 x+3 y+1=0$
Putting $\mathrm{x}=2$ and $\mathrm{y}=7$ This equation is satisfied by $(2,7)$. Thus, the bisector of acute angle between the given straight lines passes through $(2,7)$

Topic: Mathematics, Pair of Lines, NDA, NDA 2006 (Phase 1)
The blubber is formed by

Options:
A

B

C

D

The correct answer is:

Solution:
Adipose tissue is fibrous connective tissue packed with masses of fat cells. These form a thick layer under the skin and occurs around kidneys. The blubber is formed of these tissues. Blubber covers the entire body of animals such as seals, whales, and walruses.

Topic: Biology, Structural Organisation in Animals, NEET
The boiling point of water at $1 \mathrm{~atm}$ is $100^{\circ} \mathrm{C}$. Which among the following is true for vaporisation of water at $75^{\circ} \mathrm{C}$ ?

Options:
A $\Delta G_{\text {vap }}^{\circ}>0$
B $\Delta H_{\text {vap }}^{\circ} < 0$
C $K_{\text {vap }}=1$
D $\Delta S_{\text {vap }}^0 < 0$

The correct answer is:
$\Delta G_{\text {vap }}^{\circ}>0$

Solution:
According to standard Gibbs free energy equation
i.e.,
$$
\begin{aligned}
& \Delta G^{\circ}=\Delta H^{\circ}-T \Delta S^{\circ} \\
& \Delta G^{\circ}=-T \Delta S^{\circ} \\
& \Delta G^{\circ} \propto-T
\end{aligned}
$$
It means $\Delta G^{\circ}$ is inversely proportional to temperature due to negative sign present in Gibbs free energy equation.
Hence, lesser the temperature i.e., $75^{\circ} \mathrm{C}$, greater the value of $\Delta^{\circ} G_{\text {vap }}\left(\Delta G_{\text {vap }}^0>0\right)$.
At $100^{\circ} \mathrm{C}, \Delta G_{\text {vap }}^0$ value is lesser than that at temperature $75^{\circ} \mathrm{C}$.
$\therefore$ Vaporisation of water at $75^{\circ} \mathrm{C}$ has $\Delta G_{\text {vap }}^{\circ}>0$.

Topic: Chemistry, Thermodynamics (C), AP EAMCET, AP EAMCET 2021 (25 Aug Shift 1)
The bond dissociation energies for Cl2 , I2 and ICl are 242.3 , 151 and 211.3kJ/mol respectively. The enthalpy of sublimation of iodine is 62.8 kJ/mol . What is the standard enthalpy of formation of ICI (g) ?

Options:
A -211.3 kJ/mol
B -14.6kJ/mol
C -16.8 kJ/mol
D 33.5 kJ/mol

The correct answer is:
-16.8 kJ/mol

Solution:
Cl2(g)â2Cl(g) ; ÎH1=242.3kJ/mol
I2(g)â21(g) ; ÎH2=151kJ/mol
ICI(g)âI(g)+Cl(g);ÎH3=211.3kJ/mol
I2(s)âI2(g);ÎH4=62.8kJ/mol
required equation:
12I2(g)+12Cl2(g)âICl(g)î¼;ÎH=?
ÎH=62.8+151+242.32-211.3
=16.75 kJ/mol

Topic: Chemistry, Thermodynamics (C), NEET
The boundaries of which of the following kingdom is not well defined?

Options:
A Plantae
B Fungi
C Monera
D Protista

The correct answer is:
Protista

Solution:
Boundaries of kingdom Protista is not well defined as it includes all unicellular eukaryotes, which can be autotrophic in nature like euglena, heterotropic animal-like amoeba and saprophytic fungi like slime moulds/plasmodium.

Topic: Biology, Biological Classification, NEET
The breakdown in a reverse biased $\mathrm{p}-\mathrm{n}$ junction is more likely to occur due to

Options:
A large velocity of the minority charge carriers if the doping concentration is small

B
large velocity of the minority charge carries if the doping concentration is large

C
strong electric field in a depletion region if the doping concentration is small

D
strong electric field in the depletion region if the doping concentration is large

The correct answers are:
large velocity of the minority charge carriers if the doping concentration is small
,
strong electric field in the depletion region if the doping concentration is large

Solution:
In reverse biasing, minority charge carriers will be accelerated due to high electric field applied or so the minority charge carries in both junction accelerated toward depletion layer which on striking with atoms cause ionization resulting secondary electrons and So more number of charge carries more current in reverse directions. When doping concetration is large, there will be large number of ions in the depletion region, which will give rise to a strong electric field.

Topic: Physics, Semiconductors
The calculated spin-only magnetic moment values in \(\mathrm{BM}\) for \(\left[\mathrm{FeCl}_4\right]^{-}\)and \(\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-}\) are

Options:
A \(5.9 \mathrm{~BM},~ 1.732 \mathrm{~BM}~\)
B \(4.89 \mathrm{~BM},~ 1.732 \mathrm{~BM}~\)
C \(3.87 \mathrm{~BM},~ 1.732 \mathrm{~BM}~\)
D \(1.732 \mathrm{~BM},~ 2.82 \mathrm{~BM}~\)

The correct answer is:
\(5.9 \mathrm{~BM},~ 1.732 \mathrm{~BM}~\)

Solution:
Hint : \(\mathrm{Fe}^{+3}=[\mathrm{Ar}] 3 \mathrm{~d}^5 4 \mathrm{~s}^0\)
\(\left[\mathrm{FeCl}_4\right]^{-}\)
No pairing as \(\mathrm{Cl}^{-}\)is weak field ligand, hence have five unpaired electron \((n=5)\).
\(\begin{aligned} & \therefore \quad \mu=\sqrt{\mathrm{n}(\mathrm{n}+2)} \text { B.M }=\sqrt{5(5+2)} \text { B.M }=5.9 \text { B.M } \\ & {\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{-3}}\end{aligned}\)
Pairing takes place as \(\mathrm{CN}^{-}\)is strong field ligand but has one unpaired electron \((n=1)\)
\(\therefore \mu=\sqrt{n(n+2)} \text { B.M }=\sqrt{1(1+2)}=1.732 \text { B.M }\)

Topic: Chemistry, Structure of Atom, WBJEE, WBJEE 2023
The Calvin cycle proceeds in three stages:

1. Reduction, during which carbohydrate is formed at the expense of the photochemically made ATP and NADPH.
2. Regeneration, during which the carbon dioxide acceptor ribulose-1,5-biphosphate is formed.
3. Carboxylation, during which carbon dioxide combines with ribulose-1,5-biphosphate.

Identify the correct sequence.

Options:
A 3 â 1 â 2
B 3 → 2 → 1
C 1 → 2 → 3
D 2 → 1 → 3

The correct answer is:
3 â 1 â 2

Solution:
Calvin cycle (C3 cycle) occurs in three steps:
(i) Carboxylation: In the presence of Rubisco (RuBP carboxylase, most abundant protein), carbon dioxide combines with ribulose biphosphate (acceptor molecule) to form first a transitory 2-carboxy 3-keto 1,5-biphosphoribotol and then two molecules of 3-phosphoglyceric acid or PGA (first stable product).
(ii) Glycolytic reversal: With the help of ATP and triosephosphate kinase, 3-PGA is converted into 1,3-biphosphoglyceric acid. In the presence of enzyme triosephosphate dehydrogenase and NADPH, 1,3-BPGA is changed to glyceraldehyde-3-phosphate (3 phosphoglyceraldehydes). A part of glyceraldehyde-3-phosphate is changed into its isomer with the help of enzyme phophotriose isomerase. The isomer is dihydroxyacetone-3-phosphate. In the presence of enzyme aldolase, GAP and DiHAP combine giving rise to fructose-1,6-biphosphate, from which glucose and its polymers are produced.
(iii) Regeneration of RuBP: RuBP is regenerated through the formation of 4-C, 5-C and 7-C intermediates.

Topic: Biology, Photosynthesis in Higher Plants, NEET
The cartesian co-ordinates of the point on the parabola y 2 =â16x, whose parameter is 12 are

Options:
A -2,4
B 4,-1
C -1,-4
D -1,4

The correct answer is:
-1,4

Solution:
Given t= 1 2 ; y 2 =−16x
Comparing with standard parabola,
y 2 =−4ax
We get a=4
P( t )=( −a t 2 ,2at )=( −4× 1 4 ,2( 1 2 4 ) )
P( t )=( −1,4 )

Topic: Mathematics, Parabola, MHT CET, MHT CET 2018
The cell in which the following reaction occurs: \(2 \mathrm{Fe}^{3+}(a q)+2 \mathrm{I}^{-}(a q) \rightarrow 2 \mathrm{Fe}^{+}(a q)+\mathrm{I}_2(s)\) has \(\mathrm{E}_{\text {cell }}^0=\) \(0.236 \mathrm{~V}\) at \(298 \mathrm{~K}\). Calculate the standard Gibbs energy and the equilibrium constant of the cell reaction.

Options:

None

Solution:
\(2 \mathrm{Fe}^{3+}(a q)+2 \mathrm{I}^{-}(a q) \rightarrow 2 \mathrm{Fe}^{2+}(a q)+\mathrm{I}_2(s)\)
For the given cell, \(n=2\)
\(\Delta_r G^{\circ}=-n F E^{\circ}\) cell
\(=-2 \times 96487 \times 0.236\)
\(=-45.55 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
Also, \(\quad \Delta_r G^{\circ}=-2.303 R T \log K_C\)
\(\Rightarrow \quad \log K_{\mathrm{C}}=\frac{-\Delta_r G^{\circ}}{2.303 \mathrm{RT}}\)
\(=\frac{-45.55}{2.303 \times 8.314 \times 10^{-3} \times 298}\)
\(=7.983\)
\(\Rightarrow \quad K_C=\operatorname{antilog}(7.983)\)
\(=9.616 \times 10^7\)

Topic: Chemistry, Electrochemistry
The ________ cell is bigger, has abundant food reserve and a large irregularly shaped nucleus. The _________ is small and has a spindle-shaped nucleus with dense cytoplasm.

Options:
A a-vegetative cell; b-generative cell
B a-generative cell; b-vegetative cell
C a-tapetal cell; b-endothelial cell
D a-endothelial cell; b-tapetal cell

The correct answer is:
a-vegetative cell; b-generative cell

Solution:
The vegetative cell is bigger, has abundant food reserves and a large irregularly shaped nucleus. The generative cell is small and floats in the cytoplasm of the vegetative cell. It is spindle-shaped with dense cytoplasm and a nucleus.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
The cellulosic cell wall is observed in members of

Options:
A Protista
B Plantae
C Both a and b
D Monera

The correct answer is:
Both a and b

Solution:
A plant cell wall is arranged in layers and contains cellulose microfibrils, hemicellulose, pectin, lignin, and soluble protein. These components are organized into three major layers: the primary cell wall, the middle lamella, and the secondary cell wall. Beside higher plants some members of protists are also having cellulosic cell wall.

Topic: Biology, Biological Classification, NEET
The centroid of the triangle with vertices $(2,3),(-2,-5)$ and $(3,5)$ is at

Options:
A $(1,1)$
B $(2,-1)$
C $(1,-1)$
D $(1,2)$

The correct answer is:
$(1,1)$

Solution:
$\quad$ Centroid $=\left(\frac{2-2+3}{3}, \frac{3-5+5}{3}\right)$
$=(1,1)$

Topic: Mathematics, Straight Lines, NDA, NDA 2013 (Phase 2)
The centroid of the triangle with vertices $\mathrm{A}(2,-3,3), \mathrm{B}(5,-$ $3,-4$ ) and $\mathrm{C}(2,-3,-2)$ is the point

Options:
A $(-3,3,-1)$
B $(3,-3,-1)$
C $(3,1,-3)$
D $(-3,-1,-3)$

The correct answer is:
$(3,-3,-1)$

Solution:
Given vertices of triangle are $\mathrm{A}(2,-3,3), \mathrm{B}(5,-3,-4)$ and $\mathrm{C}(2,-3,-2)$
Centroid $=\left(\frac{x_{1}+x_{2}+x_{3}}{3}, \frac{y_{1}+y_{2}+y_{3}}{3}, \frac{z_{1}+z_{2}+z_{3}}{3}\right)$
$=\left(\frac{2+5+2}{3}, \frac{-3-3-3}{3}, \frac{3-4-2}{3}\right)$
$=(3,-3,-1)$

Topic: Mathematics, Straight Lines, NDA, NDA 2019 (Phase 1)
The chance of winning the race of the horse $A$ is $1 / 5$ and that of horse $B$ is $1 / 6$. What is the probability that the race will be won by $A$ or $B$ ?

Options:
A $1 / 30$
B $1 / 3$
C $11 / 30$
D $1 / 15$

The correct answer is:
$11 / 30$

Solution:
Let $\mathrm{P}(\mathrm{A})$ be the probability that the race will be won by A and $\mathrm{P}(\mathrm{B})$ be the probability that the race will be won by B.
$\therefore \quad \mathrm{P}(\mathrm{A})=\frac{1}{5} \quad$ and $\quad \mathrm{P}(\mathrm{B})=\frac{1}{6}$
$\therefore$ Probability that the race will be won by
A or $\mathrm{B}=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})=\frac{1}{5}+\frac{1}{6}=\frac{11}{30}$

Topic: Mathematics, Probability, NDA, NDA 2008 (Phase 2)
The changes that occurs in female at the onset of puberty are:

Options:
A The enlargement of breasts
B Beginning of menstrual cycle
C Stoppage of growth of long bone and height
D All the above

The correct answer is:
All the above

Solution:
The enlargement of breasts, the beginning of the menstrual cycle and the stoppage of growth of long bone and height are the changes that occurs in female at the onset of puberty and these are called as secondary sexual characters.

Topic: Biology, Human Reproduction, NEET
The characteristic feature of anamnestic response is

Options:
A a highly intensified strong response which leads to a high titre of antibodies.
B IgM is the dominant antibody in the body.
C a low intensity of antibodies which takes a longer time to establish immunity.
D Both A and C

The correct answer is:
a highly intensified strong response which leads to a high titre of antibodies.

Solution:
Anamnestic response is a highly intensified and strong response and is characterized by a high titre of antibodies. It is also known as the secondary immune response. When an individual's immune system comes in contact with the antigen for the first time, the immune system gets sensitized to this antigen, but the response is not much intensified. This is known as primary immune response. But, when the immune system comes in contact with the antigen subsequently, the immune response is highly intensified and massive, as the immune system was previously sensitized with this antigen. This is the anamnestic response.

Topic: Biology, Human Health and Diseases, NEET
The choice of a reducing agent in a particular case depends on thermodynamic factor. How far do you agree with this statement? Support your opinion with two examples.

Options:

None

Solution:
We can study the choice of a reducing agent in a particular case using Ellingham diagram.
It is evident from the diagram that metals for which the standard free energy of formation of their oxides is more negative can reduce those metal oxides for which the standard free energy of formation of their respective oxides is less negative. It means that any metal will reduce the oxides of other metals which lie above it in the Ellingham diagram. This is because the standard free energy change \(\left(\Delta_{\mathrm{r}} \mathrm{G}^{\circ}\right)\) of the combined redox reaction will be negative by an amount equal to the difference in \(\Delta_{\mathrm{f}} \mathrm{G}^{\circ}\) of the two metal oxides. Thus both \(\mathrm{Al}\) and \(\mathrm{Zn}\) can reduce \(\mathrm{FeO}\) to \(\mathrm{Fe}\) but \(\mathrm{Fe}\) cannot reduce \(\mathrm{Al}_2 \mathrm{O}_3\) to \(\mathrm{Al}\) and \(\mathrm{ZnO}\) to \(\mathrm{Zn}\). In the same way, \(\mathrm{C}\) can reduce \(\mathrm{ZnO}\) to \(\mathrm{Zn}\) but not \(\mathrm{CO}\).
Note: Only that reagent will be preferred as reducing agent which will lead to decrease in free energy value \(\left(\Delta \mathrm{G}^{\circ}\right)\) at a certain specific temperature.

Topic: Chemistry, General Principles and Processes of Isolation of Metals
The circle $x^{2}+y^{2}+4 x-4 y+4=0$ touches

Options:
A Only the $x$ -axis
B Only the $y$ -axis
C Both the axes
D Neither of the axes

The correct answer is:
Both the axes

Solution:
Given equation of circle is $x^{2}+4 x+4+y^{2}-4 y=0$
Add 4 on both side, $x^{2}+4 x+4+y^{2}-4 y+4=0+4$
$\Rightarrow(x+2)^{2}+(y-2)^{2}=2^{2}$
Here, we observe that the values of centre and radius
are same.
Hence, it touches both the axes.

Topic: Mathematics, Circle, NDA, NDA 2009 (Phase 2)
The circle $x^{2}+y^{2}+4 x-7 y+12=0$, cuts an intercept on yaxis equal to

Options:
A 1
B 3
C 4
D 7

The correct answer is:
1

Solution:
Given circle, $x^{2}+y^{2}+4 x-7 y+12=0$ Comparing with general form of circle, $a x^{2}+b y^{2}+2 g x+2 f y+c=0$
$f=\frac{-7}{2}$ and $\mathrm{c}=12 .$
$\mathrm{y}$ - intercept $=2 \sqrt{f^{2}-c}$
$=2 \sqrt{\left(\frac{-7}{2}\right)^{2}-12}=2 \sqrt{\frac{49}{4}-12}$
$=2 \sqrt{\frac{49-48}{4}}=2\left(\frac{1}{2}\right)=1$

Topic: Mathematics, Circle, NDA, NDA 2019 (Phase 1)
The coefficient of correlation when coefficients of $\begin{array}{ll}\text { regression are } 0.2 \text { and } 1.8 \text { is } & {}\end{array}$

Options:
A $0.36$
B $0.2$
C $0.6$
D $0.9$

The correct answer is:
$0.6$

Solution:
Coefficient of correlation $=\sqrt{0.2 \times 1.8}$ $=\sqrt{0.36}$
$0.6$

Topic: Mathematics, Statistics, NDA, NDA 2017 (Phase 2)
The coefficient of mutual inductance between the primary and the secondary coil of a transformer is \(0.2 \mathrm{H}\). When the current in the primary changes by \(5 \mathrm{As}^{-1}\), then the induced emf in the secondary will be

Options:
A \(5 \mathrm{~V}\)
B \(1 \mathrm{~V}\)
C \(25 \mathrm{~V}\)
D \(10 \mathrm{~V}\)

The correct answer is:
\(1 \mathrm{~V}\)

Solution:
Given, coefficient of mutual induction,
\(M=0.2 \mathrm{H}\)
The rate of change in current in primary coil.
\(\frac{d I}{d t}=5 \mathrm{As}^{-1}\)
\(\therefore\) Induced emf in the secondary coil,
\(e=M \frac{d I}{d t}=0.2 \times 5=1 \mathrm{~V}\)

Topic: Physics, Electromagnetic Induction, AP EAMCET, AP EAMCET 2020 (18 Sep Shift 1)
The coefficient of variation and standard deviation of an ungrouped data are 60 and 21 respectively. If 15 is added to every observation of the data, then the coefficient of variation of the new data is

Options:
A 30
B 42
C 40
D 20

The correct answer is:
42

Solution:
According to the given information,
Coefficient of variance $(\mathrm{CV})=\frac{\sigma}{\mu} \times 100$
Where $\sigma$ is standard deviation and $\mu$ is mean of an ungrouped data.
$\because \quad \frac{\sigma}{\mu} \times 100=60$ and $\sigma=21$
So, $\quad \mu=\frac{21 \times 10}{6}=35$
After adding 15 to each observation of the data, the new mean $\mu^{\prime}=35+15=50$ but
$\Sigma\left|\mu^{\prime}-x_i^{\prime}\right|=\Sigma\left|\mu-x_i\right|$, [where $x_i^{\prime}=x_i+15$ ]
Now, $\sigma=\sqrt{\frac{\sum\left(\mu-x_i\right)^2}{n}}=21$
$\Rightarrow \Sigma\left(\mu-x_i\right)^2=(21)^2 \times n$
So, new standard deviation
$$
\sigma^{\prime}=\sqrt{\frac{\sum\left(\mu^{\prime}-x_i^{\prime}\right)^2}{n}}=\sqrt{\frac{(21)^2 \times n}{n}}=21
$$
$\therefore$ New coefficient of variance
$$
=\frac{\sigma^{\prime}}{\mu^{\prime}} \times 100=\frac{21}{50} \times 100=21 \times 2=42
$$
Hence, option (b) is correct.

Topic: Mathematics, Statistics, AP EAMCET, AP EAMCET 2019 (20 Apr Shift 2)
The coefficient of variation for the following data is

Options:
A $9 \frac{\sqrt{23}}{2}$
B $20 \frac{\sqrt{23}}{3}$
C $15 \frac{\sqrt{23}}{4}$
D $7 \frac{\sqrt{23}}{6}$

The correct answer is:
$20 \frac{\sqrt{23}}{3}$

Solution:
Option (b), $\frac{20 \sqrt{23}}{3}$ is correct.
$\begin{aligned} & \text { Mean }(\bar{x})=\frac{\Sigma f_i x_i}{\Sigma f_i}=\frac{60}{8}=7.5 \\ & \text { Standard deviation }(\sigma)=\sqrt{\frac{\Sigma f_i x_i^2}{\Sigma f_i}-\left(\frac{\Sigma f_i x_i}{\Sigma f_i}\right)^2} \\ & =\sqrt{\frac{496}{8}-(7.5)^2}=\sqrt{6225-56.25} \\ & =\sqrt{6}=2.44\end{aligned}$
$\begin{aligned} & \therefore \text { Coefficient of variation }=\frac{\sigma}{\bar{x}} \times 100 \\ & =\frac{244}{7.5} \times 100 \\ & =3253\end{aligned}$

Topic: Mathematics, Statistics, TS EAMCET, TS EAMCET 2019 (06 May Shift 1)
The coefficient of variation of two distributions are \( 60 \) and \( 70 \). The standard deviation are \( 21 \)
and \( 16 \) respectively, then their mean is

Options:
A 35
B \( 23 \)
C \( 28.25 \)
D \( 22.85 \)

The correct answer is:
\( 28.25 \)

Solution:
Given that, $C V_{x}=60 \rightarrow(1)$
We know that, $C V=\frac{\sigma}{\chi} \times 100$
So, $\frac{21}{x} \times 100=60$
$\Rightarrow \bar{\chi}=\frac{2100}{60}=35$
Again, $C V_{y}=70$
$\frac{16}{y} \times 100=70$ $\Rightarrow \bar{y}=\frac{1600}{70}=22.86$
So, mean of two means are
$\frac{\bar{x}+\bar{y}}{2}$ $\frac{35+22.86}{2}=28.925$

Topic: Mathematics, Statistics, KCET, KCET 2016
The coefficient of $\mathrm{x}^{99}$ in the expansion of $(\mathrm{x}-1)(\mathrm{x}-2)(\mathrm{x}-3) \ldots \ldots$ $(\mathrm{x}-100)$ is

Options:
A 5050
B 5000
C $-5050$
D $-5000$

The correct answer is:
$-5050$

Solution:
Coefficient of $x^{1}$ in $\left[(x-1)(x-2)\right.$ or $\left.\left(x^{2}-3 x+2\right)\right]$ $=-3=-1-2=-(1+2)$
Coefficient of $x^{2}$ in $[(x-1)(x-2)(x-3)$ or $\begin{aligned} & \left.\left(\mathrm{x}^{3}-6 \mathrm{x}^{2}+5 \mathrm{x}-6\right)\right] \\=&-6=-[1+2+3] . \end{aligned}$
Coefficient of $x^{3}$ in $[(x-1)(x-2)(x-3)(x-4)$ or $\left(x^{4}-10 x^{3}-29 x^{2}-11 x+24\right)$
$=-10=-[1+2+3+4]$
$\therefore$ Coefficient of $x^{99}$ in $[(x-1)(x-2) \ldots \ldots . .(x-100)]$
$=-[1+2+3+\ldots . .+100]=\frac{-100(100+1)}{2}=-5050$

Topic: Mathematics, Binomial Theorem, NDA, NDA 2015 (Phase 2)
The cofactor of the element 4 in the determinant
$\left|\begin{array}{lll}1 & 2 & 3 \\ 4 & 5 & 6 \\ 5 & 8 & 9\end{array}\right|$ is

Options:
A 2
B 4
C 6
D $-6$

The correct answer is:
6

Solution:
Co-factor of $4=(-1)^{3}(2 \times 9-3 \times 8)=-(-6)=6$

Topic: Mathematics, Matrices, NDA, NDA 2013 (Phase 2)
The coils of a step-down transformer have 500 and 5000 turns. In the primary coil an \(\mathrm{AC}\) current of \(4 \mathrm{~A}\) at \(2200 \mathrm{~V}\) is sent. The value of the current and potential difference in the secondary coil is

Options:
A \(20 \mathrm{~A}, 220 \mathrm{~V}\)
B \(0.4 \mathrm{~A}, 22000 \mathrm{~V}\)
C \(40 \mathrm{~A}, 220 \mathrm{~V}\)
D \(40 \mathrm{~A}, 22000 \mathrm{~V}\)

The correct answer is:
\(40 \mathrm{~A}, 220 \mathrm{~V}\)

Solution:
For a step-down transformer,
\(\begin{aligned}
N_P & =5000, N_S=500 \\
I_P & =4 \mathrm{~A}, V_P=2200 \mathrm{~V}
\end{aligned}\)
For a transformer, we know that
\(\begin{aligned}
& \frac{N_S}{N_P}=\frac{V_S}{V_P}=\frac{I_P}{I_S} \\
& \text {Using, } \quad \frac{N_S}{N_P}=\frac{V_S}{V_P} \\
& \Rightarrow \quad V_S=\frac{N_S V_P}{N_P}=\frac{500 \times 2200}{5000}=220 \mathrm{~V} \\
& \text {Again, using } \frac{V_S}{V_P}=\frac{I_P}{I_S} \\
& I_S=\frac{I_P \times V_P}{V_S}=\frac{4 \times 2200}{220}=40 \mathrm{~A} \\
\end{aligned}\)

Topic: Physics, Alternating Current, AP EAMCET, AP EAMCET 2020 (21 Sep Shift 2)