datavorous/entrance-exam-dataset huggingface dataset:
What is the modulus of $\frac{\sqrt{2}+\mathrm{i}}{\sqrt{2}-\mathrm{i}}$ where $\mathrm{i}=\sqrt{-1}$ ?

Options:
A 3
B 1/2
C 1
D None of the above

The correct answer is:
1

Solution:
Let $z=\frac{\sqrt{2}+i}{\sqrt{2}-i}=\frac{\sqrt{2}+i}{\sqrt{2}-i} \times \frac{\sqrt{2}+i}{\sqrt{2}+i}$
$\quad=\frac{(\sqrt{2}+i)^{2}}{(\sqrt{2})^{2}-(i)^{2}}=\frac{2 \sqrt{2} i+1}{3}$
$\quad$ Now, $|z|=\sqrt{\left(\frac{2 \sqrt{2}}{3}\right)^{2}+\left(\frac{1}{3}\right)^{2}}=1$

Topic: Mathematics, Complex Number, NDA, NDA 2012 (Phase 2)
What is the molar volume of an ideal gas at standard temperature and pressure (STP), which is defined as 1 atmospheric pressure and 0 °C?

Options:
A 1 liter
B 10 liters
C 22.4 liters
D 100 liters

The correct answer is:
22.4 liters

Solution:
Correct Option is : (C)
22.4 liters

Topic: Chemistry, Kinetic Theory of Gases, NEET
What is the moment about the point $\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}$, of a
force represented by $\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}$, acting through the point
$-2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+\hat{\mathrm{k}}$ ?

Options:
A $\quad 2 \hat{\mathrm{i}}+\hat{\mathrm{j}}+2 \hat{\mathrm{k}}$
B $\hat{\mathrm{i}}-\hat{\mathrm{j}}+3 \hat{\mathrm{k}}$
C $\quad 3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}-\hat{\mathrm{k}}$
D $\quad 3 \hat{i}+\hat{j}-4 \hat{k}$

The correct answer is:
$\quad 3 \hat{i}+\hat{j}-4 \hat{k}$

Solution:
Force, $\overrightarrow{\mathrm{F}}$ is given by $\overrightarrow{\mathrm{F}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}$
$\begin{aligned} \overrightarrow{\mathrm{OA}} &=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}} \text { and } \overrightarrow{\mathrm{OB}}=-2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+\hat{\mathrm{k}} \\ \therefore \quad \overrightarrow{\mathrm{r}}=& \overrightarrow{\mathrm{AB}}=-2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+\hat{\mathrm{k}}-\hat{\mathrm{i}}-2 \hat{\mathrm{j}}-3 \hat{\mathrm{k}} \\ &=-3 \hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}} \end{aligned}$
Moment $\overrightarrow{\mathrm{M}}$ about the point $\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}$
$\overrightarrow{\mathrm{M}}=\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{F}}=(-3 \hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}}) \times(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})$
$=\left|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ -3 & 1 & -2 \\ 1 & 1 & 1\end{array}\right|=\hat{\mathrm{i}}(1+2)-\hat{\mathrm{j}}(-3+2)+\hat{\mathrm{k}}(-3-1)$
$=3 \hat{\mathrm{i}}+\hat{\mathrm{j}}-4 \hat{\mathrm{k}}$

Topic: Mathematics, Vector Algebra, NDA, NDA 2006 (Phase 1)
What is the moment about the point $\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-\hat{\mathrm{k}}$ of a force
represented by $3 \hat{i}+\hat{k}$ acting through the point $2 \hat{i}-\hat{j}+3 \hat{k}$ ?

Options:
A $\quad-3 \hat{i}+11 \hat{j}+9 \hat{k}$
B $3 \hat{i}+2 \hat{j}+9 \hat{k}$
C $\quad 3 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+9 \hat{\mathrm{k}}$
D $\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}$

The correct answer is:
$\quad-3 \hat{i}+11 \hat{j}+9 \hat{k}$

Solution:
$\begin{aligned} \overrightarrow{\mathrm{r}} &=(2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+3 \hat{\mathrm{k}})-(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-\hat{\mathrm{k}}) \\ &=(\hat{\mathrm{i}}-3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}) \\ & \text { Moment }(\tau)=\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{f}} \\ &=(\hat{\mathrm{i}}-3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}) \times(3 \hat{\mathrm{i}}+\hat{\mathrm{k}}) \\ &=-3 \hat{\mathrm{i}}+11 \hat{\mathrm{j}}+9 \hat{\mathrm{k}} \end{aligned}$

Topic: Mathematics, Vector Algebra, NDA, NDA 2018 (Phase 1)
What is the most abundant chemical in living organisms?

Options:
A Carbon
B Hydrogen
C Oxygen
D Water

The correct answer is:
Water

Solution:
The most abundant chemical in living organisms is water, not carbon, hydrogen, or oxygen.

Topic: Biology, Biomolecules (B), NEET
What is the most abundant element in the Earth's crust by mass?

Options:
A Oxygen
B Silicon
C Aluminum
D Boron

The correct answer is:
Oxygen

Solution:
Oxygen is the most abundant element in the Earth's crust by mass, with 45.5%.

Topic: Chemistry, p Block Elements (Group 13 & 14), NEET
What is the most abundant element in the universe?

Options:
A Oxygen
B Carbon
C Hydrogen
D Helium

The correct answer is:
Hydrogen

Solution:
The paragraph states that dihydrogen is the most abundant element in the universe.

Topic: Chemistry, Hydrogen, NEET
What is the most abundant isotope of nitrogen?

Options:
A 14N
B 15N
C 16N
D 17N

The correct answer is:
14N

Solution:
The paragraph states that nitrogen has two stable isotopes: 14N and 15N. 14N is the most abundant isotope of nitrogen, accounting for approximately 99.63% of naturally occurring nitrogen.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), NEET
What is the most abundant organic substance in the plant kingdom?

Options:
A Starch
B Cellulose
C Lignin
D Hemicellulose

The correct answer is:
Cellulose

Solution:
Cellulose is the most abundant organic substance in the plant kingdom.

Topic: Chemistry, Biomolecules, NEET
What is the most abundant plant pigment in the world?

Options:
A Chlorophyll b
B Xanthophylls
C Carotenoids
D Chlorophyll a

The correct answer is:
Chlorophyll a

Solution:
The ability of chlorophyll a pigment to absorb lights of different wavelengths shows that chlorophyll a is the most abundant pigment.

Topic: Biology, Photosynthesis in Higher Plants, NEET
What is the most abundant protein in animal world?

Options:
A Collagen
B Ribulose bisphosphate Carboxylase-Oxygenase (RuBisCO)
C Enzymes
D Hormones

The correct answer is:
Collagen

Solution:
Collagen is the most abundant protein in animal world.

Topic: Biology, Biomolecules (B), NEET
What is the most abundant protein in the animal world?

Options:
A Enzymes
B Hormones
C Collagen
D RuBisCO

The correct answer is:
Collagen

Solution:
Collagen is the most abundant protein in the animal world. RuBisCO is the most abundant protein in the whole of the biosphere.

Topic: Biology, Biomolecules (B), NEET
What is the most probable number of successes in 10 trials with probability of success $2 / 3$ ?

Options:
A 10
B 7
C 5
D 4

The correct answer is:
7

Solution:
This is the question of Binomial Distribution. Number of $r$ success in $n$ trial is $P_{r}={ }^{n} C_{r} p^{r} q^{n-r}$
Where $\mathrm{p}=$ prob of success $\mathrm{q}=$ prob. of failure $=1-\mathrm{p}$
Given : $n=10, p=\frac{2}{3}, q=\frac{1}{3}$
Fo $\quad r=10, P_{10}={ }^{10} C_{10} \quad\left(\frac{2}{3}\right)^{10}\left(\frac{1}{3}\right)^{0}=\frac{2^{10}}{3^{10}}$
For $\quad r=7, P_{7}=120 \cdot \frac{2^{7}}{3^{10}}$
For $\mathrm{r}=5, \mathrm{P}_{5}=252 \times \frac{2^{5}}{3^{10}}$
For $\quad r=4, P_{4}=210 \cdot \frac{2^{4}}{3^{10}}$
It is maximum for $\mathrm{r}=7$

Topic: Mathematics, Probability, NDA, NDA 2012 (Phase 2)
What is the $n^{\text {th }}$ term of the sequence $25,-125,625,-3125, \ldots ?$

Options:
A $(-5)^{2 \mathrm{n}-1}$
B $(-1)^{2 n} 5^{n+1}$
C $(-1)^{2 \mathrm{n}-1} 5^{\mathrm{n}+1}$
D $(-1)^{\mathrm{n}-1} 5^{\mathrm{n}+1}$

The correct answer is:
$(-1)^{\mathrm{n}-1} 5^{\mathrm{n}+1}$

Solution:
Given series $25,-125,625,-3125 \ldots \ldots$ is geometric progression. $a=t_{1}=25, t_{2}=-125$
$\mathrm{r}=\frac{\mathrm{t}_{2}}{\mathrm{t}_{1}}=\frac{-125}{25}=-5$
$\mathrm{n}^{\text {th }}$ term $\left(\mathrm{t}_{\mathrm{n}}\right)=\operatorname{ar}^{\mathrm{n}-1}=(25)(-5)^{\mathrm{n}-1}$
$=(5)^{2}(-1)^{\mathrm{l}-1}(5)^{\mathrm{n}-1}=(-1)^{\mathrm{n}-1}(5)^{2+\mathrm{n}-1}$
$=(-1)^{\mathrm{n}-1} 5^{\mathrm{n}+1}$

Topic: Mathematics, Sequences and Series, NDA, NDA 2019 (Phase 1)
What is the name of the process by which two male gametes fuse with the egg cell and diploid secondary nucleus, respectively?

Options:
A Syngamy
B Double fertilization
C Pollination
D Germination

The correct answer is:
Double fertilization

Solution:
Double fertilization is the unique process in angiosperms where two male gametes fuse with the egg cell and diploid secondary nucleus, respectively, resulting in the formation of a zygote and the triploid primary endosperm nucleus.

Topic: Biology, Plant Kingdom, NEET
What is the name of the small fragment that appears on a chromosome due to non-staining secondary constrictions at a constant location?

Options:
A Chromosome band
B Chromosome fragment
C Chromosome satellite
D Chromosome loop

The correct answer is:
Chromosome satellite

Solution:
A satellite is a small fragment that appears on a chromosome due to non-staining secondary constrictions at a constant location.

Topic: Biology, Cell - The Unit of Life, NEET
What is the nth term of the sequence $1,5,9,13,17, \ldots \ldots \ldots$ ?

Options:
A $\quad 2 \mathrm{n}-1$
B $2 \mathrm{n}+1$
C $4 \mathrm{n}-3$
D None of the above

The correct answer is:
$4 \mathrm{n}-3$

Solution:
Given sequence is $1,5,9,13,17, \ldots$
Which is an A.P. Here $\mathrm{a}=1, \mathrm{~d}=4$
$\therefore$ nth term $=a_{n}=a+(n-1) d=1+(n-1) 4$
$=1+4 n-4=4 n-3$

Topic: Mathematics, Sequences and Series, NDA, NDA 2012 (Phase 1)
What is the number of arbitrary constants in the particular solution of differential equation of third order?

Options:
A 0
B 1
C 2
D 3

The correct answer is:
3

Solution:
Particular solution of $\mathrm{D} . \varepsilon$. of third order have three arbitary constant.

Topic: Mathematics, Differential Equations, NDA, NDA 2014 (Phase 1)
What is the number of diagonals which can be drawn by joining the angular points of a polygon of 100 sides?

Options:
A 4850
B 4950
C 5000
D 10000

The correct answer is:
4850

Solution:
We have, $\mathrm{n}=100$
Number of diagonals $=\frac{\mathrm{n}^{2}-3 \mathrm{n}}{2}$
$=\frac{100^{2}-300}{2}=\frac{9700}{2}=4850$

Topic: Mathematics, Sequences and Series, NDA, NDA 2012 (Phase 2)
What is the number of different messages that can be represented by three 0's and two 1's?

Options:
A 10
B 9
C 8
D 7

The correct answer is:
10

Solution:
Number of different messages that can be represented by three 0 's and two 1 's is 10 .
Option (a) is correct.

Topic: Mathematics, Permutation Combination, NDA, NDA 2016 (Phase 1)
What is the number of digits in the numeral form of 8172 (Given $\left.\log _{10}{ }^{2}=0.3010\right)$

Options:
A 51
B 16
C 15
D 14

The correct answer is:
16

Solution:
$x=2^{5}$
Taking log on both sides of above equation, we get $\log x=51 \log 2$
$=51 \times 0.3010=15.381$
Number ofdigits in $8^{17}=15+1=16$

Topic: Mathematics, Basic of Mathematics, NDA, NDA 2007 (Phase 1)
What is the number of distinct solutions of the equation $\mathrm{z}^{2}+|\mathrm{z}|=0$ (where z is a complex number)?

Options:
A One
B Two
C Three
D Five

The correct answer is:
Three

Solution:
Let $z=x+i y$ $z^{2}+|z|=0$
$\Rightarrow x^{2}-y^{2}+2 i x y+\sqrt{x^{2}+y^{2}}=0$
$\Rightarrow x^{2}-y^{2}+\sqrt{x^{2}+y^{2}}+i 2 x y=0+i 0$
$\Rightarrow x^{2}-y^{2}+\sqrt{x^{2}+y^{2}}=0$
and $2 x y=0 \Rightarrow x y=0 \Rightarrow x=0$ or $y=0$
Now: For $y=0$ in eq. (1) we get :
$x^{2}+\sqrt{x^{2}}=0$
$\Rightarrow x^{2}+|x|=0$
Clearly $x^{2}+|x|$ will always be greater than 0 for all $x>0 .$
Let, $x \leq 0$
$x^{2}+|x|=0$
$\Rightarrow x^{2}-x=0 \Rightarrow x(x-1)=0$
$\Rightarrow x=0$ or $(x-1)=0$
$\Rightarrow x=0 \quad(\because x \leq 0)$
$\therefore \mathbf{z = 0}$
For $x=0$ in eq. (1) we get,
$-y^{2}+\sqrt{y^{2}}=0$
$-y^{2}+|y|=0$
If $y>0$, then $-y^{2}+|y|=0$
$\Rightarrow-y^{2}+y=0$
$\Rightarrow y=0, y=1$
$\Rightarrow y=1$
$(\because y>0)$
$\therefore \mathrm{z}=\mathbf{i}$
If $y < 0$, then $-y^{2}+|y|=0$
$\Rightarrow-y^{2}-y=0$
$\Rightarrow y=0, y=-1$
$\Rightarrow y=-1$
$(\because y < 0)$
$\therefore \mathrm{z}=-\mathbf{i}$
$\therefore \quad$ There are only 3 distinct solutions.

Topic: Mathematics, Complex Number, NDA, NDA 2016 (Phase 2)
What is the number of five-digit numbers formed with 0,1 , 2, 3, 4 without any repetition of digits?

Options:
A 24
B 48
C 96
D 120

The correct answer is:
96

Solution:
Tomakea 5 digit number, 0 can not come in the bagining. So, it can be filled in 4 ways. Rest of the places can be filled in $4 !$ ways. So total number of digit formed $=4 \times 4 !$ $=4 \times 24=96$

Topic: Mathematics, Permutation Combination, NDA, NDA 2008 (Phase 1)
What is the number of four-digit decimal numbers $( < 1)$ in which no digit is repeated?

Options:
A 3024
B 4536
C 5040
D None of the above

The correct answer is:
4536

Solution:
Let the given 4 digit decimal number is
Places after decimal can be filled in the following ways:
Total number of ways $=7 \times 8 \times 9 \times 9=4536$
#

Topic: Mathematics, Permutation Combination, NDA, NDA 2016 (Phase 1)
What is the number of moles of secondary carbon atoms in $\mathrm{n}$ mole isopentane?

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

The correct answer is:
$\mathrm{n}$

Solution:
1 molecule of isopentane has 1 secondary carbon atom.
$\therefore \quad 1 \mathrm{~mol}$ isopentane $=1 \mathrm{~mol}$ secondary carbon atoms
$\therefore \quad \mathrm{n}$ mol isopentane $=\mathrm{n}$ mol secondary carbon atoms

Topic: Chemistry, Some Basic Concepts of Chemistry, MHT CET, MHT CET 2023 (11 May Shift 2)
What is the number of natural numbers less than or equal to 1000 which are neither divisible by 10 nor 15 nor $25^{\circ}$

Options:
A 860
B 854
C 840
D 82

The correct answer is:
854

Solution:
Let $\mathrm{A}, \mathrm{B} \& \mathrm{C}$ be the sets of numbers divisible by 10,15 \& 25 respectively No. divisible by $10=100=\mathrm{n}(\mathrm{A})$ No. divisible by $15=66=\mathrm{n}(\mathrm{B})$ No. divisible by $25=40=\mathrm{n}(\mathrm{C})$
No. divisible by $(10 \& 15)=33=\mathrm{n}(\mathrm{A} \cap \mathrm{B})$
No. divisible by $(15 \& 25)=13=\mathrm{n}(\mathrm{B} \cap \mathrm{C})$
No. divisible by $(25 \& 10)=20=\mathrm{n}(\mathrm{A} \cap \mathrm{C})$
No. divisible by $(10,15 \& 25)=6=n\left(A \cap B_{\cap} C\right)$ No. divisible by 10,15 and $25=n(A \cup B \cup C)$ $=100+66+40-33-13-20+6=146$
Thus, no. which are neither divisible by 10 nor 15 nor $25=1000-146=854 .$

Topic: Mathematics, Sets and Relations, NDA, NDA 2016 (Phase 1)
What is the number of non-zero terms in the expansion of $(1+2 \sqrt{3} \mathrm{x})^{11}+(1-2 \sqrt{3} \mathrm{x})^{11}$ (after simplification)?

Options:
A 4
B 5
C 6
D 11

The correct answer is:
6

Solution:
We know, in the expansion of $(\mathrm{x}+\mathrm{y})^{\mathrm{n}}+(\mathrm{x}-\mathrm{y})^{\mathrm{n}}$, of
$\mathrm{n}=$ even, then number of non zero terms is $\frac{\mathrm{n}}{2}+1$
$\mathrm{n}=$ odd, then number of non zero terms in $\frac{\mathrm{n}+1}{2}$.
Here, $\mathrm{n}=11$ which is odd.
number of non zero terms $=\frac{11+1}{2}=6$.

Topic: Mathematics, Binomial Theorem, NDA, NDA 2018 (Phase 1)
What is the number of odd integers between 1000 and 9999 with no digit repeated?

Options:
A 2100
B 2120
C 2240
D 3331

The correct answer is:
2240

Solution:
Case I When unit digit can be $1,3,5$ or 7 \& digit at thousand's place can be $1,2,3,4,5,6,7$ or 8 . No. of ways digits can be filled are:
Total no's $=7 \times 8 \times 7 \times 4=1568$.
Case II When unit digit can be 9 \& digit at thousand's place can be $1,2,3,4,5,6,7$ or 8 . No. of ways digits can be filled are:
Total no's $=8 \times 8 \times 7 \times 1=448$.
Case III When unit digit can be $1,3,5$ or 7 \& digit at thousand's place can be $9 .$ No. of ways digits can be filled are:
Total no's $=1 \times 8 \times 7 \times 4=224$.
$\therefore \quad$ Number of odd digits between 1000 \& 9999 with no digit repeated $=1568+448+224$ $=2240 .$

Topic: Mathematics, Permutation Combination, NDA, NDA 2016 (Phase 2)
What is the number of outcomes when a coin is tossed and then a die is rolled only in case a head is shown on the coin?

Options:
A 6
B 7
C 8
D None of these

The correct answer is:
7

Solution:
Possible outcomes are (Head, 1), (Head, 2), (Head, 3), $($ Head, 4$),($ Head, 5$),($ Head, 6$)$, Tail

Topic: Mathematics, Probability, NDA, NDA 2013 (Phase 2)
What is the number of points of local maxima of the function $\mathrm{f}(\mathrm{x}) ?$
Consider the function $\mathrm{f}(\mathrm{x})=(\mathrm{x}-1)^{2}(\mathrm{x}+1)(\mathrm{x}-2)^{3}$

Options:
A None
B One
C Two
D Three

The correct answer is:
One

Solution:
Local maxima is $[\mathrm{x}=1$ ]

Topic: Mathematics, Application of Derivatives, NDA, NDA 2016 (Phase 1)
What is the number of points of local minima of the function $\mathrm{f}(\mathrm{x}) ?$
Consider the function $\mathrm{f}(\mathrm{x})=(\mathrm{x}-1)^{2}(\mathrm{x}+1)(\mathrm{x}-2)^{3}$

Options:
A None
B One
C Two
D Three

The correct answer is:
Two

Solution:
$f(x)=(x-1)^{2}(x+1)(x-2)^{3}$
$f^{\prime}(x)=2(x-1)(x+1)(x-2)^{3}+(x-1)^{2}(x-2)^{3}$
$+(x-1)^{2}(x+1) 3(x-2)^{2}$
$$
\begin{aligned}(x-1)(x-2)^{2}[2(x+1)(x-2)+(x-1)(x-2)\\+3(x-1)(x+1)] \end{aligned}
$$
$f^{\prime}(x)=(x-1)(x-2)^{2}\left[2 x^{2}-2 x-4+x^{2}-3 x+2\right.$
$\left.+3 x^{2}-3\right]$
$=(x-1)(x-2)^{2}\left[6 x^{2}-5 x-5\right]$
For maxima and minima $f^{\prime}(x)=0$
$(x-1)(x-2)^{2}\left[6 x^{2}-5 x-5\right]=0$
$x=1,2,2, \frac{5 \pm \sqrt{145}}{12}$ The change in signs of $\mathrm{f}^{\prime}(\mathrm{x})$ for
diffrent values of $x$ is shown:
$\therefore$ Local Minima are
$x=\frac{5-\sqrt{145}}{12} \& x=\frac{5+\sqrt{145}}{12}$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2016 (Phase 1)
What is the number of primary amine molecules among the following compounds?

Options:
A 1
B 2
C 3
D 4

The correct answer is:
4

Solution:
primary alcohol group $=1$
secondary alcohol group $=4$
Ans. 1, 4

Topic: Chemistry, General Organic Chemistry, MHT CET, MHT CET 2022 (07 Aug Shift 2)
What is the number of primary and secondary alcoholic groups respectively present in glucose?

Options:
A 2,3
B 3,2
C 4,2
D 1,4

The correct answer is:
1,4

Solution:

Topic: Chemistry, Biomolecules, MHT CET, MHT CET 2022 (07 Aug Shift 2)
What is the number of proper subsets of a given finite set with $n$ elements?

Options:
A $2 n-1$
B $2 n-2$
C $2^{n}-1$
D $2^{n}-2$

The correct answer is:
$2^{n}-1$

Solution:
We know that Total number of proper subsets of finite set with $n$ elements is $2^{n}-1$

Topic: Mathematics, Sets and Relations, NDA, NDA 2009 (Phase 1)
What is the number of real solutions of
$\left|x^{2}-x-6\right|=x+2 ?$

Options:
A 4
B 3
C 2
D 1

The correct answer is:
3

Solution:
Theequation is $\left|x^{2}-x-6\right|=x+2$ for $x \geq 0$
$x^{2}-x-6=x+2$
$\Rightarrow x^{2}-2 x-8=0$
$\Rightarrow(x-4)(x+2)=0$
$\Rightarrow \mathrm{x}=4,-2$
and for $x < 0$ $-x^{2}+x+6=x+2$
$\Rightarrow x^{2}=4 \Rightarrow x=\pm 2$
Thus, the number of real solutionsare, $-2,2$ and $4 .$
So, numbers of real solution of the equation $\left|x^{2}-x-6\right|=x+2$ are 3

Topic: Mathematics, Quadratic Equation, NDA, NDA 2006 (Phase 1)
What is the number of signals that can be sent by 6 flags of different colours taking one or more at a time?

Options:
A 21
B 63
C 720
D 1956

The correct answer is:
63

Solution:
Required number of ways
${ }^{6} C_{0}+6_{C_{1}}+6_{C_{2}}+\ldots . .+{ }^{6} C_{6}-1=2^{6}-1=64-1=63$

Topic: Mathematics, Permutation Combination, NDA, NDA 2010 (Phase 2)
What is the number of terms in the expansion of $\left[(2 \mathrm{x}-3 \mathrm{y})^{2}\right.$ $\left.(2 \mathrm{x}+3 \mathrm{y})^{2}\right]^{2} ?$

Options:
A 4
B 5
C 8
D 16

The correct answer is:
5

Solution:
$\left[(2 x-3 y)^{2}(2 x+3 y)^{2}\right]^{2}$
$=\left[\left(4 x^{2}-9 y^{2}\right)^{2}\right]^{2}=\left(4 x^{2}-9 y^{2}\right)^{4}$
$\therefore$ Number of terms $=4+1=5$

Topic: Mathematics, Binomial Theorem, NDA, NDA 2019 (Phase 1)
What is the number of terms in the expansion of $(a+b+c)^{n}$, $n \in N$?

Options:
A $n+1$
B $n+2$
C $n(n+1)$
D $\frac{(n+1)(n+2)}{2}$

The correct answer is:
$\frac{(n+1)(n+2)}{2}$

Solution:
Required number of terms in $(a+b+c)^{\mathrm{n}}$
$={ }^{n+2} \mathrm{C}_{2} \frac{(n+2) !}{2 ! n !}=\frac{(n+1)(n+2)}{2}$

Topic: Mathematics, Binomial Theorem, NDA, NDA 2010 (Phase 2)
What is the number of three-digit odd numbers formed by using the digits $1,2,3,4,5,6$ if repetition of digits is allowed?

Options:
A 60
B 108
C 120
D 216

The correct answer is:
108

Solution:
Total no. of digits $=6$ To form a odd numbers we have only 3 choice for the unit digits. Now, Extreme left place can be filled in 6 ways the middle place can be filled in 6 ways. $\therefore$ Required number of numbers $=6 \times 6 \times 3=108$

Topic: Mathematics, Permutation Combination, NDA, NDA 2010 (Phase 1)
What is the number of triangles that can be formed by choosing the vertices from a set of 12 points in a plane, seven of which lie on the same straight line?

Options:
A 185
B 175
C 115
D 105

The correct answer is:
185

Solution:
To form a triangle, we need 3 points. 12 points are given.
So, ${ }^{12} \mathrm{C}_{3}$ triangles can be formed. But, given that 7 points are on a straight line. selecting 3 points from this set will not form a triangle.
So, number of triangles formed ${ }^{12} \mathrm{C}_{3}-{ }^{7} \mathrm{C}_{3}$
$=\frac{12 !}{3 ! 9 !}-\frac{7 !}{3 ! 4 !}$
$=\frac{12 \times 11 \times 10}{3 \times 2 \times 1}-\frac{7 \times 6 \times 5}{3 \times 2 \times 1}=220-35=185$

Topic: Mathematics, Permutation Combination, NDA, NDA 2018 (Phase 1)
What is the number of vectors of length 5 unit perpendicular
to the vectors $\vec{a}=(1,1,0)$ and $\vec{b}=(0,1,1) ?$

Options:
A 1
B 2
C 3
D 4

The correct answer is:
2

Solution:
As given, vectors are: $\overrightarrow{\mathrm{a}}=\widehat{\mathrm{i}}+\overline{\mathrm{j}}$ and $\widehat{\mathrm{b}}=\widehat{\mathrm{j}}+\widehat{\mathrm{k}}$
So, $\vec{a} \times \vec{b}$ is perpendicular to $\vec{a}$ and $\vec{b}$
$\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}=\left|\begin{array}{ccc}\mathrm{i} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 1 & 1 & 0 \\ 0 & 1 & 1\end{array}\right|$
$=\widehat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}$
and $|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|=\sqrt{1+1+1}=\sqrt{3}$
Unit vector perpendicular to $\overrightarrow{\mathrm{a}}$ and $\overrightarrow{\mathrm{b}}$
$=\pm \frac{(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}})}{|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|}=\pm \frac{\widehat{\mathrm{i}}-\widehat{\mathrm{j}}+\widehat{\mathrm{k}}}{\sqrt{3}}$
Thus, the number of vectors perpendicular to the
vectors a and $\overrightarrow{\mathrm{b}}$ is 2 . This is true for vectors of any length. So, it is true for vector of length 5 unit.

Topic: Mathematics, Vector Algebra, NDA, NDA 2008 (Phase 1)
What is the number of ways in which 3 holiday travel tickets are to be given to 10 employees of an organization, if each employee is eligible for any one or more of the tickets?

Options:
A 60
B 120
C 500
D 1000

The correct answer is:
1000

Solution:
No. of ways in which 3 holiday travel tickets are to be given to 10 employees $=10^{3}=1000$

Topic: Mathematics, Permutation Combination, NDA, NDA 2016 (Phase 1)
What is the number of ways in which one can post 5 letters in 7 letters boxes?

Options:
A $7^{5}$
B $3^{5}$
C $5^{7}$
D 2520

The correct answer is:
$7^{5}$

Solution:
First letter can be put any 7 letters boxes $=7$ ways Similarly, 2nd, $3 \mathrm{rd}, 4$ th and 5 th letters be put in 7 ways each, respectively $\therefore 7 \times 7 \times 7 \times 7 \times 7=7^{5}$

Topic: Mathematics, Permutation Combination, NDA, NDA 2014 (Phase 2)
What is the number of ways of arranging the letters of the word 'BANANA' so that no two N's appear together?

Options:
A 40
B 60
C 80
D 100

The correct answer is:
40

Solution:
Total no. of letters in BANANA=6 No. of repeated letter $\mathrm{N}=2$ No. of repeated letter $\mathrm{A}=2$ Therefore Number of ways that can be formed by using the words
'BANANA' $=\frac{6 !}{3 ! 2 !}=\frac{6 \times 5 \times 4 \times 3 !}{3 ! \times 2 !}=60$
Number of ways in which two $\mathrm{N}$ comes together
$=\frac{5 !}{3 !}=20$
$\therefore$ Required number of ways $=60-20=40$

Topic: Mathematics, Permutation Combination, NDA, NDA 2010 (Phase 1)
What is the number of ways that 4 boys and 3 girls can be seated so that boys and girls alternate?

Options:
A 12
B 72
C 120
D 144

The correct answer is:
144

Solution:
BG BG B G B Required no. of ways $=4 ! \times 3 !=144$

Topic: Mathematics, Permutation Combination, NDA, NDA 2012 (Phase 1)