datavorous/entrance-exam-dataset huggingface dataset:
At 60°C, dinitrogen tetroxide is fifty percent dissociated. Find its standard free energy change at this temperature and one atmosphere. [ Given: log1.33=0.1239 ]

Options:
A -650J.mol -1
B -830 J.mol-1
C -790 J.mol-1
D -875 J·mol-1

The correct answer is:
-790 J.mol-1

Solution:
The reaction showing the dissociation will be:

N2O4(s)⇌2NO2(g)t=01 mol0 molt=teq0.5 mol1 mol

Clearly, from this equation we can say that the total number of moles is

0.5+1=1.5 mol

Hence, the partial pressure for the molecule will be

PN2O4 = 0.51.5×1atm = 13 atmPNO2 = 11.5×1atm =23 atm

Now, according to the law of chemical equilibrium we know that

Ksp = P2NO2PN2O4 = 23213 = 1.33 atm

We know that at equilibrium the free Gibb's energy is:

∆G = -2.303×RT×log10 KP∆G = -2.303×8.314 JK-1mol-1×333K×log10 1.33∆G = -790 Jmol-1

Topic: Chemistry, Thermodynamics (C), AP EAMCET, AP EAMCET 2021 (20 Aug Shift 1)
At 60°C, dinitrogen tetroxide is 50 per cent dissociated, what is the standard free energy change at this temperature and at one atmosphere?

Options:
A –869.4 kJ mol-1
B –763.8 kJ mol–1
C –621.5 kJ mol-1
D –597.2 kJ mol-1

The correct answer is:
–763.8 kJ mol–1

Solution:
Correct Option is : (B)
–763.8 kJ mol–1

Topic: Chemistry, Thermodynamics (C), NEET
At 298 K, the equilibrium constant of the process 1.5O2 (g)⇌O3 (g) is 3×10-29. Standard free energy change (in kJ mol-1 ) of the process is approximately R=8.314 J mol-1 K-1; log 3=0.47

Options:
A 724
B 612
C 247
D 163

The correct answer is:
163

Solution:
Free energy change of a reaction is calculated by,

∆G=∆G∘+RT lnQ

At equilibrium, the free energy change of a reaction is zero (∆G=0), and the reaction quotient (Q) is equal to the equilibrium constant (K). So, the above relation takes the form,

∆G∘=-RT lnK

∆G∘=-2.303RT log10K

where ∆G∘=standard free energy change

Therefore,

∆G∘=-2.303×8.314×298×log3×10-29 J mol-1

∆G∘=-5705.8×log3+log10-29 J mol-1

∆G∘=-5705.8×0.47-29 J mol-1

∆G∘=162787J mol-1

∆G∘≈163 kJ mol-1

Topic: Chemistry, Thermodynamics (C), TS EAMCET, TS EAMCET 2018 (04 May Shift 1)
At 298 K, the standard electrode potentials of Cu2+/Cu, Zn2+/Zn, Fe2+/Fe and Ag+/Ag are 0.34 V, -0.76 V, -0.44 V and 0.80 V, respectively.

On the basis of standard electrode potential, predict which of the following reaction cannot occur?

Options:
A CuSO4(aq)+Fe(s)→FeSO4(aq)+Cu(s)
B FeSO4(aq)+Zn(s)→ZnSO4(aq)+Fe(s)
C 2CuSO4(aq)+2Ag(s)→2Cu(s)+Ag2SO4(aq)
D CuSO4(aq)+Zn(s)→ZnSO4(aq)+Cu(s)

The correct answer is:
2CuSO4(aq)+2Ag(s)→2Cu(s)+Ag2SO4(aq)

Solution:
The given standard reduction potentials of Zn, Cu, Ag and Fe are;

E°Zn2+Zn=-0.76 V, E°Cu2+cu=0.34 VE°Ag+Ag=0.80 V, and E°Fe2+Fe=-0.44 V

The condition for reaction to occur, E°Cell=E°SRPreduction reaction-E°SRPoxidation reaction must have the positive value.

A Cu2+aq+Fe→Cu +Fe2+aq

E°Cell=0.34+0.44E°Cell=0.78 V

B Zn +Fe2+aq→Zn2+aq+Fe

E°Cell=-0.44+0.76E°Cell=+0.32 V

C Cu2+aq+2Ag→2Ag+aq+Cu

E°Cell=0.34-0.80E°Cell=-0.46 V

D Zn+Cu2+aq→Zn2+aq+Cu

E°Cell=0.34+0.76E°Cell=+1.10 V

Hence, the reaction given in C option is not possible.

Topic: Chemistry, Electrochemistry, NEET, NEET 2022 (Phase 1)
At $300 \mathrm{~K}$, the equilibrium constant for a reaction is 10 . The standard free energy change (in $\mathrm{kJ} \mathrm{mol}^{-1}$ ) for the reaction is

Options:
A -57.4
B -115.2
C +57.4
D -57.4

The correct answer is:
-57.4

Solution:
From Gibbs free energy equation,
$$
\Delta G=-2.303 R T \log K
$$
Given, $K=10, T=300 \mathrm{~K}$
$$
\begin{aligned}
& R=8.314 \mathrm{~J} / \mathrm{K} / \mathrm{mol} \\
& \therefore \Delta G=-2.303 \times 8.314 \mathrm{~J} / \mathrm{mol} / \mathrm{K} \times 300 \mathrm{~K} \times \log 10 \\
& \Delta G=-2.303 \times 8.314 \mathrm{~J} / \mathrm{mol} / \mathrm{K} \times 300 \mathrm{~K} \\
& {[\because \log 10=1]} \\
& \Delta G=-5744.14 \mathrm{~J} / \mathrm{mol}, \Delta G=-5.74 \mathrm{~kJ} / \mathrm{mol}
\end{aligned}
$$
Thus, option (d) is correct.

Topic: Chemistry, Thermodynamics (C), AP EAMCET, AP EAMCET 2019 (20 Apr Shift 2)
At a point 15 maway from the base ofa 15 mhigh house, the angle of elevation of the top is $\quad$

Options:
A $90^{\circ}$
B $60^{\circ}$
C $45^{\circ}$
D $30^{\circ}$

The correct answer is:
$45^{\circ}$

Solution:
Let $A B$ be the house of height $15 \mathrm{~m}$. Let $B$ be the base of house and $C$ be the point $15 \mathrm{~m}$ away from the base of a house. Let ' $\theta$ ' be the angle of elevation. So, in $\Delta A B C$,
we have $\tan \theta=\frac{A B}{B C}=\frac{15}{15}=1$
$\Rightarrow \tan \theta=\tan 45^{\circ} \quad\left(\because \tan 45^{\circ}=1\right)$
$\Rightarrow \theta=45^{\circ}$
Hence, the angle of elevation of the top is $45^{\circ}$

Topic: Mathematics, Heights and Distances, NDA, NDA 2011 (Phase 2)
At an extreme point of a function $f(x)$, the tangent to the curve is

Options:
A parallel to the $\mathrm{x}$ -axis
B perpendicular to the $\mathrm{x}$ -axis
C inclined at an angle $45^{\circ}$ to the $\mathrm{x}$ -axis
D inclined at an angle $60^{\circ}$ to the $\mathrm{x}$ -axis

The correct answer is:
parallel to the $\mathrm{x}$ -axis

Solution:
At an extreme point of a function $f(x)$, slope is always zero. Thus, At an extreme point of a function $f(x)$, the tangent to the curve is parallel to the $\mathrm{x}$ -axis.

Topic: Mathematics, Application of Derivatives, NDA, NDA 2011 (Phase 2)
At how many points is the function $f(x)=[x]$ discontinuous?

Options:
A 1
B 2
C 3
D Infinite

The correct answer is:
Infinite

Solution:
Given, $f(x)=[x]$ Let ' $c$ 'be any real number. $f$ is continuous at $x=c$ ifL.H.L. $=\mathrm{R}$.H.L. $=f(c)$
i.e., $\lim _{x \rightarrow c^{-}} f(x)=\lim _{x \rightarrow c^{+}} f(x)=f(c)$
L.H.L. $=\lim _{x \rightarrow c^{-}} f(x)=\lim _{x \rightarrow c^{-}}[x]=\mathrm{c}^{-1}$
R.H.L. $=\lim _{x \rightarrow c^{+}} f(x)=\lim _{x \rightarrow c^{+}}[x]=\mathrm{c}$
Since, L.H.L. $\neq$ R.H.L.
$f$ is discontinous for all $\mathrm{x} \in \mathrm{R}$.
So, $[x]$ is discontinuous at infinite points.

Topic: Mathematics, Continuity and Differentiability, NDA, NDA 2010 (Phase 2)
At standard conditions, if the change in the enthalpy for the following reaction is -109 kJmol-1

H2(g)+Br2(g)→2HBr(g)

Given that bond energy of H2 and Br2 is 435 kJmol-1 and 192 kJmol-1, respectively, what is the bond energy (in kJ mol-1) of HBr?

Options:
A 368
B 736
C 518
D 259

The correct answer is:
368

Solution:
ΔH=Σ(B·E)Reactants -Σ(B.E)Products
-109=B·E(H-H)+B·E(Br-Br)-2×B·E(H-Br)
-109=435+192-2×B·E(H-Br)
B.E(H-Br)=435+192+1092=368 KJ/mol

Topic: Chemistry, Thermodynamics (C), NEET, NEET 2020 (Phase 2)
At STP, the dissociation reaction of water is \(\mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{H}^{+}\)(aq.) \(+\mathrm{OH}^{-}\)(aq.), and the \(\mathrm{pH}\) of water is 7.0. The change of standard free energy \(\left(\Delta \mathrm{G}^{\circ}\right)\) for the above dissociation process is given by

Options:
A \(20301 \mathrm{~cal} / \mathrm{mol}\)
B \(19091 \mathrm{~cal} / \mathrm{mol}\)
C \(20096 \mathrm{~cal} / \mathrm{mol}\)
D \(21301 \mathrm{~cal} / \mathrm{mol}\)

The correct answer is:
\(19091 \mathrm{~cal} / \mathrm{mol}\)

Solution:
\(\begin{aligned}
& \text {Hint : } \Delta G^{\circ}=-2.303 R T \log \mathrm{K}_{\mathrm{w}} \\
& =-2.303 \times 1.987 \times 298 \log 10^{-14} \\
& =+2.303 \times 1.987 \times 298 \times 14 \mathrm{~cal} / \mathrm{mol} \quad \quad\left[\mathrm{K}_{\mathrm{w}}=\left[\mathrm{H}^{+}\right][\mathrm{OH}]=10^{-7} \times 10^{-7}=10^{-14} \mathrm{~as~} \mathrm{pH}=7\right] \\
& =19091.3 \mathrm{cal} / \mathrm{mol}=19091 \mathrm{cal} / \mathrm{mol}
\end{aligned}\)

Topic: Chemistry, Thermodynamics (C), WBJEE, WBJEE 2023
At what conditions is the emf of the cell equal to the standard electrode potential of a given half-cell?

Options:
A When the cell temperature is 298 K.
B When the cell concentration is unity.
C When the cell pressure is one bar.
D When the cell potential is zero.

The correct answer is:
When the cell concentration is unity.

Solution:
If the concentrations of the oxidized and reduced forms of the species in the right-hand half-cell are unity, then the cell potential is equal to the standard electrode potential of the given half-cell.

Topic: Chemistry, Electrochemistry, NEET
At what stage of the cell cycle are histone proteins synthesized in a eukaryotic cell ?

Options:
A During entire prophase
B During telophase
C During S-phase
D During $\mathrm{G}_2$ stage of prophase

The correct answer is:
During S-phase

Solution:
Histone proteins are basic proteins and are used in packing of eukaryotic (absent in prokaryotes) DNA. DNA and histones together comprise chromatin, forming bulk of the eukaryotic chromosomes. Histones are of five major kinds $H_1, H_{2 A}, H_{2 B}, H_3$ and $H_4 . H_1$ histones link neighbouring nucleosomes (fundamental packing units of an eukaryotic chromosome), while other are elements of nucleosome structure. During S-phase of cell cycle synthesis of histone proteins take place because at this stage number of chromosomes become double to that of somatic number.

Topic: Biology, Cell Cycle and Cell Division, JIPMER, JIPMER 2007
Autogamy can occur in a chasmogamous flower if:

Options:
A
pollen matures before maturity of ovule

B
ovules mature before maturity of pollen

C
both pollen and ovules mature simultaneously

D
both anther and stigma are of equal lengths

The correct answer is:
both pollen and ovules mature simultaneously

Solution:
The method of self-pollination in which the stigma of a flower receive pollens from the anther of same flower is known as autogamy. For autogamy both sex organs of a chasmogamous flower should mature at the same time. As chasmogamous flowers open at maturity, pollen release and for the process of autogamy stigma receptivity should be synchronised.
In such flowers, the length of anther and stigma plays secondary role in autogamy. e.g., in case of protandry (pollens nature early) and protogyny (stigma matures early) leads to cross-pollination.

Topic: Biology, Sexual Reproduction in Flowering Plants
Avena curvature test is a bioassay for examining the activity of

Options:
A auxins
B gibberellins
C cytokinins
D ethylene.

The correct answer is:
auxins

Solution:
Bioassay is a testing of biological activity like growth response of a substance by employing plant part. Avena curvature test is a bioassay for examining the activity of auxins. It is based upon experiments of Went. $10^{\circ}$ curvature is produced by auxin concentration of $150 \mu \mathrm{g} /$ lit at $25^{\circ} \mathrm{C}$ and $90 \%$ relative humidity. It can measure auxin upto 300 $\mu \mathrm{g}$ /itre. Auxin from shoot tip or any other plant organ is allowed to diffuse in a standard size agar block. 15 - $30 \mathrm{~mm}$ long oat coleoptile grown in dark is held vertically over water. $1 \mathrm{~mm}$ tip of coleoptile is removed without injuring the primary leaf. After 3 hours a second decapitation is carried out for $4 \mathrm{~mm}$ distance. Primary leaf is now pulled loose and agar block supported against it at the tip of decapitated coleoptile. After $90-110 \mathrm{~mm}$ the coleoptile is found to have bent. The curvature is measured.

Topic: Biology, Plant - Growth and Development, AIIMS, AIIMS 2006
B is associated with

Options:
A $2,3,5,8,9,12$
B $1,3,5,8,10,12$
C $1,3,5,8,9,12$
D None of the above

The correct answer is:
None of the above

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

Topic: Mathematics, Continuity and Differentiability, NDA, NDA 2010 (Phase 1)
Bark refers to

Options:
A phellem + phellogen + phelloderm
B periderm + cortex
C phellem + phelloderm + secondary phloem
D periderm + cortex + pericycle + secondary phloem.

The correct answer is:
periderm + cortex + pericycle + secondary phloem.

Solution:
Bark is a loose term and is used to define all the tissues outside vascular cambium. It is differentiated into outer bark or rhytidome (consisting of dead cells) and inner bark (of living cells, i.e., periderm, cortex, pericycle and secondary phloem). The outer layers of the bark are being constantly peeled off on account of the formation of new secondary vascular tissues in the interior. Bark is insect repellent, decay proof, fire-proof and acts as a heat screen.

Topic: Biology, Anatomy of Flowering Plants, AIIMS, AIIMS 2009
Based on standard electrode potential values, suggest which of the following reactions would take place? (Consult the book for E\degree value).
(i) $\mathrm{Cu}+\mathrm{Zn}^{2+} \rightarrow \mathrm{Cu}^{2+}+\mathrm{Zn}$
(ii) $\mathrm{Mg}+\mathrm{Fe}^{2+} \rightarrow \mathrm{Mg}^{2+}+\mathrm{Fe}$
(iii) $\mathrm{Br}^{2+} 2 \mathrm{Cl}^{-} \rightarrow \mathrm{Cl}^2+2 \mathrm{B}^{\mathrm{r}-}$
(iv) $\mathrm{Fe}+\mathrm{Cd}^{2+} \rightarrow \mathrm{Cd}+\mathrm{Fe}^{2+}$

Options:

None

Solution:
(i) $\mathrm{Cu}+\mathrm{Zn}^{2+} \rightarrow \mathrm{Cu}^{2+}+\mathrm{Zn}$
Here $\mathrm{Cu}$ undergoes oxidation so it acts as anode and $\mathrm{Zn}$ acts as the cathode. So from the table. For example $\mathrm{E}^{\circ}$ anode $=-0.76 \mathrm{~V}$
For anode $\mathrm{E}^{\circ}$ anode $=0.52 \mathrm{~V}$
$\mathrm{E}^{\circ}$ cell $=-0.24 \mathrm{~V}$
As the EMF of the cell is negative the given reaction will not occur spontaneously if they were to form a cell placed as electrodes.
(ii) $\mathrm{Mg}+\mathrm{Fe}^{2+} \rightarrow \mathrm{Mg}^{2+}+\mathrm{Fe}$
Similarly, we can say that $\mathrm{Mg}$ undergoes oxidation and Fe undergoes reduction,
$\mathrm{E}^{\circ}$ cathode $=-0.44 \mathrm{~V}$
$\mathrm{E}^{\circ}$ anode $=-2.36 \mathrm{~V}$
$\mathrm{E}^{\circ}$ cell $=+1.92 \mathrm{~V}$
Positive EMF implies that the reaction will give out energy and attain stability, thus it will occur spontaneously. So the given redox reaction will occur.
(iii) $\mathrm{Br}_2+2 \mathrm{Cl}^{-} \rightarrow \mathrm{Cl}_2+2 \mathrm{Br}^{-}$
Here $\mathrm{Br}$ undergoes reduction thus acting as cathode and $\mathrm{Cl}$ acting as the anode.
For cathode $\mathrm{E}^{\circ}$ cathode $=1.09 \mathrm{~V}$
For anode $\mathrm{E}^{\circ}$ cathode $=1.36 \mathrm{~V}$
$\mathrm{E}^{\circ}$ cell $=-0.25$
The negative potential prevents easy reaction, so the redox reaction will not occur.
(iv) $\mathrm{Fe}+\mathrm{Cd}^{2+} \rightarrow \mathrm{Cd}+\mathrm{Fe}^{2+}$
$\mathrm{Fe}$ is the cathode and $\mathrm{Cd}$ is the anode
For cathode $\mathrm{E}^{\circ}$ cathode $=-0.44 \mathrm{~V}$
For anode $\mathrm{E}^{\circ}$ anode $=-0.40 \mathrm{~V}$
$\mathrm{E}^{\circ}$ cell $=-0.04 \mathrm{~V}$
The negative potential prevents easy reaction, so the redox reaction will not occur.

Topic: Chemistry, Redox Reactions
Bast fibres are made up of

Options:
A Parenchyma cells
B Collenchyma cells
C Sclerenchyma cells
D Chlorenchyma cells

The correct answer is:
Sclerenchyma cells

Solution:
Phloem fibres (bast fibres) are made up of sclerenchymatous cells. These are generally absent in the primary phloem but are found in the secondary phloem. These are much elongated, unbranched and have pointed, needle-like apices. The cell wall of phloem fibres is quite thick. At maturity, these fibres lose their protoplasm and become dead. Phloem fibres of jute, flax and hemp are used commercially.

Topic: Biology, Anatomy of Flowering Plants, NEET
Beads on string like structures of $A$ are seen in $B$, which further condense to form chromosomes in C stage of cell division.
A $\quad$B $\quad$ C

Options:
A Chromonema Chromatin Metaphase
B Chromatin Chromatid Metaphase
C Chromonema Chromosome Anaphase
D Chromonema Chromatid Anaphase.

The correct answer is:
Chromonema Chromatin Metaphase

Solution:
The "beads-on-a-string" structure is seen in electron microscope of isolated metaphase chromosomes. The chromonema form the gene bearing portions of the chromosome. Basically chromonema is made up of nucleosome chains. Nucleosome chain gives a beads on string appearance under electron microscope. Nucleosome is the fundamental packaging unit in eukaryotic chromosomes.

Topic: Biology, Cell Cycle and Cell Division, AIIMS, AIIMS 2012
\begin{array}{|c|c|c|c|c|}
\hline Class Interval & 1-5 & 6-10 & 11-15 & 16-20 \\
\hline Frequency & 3 & 7 & 6 & 5 \\
\hline
\end{array} Consider the following statements in respect of the above frequency distribution.
I. The median is contained in the modal class.
II. The distribution is bell-shaped. Which of the above statements is/are correct?

Options:
A Only I
B Only II
C Both I and II
D Neither I nor II

The correct answer is:
Neither I nor II

Solution:
\begin{array}{|c|c|c|}
\hline Class Interval & \mathrm{f} & cf \\
\hline 0.5-5.5 & 3 & 3 \\
\hline 5.5-10.5 & 7 & 10 \\
\hline 10.5-15.5 & 6 & 16 \\
\hline 15.5-20.5 & 5 & 21 \\
\hline Total & 21 & 50 \\
\hline
\end{array} $N=21$
$\frac{N}{2}=\frac{21}{2}=10.5$
Median class is $10.5-15.5$
Hence, Median $=10.5+\frac{10.5-10}{6} \times 5$
$=10.5+0.417=10.917$
Thus, median is not contained in the modal class and the distribution is not bell-shaped.

Topic: Mathematics, Statistics, NDA, NDA 2011 (Phase 1)
\begin{array}{|c|c|c|c|c|}
\hline \mathrm{X} & 1 & 2 & 3 & 4 \\
\hline Frequency & 2 & 3 & f & 5 \\
\hline
\end{array} The frequency distribution of a discrete variable $X$ with one missing frequency $f$ is given above. If the arithmetic mean of
$X$ is $\frac{23}{8}$, what is the value of the missing frequency?

Options:
A 5
B 6
C 8
D 10

The correct answer is:
6

Solution:
Arithmetic mean $=\frac{2 \times 1+3 \times 2+3 f+4 \times 5}{2+3+f+5}$
$\Rightarrow \frac{23}{8}=\frac{28+3 f}{10+f}$
$230+23 f=224+24 f$
$\Rightarrow f=6$

Topic: Mathematics, Statistics, NDA, NDA 2008 (Phase 2)
\begin{array}{|c|c|c|c|c|c|}
\hline Class Interval & 0-10 & 10-20 & 20-30 & 30-40 & 40-50 \\
\hline Frequency & 5 & 10 & 20 & 5 & 10 \\
\hline
\end{array}
What is the mean of the above frequency distribution?

Options:
A 25
B 26
C 27
D 28

The correct answer is:
26

Solution:
\begin{array}{|c|c|c|c|c|}
\hline Class Interval & \mathbf{f} & \mathbf{e f} & \mathbf{x} & \mathbf{f x} \\
\hline 0-10 & 5 & 5 & 5 & 25 \\
\hline 10-20 & 10 & 15 & 15 & 150 \\
\hline 20-30 & 20 & 35 & 25 & 500 \\
\hline 30-40 & 5 & 40 & 35 & 175 \\
\hline 40-50 & 10 & 50 & 45 & 450 \\
\hline Total & 50 & 145 & 125 & 1300 \\
\hline
\end{array} $\frac{N}{2}=\frac{50}{2}=25$
Mean $=\frac{\sum f x}{\sum f}=\frac{1300}{50}=26$

Topic: Mathematics, Statistics, NDA, NDA 2011 (Phase 1)
\begin{array}{|c|c|c|c|c|c|}
\hline Class Interval & 0-10 & 10-20 & 20-30 & 30-40 & 40-50 \\
\hline Frequency & 5 & 10 & 20 & 5 & 10 \\
\hline
\end{array}
What is the median of the above frequency $\begin{array}{ll}\text { distribution? } & {}\end{array}$

Options:
A 23
B 24
C 25
D 26

The correct answer is:
25

Solution:
\begin{array}{|c|c|c|c|c|}
\hline Class Interval & \mathbf{f} & \mathbf{e f} & \mathbf{x} & \mathbf{f x} \\
\hline 0-10 & 5 & 5 & 5 & 25 \\
\hline 10-20 & 10 & 15 & 15 & 150 \\
\hline 20-30 & 20 & 35 & 25 & 500 \\
\hline 30-40 & 5 & 40 & 35 & 175 \\
\hline 40-50 & 10 & 50 & 45 & 450 \\
\hline Total & 50 & 145 & 125 & 1300 \\
\hline
\end{array} $\frac{N}{2}=\frac{50}{2}=25$
Median group is 20-30. $\Rightarrow$ Median $=20+\frac{25-15}{20} \times 10=20+5=25$

Topic: Mathematics, Statistics, NDA, NDA 2011 (Phase 1)
\begin{array}{|c|c|c|c|c|c|c|c|}
\hline \multirow{2}{*} { Year } & \multicolumn{3}{|c|} { Male } & \multicolumn{2}{|c|} { Female } & \multirow{2}{*} { Total } \\
\cline { 2 - 7 } & & Urban & Rural & Total & Urban & Rural & Total & \\
\hline 1995 & 280 & 350 & & & 310 & & 1350 \\
\hline 1996 & 370 & & 670 & 180 & & 450 & \\
\hline 1997 & & 130 & 440 & & 190 & & \\
\hline 1998 & 400 & 280 & & 290 & & & \\
\hline Total & & & & 1060 & 850 & & \\
\hline
\end{array}
In which year is the male population minimum?

Options:
A 1995
B 1996
C 1997
D 1998

The correct answer is:
1997

Solution:
Complete table is
\begin{array}{|c|c|c|c|c|c|c|c|}
\hline \multirow{2}{*} { Year } & \multicolumn{3}{|c|} { Male } & \multicolumn{3}{|c|} { Female } & \multirow{2}{*} { Total } \\
\cline { 2 - 8 } & Urban & Rural & Total & Urban & Rural & Total & \\
\hline 1995 & 280 & 350 & 630 & 410 & 310 & 720 & 1350 \\
\hline 1996 & 370 & 300 & 670 & 180 & 270 & 450 & 1120 \\
\hline 1997 & 310 & 130 & 440 & 180 & 190 & 370 & 810 \\
\hline 1998 & 400 & 280 & 680 & 290 & 80 & 370 & 1050 \\
\hline Total & 1360 & 1060 & 2420 & 1060 & 850 & 1910 & 4330 \\
\hline
\end{array}
In 1997 , the male population is minimum.

Topic: Mathematics, Statistics, NDA, NDA 2011 (Phase 2)
\begin{array}{|c|c|c|c|c|c|c|c|}
\hline \multirow{2}{*} { Year } & \multicolumn{3}{|c|} { Male } & \multicolumn{2}{|c|} { Female } & \multirow{2}{*} { Total } \\
\cline { 2 - 7 } & & Urban & Rural & Total & Urban & Rural & Total & \\
\hline 1995 & 280 & 350 & & & 310 & & 1350 \\
\hline 1996 & 370 & & 670 & 180 & & 450 & \\
\hline 1997 & & 130 & 440 & & 190 & & \\
\hline 1998 & 400 & 280 & & 290 & & & \\
\hline Total & & & & 1060 & 850 & & \\
\hline
\end{array}
What is the difference between the number of females and the number of males in the year 1995 ?

Options:
A 90
B 100
C 110
D 120

The correct answer is:
90

Solution:
Complete table is
\begin{array}{|c|c|c|c|c|c|c|c|}
\hline \multirow{2}{*} { Year } & \multicolumn{3}{|c|} { Male } & \multicolumn{3}{|c|} { Female } & \multirow{2}{*} { Total } \\
\cline { 2 - 8 } & Urban & Rural & Total & Urban & Rural & Total & \\
\hline 1995 & 280 & 350 & 630 & 410 & 310 & 720 & 1350 \\
\hline 1996 & 370 & 300 & 670 & 180 & 270 & 450 & 1120 \\
\hline 1997 & 310 & 130 & 440 & 180 & 190 & 370 & 810 \\
\hline 1998 & 400 & 280 & 680 & 290 & 80 & 370 & 1050 \\
\hline Total & 1360 & 1060 & 2420 & 1060 & 850 & 1910 & 4330 \\
\hline
\end{array}
Required difference $=720-630=90$.

Topic: Mathematics, Statistics, NDA, NDA 2011 (Phase 2)
\begin{array}{|c|c|c|c|c|c|c|c|}
\hline \multirow{2}{*} { Year } & \multicolumn{3}{|c|} { Male } & \multicolumn{2}{|c|} { Female } & \multirow{2}{*} { Total } \\
\cline { 2 - 7 } & & Urban & Rural & Total & Urban & Rural & Total & \\
\hline 1995 & 280 & 350 & & & 310 & & 1350 \\
\hline 1996 & 370 & & 670 & 180 & & 450 & \\
\hline 1997 & & 130 & 440 & & 190 & & \\
\hline 1998 & 400 & 280 & & 290 & & & \\
\hline Total & & & & 1060 & 850 & & \\
\hline
\end{array}
What is the female urban population in the year $1995 ?$

Options:
A 390
B 410
C 430
D 470

The correct answer is:
410

Solution:
Complete table is
\begin{array}{|c|c|c|c|c|c|c|c|}
\hline \multirow{2}{*} { Year } & \multicolumn{3}{|c|} { Male } & \multicolumn{3}{|c|} { Female } & \multirow{2}{*} { Total } \\
\cline { 2 - 8 } & Urban & Rural & Total & Urban & Rural & Total & \\
\hline 1995 & 280 & 350 & 630 & 410 & 310 & 720 & 1350 \\
\hline 1996 & 370 & 300 & 670 & 180 & 270 & 450 & 1120 \\
\hline 1997 & 310 & 130 & 440 & 180 & 190 & 370 & 810 \\
\hline 1998 & 400 & 280 & 680 & 290 & 80 & 370 & 1050 \\
\hline Total & 1360 & 1060 & 2420 & 1060 & 850 & 1910 & 4330 \\
\hline
\end{array}
Female urban population in the year $1995=410$.

Topic: Mathematics, Statistics, NDA, NDA 2011 (Phase 2)
\begin{array}{|c|c|c|c|c|c|c|c|}
\hline \multirow{2}{*} { Year } & \multicolumn{3}{|c|} { Male } & \multicolumn{2}{|c|} { Female } & \multirow{2}{*} { Total } \\
\cline { 2 - 7 } & & Urban & Rural & Total & Urban & Rural & Total & \\
\hline 1995 & 280 & 350 & & & 310 & & 1350 \\
\hline 1996 & 370 & & 670 & 180 & & 450 & \\
\hline 1997 & & 130 & 440 & & 190 & & \\
\hline 1998 & 400 & 280 & & 290 & & & \\
\hline Total & & & & 1060 & 850 & & \\
\hline
\end{array}
What is the total population for the year 1997?

Options:
A 810
B 830
C 970
D 1030

The correct answer is:
810

Solution:
Complete table is
\begin{array}{|c|c|c|c|c|c|c|c|}
\hline \multirow{2}{*} { Year } & \multicolumn{3}{|c|} { Male } & \multicolumn{3}{|c|} { Female } & \multirow{2}{*} { Total } \\
\cline { 2 - 8 } & Urban & Rural & Total & Urban & Rural & Total & \\
\hline 1995 & 280 & 350 & 630 & 410 & 310 & 720 & 1350 \\
\hline 1996 & 370 & 300 & 670 & 180 & 270 & 450 & 1120 \\
\hline 1997 & 310 & 130 & 440 & 180 & 190 & 370 & 810 \\
\hline 1998 & 400 & 280 & 680 & 290 & 80 & 370 & 1050 \\
\hline Total & 1360 & 1060 & 2420 & 1060 & 850 & 1910 & 4330 \\
\hline
\end{array}
Total population for the year $1997=440+370=810$

Topic: Mathematics, Statistics, NDA, NDA 2011 (Phase 2)