datavorous/entrance-exam-dataset huggingface dataset:
A complex compound of $\mathrm{Co}^{3+}$ with molecular formula $\mathrm{CoCl}_{\mathrm{x}} \cdot \mathrm{yNH}_3$ gives a total of 3 ions when dissolved in water. How many $\mathrm{Cl}^{-}$ions satisfy both primary and secondary valencies in this complex?

Options:
A 3
B 1
C 4
D zero

The correct answer is:
1

Solution:
As cobalt is present as $\mathrm{Co}^{3+}$ and co-ordination number of cobalt is 6 , the molecular formula of compound should be $\mathrm{CoCl}_3 \cdot \mathrm{yNH}_3$. Now, as it gives a total of 3 ions when dissolved in water, its structural formula must be $\left[\mathrm{CoCl}\left(\mathrm{NH}_3\right)_5\right] \mathrm{Cl}_2$
$\left[\mathrm{CoCl}\left(\mathrm{NH}_3\right)_5\right] \mathrm{Cl}_2 \rightleftharpoons\left[\mathrm{CoCl}\left(\mathrm{NH}_3\right)_5\right]^{2+}+2 \mathrm{Cl}^{-}$
Thus only one $\mathrm{Cl}^{-}$ion is satisfying both primary and secondary valency of $\mathrm{Co}^{3+}$ in this compound.

Topic: Chemistry, Coordination Compounds, AP EAMCET, AP EAMCET 2005
A complex compound of $\mathrm{Co}^{3+}$ with molecular formula $\mathrm{CoCl}_{\mathrm{x}} \cdot \mathrm{yNH}_3$ gives a total of 3 ions when dissolved in water. How many $\mathrm{Cl}^{-}$ions satisfy both primary and secondary valencies in this complex?

Options:
A 3
B 1
C 4
D zero

The correct answer is:
1

Solution:
As cobalt is present as $\mathrm{Co}^{3+}$ and co-ordination number of cobalt is 6 , the molecular formula of compound should be $\mathrm{CoCl}_3 \cdot \mathrm{yNH}_3$. Now, as it gives a total of 3 ions when dissolved in water, its structural formula must be $\left[\mathrm{CoCl}\left(\mathrm{NH}_3\right)_5\right] \mathrm{Cl}_2$
$\left[\mathrm{CoCl}\left(\mathrm{NH}_3\right)_5\right] \mathrm{Cl}_2 \rightleftharpoons\left[\mathrm{CoCl}\left(\mathrm{NH}_3\right)_5\right]^{2+}+2 \mathrm{Cl}^{-}$
Thus only one $\mathrm{Cl}^{-}$ion is satisfying both primary and secondary valency of $\mathrm{Co}^{3+}$ in this compound.

Topic: Chemistry, Coordination Compounds, TS EAMCET, TS EAMCET 2005
A complex number is given by $z=\frac{1+2 i}{1-(1-i)^{2}}$
What is the modulus of $z$ ?

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

The correct answer is:
1

Solution:
$z=\frac{1+2 i}{1-(1-i)^{2}}=\frac{1+2 i}{1-(1-1-2 i)}$
$=\frac{1+2 i}{1+2 i}=1$
$\therefore|z|=1$

Topic: Mathematics, Complex Number, NDA, NDA 2019 (Phase 1)
A complex number is given by $z=\frac{1+2 i}{1-(1-i)^{2}}$
What is the principal argument of $z$ ?

Options:
A 0
B $\frac{\pi}{4}$
C $\frac{\pi}{2}$
D $\pi$

The correct answer is:
0

Solution:
$\begin{aligned} & z=\frac{1+2 i}{1-(1-i)^{2}}=1=1+0 . i \\ & \therefore \text { Principal argument of } z=\tan \theta \\ &=\frac{0}{1}=0 \\ & \therefore \theta=0^{\circ} \end{aligned}$

Topic: Mathematics, Complex Number, NDA, NDA 2019 (Phase 1)
A compound (X) with molecular formula C3H9N reacts with C6H5SO2Cl to give a solid which is insoluble in alkali (X) is

Options:
A CH3CH2CH2NH2
B
C CH3-NH-CH2CH3
D

The correct answer is:
CH3-NH-CH2CH3

Solution:
Since the compound reacts with benzenesulphonyl chloride to give a product which is insoluble in alkali, it shows there is no H attached to N in the product.
Hence, the compound X is a secondary amine.

Topic: Chemistry, Amines, NEET
A compound $\mathrm{Z}$ with molecular formula $\mathrm{C}_3 \mathrm{H}_9 \mathrm{~N}$ reacts with $\mathrm{C}_6 \mathrm{H}_5 \mathrm{SO}_2 \mathrm{Cl}$ to give a solid, insoluble in alkali. Idenfity $\mathrm{Z}$.

Options:

None

Solution:
Since, $\mathrm{Z}$ on reaction with $\mathrm{C}_6 \mathrm{H}_5 \mathrm{SO}_2 \mathrm{Cl}$ gives a product insoluble in alkali. Therefore, the product does not have a replaceable shows that $\mathrm{Z}$ is a secondary amine.
![rendered image](https://cdn.mathpix.com/snip/images/kY6f1uLbfmvn5fXFwz0vZHkPniN4tEBwFKmyZNHPX5g.rendered.fullsize.png)

Topic: Chemistry, Amines
A condition where the protoxylem lies towards periphery and metaxylem lies towards the center is

Options:
A Exarch
B Endarch
C Mesarch
D Diarch

The correct answer is:
Exarch

Solution:
In exarch, the development of protoxylem occurs from the outermost part of the procambium, i.e., the development is from outside to inside. The protoxylem is found outside of the metaxylem (centripetal pattern). It is found in roots.

Topic: Biology, Anatomy of Flowering Plants, NEET
A conjoint and open vascular bundle will be observed in the transverse section of

Options:
A Monocot root
B Monocot stem
C Dicot root
D Dicot stem

The correct answer is:
Dicot stem

Solution:
Each vascular bundle is conjoint, collateral, open, and with endarch in protoxylem dicot stem

Topic: Biology, Anatomy of Flowering Plants, NEET
A coordination compound is made of $\mathrm{Co}^{3+}$, $\mathrm{NH}_3$, and $\mathrm{Cl}^{-}, 0.1 \mathrm{M}$ solution of this complex when treated with excess silver nitrate gave no precipitate. The formula of the complex and secondary valency of metal are respectively.

Options:
A $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_3 \mathrm{Cl}_3\right], 6$
B $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5 \mathrm{Cl}_3 \mathrm{Cl}_2, 6\right.$
C $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_3 \mathrm{Cl}_3\right], 3$
D $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_4 \mathrm{Cl}_2\right] \mathrm{Cl}, 6$

The correct answer is:
$\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_3 \mathrm{Cl}_3\right], 6$

Solution:
Since, no precipitate of $\mathrm{AgCl}$ is obtained on treating the compound with silver nitrate that means no $\mathrm{Cl}^{-}$ions are available outside the coordination sphere. So, correct formula of coordination compound is $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_3 \mathrm{Cl}_3\right.$. Since, central metal $\mathrm{Co}$ forms six coordination bonds thus the secondary valency is 6 .

Topic: Chemistry, Coordination Compounds, AP EAMCET, AP EAMCET 2018 (23 Apr Shift 2)
A corner point of a feasible region is a point in the region which is the of two boundary lines.

Options:

None

Solution:
Intersection

Topic: Mathematics, Linear Programming
$A_{(\alpha)}=\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha\end{array}\right], A_{(\beta)}=\left[\begin{array}{cc}\cos \beta & -\sin \beta \\ \sin \beta & \cos \beta\end{array}\right]$
Which one of the following is correct?

Options:
A $\mathrm{A}_{(-\alpha)} \mathrm{A}_{(-\beta)}=\mathrm{A}_{(\alpha+\beta)}$
B $\mathrm{A}_{(-\alpha)} \mathrm{A}_{(\beta)}=\mathrm{A}_{(\beta-\alpha)}$
C $A_{(\alpha)}+A_{(-\beta)}=A_{\{-(\beta-\alpha)\}}$
D $A_{(\alpha)}+A_{(\beta)}=A_{(\alpha+\beta)}$

The correct answer is:
$A_{(\alpha)}+A_{(\beta)}=A_{(\alpha+\beta)}$

Solution:
As given $\mathrm{A}(\alpha)=\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha\end{array}\right]$
and $\quad A(\beta)=\left[\begin{array}{cc}\cos \beta & -\sin \beta \\ \sin \beta & \cos \beta\end{array}\right]$
Hence,
$\mathrm{A}(\alpha) \cdot \mathrm{A}(\beta)=\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha\end{array}\right]\left[\begin{array}{cc}\cos \beta & -\sin \beta \\ \sin \beta & \cos \beta\end{array}\right]$
$=\left[\begin{array}{cc}\cos \alpha \cos \beta-\sin \alpha \sin \beta & -\cos \alpha \sin \beta-\sin \alpha \cos \beta \\ \sin \alpha \cos \beta+\cos \alpha \sin \beta & \sin \alpha \sin \beta+\cos \alpha \cos \beta\end{array}\right]$
$=\left|\begin{array}{cc}\cos (\alpha+\beta) & -\sin (\alpha+\beta) \\ \sin (\alpha+\beta) & \cos (\alpha+\beta)\end{array}\right|=\mathbf{A}_{(\alpha+\beta)}$

Topic: Mathematics, Matrices, NDA, NDA 2006 (Phase 1)
A current of 5 A is flowing at 220 V in the primary coil of a transformer. If the voltage produced in the secondary coil is 2200 V and 50% of power is lost, then the current in the secondary coil is

Options:
A 0.25A
B 2.5A
C 0.5A
D 5 A

The correct answer is:
0.25A

Solution:
Given, power output
=12Ã power input (or 50% of input power)
â´IsVs=12IpâVp âµP=VI
Here, Ip=5 A, Vp=220 V
and Vs=2200 V
â´Is=IpâVp2Vs=5Ã2202Ã2200=0.25 A

Topic: Physics, Alternating Current, NEET
A current of $5 \mathrm{~A}$ is flowing at $220 \mathrm{~V}$ in the primary coil of a transformer. If the voltage produced in the secondary coil is $2200 \mathrm{~V}$ and $50 \%$ of power is lost, then the current in the secondary will be

Options:
A $2.5 \mathrm{~A}$
B $5 \mathrm{~A}$
C $0.25 \mathrm{~A}$
D $0.5 \mathrm{~A}$

The correct answer is:
$0.25 \mathrm{~A}$

Solution:
For a transformer,
$\begin{aligned} I_{P}=5 \mathrm{~A}, V_{P} &=220 \mathrm{~V} \\ V_{S}=2200 \mathrm{~V} \end{aligned}$
Since, $50 \%$ power is lost, hence power at secondary side,
$$
\begin{array}{cl}
& P_{0}=50 \% \text { of } P_{i}=\frac{50}{100} \times 1100 \\
\Rightarrow & P_{0}=550 \mathrm{~W} \\
\Rightarrow & V_{S} I_{S}=550 \\
\Rightarrow & 2200 \times I_{S}=550 \\
\Rightarrow & I_{S}=\frac{550}{2200}=\frac{1}{4}=0.25 \mathrm{~A}
\end{array}
$$

Topic: Physics, Alternating Current, COMEDK, COMEDK 2018
A current of $5 \mathrm{~A}$ is flowing at $220 \mathrm{~V}$ in the primary coil of a transformer. If the voltage produced in the secondary coil is $2200 \mathrm{~V}$ and $50 \%$ of power is lost, then the current in secondary will be

Options:
A $2.5 \mathrm{~A}$
B $5 \mathrm{~A}$
C $0.25 \mathrm{~A}$
D $0.5 \mathrm{~A}$

The correct answer is:
$0.25 \mathrm{~A}$

Solution:
$\mathrm{V}_{\mathrm{p}}=220 \mathrm{~V}, \mathrm{~V}_{\mathrm{s}}=2200 \mathrm{~V}, \mathrm{I}_{\mathrm{p}}=5 \mathrm{~A}, \mathrm{I}_{\mathrm{s}}=$ ?
Power loss $=50 \%$
Efficiency of transformer ( $\eta$ ) is defined as the ratio of output power and input power
$$
\eta \%=\frac{\mathrm{P}_{\mathrm{out}}}{\mathrm{P}_{\mathrm{in}}} \times 100=\frac{\mathrm{V}_{\mathrm{s}} \mathrm{I}_{\mathrm{s}}}{\mathrm{V}_{\mathrm{p}} \mathrm{I}_{\mathrm{p}}} \times 100
$$
$$
\begin{aligned}
50 &=\frac{2200 \times \mathrm{I}_{\mathrm{s}}}{220 \times 5} \times 100 \\
\mathrm{I}_{\mathrm{s}} &=0.25 \mathrm{~A}
\end{aligned}
$$

Topic: Physics, Alternating Current, KCET, KCET 2008
A current of 5 A is flowing at 220 V in the primary coil of a transformer. If the voltage produced in the secondary coil is 2200 V and 50% of power is lost, then the current in the secondary coil is

Options:
A 0.25 A
B 2.5 A
C 0.5 A
D 5 A

The correct answer is:
0.25 A

Solution:
Given, power output
=12× power input (or 50% of input power)
∴IsVs=12Ip⋅Vp ∵P=VI
Here, Ip=5 A, Vp=220 V
and Vs=2200 V
∴Is=Ip⋅Vp2Vs=5×2202×2200=0.25 A

Topic: Physics, Alternating Current, NEET
A current of $5 \mathrm{~A}$ is flowing at $220 \mathrm{~V}$ in the primary coil of transformer. If the voltage produced in the secondary coil is $2200 \mathrm{~V}$ and $50 \%$ of power is lost, then the current in secondary will be

Options:
A 2.5 A
B 0.5 A
C 0.25 A
D 5 A

The correct answer is:
0.25 A

Solution:
Let $I$ be the current flowing through secondary coil.
Current on primary (input) coil $i=5 \mathrm{~A}$
Voltage in primary coil $V=220 \mathrm{~V}$
Voltage in secondary coil $V^{\prime}=2200 \mathrm{~V}$
Considering only $50 \%$ power is transmitted:
$\begin{aligned}
& 0.5(\mathrm{Vi})=V^{\prime} I \\
& \therefore 0.5 \times 220 \times 5=2200 \times I \\
& \Rightarrow I=0.25 \mathrm{~A}
\end{aligned}$

Topic: Physics, Alternating Current, MHT CET, MHT CET 2022 (08 Aug Shift 1)
A current of 6 A is flowing at 220 V in the primary coil of a transformer. If the voltage produced in the secondary coil is 1100 V and 40% of power is lost, then the current in the secondary coil will be:

Options:
A 0.28 A
B 0.36 A
C 0.48 A
D 0.42 A

The correct answer is:
0.48 A

Solution:
Given: Voltage in the primary coil Vi=220 V, voltage in the secondary coil VO=1100 V, current in the primary coil, Ii=6 A, Power lost=40%

Let the current in the secondary coil be IO and η be the efficiency of the transformer.

Efficiency η=POUTPIN

Then, POUT=PIN×η

VO×IO=Vi×Ii×η

IO=Vi×Ii×ηVO

IO=220×6×40110×100=0.48 A

Topic: Physics, Dual Nature of Matter, TS EAMCET, TS EAMCET 2020 (09 Sep Shift 1)
A current of 10 A in the primary coil of a circuit is reduced to zero. If the coefficient of mutual inductance is 3 H and emf induced in the secondary coil is 30 kV, time taken for the change of current is

Options:
A 103 s
B 102 s
C 10-3 s
D 10-2 s

The correct answer is:
10-3 s

Solution:
As e=Mdidt
â´ 30Ã103=3Ã10dt,
dt= 3030Ã103=10-3 s

Topic: Physics, Electromagnetic Induction, NEET
A cylinder with a movable piston contains 3 moles of hydrogen at standard temperature and pressure. The walls of the cylinder are made of a heat insulator, and the piston is insulated by having a pile of sand on it. By what factor does the pressure of the gas increase, if the gas is compressed to half its original volume?

Options:

None

Solution:
The process is adiabatic, as no heat in exchanged
$$
\therefore \quad P_2 V_2^\gamma=P_1 V_1^\gamma \quad \therefore \frac{P_2}{P_1}=\left(\frac{V_1}{V_2}\right)^\gamma
$$
Given $V_2=1 / 2 V_1$
$$
\therefore \quad \frac{P_2}{P_1}=\left(\frac{V_1}{V_1 / 2}\right)^{1.4}=2.64
$$

Topic: Physics, Thermodynamics
A cylindrical jar without a lid has to be constructed using a given surface area of a metal sheet. If the capacity of the jar is to be maximum, then the diameter of the jar must be $\mathrm{k}$ times the height of the jar. The value of k is

Options:
A 1
B 2
C 3
D 4

The correct answer is:
2

Solution:
The capacity of the jar will be maximum if height and radius of the cylinder are equal. $\therefore$ height $=$ radius
(1) Given diameter is $\mathrm{k}$ times the height of the jar. Diameter $=2 \times$ radius $=2 \times$ height $\quad \ldots .($ from $(1))$ $\therefore \mathrm{k}=2$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2017 (Phase 2)
A cylindrical tube open at both ends has a fundamental frequency f in air. The tube is dipped vertically in water so that 60 % of the tube is in water. Then the fundamental frequency of air column is

Options:
A f2
B 5f4
C 3f4
D 2f

The correct answer is:
5f4

Solution:
Consider the length of the tube be L

The fundamental frequency for the open organ pipe is,

f=V2 L

The figure below represents the 60 % of tube is dipped
vertically in water.

The length of the closed organ pipe is,

l'=40100 L

l'=2 L5

The fundamental frequency of air column is,

f'=V4l'

Substitute the values in the above equation.

f'=V4 2 L5

f'=54V2 L

f'=54f

Topic: Physics, Waves and Sound, AP EAMCET, AP EAMCET 2019 (21 Apr Shift 2)
A cylindrical tube open at both ends has a fundamental
frequency 'f' in air. The tube is dipped vertically in water so
that half of it is in water. The new fundamental frequency is

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

The correct answer is:
f

Solution:
Initially
$$
\mathrm{f}=\frac{\mathrm{V}}{2 \ell}
$$
Finally
From diagram, we can see
$$
\begin{aligned}
& \frac{\lambda}{4}=\frac{\ell}{2} \\
& \Rightarrow \lambda=21
\end{aligned}
$$
Now, fundamental frequency
$$
=\frac{\mathrm{V}}{\lambda}=\frac{\mathrm{V}}{2 \ell}=\mathrm{f}
$$
So, there will be no change in fundamental frequency.

Topic: Physics, Waves and Sound, TS EAMCET, TS EAMCET 2022 (20 Jul Shift 2)
A cylindrical tube open at both ends has fundamental frequency $f$ in air. When the tube is dipped vertically in water so that one-third part of the tube is in water, the fundamental frequency of air column becomes (neglect end correction)

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

The correct answer is:
$\frac{3 f}{2}$

Solution:
We know. Frequency is ratio of velocity of sound and wavelength:
$f=\frac{v}{\lambda}$
For tube of length , in fundamental mode wavelength is given by $\frac{\lambda}{2}=L$
$\frac{\lambda}{2}=L$.
$\therefore f=\frac{v}{2 L}$
Now, if tube is dipped one third, then for fundamental mode in air column has:
$\begin{aligned} & \frac{\lambda^{\prime}}{2}=\left(\frac{2}{3} L\right) \\ & \therefore f^{\prime}=\frac{3 v}{4 L}=\frac{3}{2} f\end{aligned}$

Topic: Physics, Waves and Sound, MHT CET, MHT CET 2022 (10 Aug Shift 1)
A cylindrical tube open at both ends has fundamental frequency ' $n$ ' in air. The tube is dipped vertically in water so that one-fourth of it is in water. The fundamental frequency of the air column becomes

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

The correct answer is:
$\frac{2 n}{3}$

Solution:
The fundamental frequency of open tube is
$\mathrm{n}_1=\frac{\mathrm{v}}{2 \ell_1}$
When tube is dipped in water, one-fourth of it is in water and three-fourth is in air.
Hence, it becomes a tube closed at one end with length $\ell_2=\frac{3}{4} \ell_1$ The fundamental frequency of closed tube is
$\begin{aligned}
& \mathrm{n}_2=\frac{\mathrm{v}}{4 \ell_2} \\
& \therefore \frac{\mathrm{n}_2}{\mathrm{n}_1}=\frac{1}{4 \ell_2} \times 2 \ell_1=\frac{\ell_1}{2 \ell_2}=\frac{4}{2 \times 3} \quad\left[\because \frac{\ell_1}{\ell_2}=\frac{4}{3}\right] \\
& \therefore \mathrm{n}_2=\frac{2}{3} \mathrm{n}_1=\frac{2}{3} \mathrm{n}
\end{aligned}$

Topic: Physics, Waves and Sound, MHT CET, MHT CET 2021 (24 Sep Shift 2)
A cylindrical tube open at both the ends has a fundamental frequency of $390 \mathrm{~Hz}$ in air. If $\frac{1}{4}$ th of the tube is immersed vertically in water the fundamental frequency of air column is

Options:
A $260 \mathrm{~Hz}$
B $130 \mathrm{~Hz}$
C $390 \mathrm{~Hz}$
D $520 \mathrm{~Hz}$

The correct answer is:
$260 \mathrm{~Hz}$

Solution:
Fundamental frequency of cylindrical open tube
$$
\mathrm{n}=\frac{\mathrm{v}}{2 \mathrm{~L}}=390 \mathrm{~Hz}
$$
When it is immersed in water it becomes a closed tube of length $\frac{3}{4}$ th of the initial length.
Therefore, its fundamental frequency is
$$
\begin{aligned}
\mathrm{n}^{\prime} &=\frac{\mathrm{V}}{4\left(\frac{3}{4} \mathrm{~L}\right)}=\frac{\mathrm{v}}{3 \mathrm{~L}}=\frac{2}{3}\left(\frac{\mathrm{V}}{2 \mathrm{~L}}\right) \\
&=\frac{2}{3} \times 390 \mathrm{~Hz}=260 \mathrm{~Hz}
\end{aligned}
$$

Topic: Physics, Waves and Sound, KCET, KCET 2009
A cylindrical tube open at both the ends has a fundamental frequency of 390 Hz in the air. If 14 of the tube is immersed vertically in water, the fundamental frequency of air column is

Options:
A 260 Hz

B 130 Hz

C 390 Hz

D 520 Hz

The correct answer is:
260 Hz

Solution:
The fundamental frequency of cylindrical open tube
n=v2L=390 Hz
When it is immersed in water it becomes a closed tube of length, 34th of the initial length.
Therefore, its fundamental frequency is
n′=v434L=v3L=23v2L
=23×390 Hz=260 Hz

Topic: Physics, Waves and Sound, NEET
A cylindrical tube, open at both the ends has fundamental frequency n . If one of the ends is closed, the fundamental frequency will become

Options:
A n2
B 2n
C n
D 3n

The correct answer is:
n2

Solution:
For a cylindrical tube
If both the ends is closed then
Fundamental frequency, n=v2l .....(i)
If one of the ends is closed then fundamental frequency, n1=v4l ....(ii)
from equation (i) and (ii)
n1=n2

Topic: Physics, Oscillations, NEET
A die is thrown 100 times, then the standard deviation of getting an even number is

Options:
A 10
B 5
C 20
D 15

The correct answer is:
5

Solution:
Let $x$ denote the number of succeses in 100 Then $x$ follows binomial distribution with $n=1008$ $p=y_{2}, q=\frac{1}{2}$
$\therefore$ variance of $x=n p q=100 \times \frac{1}{2} \times \frac{1}{2}=25$
$\begin{array}{l}\text { Standard } \\ \text { Deviation }\end{array}=\sqrt{\text { vatiance }}=\sqrt{25}$
$=5$

Topic: Mathematics, Statistics, MHT CET, MHT CET 2020 (14 Oct Shift 1)
A die is tossed twice. What is the probability of getting a sum of $10 ?$

Options:
A $\frac{1}{18}$
B $\frac{1}{6}$
C $\frac{1}{12}$
D $\frac{5}{12}$

The correct answer is:
$\frac{1}{12}$

Solution:
Number of possible outcomes $=36$ When sum is 10 , samples are $(5,5),(4,6)$ and $(6,4)$ Required probability $=\frac{3}{36}=\frac{1}{12}$

Topic: Mathematics, Probability, NDA, NDA 2013 (Phase 2)
A fair coin is tossed 100 times. What is the probability of getting tails an odd number of times?

Options:
A $\frac{1}{2}$
B $\frac{3}{8}$
C $\frac{1}{4}$
D $\frac{1}{8}$

The correct answer is:
$\frac{1}{2}$

Solution:
Let $\mathrm{x}$ denote number of tails. Then, $\mathrm{X}$ is a binomial variate with parameters:
$\mathrm{x}=100 \& \mathrm{p}=\frac{1}{2}$
$\therefore \mathrm{p}(\mathrm{x}=\mathrm{r})={ }^{100} \mathrm{C}_{\mathrm{r}}\left(\frac{1}{2}\right)^{100} ;(\mathrm{r}=0,1,2, \ldots \ldots .100)$
Req. probability $=\mathrm{P}(\mathrm{x}=1)+\mathrm{P}(\mathrm{x}=3)+\ldots .+\mathrm{P}(\mathrm{x}=99)$
$=\left[\frac{1}{2}\right]^{100}\left[{ }^{100} \mathrm{C}_{1}+{ }^{100} \mathrm{C}_{3}+\ldots .+{ }^{100} \mathrm{C}_{99}\right]$
$=\frac{1}{2^{100}} 2^{99}=\frac{1}{2}$

Topic: Mathematics, Probability, NDA, NDA 2016 (Phase 1)
A fair coin is tossed and an unbiased dice is rolled together. What is the probability of getting a 2 or 4 or 6 along with head?

Options:
A $\frac{1}{2}$
B $\frac{1}{3}$
C $\frac{1}{4}$
D $\frac{1}{6}$

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

Solution:
Total number of events with dice and coin $=6 \times 2=12$ Number of possibilities $=(2, \mathrm{H}),(4, \mathrm{H})$ and $(6, \mathrm{H})$ i.e., 3
$\therefore$ Probability $=\frac{3}{12}=\frac{1}{4}$

Topic: Mathematics, Probability, NDA, NDA 2019 (Phase 1)
A fair coin is tossed four times. What is the probability that at most three tails occur ?

Options:
A $7 / 9$
B $15 / 16$
C $13 / 16$
D $3 / 4$

The correct answer is:
$15 / 16$

Solution:
$\begin{array}{l}\quad \mathrm{n}(\mathrm{S})=2^{4}=16 \\ \text { and } \mathrm{n}(\mathrm{E})={ }^{4} \mathrm{C}_{0}+{ }^{4} \mathrm{C}_{1}+{ }^{4} \mathrm{C}_{2}+{ }^{4} \mathrm{C}_{3} \\ & =1+4+\frac{4 \times 3}{2 \times 1}+4=1+4+6+4=15 \\ & \therefore \mathrm{P}(\mathrm{E})=\frac{\mathrm{n}(\mathrm{E})}{\mathrm{n}(\mathrm{S})}=\frac{15}{16}\end{array}$

Topic: Mathematics, Probability, NDA, NDA 2014 (Phase 2)
A fair coin is tossed repeatedly. The probability of getting a result in the fifth toss different from those obtained in the first four tosses is:

Options:
A $\frac{1}{2}$
B $\frac{1}{32}$
C $\frac{31}{32}$
D $\frac{1}{16}$

The correct answer is:
$\frac{1}{16}$

Solution:
P(HHHHT or TTTTH) = P (HHHHT) + P (TTTTH)
$=\left(\frac{1}{2}\right)^{5}+\left(\frac{1}{2}\right)^{5}=\frac{1}{16}$

Topic: Mathematics, Probability, NDA, NDA 2013 (Phase 1)
A firm employing 30 workers and paying on an average Rs 500 is combined with another firm employing 20 workers paying on an average Rs 600 . What is the average pay of the workers of the combined firm?

Options:
A $\operatorname{Rs} 540$
B $\mathrm{Rs} 550$
C $\operatorname{Rs} 560$
D $\operatorname{Rs} 580$

The correct answer is:
$\operatorname{Rs} 540$

Solution:
Let $\mathrm{n}$ denote number of workers and $\mathrm{x}$, the pay. Let $\mathrm{n}$ denote number of workers ar
$\mathrm{n}_{1}=30, \mathrm{n}_{2}=20, \mathrm{x}_{1}=₹ 500, \mathrm{x}_{2}=₹ 600$
Combined average $=\frac{\mathrm{n}_{1} \mathrm{x}_{1}+\mathrm{n}_{2} \mathrm{x}_{2}}{\mathrm{n}_{1}+\mathrm{n}_{2}}$
$=\frac{30 \times 500+20 \times 600}{30+20}=\frac{15000+12000}{50}$
$=\frac{27000}{50}=540$
Combined average pay $=₹ 540$

Topic: Mathematics, Statistics, NDA, NDA 2006 (Phase 2)
A five-digit number divisible by 3 is to be formed using the digits $0,1,2,3$ and 4 without repetition of digits. What is the number of ways this can be done?

Options:
A 96
B 48
C 32
D Nonumber can be formed

The correct answer is:
Nonumber can be formed

Solution:
Since sum of digits $=10$ (which is not divisible by 3 ) $\therefore$ No numbers can be formed.

Topic: Mathematics, Permutation Combination, NDA, NDA 2016 (Phase 2)