datavorous/entrance-exam-dataset huggingface dataset:
If $\overrightarrow{\mathrm{a}}=2 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}-\hat{\mathrm{k}}, \overrightarrow{\mathrm{b}}=\hat{\mathrm{i}}+4 \hat{\mathrm{j}}-2 \hat{\mathrm{k}} ; \quad$ then what is
$(\vec{a}+\vec{b}) \times(\vec{a}-\vec{b})$ equal to ?

Options:
A $2(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}})$
B $-2(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}})$
C $\quad(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}})$
D $-(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}})$

The correct answer is:
$-2(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}})$

Solution:
Given vectors are:
$\begin{aligned} & \overrightarrow{\mathrm{a}}=2 \hat{\mathrm{i}}-3 \overline{\mathrm{j}}-\hat{\mathrm{k}} \text { and } \overrightarrow{\mathrm{b}}=\overline{\mathrm{i}}+4 \overline{\mathrm{j}}-2 \overrightarrow{\mathrm{k}} \\ \Rightarrow \quad & \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}=(2 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}-\hat{\mathrm{k}})+(\hat{\mathrm{i}}+4 \hat{\mathrm{j}}-2 \hat{\mathrm{k}}) \\ &=3 \overline{\mathrm{i}}+\overline{\mathrm{j}}-3 \hat{\mathrm{k}} \end{aligned}$
and $\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}=(2 \hat{\mathrm{i}}-3 \overline{\mathrm{j}}-\hat{\mathrm{k}})-(\overline{\mathrm{i}}+4 \hat{\mathrm{j}}-2 \hat{\mathrm{k}})$
$=\overline{\mathrm{i}}-7 \overline{\mathrm{j}}+\hat{\mathrm{k}}$
$\therefore(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}) \times(\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}})=\left|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 3 & 1 & -3 \\ 1 & -7 & 1\end{array}\right|$
$=\hat{\mathrm{i}}\left|\begin{array}{cc}1 & -3 \\ -7 & 1\end{array}\right|-\hat{\mathrm{j}}\left|\begin{array}{cc}3 & -3 \\ 1 & 1\end{array}\right|+\widehat{k}\left|\begin{array}{cc}3 & 1 \\ 1 & -7\end{array}\right|$
$=\hat{\mathrm{i}}(1-21)-\hat{\mathrm{j}}(3+3)+\mathrm{k}(-21-1)$
$=-20 \hat{\mathrm{i}}-6 \widehat{\mathrm{j}}-22 \hat{\mathrm{k}}$
$=-2(10 \hat{\mathrm{i}}+3 \widehat{\mathrm{j}}+11 \widehat{\mathrm{k}})$
Now, $\vec{a} \times \vec{b}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 2 & -3 & -1 \\ 1 & 4 & -2\end{array}\right|$
$=\hat{i}\left|\begin{array}{cc}-3 & -1 \\ 4 & -2\end{array}\right|-\hat{j}\left|\begin{array}{cc}2 & -1 \\ 1 & -2\end{array}\right|+\hat{k}\left|\begin{array}{cc}2 & -3 \\ 1 & 4\end{array}\right|$
$=\widehat{\mathrm{i}}(6+4)-\hat{\mathrm{j}}(-4+1)+\mathrm{k}(8+3)$
$=10 \hat{\mathrm{i}}+3 \widehat{\mathrm{j}}+11 \hat{\mathrm{k}}$
Hence, $(\vec{a}+\vec{b}) \times(\vec{a}-\vec{b})=-2(\vec{a} \times \vec{b})$

Topic: Mathematics, Vector Algebra, NDA, NDA 2008 (Phase 1)
If $\mathrm{A}=\left[\begin{array}{cc}2 \mathrm{x} & 0 \\ \mathrm{x} & \mathrm{x}\end{array}\right]$ and $\mathrm{A}^{-1}=\left[\begin{array}{cc}1 & 0 \\ -1 & 2\end{array}\right]$, then what is the value of $\mathrm{x}$?

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

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

Solution:
Given matrices are :
$\mathrm{A}=\left[\begin{array}{cc}2 \mathrm{x} & 0 \\ \mathrm{x} & \mathrm{x}\end{array}\right]$ and $\mathrm{A}^{-1}=\left[\begin{array}{cc}1 & 0 \\ -1 & 2\end{array}\right]$
$\mathrm{AA}^{-1}=\mathrm{I}$
$\Rightarrow\left[\begin{array}{cc}2 \mathrm{x} & 0 \\ \mathrm{x} & \mathrm{x}\end{array}\right]\left[\begin{array}{cc}1 & 0 \\ -1 & 2\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
$\Rightarrow\left[\begin{array}{cc}2 x & 0 \\ 0 & 2 x\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right] \Rightarrow 2 x=1$

Topic: Mathematics, Matrices, NDA, NDA 2007 (Phase 2)
If $A=\left[\begin{array}{ll}3 & 2 \\ 1 & 4\end{array}\right]$, then what is $A(\operatorname{adj} A)$ equal to?

Options:
A $\left[\begin{array}{cc}0 & 10 \\ 10 & 0\end{array}\right]$
B $\left[\begin{array}{cc}10 & 0 \\ 0 & 10\end{array}\right]$
C $\left[\begin{array}{cc}1 & 10 \\ 10 & 1\end{array}\right]$
D $\left[\begin{array}{cc}10 & 1 \\ 1 & 10\end{array}\right]$

The correct answer is:
$\left[\begin{array}{cc}10 & 0 \\ 0 & 10\end{array}\right]$

Solution:
Let $A=\left[\begin{array}{ll}3 & 2 \\ 1 & 4\end{array}\right]$
We have If $A$ is a square matrix of order $n$ then $A(\operatorname{adj} A)=|A| . \mathrm{I}_{\mathrm{n}}$
Here, $n=2$ $\therefore A(\operatorname{adj} A)=\mathrm{I}_{2}|A|$
$=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]\left|\begin{array}{ll}3 & 2 \\ 1 & 4\end{array}\right|=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right](12-2)=10\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
$=\left[\begin{array}{ll}10 & 0 \\ 0 & 10\end{array}\right]$

Topic: Mathematics, Matrices, NDA, NDA 2009 (Phase 1)
If $\mathrm{A}=\left[\begin{array}{ll}3 & 4 \\ 5 & 6 \\ 7 & 8\end{array}\right]$ and $\mathrm{B}=\left[\begin{array}{lll}3 & 5 & 7 \\ 4 & 6 & 8\end{array}\right]$, then which one of the following is correct?

Options:
A $\mathrm{B}$ is the inverse of $\mathrm{A}$
B $\mathrm{B}$ is the adjoint of $\mathrm{A}$
C $\mathrm{B}$ is the transpose of A
D None of the above

The correct answer is:
$\mathrm{B}$ is the transpose of A

Solution:
We have,
$A=\left[\begin{array}{ll}3 & 4 \\ 5 & 6 \\ 7 & 8\end{array}\right]$
$\mathrm{A}^{\mathrm{T}}=\left[\begin{array}{lll}3 & 5 & 7 \\ 4 & 6 & 8\end{array}\right]=\mathrm{B}$
$\therefore \mathrm{B}$ is the transpose of $\mathrm{A}$

Topic: Mathematics, Matrices, NDA, NDA 2012 (Phase 2)
If $|\overrightarrow{\mathrm{a}}|=3,|\overrightarrow{\mathrm{b}}|=4$ and $|\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}|=5$, then what is the value of
$|\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}|=$?

Options:
A 8
B 6
C $5 \sqrt{2}$
D 5

The correct answer is:
5

Solution:
$\quad|\vec{a}+\vec{b}|^{2}+|\vec{a}-\vec{b}|^{2}=2\left\{|\vec{a}|^{2}+|\vec{b}|^{2}\right\}=2 \times 25=50$
$\Rightarrow \quad|\vec{a}+\vec{b}|^{2}+25=50$
$\Rightarrow \quad|\vec{a}+\vec{b}|^{2}=25$
$\Rightarrow \quad|\vec{a}+\vec{b}|=5$

Topic: Mathematics, Vector Algebra, NDA, NDA 2018 (Phase 2)
If $|\vec{a}|=3,|\vec{b}|=4$ and $|\vec{a}-\vec{b}|=7$, then what is the value of
$|\vec{a}+\vec{b}| ?$

Options:
A 3
B 2
C 1
D 0

The correct answer is:
1

Solution:
Given, $|\vec{a}|=3,|\vec{b}|=4$ and $|\vec{a}-\vec{b}|=7$
Since, $|\vec{a}+\vec{b}|^{2}+|\vec{a}-\vec{b}|^{2}=2\left[|\vec{a}|^{2}+|\vec{b}|^{2}\right]$
$\therefore$ By putting the values of $|\vec{a}|,|\vec{b}|$ and $|\vec{a}-\vec{b}|$ we get
$|\vec{a}+\vec{b}|^{2}+7^{2}=2\left[3^{2}+4^{2}\right]$
$|\vec{a}+\vec{b}|^{2}=50-49 \Rightarrow|\vec{a}+\vec{b}|^{2}=1 \Rightarrow|\vec{a}+\vec{b}|=1$

Topic: Mathematics, Vector Algebra, NDA, NDA 2009 (Phase 2)
If $|\overrightarrow{\mathbf{a}}|=3,|\overrightarrow{\mathbf{b}}|=4$, then for what value of 1 is $(\overrightarrow{\mathbf{a}}+\lambda \overrightarrow{\mathbf{b}})$
perpendicular to $(\overrightarrow{\mathbf{a}}-\lambda \overrightarrow{\mathbf{b}})$ ?

Options:
A $\frac{3}{4}$
B $\frac{4}{3}$
C $\frac{9}{16}$
D $\frac{3}{5}$

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

Solution:
$\because(\overrightarrow{\mathrm{a}}+\lambda \overrightarrow{\mathrm{b}})$ is perpendicular to $(\overrightarrow{\mathrm{a}}-\lambda \overrightarrow{\mathrm{b}})$, their dot
product is zero, so, $(\vec{a}+\lambda \vec{b}) \cdot(\vec{a}-\lambda \vec{b})=0$
$\Rightarrow|\overrightarrow{\mathrm{a}}|^{2}-\lambda^{2}|\overrightarrow{\mathrm{b}}|^{2}-\lambda \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}+\lambda \overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{a}}=0$
$\Rightarrow|\overrightarrow{\mathrm{a}}|^{2}-\lambda^{2}|\overrightarrow{\mathrm{b}}|^{2}=0 \quad(\because \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}=\overrightarrow{\mathrm{b} \cdot \mathrm{a}})$
$\Rightarrow 9-16 \lambda^{2}=0$
$\Rightarrow \lambda=\pm \frac{3}{4} \quad \lambda=\frac{3}{4}$ matches with the given option.

Topic: Mathematics, Vector Algebra, NDA, NDA 2007 (Phase 1)
If $A=\left(\begin{array}{cc}4 & x+2 \\ 2 x-3 & x+1\end{array}\right)$ is symmetric, then what is $x$ equal
to ?

Options:
A 2
B 3
C $-1$
D 5

The correct answer is:
5

Solution:
$\because \mathrm{A}=\mathrm{A}^{\prime}$
$\Rightarrow\left(\begin{array}{cc}4 & x+2 \\ 2 x-3 & x+1\end{array}\right)=\left(\begin{array}{cc}4 & 2 x-3 \\ x+2 & x+1\end{array}\right)$
$\Rightarrow 2 x-3=x+2$
$\therefore x=5$

Topic: Mathematics, Matrices, NDA, NDA 2014 (Phase 2)
If $A=\left[\begin{array}{cc}4 i-6 & 10 i \\ 14 i & 6+4 i\end{array}\right]$ and $k=\frac{1}{2 i}$ where $i=\sqrt{-1}$, then
$\mathrm{kA}$ is equal to

Options:
A $\left[\begin{array}{cc}2+3 \mathrm{i} & 5 \\ 7 & 2-3 \mathrm{i}\end{array}\right]$
B $\left[\begin{array}{cc}2-3 \mathrm{i} & 5 \\ 7 & 2+3 \mathrm{i}\end{array}\right]$
C $\left[\begin{array}{cc}2-3 \mathrm{i} & 7 \\ 5 & 2+3 \mathrm{i}\end{array}\right]$
D $\left[\begin{array}{cc}2+3 \mathrm{i} & 5 \\ 7 & 2+3 \mathrm{i}\end{array}\right]$

The correct answer is:
$\left[\begin{array}{cc}2+3 \mathrm{i} & 5 \\ 7 & 2-3 \mathrm{i}\end{array}\right]$

Solution:
$A=\left[\begin{array}{cc}4 i-6 & 10 i \\ 14 i & 6+4 i\end{array}\right]$ and $K=\frac{1}{2 i}$
$\mathrm{K}=\frac{1}{2 \mathrm{i}}=\frac{\mathrm{i}}{2 \mathrm{i}(\mathrm{i})}=\frac{\mathrm{i}}{2 \mathrm{i}^{2}}=\frac{-\mathrm{i}}{2} .$
$\therefore \mathrm{KA}=\frac{-\mathrm{i}}{2}\left[\begin{array}{cc}4 \mathrm{i}-6 & 10 \mathrm{i} \\ 14 \mathrm{i} & 6+4 \mathrm{i}\end{array}\right]$
$=\left[\begin{array}{cc}(4 \mathrm{i}-6)\left(\frac{-\mathrm{i}}{2}\right) & 10 \mathrm{i}\left(\frac{-\mathrm{i}}{2}\right) \\ 14 \mathrm{i}\left(\frac{-\mathrm{i}}{2}\right) & \left.(6+4 \mathrm{i})\left(\frac{-\mathrm{i}}{2}\right)\right]\end{array}\right.$
$=\left[\begin{array}{cc}-2 \mathrm{i}^{2}+3 \mathrm{i} & -5 \mathrm{i}^{2} \\ -7 \mathrm{i}^{2} & -3 \mathrm{i}-2 \mathrm{i}^{2}\end{array}\right]=\left[\begin{array}{cc}2+3 \mathrm{i} & 5 \\ 7 & 2-3 \mathrm{i}\end{array}\right]$

Topic: Mathematics, Complex Number, NDA, NDA 2017 (Phase 2)
If $A=\{4 n+2 \mid n$ is a natural number $\}$ and $\mathrm{B}=\{3 n\rfloor n$ is a natural number\}, then what is $(A \cap B)$ equal to?

Options:
A $\left\{12 n^{2}+6 n \mid n\right.$ is a natural number}
B $\{24 n-12 \mid n$ is a natural number}
C $\{60 n+30 \mid n$ is a natural number}
D $\{12 n-6 \mid n$ is a natural number}

The correct answer is:
$\{12 n-6 \mid n$ is a natural number}

Solution:
Let $A=\{4 n+2: n \in N\}$
and $B=\{3 n: n \in N\}$
$\Rightarrow \quad A=\{6,10,14,18,22,26,30,34,38,42, \ldots \ldots$
and $\mathrm{B}=\{3,6,9,12,15,18,21,24,27,30, \ldots .\}$
$\mathrm{A} \cap \mathrm{B}=\{6,18,30,42, \ldots . .\}$
$=6+12 n-12=12 n-6 .$
Hence, $\mathrm{A} \cap \mathrm{B}=\{12 n-6: n$ is a natural number $\}$.

Topic: Mathematics, Sets and Relations, NDA, NDA 2011 (Phase 1)
If $|\vec{a}|=7,|\vec{b}|=11$ and $|\vec{a}+\vec{b}|=10 \sqrt{3}$, then $|\vec{a}-\vec{b}|$ is equal
to

Options:
A 40
B 10
C $4 \sqrt{10}$
D $2 \sqrt{10}$

The correct answer is:
$2 \sqrt{10}$

Solution:
$|\overrightarrow{\mathrm{a}}|=7,|\overrightarrow{\mathrm{b}}|=11$ and $|\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}|=10 \sqrt{3}$
Now $|\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}|^{2}=|\overrightarrow{\mathrm{a}}|^{2}+|\overrightarrow{\mathrm{b}}|^{2}+2|\overrightarrow{\mathrm{a}}||\overrightarrow{\mathrm{b}}| \cos \theta$
$\begin{array}{ll}\therefore & (10 \sqrt{3})^{2}=49+121+2 \times 7 \times 7 \cos \theta \\ \therefore & 300=170+154 \cos \theta \\ & \frac{300-170}{154}=\cos \theta \\ \therefore & \frac{65}{77}=\cos 0 \\ & \quad \text { Now, }|\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}|^{2}=|\overrightarrow{\mathrm{a}}|^{2}+|\overrightarrow{\mathrm{b}}|^{2}-2|\overrightarrow{\mathrm{a}}||\overrightarrow{\mathrm{b}}| \cos \theta \\ & =(7)^{2}+(11)^{2}-2 \times 7 \times 11 \times \frac{65}{77} \\ & =49+121-2 \times 65 \\ & \equiv 170-130 \equiv 40 \\ \therefore & |\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}|=\sqrt{40}=2 \sqrt{10}\end{array}$
$\therefore \quad$ Option (d) is correct.

Topic: Mathematics, Vector Algebra, NDA, NDA 2015 (Phase 1)
If $|A|=8$, where $A$ is square matrix of order 3 , then what is $|\operatorname{adj} A|$ equal to?

Options:
A 16
B 24
C 64
D 512

The correct answer is:
64

Solution:
Let $|\mathrm{A}|=8$ and $\mathrm{A}$ is a square matrix of order $3 .$ We know that $|\operatorname{adj} \mathrm{A}|=|\mathrm{A}|^{n-1} . \mathrm{I}$ where
$n^{\prime}$ is the order of the matrix $\mathrm{A}$.
$\therefore|\operatorname{adj} \mathrm{A}|=8^{3-1}=8^{2}=64$

Topic: Mathematics, Matrices, NDA, NDA 2010 (Phase 2)
If $|\vec{a}|=10,|\overrightarrow{\mathrm{b}}|=2$ and $\overrightarrow{\mathrm{a}} \bullet \overrightarrow{\mathrm{b}}=12$, then what is the value of
$|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}| ?$

Options:
A 12
B 16
C 20
D 24

The correct answer is:
16

Solution:
We have
$\cos \theta=\frac{\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}}{|\overrightarrow{\mathrm{a}}||\overrightarrow{\mathrm{b}}|}$ and $\sin \theta=\frac{|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|}{|\overrightarrow{\mathrm{a}}||\overrightarrow{\mathrm{b}}|}$
Given: $|\overrightarrow{\mathrm{a}}|=10,|\overrightarrow{\mathrm{b}}|=2, \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}=12$
$\therefore \cos \theta=\frac{12}{20}$ and $\sin \theta=\frac{|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|}{20}$
Now, By squaring and adding, we get $\sin ^{2} \theta+\cos ^{2} \theta=1$
$\Rightarrow \frac{|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|^{2}}{400}+\frac{144}{400}=1 \Rightarrow|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|^{2}=256$
$\Rightarrow|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|=16$

Topic: Mathematics, Vector Algebra, NDA, NDA 2012 (Phase 1)
If $A=\frac{41 \pi}{12}$, then what is the value of $\frac{1-3 \tan ^{2} A}{3 \tan A-\tan ^{3} A}$ ?

Options:
A $-1$
B 1
C $1 / 3$
D 3

The correct answer is:
1

Solution:
Consider $\frac{1-3 \tan ^{2} A}{3 \tan A-\tan ^{3} A}=\frac{1}{\tan 3 A}$
$\left(\because \tan 3 A=\frac{3 \tan A-\tan ^{3} A}{1-3 \tan ^{2} A}\right)$
$=\frac{1}{\tan \frac{41 \pi}{4}}\left(\because A=\frac{4 \pi}{12}\right)$
$=\frac{1}{\tan \left(10 \pi+\frac{\pi}{4}\right)}=\frac{1}{\tan \frac{\pi}{4}}=1$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2009 (Phase 2)
If $A=\{a, b, c\}$ and $R=\{(a, a),(a, b),(b, c),(b, b),(c, c),(c, a)\}$
is a binary relation of $A$, then which one of the following is
correct?

Options:
A $R$ is reflexive and symmetric, but not transitive
B $R$ is reflexive and transitive, but not symmetric
C $R$ is reflexive, but neither symmetric nor transitive
D $R$ is reflexive, symmetric and transitive

The correct answer is:
$R$ is reflexive, but neither symmetric nor transitive

Solution:
Let $\mathrm{A}=\{a, b, c\}$ and
$\mathrm{R}=\{(a, a),(a, b),(b, c),(b, b),(c, c),(c, a)\}$
Since, $(a, a),(b, b),(c, c) \in k$
$R$ is reflexive relation But $(a, b) \in R$ and $(b, a) \notin R$ $R$ is not symmetric relation. Also, $(a, b),(b, c) \in \bar{k}$
$\Rightarrow(c, a) \in R$ But $(a, c) \in R$
$R$ isnot transitive relations.

Topic: Mathematics, Sets and Relations, NDA, NDA 2009 (Phase 1)
If $A=\{a, b, c, d\}$, then what is the number of proper subsets
of $A$ ?

Options:
A 16
B 15
C 14
D 12

The correct answer is:
15

Solution:
Let $\mathrm{A}=\{a, b, c, d\}$
Let $n=$ no. of elementsin $\mathrm{A}=n(\mathrm{~A})=4$
Now, number of subsets $=2^{n}=2^{4}=16$ As we know that no. of proper subsets $=2^{n}-1$ Number of proper subsets $=2^{4}-1=16-1=15$

Topic: Mathematics, Sets and Relations, NDA, NDA 2010 (Phase 1)
If $\mathrm{A}=\{\mathrm{a}, \mathrm{b}, \mathrm{c}\}$, then what is thenumber of proper subsets A?

Options:
A 5
B 6
C 7
D 8

The correct answer is:
7

Solution:
Let $A=\{a, b, c\} \Rightarrow O(A)=3$
Now, number of proper subsets of $\mathrm{A}=2^{O(A)}-1=2^{3}-1=7$

Topic: Mathematics, Sets and Relations, NDA, NDA 2011 (Phase 2)
If $\mathrm{A}$ and $\mathrm{B}$ are any two events such that
$\mathrm{P}(\overline{\mathrm{A}})=0.4, \mathrm{P}(\overline{\mathrm{B}})=0.3, \mathrm{P}(\mathrm{A} \cup \mathrm{B})=0.9$, then what is the
value of $\mathrm{P}(\overline{\mathrm{A}} \cup \overline{\mathrm{B}})$ equal to ?

Options:
A $0.2$
B $0.5$
C $0.6$
D $0.7$

The correct answer is:
$0.6$

Solution:
We have $\mathrm{P}(\overline{\mathrm{A}} \cup \overline{\mathrm{B}})=\mathrm{P}(\overline{\mathrm{A}})+\mathrm{P}(\overline{\mathrm{B}})-\mathrm{P}(\overline{\mathrm{A}} \cap \overline{\mathrm{B}})$
$=\mathrm{P}(\overline{\mathrm{A}})+\mathrm{P}(\overline{\mathrm{B}})-\mathrm{P}(\overline{\mathrm{A} \cup \mathrm{B}})$
By (De-Morgan's law)
$=\mathrm{P}(\overline{\mathrm{A}})+\mathrm{P}(\overline{\mathrm{B}})-(1-\mathrm{P}(\mathrm{A} \cup \mathrm{B}))=.4+.3-(1-.9)=.6$

Topic: Mathematics, Probability, NDA, NDA 2013 (Phase 1)
If $\mathrm{A}$ and $\mathrm{B}$ are any two events such that $\mathrm{P}(\mathrm{A} \cup \mathrm{B})=\frac{3}{4}$,
$(A \cap B)=\frac{1}{4}$ and $P(\bar{A})=\frac{2}{3}$, where $\bar{A}$ stands for the complementary event of $A$, then what is $\mathrm{P}(\mathrm{B})$ ?

Options:
A $\frac{1}{3}$
B $\frac{2}{3}$
C $\frac{1}{9}$
D $\frac{2}{9}$

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

Solution:
As given, $\mathrm{P}(\mathrm{A} \cup \mathrm{B})=\frac{3}{4}, \mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{1}{4}$
and $\mathrm{P}(\overline{\mathrm{A}})=\frac{2}{3}$
$\mathrm{P}(\overline{\mathrm{A}})=1-\mathrm{P}(\mathrm{A})=\frac{2}{3} \Rightarrow \mathrm{P}(\mathrm{A})=\frac{1}{3}$
We know that
$\mathrm{P}(\mathrm{A} \cup \mathrm{B})=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})$
$\Rightarrow \frac{3}{4}=\frac{1}{3}+\mathrm{P}(\mathrm{B})-\frac{1}{4}$
$\Rightarrow 1-\frac{1}{3}=\mathrm{P}(\mathrm{B}) \Rightarrow \frac{2}{3}=\mathrm{P}(\mathrm{B})$

Topic: Mathematics, Probability, NDA, NDA 2007 (Phase 1)
If $\mathrm{A}$ and $\mathrm{B}$ are any two sets, then what is $\mathrm{A} \cap(\mathrm{A} \cup \mathrm{B})$ equal to?

Options:
A Complement of $\mathrm{A}$
B Complement of $\mathrm{B}$
C $\mathrm{B}$
D $\bar{A}$

The correct answer is:
$\bar{A}$

Solution:
$\mathrm{A} \cap(\mathrm{A} \cup \mathrm{B})=(\mathrm{A} \cap \mathrm{A}) \cup(\mathrm{A} \cap \mathrm{B})=\mathrm{A} \cup(\mathrm{A} \cap \mathrm{B})$
$=\mathrm{A}($ By diagram $)$
Thus, $A \cap(A \cup B)=A$

Topic: Mathematics, Sets and Relations, NDA, NDA 2012 (Phase 1)
If $\mathrm{A}$ and $\mathrm{B}$ are disjoint sets, then $A \cap\left(A^{\prime} \cup B\right)$ is equal to which one of the following?

Options:
A $\phi$
B $\mathrm{A}$
C $A \cup B$
D $A-B$

The correct answer is:
$\phi$

Solution:
$A \cap\left(A^{\prime} \cup B\right)=A \cap A^{\prime}=\phi$

Topic: Mathematics, Sets and Relations, NDA, NDA 2008 (Phase 2)
If $A$ and $B$ are events such that $P(A \cup B)=0.5, P(\bar{B})=0.8$
and $P(A / B)=0.4$, then what is $P(A \cap B)$ equal to?

Options:
A $0.08$
B 0.02
C $0.8$
D $0.2$

The correct answer is:
$0.08$

Solution:
Let $P(A \cup B)=0.5, P(\bar{B})=0.8, P\left(\frac{A}{B}\right)=0.4$
$\quad P(\bar{B})=1-P(B)$
$\Rightarrow 0.8=1-P(B)$
$\Rightarrow P(B)=1-0.8=0.2$
Now, $P\left(\frac{A}{B}\right)=\frac{P(A \cap B)}{P(B)}$
$\Rightarrow P(B) \times P\left(\frac{A}{B}\right)=P(A \cap B)$
$\Rightarrow 0.4 \times 0.2=P(A \cap B)$
$\Rightarrow 0.08=P(A \cap B)$

Topic: Mathematics, Probability, NDA, NDA 2011 (Phase 1)
If $A$ and $B$ are independent events such that
$P(A)=\frac{1}{5}, P(A \cup B)=\frac{7}{10}$, then what is $P(\bar{B})$ equal to ?

Options:
A $\frac{2}{7}$
B $\frac{3}{7}$
C $\frac{3}{8}$
D $\frac{7}{9}$

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

Solution:
As A and B are independent event So, $\mathrm{P}(\mathrm{A} \cap \mathrm{B})=\mathrm{P}(\mathrm{A}) \cdot \mathrm{P}(\mathrm{B})$
$\mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{1}{5} \cdot \mathrm{P}(\mathrm{B})$
Now, $\mathrm{P}(\mathrm{A} \cup \mathrm{B})=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})$
$\frac{7}{10}=\frac{1}{5}+\mathrm{P}(\mathrm{B})-\frac{1}{5} \cdot \mathrm{P}(\mathrm{B})$
$\mathrm{P}(\mathrm{B})\left(1-\frac{1}{5}\right)=\frac{7}{10}-\frac{1}{5}$
$\frac{4}{5} \cdot \mathrm{P}(\mathrm{B})=\frac{1}{2} \Rightarrow \mathrm{P}(\mathrm{B})=\frac{5}{8}$
Now, $\mathrm{P}(\overline{\mathrm{B}})=1-\mathrm{P}(\mathrm{B})=1-\frac{5}{8}=\frac{3}{8}$.

Topic: Mathematics, Probability, NDA, NDA 2013 (Phase 2)
If $\vec{a}$ and $\overrightarrow{\mathrm{b}}$ are position vectors of the points $\mathrm{A}$ and $\mathrm{B}$ respectively, then what is the position vector of a point $\mathrm{C}$ on $\mathrm{AB}$ produced such that $\overrightarrow{\mathrm{AC}}=\overrightarrow{2 \mathrm{AB}}$ ?

Options:
A $\quad 2 \overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}$
B $\quad 2 \overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{a}}$
C $\vec{a}-2 \vec{b}$
D $\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}$

The correct answer is:
$\quad 2 \overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{a}}$

Solution:
Let $\vec{c}$ be the position vector of point $C$ on AB produced. From the laws of vector addition,
$\overrightarrow{\mathrm{OA}}+\overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{OB}}$
$\Rightarrow \overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{OB}}-\overrightarrow{\mathrm{OA}}=\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{a}}$
and similarly in $\Delta \mathrm{AOC}, \overrightarrow{\mathrm{OA}}+\overrightarrow{\mathrm{AC}}=\overrightarrow{\mathrm{OC}}$
$\Rightarrow \overrightarrow{\mathrm{AC}}=\overrightarrow{\mathrm{OC}}-\overrightarrow{\mathrm{OA}}=\overrightarrow{\mathrm{c}}-\overrightarrow{\mathrm{a}}$
As given, $\overrightarrow{\mathrm{AC}}=2 \overrightarrow{\mathrm{AB}}=2 \overrightarrow{\mathrm{b}}-2 \overrightarrow{\mathrm{a}}$
$\Rightarrow 2 \mathrm{~b}-2 \mathrm{a}=\mathrm{c}-\mathrm{a}$
$\Rightarrow \overrightarrow{\mathrm{c}}=2 \overrightarrow{\mathrm{b}}-2 \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{a}}=2 \overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{a}}$

Topic: Mathematics, Vector Algebra, NDA, NDA 2007 (Phase 1)
If $a$ and $b$ are rational and $b$ is not perfect square, then the quadratic equation with rational coefficients whose one root is $3 a+\sqrt{b}$ is

Options:
A $x^{2}-6 a x+9 a^{2}-b=0$
B $3 a x^{2}+x-\sqrt{b}=0$
C $x^{2}+3 a x+\sqrt{b}=0$
D $\sqrt{b} x^{2}+x-3 a=0$

The correct answer is:
$x^{2}-6 a x+9 a^{2}-b=0$

Solution:
Sinceb is not a perfect square, therefore other root will be $3 a-\sqrt{b}$
Required quadratic equation is $x^{2}-[(3 a+\sqrt{b})+(3 a-\sqrt{b})] x+(3 a+\sqrt{b})(3 a-\sqrt{b})=0$
$\Rightarrow x^{2}-6 a x+9 a^{2}-b=0$

Topic: Mathematics, Quadratic Equation, NDA, NDA 2013 (Phase 2)
If $\mathrm{A}$ and $\mathrm{B}$ are square matrices of second order such that $|\mathrm{A}|=-1,|\mathrm{~B}|=3$, then what is $|3 \mathrm{AB}|$ equal to?

Options:
A 3
B $-9$
C $-27$
D None of these

The correct answer is:
$-27$

Solution:
We know that,$|\mathrm{kA}|=\mathrm{k}^{\mathrm{n}}|\mathrm{A}|$, where $\mathrm{n}$ is order of matrix $\mathrm{A}$.
$\therefore|3 \mathrm{AB}|=3^{2}|\mathrm{~A}||\mathrm{B}| \quad(\because|\mathrm{AB}|=|\mathrm{A}||\mathrm{B}|)$
$=9(-1)(3)$
$=-27 \quad(\because|\mathrm{A}|=-1,|\mathrm{~B}|=3)$

Topic: Mathematics, Determinants, NDA, NDA 2014 (Phase 2)
If $A$ and $B$ are subsets of a set $X$, then what is $\{A \cap(X-B)\} \cup B$ equal to?

Options:
A $A \cup B$
B $A \cap B$
C $A$
D $\bar{B}$

The correct answer is:
$A \cup B$

Solution:
Since, $A$ and $B$ are subsets of set $X$ therefore $A \subseteq X$ and $B \subseteq X$
Consider $\{(A \cap(X-B)\} \cup B$
$=\left(A \cap B^{\prime}\right) \cup B$
$=A \cup B$
$\left(\because B^{\prime} \cap B=B\right)$

Topic: Mathematics, Sets and Relations, NDA, NDA 2009 (Phase 1)