datavorous/entrance-exam-dataset huggingface dataset:
In any discrete series (when all values are not same) if $x$ represents mean deviation about mean and $y$ represents standard deviation, then which one of the following is correct?

Options:
A $y \geq x$
B $y \leq x$
C $x=y$
D $x < y$

The correct answer is:
$x < y$

Solution:
$x=\mathrm{M}, \mathrm{D}_{-,}, y=\mathrm{S} . \mathrm{D}$.
$$
\mathrm{M}, \mathrm{D},=\frac{4}{5} \mathrm{S.D.} \Rightarrow x < y
$$

Topic: Mathematics, Statistics, NDA, NDA 2018 (Phase 2)
In any triangle $A B C$, the sides are $6 \mathrm{~cm}, 10 \mathrm{~cm}$ and $14 \mathrm{~cm}$. Then the triangle is obtuse angled with the obtuse angle equal to

Options:
A $150^{\circ}$
B $135^{\circ}$
C $120^{\circ}$
D $105^{\circ}$

The correct answer is:
$120^{\circ}$

Solution:
Since, $\mathrm{c}=14$ is the largest side $\therefore$ Angle $\mathrm{C}$ will be obtuse
$\cos C=\frac{a^{2}+b^{2}-c^{2}}{2 a b}=\frac{(6)^{2}+(10)^{2}-(14)^{2}}{2(6)(10)}$
$=\frac{36+100-196}{2 \times 6 \times 10}=\frac{-1}{2}$
$\Rightarrow C=\cos ^{-1}\left(\frac{-1}{2}\right)=\pi-\frac{\pi}{3}=\frac{2 \pi}{3}=\frac{2 \times 180}{3}=120$

Topic: Mathematics, Properties of Triangles, NDA, NDA 2011 (Phase 2)
In any triangle $\mathrm{ABC}, \mathrm{a}=18, \mathrm{~b}=24$ and $\mathrm{c}=30$. Then what is $\begin{array}{ll}\text { sin C equal to : } & {}\end{array}$

Options:
A $\frac{1}{4}$
B $\frac{1}{3}$
C $\frac{1}{2}$
D 1

The correct answer is:
1

Solution:
$\cos C=\frac{a^{2}+b^{2}-c^{2}}{2 a b}$
$\Rightarrow \cos C=\frac{(18)^{2}+(24)^{2}-(30)^{2}}{2 \times 18 \times 24}=\frac{9+16-5^{2}}{2 \times 3 \times 4}=0$
Now, $\sin C=\sqrt{1-\cos ^{2} C}=\sqrt{1-0}=1$
Hence $\sin C=1$

Topic: Mathematics, Properties of Triangles, NDA, NDA 2013 (Phase 1)
In Basidiomycetes, meiosis will take place in

Options:
A Primary mycelium
B Secondary mycelium
C Basidiocarp
D Basidium

The correct answer is:
Basidium

Solution:
Basidium is a club-shaped structure. Karyogamy and meiosis both will occur in basidium, and four basidiospores will form.

Topic: Biology, Biological Classification, NEET
In binary system the decimal number $0.3$ can be expressed as

Options:
A $(0.01001 \ldots \ldots)_{2}$
B $(0.10110 \ldots \ldots)_{2}$
C $(0.11001 \ldots \ldots)_{2}$
D $(0.10101 \ldots \ldots)$

The correct answer is:
$(0.01001 \ldots \ldots)_{2}$

Solution:
Consider $0.3 \times 2=0.6 \times 2=1.2 \times 2$
Now, treated 1 as 0
So, $0.2 \times 2=0.4 \times 2=0.8 \times 2=1.6 \times 2$
Again, $1.6 \times 2=0.6 \times 2=1.2$
Thus,
$0.3 \times 2=0.6 \times 2=1.2 \times 2=0.4 \times 2$
Hence, $0.3=(0.01001 \ldots-)_{2}$

Topic: Mathematics, Sets and Relations, NDA, NDA 2011 (Phase 2)
In Calotropis, the leaves show (X) phyllotaxy while the petals show (Y) aestivation.

Options:
A (X): opposite and (Y): valvate
B (X): alternate and (Y): valvate
C (X): opposite and (Y): twisted.
D (X): alternate and (Y): twisted

The correct answer is:
(X): opposite and (Y): valvate

Solution:
When sepals or petals in a whorl just touch one another at the margin, without overlapping, it is said to be valvate.
If one margin of the appendage overlaps that of the next one and so on, it is called twisted. Calotropis has valvate aestivation. In alternate type of phyllotaxy, a single leaf arises at each node in alternate manner. In opposite type, a pair of leaves arise at each node and lie opposite to each other as in Calotropis.

Topic: Biology, Morphology of Flowering Plants, NEET
In CGS system the magnitude of the force is 100 dynes. In another system where the fundamental physical. quantities are kilogram, meter and minute, the magnitude of the force is

Options:
A $0.036$
B $0.36$
C $3.6$
D $36$

The correct answer is:
$3.6$

Solution:
$n_2=n_1\left[\left(\frac{m_1}{m_2}\right)^a\left(\frac{L_1}{L_2}\right)^b\left(\frac{T_1}{T_2}\right)^c\right]$
Dimensions of force is [ $\mathrm{M} \mathrm{L} \mathrm{T}^{-2}$ ]
$a=1, b=1, c=-2$
$\begin{aligned} n_2 & =100\left[\left(\frac{\mathrm{g}}{\mathrm{kg}}\right)^1\left(\frac{\mathrm{cm}}{\text { metre }}\right)^1\left(\frac{\mathrm{sec}}{\mathrm{min}}\right)^{-3}\right] \\ & =100\left[\frac{\mathrm{g}}{100 \mathrm{~g}} \times \frac{\mathrm{cm}}{100 \mathrm{~cm}} \times\left(\frac{\mathrm{sec}}{60 \mathrm{sec}}\right)^{-2}\right] \\ & =100\left[\frac{1}{1000} \times \frac{1}{100} \times(60)^2\right] \\ & =100 \times \frac{1}{1000} \times \frac{1}{100} \times 3600=3.6\end{aligned}$

Topic: Physics, Units and Dimensions, AP EAMCET, AP EAMCET 2001
In CGS system the magnitude of the force is 100 dynes. In another system where the fundamental physical. quantities are kilogram, meter and minute, the magnitude of the force is

Options:
A $0.036$
B $0.36$
C $3.6$
D $36$

The correct answer is:
$3.6$

Solution:
$n_2=n_1\left[\left(\frac{m_1}{m_2}\right)^a\left(\frac{L_1}{L_2}\right)^b\left(\frac{T_1}{T_2}\right)^c\right]$
Dimensions of force is [ $\mathrm{M} \mathrm{L} \mathrm{T}^{-2}$ ]
$a=1, b=1, c=-2$
$\begin{aligned} n_2 & =100\left[\left(\frac{\mathrm{g}}{\mathrm{kg}}\right)^1\left(\frac{\mathrm{cm}}{\text { metre }}\right)^1\left(\frac{\mathrm{sec}}{\mathrm{min}}\right)^{-3}\right] \\ & =100\left[\frac{\mathrm{g}}{100 \mathrm{~g}} \times \frac{\mathrm{cm}}{100 \mathrm{~cm}} \times\left(\frac{\mathrm{sec}}{60 \mathrm{sec}}\right)^{-2}\right] \\ & =100\left[\frac{1}{1000} \times \frac{1}{100} \times(60)^2\right] \\ & =100 \times \frac{1}{1000} \times \frac{1}{100} \times 3600=3.6\end{aligned}$

Topic: Physics, Units and Dimensions, TS EAMCET, TS EAMCET 2001
In China rose the flowers are

Options:
A actinomorphic, hypogynous with twisted aestivation
B actinomorphic, epigynous with valvate aestivation
C zygomorphic, hypogynous with imbricate aestivation
D zygomorphic, epigynous with twisted aestivation

The correct answer is:
actinomorphic, hypogynous with twisted aestivation

Solution:
Actinomorphic (star-shaped) can be divided into 3 or more identical sectors, which are related to each other by rotation about the centre of the flower, e.g., China rose. Zygomorphic flowers can be divided by only a single plane into two mirror image halves, e.g., orchids.
Valvate aestivation The sepals or petals in a whorl just touch one another at the margin, e.g., Calotropis.
Twisted aestivation One margin of the appendage overlaps that of the next one, e.g., China rose. Imbricate aestivation the margins of sepals or petals overlap but not necessarily in specific direction, e.g., Cassia. For further detail refer to ans. 97

Topic: Biology, Morphology of Flowering Plants, NEET, NEET 2013 (All India)
In chloroethane, the carbon bearing halogen is bonded to ... hydrogen $(s)$. It is called ...... alkyl halide.

Options:
A Two, primary
B Three, primary
C Two, secondary
D One, tertiary

The correct answer is:
Two, primary

Solution:

Topic: Chemistry, Haloalkanes and Haloarenes, AP EAMCET, AP EAMCET 2002
In chloroethane, the carbon bearing halogen is bonded to ... hydrogen $(s)$. It is called ...... alkyl halide.

Options:
A Two, primary
B Three, primary
C Two, secondary
D One, tertiary

The correct answer is:
Two, primary

Solution:

Topic: Chemistry, Haloalkanes and Haloarenes, TS EAMCET, TS EAMCET 2002
In computing a measure of the central tendency for any set of 51 numbers, which one of the following measures is welldefined but uses only very few of the numbers of the set?

Options:
A Arithmetic mean
B Geometric mean
C Median
D Mode

The correct answer is:
Mode

Solution:
Mode is the required measure.

Topic: Mathematics, Statistics, NDA, NDA 2007 (Phase 2)
In conifers fibres are likely to be absent in

Options:
A
secondary phloem

B
secondary xylem

C
primaryphloem

D
leaves

The correct answer is:
secondary xylem

Solution:
Fibres occur in all those parts of plants where mechanical strength is required i.e., leaves, petioles, cortex, xylem, phloem, etc. In conifers, they are likely to be absent in secondary xylem.

Topic: Biology, Anatomy of Flowering Plants
In dehydrohalogenation, the rate of the reaction is influenced by the nature of the alkyl group. Which type of alkyl group exhibits the highest rate of reaction?

Options:
A Primary alkyl group
B Secondary alkyl group
C Tertiary alkyl group
D Quaternary alkyl group

The correct answer is:
Tertiary alkyl group

Solution:
Tertiary alkyl groups exhibit the highest rate of dehydrohalogenation reactions.

Topic: Chemistry, Hydrocarbons, NEET
In dicot root, the vascular cambium is

Options:
A Completely primary in origin
B Partly primary and partly secondary
C Completely secondary in origin
D None of these

The correct answer is:
Completely secondary in origin

Solution:
In dicot roots cambium is absent, thus during secondary growth parenchyma cells of pericycle and parenchyma cells of conjuctive tissue lying inner to phloem becomes meristematic and forms wavy ring of vascular cambium which is completely secondary in origin.

Topic: Biology, Anatomy of Flowering Plants, NEET
In dicot stem, vascular bundles are:

Options:
A Numerous and scattered
B Arranged in a ring
C Without cambium
D Surrounded by bundle sheath

The correct answer is:
Arranged in a ring

Solution:
The characteristic features of vascular bundles of dicot stem are as follows:
1. Vascular bundles are arranged in a ring.
2. They are conjoint, i.e. xylem and phloem are present on the same radius.
3. They are open, i.e. cambium layer is found between xylem and phloem.
4. They are not surrounded by bundle sheath.
5. The position of protoxylem is towards the centre, i.e. endarch.

Topic: Biology, Anatomy of Flowering Plants, NEET
In dicotyledonous stems, what is present between phloem and xylem?

Options:
A Cambium
B Secondary xylem
C Secondary phloem
D Closed vascular bundles

The correct answer is:
Cambium

Solution:
In dicotyledonous stems, cambium is present between phloem and xylem.

Topic: Biology, Anatomy of Flowering Plants, NEET
In eight throws of a die, 5 or 6 is considered a success. The mean and standard deviation of total number of successes
is respectively given by

Options:
A $\frac{8}{3}, \frac{16}{9}$
B $\frac{8}{3}, \frac{4}{3}$
C $\frac{4}{3}, \frac{4}{3}$
D $\frac{4}{3}, \frac{16}{9}$

The correct answer is:
$\frac{8}{3}, \frac{4}{3}$

Solution:
5 or 6 is success
$\therefore p=\frac{2}{6}=\frac{1}{3}$
$\therefore q=1-\frac{1}{3}=\frac{2}{3}$
$\mathrm{n}=8$
$\therefore$ Mean $=\mathrm{n} \mathrm{p}=8\left(\frac{1}{3}\right)=\frac{8}{3}$
Standard deviation $=\sqrt{n p q}=\sqrt{8\left(\frac{1}{3}\right)\left(\frac{2}{3}\right)}$
$=\sqrt{\frac{16}{9}}=\frac{4}{3}$

Topic: Mathematics, Probability, NDA, NDA 2019 (Phase 1)
In Ellingham diagram, the plot is drawn between

Options:
A temperature, $\Delta H^{\circ}$
B temperature, $\Delta G^{\circ}$
C pressure, $\Delta S^{\circ}$
D temperature, $\Delta E^{\circ}$

The correct answer is:
temperature, $\Delta G^{\circ}$

Solution:
Standard Gibbs free energy ∆G° is plotted
against temperature (T)

Topic: Chemistry, General Principles and Processes of Isolation of Metals, AP EAMCET, AP EAMCET 2018 (22 Apr Shift 1)
In endarch condition of xylem, metaxylem lies _____ of protoxylem

Options:
A on outer side
B on inner side
C both on inner and outer side
D in centre

The correct answer is:
on outer side

Solution:
The endarch condition of xylem refers to the condition when the metaxylem lies towards the periphery and the protoxylem towards the pith while in exarch condition the metaxylem lies towards the centre and the protoxylem is situated towards the periphery. Endarch condition is seen in stem whereas exarch condition is found in roots of all vascular plants.

Topic: Biology, Anatomy of Flowering Plants, NEET
In flowering plants, arrange the following steps in their correct sequence.

a Formation of the embryo sac

b Megasporogenesis

c Formation of endosperm

d Fertilization

e Development of embryo

Options:
A b→ a→e→c→d
B b→a→c→d→e
C a→b→ c→d →e
D b→a→d→c→e

The correct answer is:
b→a→d→c→e

Solution:
Formation of megaspore is called megasporogenesis and megaspore germinates to form female gametophyte i.e. embryo sac. Embryo sac has egg that participates in fertilization and secondary nucleus of embryo sac leads to formation of PEN after double fertilization. Thus endosperm formation is followed by embryo formation.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
In flowering plants, what is the correct sequence of events that takes place in the integumented ovule?

a. Embryo sac formation
b. Megasporogenesis
c. Endosperm formation
d. Fertilization
e. Embryogenesis

Options:
A a→b→c→d→e
B b→a→d→c→e
C b→a→e→c→d
D b→a→c→d→e

The correct answer is:
b→a→d→c→e

Solution:
Formation of megaspore is called megasporogenesis, and megaspore germinates to form female gametophyte, i.e., embryosac. Embryosac has an egg that participates in fertilization, and secondary nucleus of embryo sac leads to the formation of PEN after double fertilization. Thus, endosperm formation is followed by embryo formation.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
In frog chromosome number is reduced to half

Options:
A when 2nd polar body is separated
B when 2nd polar body is divided
C when 3rd polar body is separated
D when 1st polar body is separated

The correct answer is:
when 1st polar body is separated

Solution:
Each primary oocyte undergo two maturation divisions, first meiotic and the second meiotic division. In the I-meiotic division, the primary oocyte divides into two unequal haploid daughter cells-a large secondary oocyte and very small first polar body. In II-maturation division, the first polar body may divide to form two second polar bodies. The secondary oocyte again divides into two unequal daughter cells, a large ootid and a very small second polar body. The ootid grows into a functional haploid ovum. The I meiotic division is reductional and II meiotic division is equational division.

Topic: Biology, Structural Organisation in Animals, JIPMER, JIPMER 2005
In fundamental mode, the time required for the sound wave to each upto the closed end of a pipe filled with air is ' $t$ ' second. The frequency of vibration of air column is

Options:
A $\frac{2}{t}$
B $\frac{0.5}{t}$
C $\frac{1}{t}$
D $\frac{0.25}{t}$

The correct answer is:
$\frac{0.25}{t}$

Solution:
$\begin{aligned}
& \Rightarrow V=\frac{L}{y}=(4 L) g \\
& \Rightarrow f=\frac{0.25}{t}
\end{aligned}$
Option (D) is correct

Topic: Physics, Waves and Sound, MHT CET, MHT CET 2022 (08 Aug Shift 2)
In fundamental mode, the time required for the sound wave to reach up to the closed end of a pipe ﬁlled with air is ‘ t ’ second. The frequency of vibration of air column is

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

The correct answer is:
4t-1

Solution:
∵ f0= v4l
f0= 14lDistancetime
f0=14llt
f0=14t

Topic: Physics, Waves and Sound, MHT CET, MHT CET 2019 (02 May Shift 1)
In fundamental mode, the time required for the sound wave to reach up to the closed end of pipe filled with air is $t$ second. The frequency of vibration of air column is

Options:
A $\frac{1}{\mathrm{t}}$
B $\frac{2}{\mathrm{t}}$
C $\frac{3}{\mathrm{t}}$
D $\frac{0.25}{\mathrm{t}}$

The correct answer is:
$\frac{0.25}{\mathrm{t}}$

Solution:
In fundamental mode the length of the closed tube is $\frac{\lambda}{4}$
Time required to travel a distance $\frac{\lambda}{4}$ is $\mathrm{t}$.
Hence time required to travel a distance $\lambda$ will be $4 \mathrm{t}$.
$\therefore$ Time period $\mathrm{T}=4 \mathrm{t}$
Frequency, $\mathrm{n}=\frac{1}{\mathrm{~T}}=\frac{1}{4 \mathrm{t}}=\frac{0.25}{\mathrm{t}}$

Topic: Physics, Waves and Sound, MHT CET, MHT CET 2021 (21 Sep Shift 1)
In fundamental mode, the time required for the sound wave to reach upto the closed end of a pipe filled with air is 't' second. The frequency of vibration of air column is

Options:
A $\frac{1}{3 t}$
B $\frac{1}{t}$
C $\frac{1}{4 t}$
D $\frac{1}{2 t}$

The correct answer is:
$\frac{1}{4 t}$

Solution:
The distance between the open end and the closed end in the fundamental mode is $\frac{\lambda}{4}$. If the time required to cover the distance is $\mathrm{t}$, then the time required to travel distance equal to $\lambda$ will be $4 \mathrm{t}$. Therefore the period of the wave is $4 \mathrm{t}$. Hence the frequency is $\frac{1}{4 \mathrm{t}}$.

Topic: Physics, Waves and Sound, MHT CET, MHT CET 2020 (14 Oct Shift 2)
In how many of the following, primary succession will occur?
(Abandoned farmland, newly created pond, bare rock, burnt forest)

Options:
A Newly created ponds, bare rock, bandoned farmlands and burnt forests
B Newly created ponds and bare rock
C Abandoned farmlands, bare rocks and burnt forests
D Bare rocks

The correct answer is:
Newly created ponds and bare rock

Solution:
When succession occurs in an area where no living organism ever existed; it is called primary succession and when it takes place in areas that have lost all living organisms that had existed there; it is called secondary succession. Newly created ponds, bare rock are examples of primary succession. Abandoned farmlands and burnt forests are examples of secondary succession.

Topic: Biology, Ecosystem, NEET
In how many points do the ellipse $\frac{x^{2}}{4}+\frac{y^{2}}{8}=1$ and the circle $\mathrm{x}^{2}+\mathrm{y}^{2}=9$ intersect ?

Options:
A One
B Two
C Four
D None of the above

The correct answer is:
None of the above

Solution:
The given equation of circle is: $x^{2}+y^{2}=9$...(i)
and ellipse is : $\frac{x^{2}}{4}+\frac{y^{2}}{8}=1$...(ii)
From eqn. (1) and (2), we get
$\frac{x^{2}}{4}+\frac{9-x^{2}}{8}=1$
$\Rightarrow 2 x^{2}+9-x^{2}=8 \Rightarrow \quad x^{2}=-1$
$x$ is not real Hence, circle and ellipse do not intersect.

Topic: Mathematics, Ellipse, NDA, NDA 2007 (Phase 2)
In how many trophic levels can humans function in a food chain?

Options:
A 4
B 3
C 2
D 1

The correct answer is:
3

Solution:
Humans feed on vegetables and fruits. So they can be called primary consumers (Trophic level T2). Some humans also consume herbivores animals such as goat and lamb. So humans can be called secondary consumers (Trophic level T3). Humans also consume fishes such as salmon which are carnivorous animals. So humans can be tertiary consumers (Trophic level T4)

Topic: Biology, Ecosystem, NEET
In how many ways 6 girls can be seated in two chairs?

Options:
A 10
B 15
C 24
D 30

The correct answer is:
30

Solution:
Required number of ways $=6 \times 5=30$

Topic: Mathematics, Permutation Combination, NDA, NDA 2011 (Phase 1)
In how many ways can 3 books on Hindi and 3 books on English be arranged in a row on a shelf, so that not all the Hindi books are together?

Options:
A 144
B 360
C 576
D 720

The correct answer is:
576

Solution:
Total number of arrangement $=6 !=720$ Total number of arrangement while all the Hindi books are together $=4 ! \times 3 !=24 \times 6=144$ $\therefore$ The number of ways, in which books are arranged, while all the Hindi books are not together $=720-144=576$

Topic: Mathematics, Permutation Combination, NDA, NDA 2008 (Phase 2)
In how many ways can 7 persons stand in the form of a ring?

Options:
A $\mathrm{P}(7,2)$
B $7 !$
C $6 !$
D $\frac{7 !}{2}$

The correct answer is:
$6 !$

Solution:
Number of ways in which 7 persons can stand in the form of a ring $=(7-1) !=6 !$

Topic: Mathematics, Permutation Combination, NDA, NDA 2006 (Phase 2)
In how many ways can a committee consisting of 3 men and 2 women be formed from 7 men and 5 women?

Options:
A 45
B 350
C 700
D 4200

The correct answer is:
350

Solution:
Total no. of Men $=7$ Total no. of women $=5$ Required number of ways $={ }^{7} \mathrm{C}_{3} \times{ }^{5} \mathrm{C}_{2}$
$=\frac{7 !}{3 ! 4 !} \times \frac{5 !}{2 ! 3 !}=\frac{7 \times 6 \times 5}{3 \times 2} \times \frac{5 \times 4}{2}$
$=7 \times 5 \times 10=35 \times 10=350$

Topic: Mathematics, Permutation Combination, NDA, NDA 2010 (Phase 2)
In how many ways can be letters of the word 'CABLE' be arranged so that the vowels should always occupy odd positions?

Options:
A 12
B 18
C 24
D 36

The correct answer is:
36

Solution:
There are two vowels A and E. There are total 5 places out of which two places are to be occupied by vowels. So, 3 places can be occupied by 2 vowels in ${ }^{3} \mathrm{P}_{2}$ ways and after two vowels occupy two places, 3 consonants will occupy 3 places in ${ }^{3} \mathrm{P}_{3}=3$ ! way, hence, Required number of ways $={ }^{3} \mathrm{P}_{2} \times 3 !=6 \times 6=36$

Topic: Mathematics, Permutation Combination, NDA, NDA 2007 (Phase 1)
In how many ways can the letters of the word CORPORATION be arranged so that vowels always occupy even places?

Options:
A 120
B 2700
C 720
D 7200

The correct answer is:
7200

Solution:
CORPORATION is 11 letter word. It has 5 vowels $(\mathrm{O}, \mathrm{O}, \mathrm{O}, \mathrm{A}, \mathrm{I})$ and 6 consonants $(\mathrm{C}, \mathrm{R}$
$\mathrm{P}, \mathrm{R}, \mathrm{T}, \mathrm{N})$
In 11 letters, there are 5 even places $\left(2^{\text {nd }}, 4^{\text {th }}, 6^{\text {th }}, 8^{\text {th }}\right.$ and $10^{\text {th }}$ positions)
5 vowels can take 5 even places in $\frac{5 !}{3 !}$ ways
$(\because$ Since $\mathrm{O}$ is repeated thrice)
Similarly, 6 consonasts can take 6 odd places in $\frac{6 !}{2 !}$
ways. $(\because \mathrm{R}$ is repeated twice)
$\therefore \quad$ Total number of ways $=\frac{5 !}{3 !} \times \frac{6 !}{2 !}=20 \times 360=7200$

Topic: Mathematics, Permutation Combination, NDA, NDA 2006 (Phase 1)
In how many ways can the letters of the word 'GLOOMY' be arranged so that the two $\mathrm{O}$ 's should not be together?

Options:
A 240
B 480
C 600
D 720

The correct answer is:
240

Solution:
$\frac{6 !}{2 !}-5 !=240$

Topic: Mathematics, Permutation Combination, NDA, NDA 2013 (Phase 1)
In human female, second polar body is formed in the ovum

Options:
A within ovary
B within oviducts
C within uterine cavity
D within uterine endometrium

The correct answer is:
within oviducts

Solution:
During oogenesis, the primary oocyte undergoes meiosis I to form the secondary oocyte and the first polar body during the formation of the Graafian follicle. This occurs in the ovary. This secondary oocyte is released during ovulation and reaches the fallopian tubes (oviducts). Fertilization occurs here. When the sperm enters the ovum, the secondary oocyte undergoes the second meiotic division to release the second polar body. Also, the first polar body at the same time completes second meiotic division to form two secondary polar bodies.

Topic: Biology, Human Reproduction, NEET
In human females, meiosis-II is not completed until?

Options:
A Fertilization
B Uterine implantation
C Birth
D Puberty

The correct answer is:
Fertilization

Solution:
In human females, meiosis-II is completed after the entry of sperm into the cytoplasm of the secondary oocyte at the time of fertilization leading to the formation of ovum and secondary polar body.

Topic: Biology, Human Reproduction, NEET, NEET 2015 (Phase 2)
In humans at the end of first meiotic division, the male germ cells differentiate into the

Options:
A primary spermatocytes
B secondary spermatocytes
C spermatids
D spermatogonia.

The correct answer is:
secondary spermatocytes

Solution:
At sexual maturity, the undifferentiated primordial germ cells divide several times by mitosis to produce a large number of spermatogonia. Each spermatogonium then actively grows to a larger primary spermatocyte. Each primary spermatocyte undergoes two successive divisions, called maturation divisions. The first maturation division is reductional or meiotic. Hence, the primary spermatocyte divides into the haploid daughter cells called secondary spermatocytes.

Topic: Biology, Human Reproduction, AIIMS, AIIMS 2009