datavorous/entrance-exam-dataset huggingface dataset:
The most abundant metal in the earth crust is

Options:
A Na
B Â Mg
C Al
D Â Fe

The correct answer is:
Al

Solution:
Aluminium is the most abundant metal in the Earth’s crust, and the third most abundant element there is after oxygen and silicon. It makes up about 8% by weight of the Earth’s solid surface. Earth’s crust occupies less than 1% of Earth’s volume. Oxygen is 46.6%, Silicon 27.77%, Aluminum is 8.09% and Iron is 5%.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, NEET
The most abundant noble gas in the atmosphere is

Options:
A Neon
B Argon
C Xenon
D Krypton

The correct answer is:
Argon

Solution:
Argon is the most abundant noble gas present in the atmosphere.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), MHT CET, MHT CET 2016
The most abundant prokaryotes helpful to human in making curd from milk and in production of antibiotics are the ones categorized as

Options:
A cyanobacteria
B archaebacteria
C chemosynthetic autotrophs
D heterotrophic bacteria

The correct answer is:
heterotrophic bacteria

Solution:
Heterotrophic bacteria are most abundant in nature. Many of them have a significant impact on human affairs. These are helpful in making curd from milk (e.g., Lactobacillus sp.), production of antibiotics (e.g., Streptomyces sp.) and fixing nitrogen in legume roots (e.g., Rhizobium spe).

Topic: Biology, Microbes in Human Welfare, NEET, NEET 2012 (Screening)
The most abundant prokaryotes helpful to humans in making curd from milk and in production of antibiotics are the ones categorised as

Options:
A Cyanobacteria
B Archaebacteria
C Chemosynthetic autotrophs
D Heterotrophic bacteria

The correct answer is:
Heterotrophic bacteria

Solution:
A maximum number of antibiotics are produced by mycelial bacteria known as actinomycetes, and most of the actinomycetes are saprotrophic (heterotrophic). Lactic acid bacteria that are used in the preparation of curd are also heterotrophic.

Topic: Biology, Microbes in Human Welfare, NEET
The most abundant protein in the animal world is:

Options:
A Trypsin
B Haemoglobin
C Collagen
D Insulin

The correct answer is:
Collagen

Solution:
Collagen is an insoluble fibrous protein found extensively in the connective tissue of skin, tendons and bone. Collagen accounts for over 30% of the total body proteins of mammals and it is the most abundant animal protein.

Topic: Chemistry, Biomolecules, NEET
The most abundant protein of the biosphere is found in:

Options:
A Chloroplasts
B Mitochondria
C Ribosomes
D Dictyosomes

The correct answer is:
Chloroplasts

Solution:
Collagen is the most abundant protein in the animal world, and Ribulose bisphosphate Carboxylase-Oxygenase (RuBisCO) is the most abundant protein in the whole of the biosphere.
Carboxylation is the fixation of CO2 into a stable organic intermediate. Carboxylation is the most crucial step of the Calvin cycle where CO2 is utilised for the carboxylation of RuBP. This reaction is catalysed by the enzyme RuBP carboxylase which results in the formation of two molecules of 3-PGA. Since this enzyme also has an oxygenation activity, it would be correct to call it RuBP carboxylase-oxygenase or RuBisCO.

Topic: Chemistry, Biomolecules, NEET
The most abundant transition metal in human body is

Options:
A copper
B iron
C zinc
D manganese

The correct answer is:
iron

Solution:
Most abundant element (transition) in human body is Iron.

Topic: Chemistry, d and f Block Elements, KVPY, KVPY 2015 (SB/SX)
The most active phagocytic white blood cells are

Options:
A neutrophils and eosinophils
B lymphocytes and macrophages
C eosinophils and lymphocytes
D neutrophils and monocytes

The correct answer is:
neutrophils and monocytes

Solution:
Neutrophils are most abundant granulecytes which engulf microbes by phagocytosis. These have 2-7 lobulated nucleus and do not stain any basic or acidic dye monocytes are large sized leucocytes, which are highly motile and phagocytic in nature.

Topic: Biology, Body Fluids and Circulation, JIPMER, JIPMER 2012
The most reactive molecules towards SN1 reaction is

Options:
A (I), (IV) and (VI)
B (I), (II) and (IV)
C (II), (III) and (V)
D (IV), (V) and (VI)

The correct answer is:
(I), (IV) and (VI)

Solution:
According to the question, The rate of the SN1 (nucleophilic substitution unimolecular reaction) reaction depends only on the concentration of alkyl halide. Relative reactivity of different alkyl halides in SN1 Reaction is as follows: Tertiary alkyl halide> secondary alkyl halide > Primary alkyl halide.

Product formation takes place by the formation of carbocation as reaction intermediate and their stability. Stability of various alkyl carbocation are-

Tertiary alkyl carbocation >secondary alkyl carbocation > Primary alkyl carbocation.

In option (a) carbocation is formed, due to hetrolytic bond fission of C-Cl bond is present in the conjugation with double bond of two benzene rings, here methoxy group show +R (maximum) and -I (minimum) effect. Due to this stability of carbocation is increases. So it is capable to show SN1 Reaction.

Option (d) is tertiary butyl chloride (tertiary alkyl halide, which show SN1 Reaction in maximum extant.

In option (f), due to hetrolytic bond fission of C-Cl bond. Tertiary carbocation is formed, which is stabilised by resonance effect of phenyl ring and +I effect of methyl group. So it also capable to showSN1Reaction.

Topic: Chemistry, Haloalkanes and Haloarenes, TS EAMCET, TS EAMCET 2021 (05 Aug Shift 2)
The most suitable reagent for the conversion of $R \mathrm{CH}_2 \mathrm{OH} \rightarrow R \mathrm{CHO}$ is

Options:
A $\mathrm{KMnO}_4$
B $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$
C $\mathrm{CrO}_3$
D PCC (pyridine chloro chromate).

The correct answer is:
PCC (pyridine chloro chromate).

Solution:
Pyridine chloro chromate (PCC) $\left[\mathrm{C}_5 \mathrm{H}_5 \mathrm{NH}^{+} \mathrm{ClCrO}_3^{-}\right]$selectively oxidizes primary alcohols to aldehydes. Secondary alcohols can be oxidized to ketones with PCC.

Topic: Chemistry, Aldehydes and Ketones, AIIMS, AIIMS 2004
The motion of a particle is described as $\mathrm{s}=2-3 \mathrm{t}+4 \mathrm{t}^{3}$. What is the acceleration of the particle at the point where its velocity is zero?

Options:
A 0
B 4 unit
C 8 unit
D 12 unit

The correct answer is:
12 unit

Solution:
Given rule is:
Distance, $\mathrm{s}=2-3 \mathrm{t}+4 \mathrm{t}^{3}$
$\Rightarrow$ Velocity $=\frac{\mathrm{ds}}{\mathrm{dt}}=-3+12 \mathrm{t}^{2}$
$\Rightarrow$ Acceleration $=\frac{\mathrm{d}^{2} \mathrm{~s}}{\mathrm{dt}^{2}}=24 \mathrm{t}$
Since, velocity is zero
$\therefore \quad \frac{\mathrm{ds}}{\mathrm{dt}}=0$
$\Rightarrow 0=-3+12 \mathrm{t}^{2} \Rightarrow \mathrm{t}=\sqrt{\frac{3}{12}}=\frac{1}{2}$
Acceleration (when velocity is zero)
$\Rightarrow \quad \frac{\mathrm{d}^{2} \mathrm{~s}}{\mathrm{dt}^{2}}=24 \mathrm{t}=24 \times \frac{1}{2}=12 \mathrm{unit}$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2007 (Phase 2)
The multiplication of the number $(10101)_{2}$ by $(1101)_{2}$ yields which one of the following ?

Options:
A $(100011001)_{2}$
B $(100010001)_{2}$
C $(110010011)_{2}$
D $(100111001)_{2}$

The correct answer is:
$(100010001)_{2}$

Solution:
$\begin{aligned} \because(10101)_{2} &=2^{4} \times 1+0 \times 2^{3}+1 \times 2^{2}+0 \times 2+1 \times 2^{0} \\ &=16+4+1=21 \end{aligned}$
and $\begin{aligned}(1101)_{2} &=1 \times 2^{3}+1 \times 2^{2}+0 \times 2^{1}+1 \times 2^{0} \\ &=8+4+1=13 \end{aligned}$
$\therefore(10101)_{2} \times(1101)_{2}=21 \times 13$
$=273=256+16+1=2^{8}+2^{4}+2^{0}$
So, there will be 1 at $9^{\text {th }}, 5^{\text {th }}$ and first place from right and zero at other places So, $(273)_{10}=(100010001)_{2}$

Topic: Mathematics, Sets and Relations, NDA, NDA 2007 (Phase 2)
The municipal waste material which is carried away in drains and sewers is called ___.

Options:
A Bacteria
B Secondary sewage treatment
C Sewage
D Primary sewage treatment

The correct answer is:
Sewage

Solution:
The municipal waste material which is carried away in drains and sewers is called sewage.

Topic: Biology, Microbes in Human Welfare, NEET
The $n$ th term of an AP. is $\frac{3+\mathrm{n}}{4}$, then the sum of first 105 terms is

Options:
A 270
B 735
C 1409
D 1470

The correct answer is:
1470

Solution:
$\mathrm{T}_{\mathrm{n}}=\frac{3+\mathrm{n}}{4}$
$\mathrm{~S}_{\mathrm{n}}=\sum_{\mathrm{n}=0}^{\infty} \mathrm{T}_{\mathrm{n}}=\sum\left(\frac{3}{4}+\frac{\mathrm{n}}{4}\right)$
$=\frac{3}{4} \mathrm{n}+\frac{1}{4} \times \frac{\mathrm{n}(\mathrm{n}+1)}{2}=\frac{7}{8} \mathrm{n}+\frac{\mathrm{n}^{2}}{8}$
$\mathrm{~S}_{105}=\frac{7}{8} \times 105+\frac{(105)^{2}}{8}=1470$

Topic: Mathematics, Sequences and Series, NDA, NDA 2015 (Phase 2)
The natural relative abundance of isotopes of hydrogen is

Options:
A ${ }_1^1 \mathrm{H}=99.985 \% ;{ }_1^2 \mathrm{D}=0.015 \%$
B ${ }_1^1 \mathrm{H}=99.985 \% ;{ }_1^2 \mathrm{D}=0.015 \% ;{ }_1^3 \mathrm{~T}=10^{-16} \%$
C ${ }_1^1 \mathrm{H}=99.100 \% ;{ }_1^2 \mathrm{D}=0.900 \%$
D ${ }_1^1 \mathrm{H}=99.900 \% ;{ }_1^2 \mathrm{D}=0.10 \% ;{ }_1^3 \mathrm{~T}=10^{-15} \%$

The correct answer is:
${ }_1^1 \mathrm{H}=99.985 \% ;{ }_1^2 \mathrm{D}=0.015 \% ;{ }_1^3 \mathrm{~T}=10^{-16} \%$

Solution:
Isotope of hydrogen is hydrogen -2 (also called) deuterium abbreviated as D.
Which contains one proton and one neutron and has a natural abundance of only $0.015 \%$.
$$
{ }_1^1 \mathrm{H}=99.985 \% ;{ }_1^2 \mathrm{D}=0.015 \% ;{ }_1^3 \mathrm{~T}=10^{-16} \%
$$

Topic: Chemistry, Hydrogen, TS EAMCET, TS EAMCET 2018 (04 May Shift 2)
The nitrogen-fixing bacteria Rhizobium is ________________ in nature.

Options:
A chemoautotrophic
B photoautotrophic
C chemosynthetic
D heterotrophic

The correct answer is:
heterotrophic

Solution:
Heterotrophic bacteria are the most abundant in nature. The majority are important decomposers. Many of them have a significant impact on human affairs. They are helpful in making curd from milk, production of antibiotics, fixing nitrogen in legume roots like Rhizobium, etc.

Topic: Biology, Biological Classification, NEET
The number $(2+\sqrt{2})^{2}$ is

Options:
A a natural number
B an irrational number
C a rational number
D a whole number

The correct answer is:
an irrational number

Solution:
$(2+\sqrt{2})^{2}=4+2+4 \sqrt{2}=6+4 \sqrt{2}$
So, it is an irrational number

Topic: Mathematics, Sets and Relations, NDA, NDA 2008 (Phase 2)
The number $0.0011$ in binary system represents

Options:
A rational number $3 / 8$ in decimal system
B rational number $1 / 8$ in decimal system
C rational number $3 / 16$ in decimal system
D rational number $5 / 16$ in decimal system

The correct answer is:
rational number $3 / 16$ in decimal system

Solution:
$(0.0011)=0 \times \frac{1}{2}+0 \times \frac{1}{2^{2}}+1 \times \frac{1}{2^{3}}+1 \times \frac{1}{2^{4}}$
$=0+0+\frac{1}{8}+\frac{1}{16}=\frac{3}{16}$
Hence, option (c) is correct.

Topic: Mathematics, Sets and Relations, NDA, NDA 2009 (Phase 2)
The number 25 l in decimal system is expressed in binary system by

Options:
A 11110111
B 1111101
C 11111101
D 11111110

The correct answer is:
1111101

Solution:
\begin{array}{c|c|c}
2 & 251 & 1 \\
\hline 2 & 125 & 1 \\
\hline 2 & 62 & 0 \\
\hline 2 & 31 & 1 \\
\hline 2 & 15 & 1 \\
\hline 2 & 7 & 1 \\
\hline 2 & 3 & 1 \\
\hline & 1 &
\end{array}
Therefore, $(251)_{10}=(11111011)_{2}$
Sol. (197-199)

Topic: Mathematics, Sets and Relations, NDA, NDA 2014 (Phase 1)
The number 83 is written in the binary system as

Options:
A 100110
B 101101
C 1010011
D 110110

The correct answer is:
1010011

Solution:
\begin{array}{c|c|c}
2 & 83 & 1 \\
\hline 2 & 41 & 1 \\
\hline 2 & 20 & 0 \\
\hline 2 & 10 & 0 \\
\hline 2 & 5 & 1 \\
\hline 2 & 2 & 0 \\
\hline & 1 &
\end{array}
Therefore, $(83)_{10}=(1010011)_{2}$

Topic: Mathematics, Sets and Relations, NDA, NDA 2013 (Phase 2)
The number 10101111 in binary system is represented in decimal system by which one of the following numbers?

Options:
A 157
B 175
C 57
D 75

The correct answer is:
175

Solution:
The number 10101111 can be rewritten as $10101111=2^{7} \times 1+2^{6} \times 0+2^{5} \times 1+2^{4} \times 0+2^{3} \times 1$
$+2^{2} \times 1+2^{1} \times 1+2^{0} \times 1$
$=128+32+8+4+2+1=175$

Topic: Mathematics, Sets and Relations, NDA, NDA 2011 (Phase 1)
The number of 3 -digit even numbers that can be formed from the digits $0,1,2,3,4$ and 5 , repetition of digits being not allowed, is

Options:
A 60
B 56
C 52
D 48

The correct answer is:
52

Solution:
No. of digits to be filled at one's place $=3$ No. of digits to be filled at 10 's place $=5$ No. of digits to be filled at 100 's place $=4$ $\therefore$ Total no. of digits formed $=3 \times 5 \times 4=60$ If zero is at 100 's place; Then; no. of digits to be filled at one's place $=2$ \& no. of digits to be filled at 10 's place $=4$ $\therefore$ No. of digits formed with zero at 100 's place $=1 \times 2 \times 4=8$
$\therefore$ Required no. of digits formed $=60-8=52$.

Topic: Mathematics, Permutation Combination, NDA, NDA 2015 (Phase 2)
The number of air molecules in a $(5 \mathrm{~m} \times 5 \mathrm{~m} \times 4 \mathrm{~m})$ room at standard temperature and pressure is of the order of

Options:
A $6 \times 10^{23}$
B $3 \times 10^{24}$
C $3 \times 10^{27}$
D $6 \times 10^{30}$

The correct answer is:
$3 \times 10^{27}$

Solution:
$\mathrm{PV}=\mathrm{N} \times \mathrm{K} \times \mathrm{T}$
where $\mathrm{K}$ is Boltzmann constant
$\begin{array}{l}
10^{5} \times 100=\mathrm{N} \times 1.38 \times 10^{-23} \times 273 \\
\mathrm{~N} \approx 3 \times 10^{27}
\end{array}$

Topic: Physics, Kinetic Theory of Gases, KVPY, KVPY 2011 (SA)
The number of chromosomes in a primary spermatocyte is

Options:
A same as in spermatid
B same as in spermatogonium
C help of that in spermatogonium
D same as in secondary spermatocyte

The correct answer is:
same as in spermatogonium

Solution:
Both spermatogonium and primary spermatocyte in human testes are diploid cells (2n).

Topic: Biology, Human Reproduction, JIPMER, JIPMER 2015
The number of common characters among organisms

Options:
A Decreases (Angiospermae to Triticum)
B Increases (Monocotyledone to Plantae)
C Increases (Sapindales to Angiospermae)
D Decreases (Poaceae to Plantae)

The correct answer is:
Decreases (Poaceae to Plantae)

Solution:
Maximum similarity occurs in species and this decreases with the increasing rank in the hierarchy. Plantae is the kingdom and is the highest category in the taxonomy that includes all the organisms which share a set of distinguishing characters. Poaceae is the category family and is of higher rank than that of the genus. It consists of one to many related genera, which are more similar to one another than to genera of other families.

Topic: Biology, The Living World, NEET
The number of male gametes and the number of meiotic divisions required for the formation of 100 seeds in a typical angiosperm is

Options:
A 100 male gametes and 125 meiotic divisions
B 200 male gametes and 125 meiotic divisions
C 100 male gametes and 25 meiotic divisions
D 200 male gametes and 25 meiotic divisions

The correct answer is:
200 male gametes and 125 meiotic divisions

Solution:
Angiosperms undergo double fertilization in which one male gamete fuses with the egg, whereas the second male gamete fuses with the secondary nucleus. Thus, for the formation of one seed, two male gametes are required. One egg is produced by one meiotic division. Four pollen grains are produced by one meiotic division. Thus, for 100 seeds, 100+25 = 125 meiosis are required with 200 male gametes [x+(x/4), 100+(100/4)=100+25=125].

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
The number of moles of oxygen in one litre of air containing 21% oxygen by volume in standard condition is

Options:
A 0.186 mole
B 2.10 mole
C 0.210 mole
D 0.0093 mole

The correct answer is:
0.0093 mole

Solution:
Air containing 21% oxygen in one litre =21100=0.21 litre
1 mole occupies a volume = 22.4 litre
x mole occupies a volume = 0.21 litre
So x=V22.4=0.2122.4=0.0093 mole of O2 .

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET
The number of non-zero integral solutions of the equation $|1-2 i|^{x}=5^{x}$ is

Options:
A Zero (No solution)
B One
C Two
D Three

The correct answer is:
Zero (No solution)

Solution:
$|1-2 i|^{x}=5^{x}$
$\Rightarrow\left(\left((1)^{2}+(-2)^{2}\right)^{\frac{1}{2}}\right)^{\mathrm{x}}=5^{\mathrm{x}} \quad \because|\mathrm{x}+\mathrm{iy}|=\sqrt{\mathrm{x}^{2}+\mathrm{y}^{2}}$
$\Rightarrow(5)^{\frac{x}{2}}=5^{x}$
$\Rightarrow \frac{x}{2}=x \Rightarrow x=2 x$
$\Rightarrow x=0$
There is no non zero integral solution.

Topic: Mathematics, Complex Number, NDA, NDA 2018 (Phase 1)
The number of $\mathrm{p} \pi-\mathrm{d} \pi$ 'pi' bonds present in $\mathrm{XeO}_{3}$ and $\mathrm{XeO}_{4}$ molecules, respectively are

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

The correct answer is:
3,4

Solution:

Topic: Chemistry, Chemical Bonding and Molecular Structure, VITEEE, VITEEE 2009
The number of permutations that can be formed from all the letters of the word 'BASEBALL' is:

Options:
A 540
B 1260
C 3780
D 5040

The correct answer is:
5040

Solution:
There are total 8 letters in the word BASEBALL, in which we have 2 B's, 2 A's and 2 L's.
$\therefore \quad$ Required No. of permutations $=\frac{8 !}{2 ! \times 2 ! \times 2 !}$
$=\frac{8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{8}=5040$

Topic: Mathematics, Permutation Combination, NDA, NDA 2012 (Phase 2)
The number of possible natural oscillations of air column in a pipe closed at one end of length 85 cm whose frequencies lies below 1250 Hz are: (velocity of sound =340 ms-1)

Options:
A 4
B 5
C 7
D 6

The correct answer is:
6

Solution:
Fundamental frequency of a closed organ pipe is f1=V4l=3404×0.85=100 Hz
The natural frequencies of the organ pipe will be f=100 Hz, 300 Hz, 500 Hz, 700 Hz, 900 Hz, 1100 Hz which are below 1250 Hz

Topic: Physics, Waves and Sound, NEET, NEET 2014
The number of real ro the equation $x^{2}-3|x|+2=0$ is

Options:
A 4
B 3
C 2
D 1

The correct answer is:
4

Solution:
$x^{2}-3|x|+2=0$
Case (i) when $x \geq 0$
$x^{2}-3 x+2=0$
$(x-1)(x-2)=0$
$\mathrm{x}=1,2($ both roots satisfy the condition $\mathrm{x} \geq 0$ )
Case (ii) when $\mathrm{x} < 0$ $x^{2}+3 x+2=0$
$(x+1)(x+2)=0$
$\mathrm{x}=-1,-2$ (both roots satisfy the condition $\mathrm{x} < 0$ ) So no. of real roots is 4

Topic: Mathematics, Quadratic Equation, NDA, NDA 2015 (Phase 2)
The number of real roots for the equation $x^{2}+9|x|+20=0$ is

Options:
A Zero
B One
C Two
D Three

The correct answer is:
Zero

Solution:
$x^{2}+9|x|+20=0$
The sum of three positive quantities can never be zero. So, the equation has no solution.

Topic: Mathematics, Quadratic Equation, NDA, NDA 2019 (Phase 1)
The number of roots of the equation $z^{2}=2 \bar{z}$ is

Options:
A 2
B 3
C 4
D zero

The correct answer is:
4

Solution:
$\mathrm{z}^{2}=2 \overline{\mathrm{z}}$
Let $\mathrm{z}=\mathrm{x}+\mathrm{iy} \Rightarrow \mathrm{z}^{2}=(\mathrm{x}+\mathrm{iy})^{2}=\mathrm{x}^{2}-\mathrm{y}^{2}+2 \mathrm{xyi}$
$\therefore \quad \mathrm{z}^{2}=2 \overline{\mathrm{z}} \Rightarrow \mathrm{x}^{2}-\mathrm{y}^{2}+2 \mathrm{xyi}=2(\mathrm{x}-\mathrm{iy})$
$\Rightarrow \mathrm{x}^{2}-\mathrm{y}^{2}=2 \mathrm{x} ; 2 \mathrm{xy}=-2 \mathrm{y} \Rightarrow 2(\mathrm{x}+1) \mathrm{y}=0 .$
$\mathrm{x}=-1$ and $\mathrm{y}=0$
$\Rightarrow(-1)^{2}-\mathrm{y}^{2}=2(-1)$
$\Rightarrow 1-\mathrm{y}^{2}=-2$
$\Rightarrow \mathrm{y}^{2}=3 \Rightarrow \mathrm{y}=\pm \sqrt{3}$
$\therefore$ Roots are $-1+\sqrt{3} \mathrm{i}$ and $-1-\sqrt{3} \mathrm{i} .$
and for $\mathrm{y}=0, \mathrm{x}^{2}-0=2 \mathrm{x}$
$\mathrm{x}(\mathrm{x}-2)=0$
$\mathrm{x}=0$ and $2 .$
Hence, roots are 0,2

Topic: Mathematics, Complex Number, NDA, NDA 2017 (Phase 1)
The number of rows in a lecture hall equals the number of seats in a row. If the number of rows is doubled and the
number of seats in every row is reduced by 10 , the number ofseats is increased by 300 . If $x$ denotes the number of rows in the lecture hall, then what is the value of $x$ ?

Options:
A 10
B 15
C 20
D 30

The correct answer is:
30

Solution:
As given $\Rightarrow$ Number of seats in each row $=x$ Total number of seats in the hall $=x$ Revised number of rows $=2 x$ Revised number in each row $=x-10$ Thus Revised number of seats $=2 x(x-10)=2 x^{2}-20 x$ According to question,
$\Rightarrow x^{2}-20 x-300=0$
$\Rightarrow x^{2}-30 x+10 x-300=0$
$(x-30)(x+10)=0$
$\Rightarrow x=30$
$(\because \quad x \neq-10)$

Topic: Mathematics, Quadratic Equation, NDA, NDA 2007 (Phase 2)
The number of structural isomers possible from the molecular formula C3H9N is:

Options:
A 4
B 5
C 2
D 3

The correct answer is:
4

Solution:
The given formula is C3H9N which have degree of usaturation is zero i.e. compounds will be saturated.

Total possible primary amine is 2 , secondary amine 1 ans tertiary amine also 1.

CH3-CH2-CH2-NH2 ; CH3-NH-CH2-CH3

Topic: Chemistry, General Organic Chemistry, NEET, NEET 2015 (Phase 2)
The number of terms in the expansion of $(x+a)^{100}+$ $(x-a)^{100}$ after simplification is

Options:
A 202
B 101
C 51
D 50

The correct answer is:
51

Solution:
$(x+a)^{100}+(x-a)^{100}$
Simple logic is we get ${ }^{\mathrm{n}} \mathrm{c}_{0},{ }^{\mathrm{n}} \mathrm{c}_{2},{ }^{\mathrm{n}} \mathrm{c}_{4} \ldots{ }^{\mathrm{n}} \mathrm{C}_{100}$ in this
expansion.
The number of terms from ${ }^{\mathrm{n}} \mathrm{c}_{0}$ to ${ }^{\mathrm{n}} \mathrm{c}_{100}$ are 51

Topic: Mathematics, Binomial Theorem, NDA, NDA 2017 (Phase 2)
The number of turns in primary and secondary coils of a transformer is 50 and 200 respectively. If the current in the primary coil is $4 \mathrm{~A}$, then the current in the secondary coil is

Options:
A $1 \mathrm{~A}$
B $2 \mathrm{~A}$
C $4 \mathrm{~A}$
D $5 \mathrm{~A}$

The correct answer is:
$1 \mathrm{~A}$

Solution:
In a transformer
$$
\begin{array}{rlrl}
& & \frac{N_P}{N_S} & =\frac{I_S}{I_P} \\
& \frac{50}{200} & =\frac{I_S}{4} \\
& I_S & =1 \mathrm{~A}
\end{array}
$$

Topic: Physics, Alternating Current, AP EAMCET, AP EAMCET 2008
The number of turns in primary and secondary coils of a transformer is 50 and 200 respectively. If the current in the primary coil is $4 \mathrm{~A}$, then the current in the secondary coil is

Options:
A $1 \mathrm{~A}$
B $2 \mathrm{~A}$
C $4 \mathrm{~A}$
D $5 \mathrm{~A}$

The correct answer is:
$1 \mathrm{~A}$

Solution:
In a transformer
$$
\begin{array}{rlrl}
& & \frac{N_P}{N_S} & =\frac{I_S}{I_P} \\
& \frac{50}{200} & =\frac{I_S}{4} \\
& I_S & =1 \mathrm{~A}
\end{array}
$$

Topic: Physics, Alternating Current, TS EAMCET, TS EAMCET 2008
The number of ways in which 3 holiday tickets can be given to 20 employees of an organization if each employee is eligible for any one or more of the tickets, is

Options:
A 1140
B 3420
C 6840
D 8000

The correct answer is:
8000

Solution:
$\because$ Each employee is eligible for 1 or more of the tickets. $\therefore$ No. of ways $=20 \times 20 \times 20=8000$.

Topic: Mathematics, Permutation Combination, NDA, NDA 2015 (Phase 2)
The numbers 1,5 and 25 can be three terms (not necessarily consecutive) of

Options:
A only one AP
B more than one but unite numbers of APs
C inunite number of APs
D unite number of GPs

The correct answer is:
inunite number of APs

Solution:
Given, numbers $1,5,25$ Let $p^{\text {th }}$ term $=1 \Rightarrow a+(p-1) d=1$ ...(1)
Let $q^{\text {th }}$ term $=5 \Rightarrow a+(q-1) d=5$ ...(2)
Let $\mathrm{r}^{\text {th }}$ term $=25 \Rightarrow \mathrm{a}+(\mathrm{r}-1) \mathrm{d}=25$...(3)
(3),$(2) \Rightarrow \mathrm{r}-\mathrm{q}=25-5=20$
(2), $(1) \Rightarrow \mathrm{q}-\mathrm{p}=5-1=4$
$\frac{r-q}{q-p}=\frac{20}{4}=5$ which is an integer.
So, the given series forms an $\mathrm{AP}$.
Infinite AP's are possible

Topic: Mathematics, Sequences and Series, NDA, NDA 2019 (Phase 1)