datavorous/entrance-exam-dataset huggingface dataset:
A and B are two events such that $\bar{A}$ and $\bar{B}$ are mutually exclusive. If $\mathrm{P}(\mathrm{A})=0.5$ and $\mathrm{P}(\mathrm{B})=0.6$, then what is the value of $\mathrm{P}(\mathrm{A} \mid \mathrm{B})$ ?

Options:
A $\frac{1}{5}$
B $\frac{1}{6}$
C $\frac{2}{5}$
D $\frac{1}{3}$

The correct answer is:
$\frac{1}{6}$

Solution:
$\bar{A}$ and $\bar{B}$ are mutually exclusive $\therefore P(\bar{A} \cap \bar{B})=0$
Given, $\mathrm{P}(\mathrm{A})=0.5 \Rightarrow P(\bar{A})=1-0.5=0.5$
$\mathrm{P}(\mathrm{B})=0.6 \Rightarrow P(\bar{B})=1-0.6=0.4$
$P(A \mid B)=\frac{P(A \cap B)}{P(B)}=\frac{1-P(\bar{A} \cup \bar{B})}{P(B)}$
$=\frac{1-(P(\bar{A})+P(\bar{B}))}{P(B)}=\frac{1-(0.5+0.4)}{0.6}$
$=\frac{1-0.9}{0.6}=\frac{0.1}{0.6}=\frac{1}{6}$

Topic: Mathematics, Probability, NDA, NDA 2019 (Phase 1)
A and B are two metals. The standard reduction potential of $\mathrm{A}^{+}(\mathrm{aq}) / \mathrm{A}(\mathrm{s})$ and $\mathrm{B}^{+}(\mathrm{aq}) / \mathrm{B}(\mathrm{s})$ are $-0.5 \mathrm{~V}$ and $+0.5 \mathrm{~V}$ respectively. What is the $\log K_C$ value for the following reaction at $298 \mathrm{~K}$ ?
$$
\mathrm{A}(\mathrm{s})+\mathrm{B}^{+}(\mathrm{aq}) \rightleftharpoons \mathrm{A}^{+}(\mathrm{aq})+\mathrm{B}(\mathrm{s})
$$
$\left(\right.$ Given : $\left.\frac{2.303 R T}{F}=0.06 \mathrm{~V}\right)$

Options:
A $\frac{6}{100}$
B $\frac{100}{6}$
C $\frac{}{200}$
D $\frac{200}{6}$

The correct answer is:
$\frac{100}{6}$

Solution:
$\begin{aligned} & \text { } \Delta \mathrm{G}^{\circ}=\mathrm{nFE}_{\text {cell }}^{\circ} \\ & \Delta \mathrm{G}^{\circ}=-\mathrm{RT} \ln \mathrm{K}_{\mathrm{c}} \\ & \therefore \quad \mathrm{nFE}_{\text {cell }}^{\circ}=-\mathrm{RT} \ln \mathrm{K}_{\mathrm{c}}=\frac{-2.303 \mathrm{RT}}{\mathrm{nF}} \log \mathrm{K}_{\mathrm{c}} \\ & \mathrm{n}=\text { no. of electrons involved }=1 \\ & \mathrm{E}_{\text {cell }}^{\circ}=\mathrm{E}_{\text {cathode }}^{\circ}-\mathrm{E}_{\text {Anode }}^{\circ}=0.5-(-0.5)=1 \text {. } \\ & \therefore \quad \mathrm{E}_{\text {cell }}^{\circ}=\frac{-2.303 \mathrm{RT} \log \mathrm{K}_{\mathrm{c}}}{\mathrm{nF}} \\ & 1=0.06 \log \mathrm{K}_{\mathrm{c}} \text {. } \\ & \therefore \log \mathrm{K}_{\mathrm{c}}=\frac{100}{6} \\ & \end{aligned}$

Topic: Chemistry, Electrochemistry, TS EAMCET, TS EAMCET 2023 (13 May Shift 2)
A and $\mathrm{B}$ are two sets having 3 elements in common $\operatorname{lfn}(\mathrm{A})=5, \mathrm{n}(\mathrm{B})=4$, then what is $\mathrm{n}(\mathrm{A} \times \mathrm{B})$ equal to

Options:
A 0
B 9
C 15
D 20

The correct answer is:
20

Solution:
Here, $\mathrm{n}(\mathrm{A})=5$ and $\mathrm{n}(\mathrm{B})=$
$\cdot n(A \times B)=5 \times 4=20$
$[\because \mathrm{n}(\mathrm{A})=\mathrm{m}, \mathrm{n}(\mathrm{B})=\mathrm{n} \Rightarrow \mathrm{n}(\mathrm{A} \times \mathrm{B})=\mathrm{mn}]$

Topic: Mathematics, Sets and Relations, NDA, NDA 2014 (Phase 2)
A, B are two events and $\bar{A}$ denotes the complements of A. Consider the following statements
$1 \quad P(A \cup B) \leq P(B)+P(A)$
$2 \quad \mathrm{P}(\mathrm{A})+\mathrm{P}(\overline{\mathrm{A}} \cup \mathrm{B}) \leq 1+\mathrm{P}(\mathrm{B})$
Which of the above statements is/are correct?

Options:
A 1 only
B 2 only
C Both 1 and 2
D Neither 1 nor 2

The correct answer is:
Both 1 and 2

Solution:
(1) $\mathrm{P}(\mathrm{A} \cup \mathrm{B}) \leq \mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})$ is correct
and (2) $\mathrm{P}(\mathrm{A})+\mathrm{P}(\overline{\mathrm{A}} \cup \mathrm{B}) \leq 1+\mathrm{P}(\mathrm{B})$ is also correct
$\Rightarrow$ Both the given statement are correct.

Topic: Mathematics, Probability, NDA, NDA 2007 (Phase 2)
$\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}=\overrightarrow{0}$ such that $|\vec{a}|=3,|\overrightarrow{\mathrm{b}}|=5$ and $|\overrightarrow{\mathrm{c}}|=7$
What is $\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c}$ a equal to ?

Options:
A $-83$
B $-83 / 2$
C 75
D $-75 / 2$

The correct answer is:
$-83 / 2$

Solution:
We have, $\overline{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overline{\mathrm{c}}=0$ ...(i)
On squaring both sides request.
$\bar{a}^{2}+\vec{b}^{2}+\bar{c}^{2}+2 \bar{a} \vec{b}+2 \vec{b} \bar{c}+2 \vec{c} \bar{a}=0$
$\quad(\because \vec{a} \cdot \vec{b}=\vec{b} \cdot \vec{a}, \vec{b} \cdot \vec{c}=\bar{c} \vec{b}$ and $\bar{c} \cdot \vec{a}=\bar{a} \cdot \vec{c})$
$\Rightarrow|a|^{2}+|b|^{2}+|c|^{2}=-2[a \cdot b+b . c+c . a]$
$\Rightarrow(3)^{2}+(5)^{2}+(7)^{2}=-2[a \cdot b+b \cdot c+c . a]$
$\Rightarrow a \cdot b+b . c+c . a=\frac{9+25+49}{-2}=-\frac{83}{2}$
Now $a+b+c=0 \quad$ [using eq. (i)]
$\Rightarrow \mathrm{a}+\mathrm{b}=-\mathrm{c}$
On squaring both sides, we get $\Rightarrow \mathrm{a}^{2}+\mathrm{b}^{2}+2 \mathrm{a} \cdot \mathrm{b}=\mathrm{c}^{2}$
$\Rightarrow(3)^{2}+(5)^{2}+2 \mathrm{ab}=(7)^{2}$
$\Rightarrow \overline{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}=\frac{15}{2}$
$\Rightarrow|\overline{\mathrm{a}}||\overrightarrow{\mathrm{b}}| \cos \theta=\frac{15}{2} \Rightarrow 3.5 \cos \theta=\frac{15}{2}$
$\Rightarrow \cos \theta=\frac{1}{2}=\cos \frac{\pi}{3}$
$\theta=\frac{\pi}{3}$
From eq. (i), $\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}=-\overline{\mathrm{a}}$
$\Rightarrow b^{2}+c^{2}+2 b \cdot c=a^{2}$
$\Rightarrow 2 \mathrm{~b} \cdot \mathrm{c}=\mathrm{a}^{2}-\mathrm{b}^{2}-\mathrm{c}^{2}=9-25-49=-65$
$\Rightarrow \mathrm{b.c}=-\frac{65}{2} \Rightarrow|\mathrm{b}||\mathrm{c}| \cos \theta=-\frac{65}{2}$
$\Rightarrow \cos \theta=\frac{65}{2} \times \frac{1}{5} \times \frac{1}{7}=-\frac{13}{14}$
Also, $|\vec{a}+\vec{b}|=|-\vec{c}|=|\vec{c}|=7$

Topic: Mathematics, Vector Algebra, NDA, NDA 2014 (Phase 2)
$\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}=\overrightarrow{0}$ such that $|\vec{a}|=3,|\overrightarrow{\mathrm{b}}|=5$ and $|\overrightarrow{\mathrm{c}}|=7$
What is $|\vec{a}+\vec{b}|$ equal to ?

Options:
A 7
B 8
C 10
D 11

The correct answer is:
7

Solution:
We have, $\overline{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overline{\mathrm{c}}=0$ ...(i)
On squaring both sides request.
$\bar{a}^{2}+\vec{b}^{2}+\bar{c}^{2}+2 \bar{a} \vec{b}+2 \vec{b} \bar{c}+2 \vec{c} \bar{a}=0$
$\quad(\because \vec{a} \cdot \vec{b}=\vec{b} \cdot \vec{a}, \vec{b} \cdot \vec{c}=\bar{c} \vec{b}$ and $\bar{c} \cdot \vec{a}=\bar{a} \cdot \vec{c})$
$\Rightarrow|a|^{2}+|b|^{2}+|c|^{2}=-2[a \cdot b+b . c+c . a]$
$\Rightarrow(3)^{2}+(5)^{2}+(7)^{2}=-2[a \cdot b+b \cdot c+c . a]$
$\Rightarrow a \cdot b+b . c+c . a=\frac{9+25+49}{-2}=-\frac{83}{2}$
Now $a+b+c=0 \quad$ [using eq. (i)]
$\Rightarrow \mathrm{a}+\mathrm{b}=-\mathrm{c}$
On squaring both sides, we get $\Rightarrow \mathrm{a}^{2}+\mathrm{b}^{2}+2 \mathrm{a} \cdot \mathrm{b}=\mathrm{c}^{2}$
$\Rightarrow(3)^{2}+(5)^{2}+2 \mathrm{ab}=(7)^{2}$
$\Rightarrow \overline{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}=\frac{15}{2}$
$\Rightarrow|\overline{\mathrm{a}}||\overrightarrow{\mathrm{b}}| \cos \theta=\frac{15}{2} \Rightarrow 3.5 \cos \theta=\frac{15}{2}$
$\Rightarrow \cos \theta=\frac{1}{2}=\cos \frac{\pi}{3}$
$\theta=\frac{\pi}{3}$
From eq. (i), $\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}=-\overline{\mathrm{a}}$
$\Rightarrow b^{2}+c^{2}+2 b \cdot c=a^{2}$
$\Rightarrow 2 \mathrm{~b} \cdot \mathrm{c}=\mathrm{a}^{2}-\mathrm{b}^{2}-\mathrm{c}^{2}=9-25-49=-65$
$\Rightarrow \mathrm{b.c}=-\frac{65}{2} \Rightarrow|\mathrm{b}||\mathrm{c}| \cos \theta=-\frac{65}{2}$
$\Rightarrow \cos \theta=\frac{65}{2} \times \frac{1}{5} \times \frac{1}{7}=-\frac{13}{14}$
Also, $|\vec{a}+\vec{b}|=|-\vec{c}|=|\vec{c}|=7$

Topic: Mathematics, Vector Algebra, NDA, NDA 2014 (Phase 2)
$\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}=\overrightarrow{0}$ such that $|\vec{a}|=3,|\overrightarrow{\mathrm{b}}|=5$ and $|\overrightarrow{\mathrm{c}}|=7$
What is cosine of the angle between $\overrightarrow{\mathrm{b}}$ and $\overrightarrow{\mathrm{c}}$ ?

Options:
A $11 / 12$
B $13 / 14$
C $-11 / 12$
D $-13 / 14$

The correct answer is:
$-13 / 14$

Solution:
We have, $\overline{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overline{\mathrm{c}}=0$ ...(i)
On squaring both sides request.
$\bar{a}^{2}+\vec{b}^{2}+\bar{c}^{2}+2 \bar{a} \vec{b}+2 \vec{b} \bar{c}+2 \vec{c} \bar{a}=0$
$\quad(\because \vec{a} \cdot \vec{b}=\vec{b} \cdot \vec{a}, \vec{b} \cdot \vec{c}=\bar{c} \vec{b}$ and $\bar{c} \cdot \vec{a}=\bar{a} \cdot \vec{c})$
$\Rightarrow|a|^{2}+|b|^{2}+|c|^{2}=-2[a \cdot b+b . c+c . a]$
$\Rightarrow(3)^{2}+(5)^{2}+(7)^{2}=-2[a \cdot b+b \cdot c+c . a]$
$\Rightarrow a \cdot b+b . c+c . a=\frac{9+25+49}{-2}=-\frac{83}{2}$
Now $a+b+c=0 \quad$ [using eq. (i)]
$\Rightarrow \mathrm{a}+\mathrm{b}=-\mathrm{c}$
On squaring both sides, we get $\Rightarrow \mathrm{a}^{2}+\mathrm{b}^{2}+2 \mathrm{a} \cdot \mathrm{b}=\mathrm{c}^{2}$
$\Rightarrow(3)^{2}+(5)^{2}+2 \mathrm{ab}=(7)^{2}$
$\Rightarrow \overline{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}=\frac{15}{2}$
$\Rightarrow|\overline{\mathrm{a}}||\overrightarrow{\mathrm{b}}| \cos \theta=\frac{15}{2} \Rightarrow 3.5 \cos \theta=\frac{15}{2}$
$\Rightarrow \cos \theta=\frac{1}{2}=\cos \frac{\pi}{3}$
$\theta=\frac{\pi}{3}$
From eq. (i), $\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}=-\overline{\mathrm{a}}$
$\Rightarrow b^{2}+c^{2}+2 b \cdot c=a^{2}$
$\Rightarrow 2 \mathrm{~b} \cdot \mathrm{c}=\mathrm{a}^{2}-\mathrm{b}^{2}-\mathrm{c}^{2}=9-25-49=-65$
$\Rightarrow \mathrm{b.c}=-\frac{65}{2} \Rightarrow|\mathrm{b}||\mathrm{c}| \cos \theta=-\frac{65}{2}$
$\Rightarrow \cos \theta=\frac{65}{2} \times \frac{1}{5} \times \frac{1}{7}=-\frac{13}{14}$
Also, $|\vec{a}+\vec{b}|=|-\vec{c}|=|\vec{c}|=7$

Topic: Mathematics, Vector Algebra, NDA, NDA 2014 (Phase 2)
$\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}=\overrightarrow{0}$ such that $|\vec{a}|=3,|\overrightarrow{\mathrm{b}}|=5$ and $|\overrightarrow{\mathrm{c}}|=7$
What is the angle between $|\vec{a}|$ and $|\vec{b}|$ ?

Options:
A $\pi / 6$
B $\pi / 4$
C $\pi / 3$
D $\pi / 2$

The correct answer is:
$\pi / 3$

Solution:
We have, $\overline{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overline{\mathrm{c}}=0$ ...(i)
On squaring both sides request.
$\bar{a}^{2}+\vec{b}^{2}+\bar{c}^{2}+2 \bar{a} \vec{b}+2 \vec{b} \bar{c}+2 \vec{c} \bar{a}=0$
$\quad(\because \vec{a} \cdot \vec{b}=\vec{b} \cdot \vec{a}, \vec{b} \cdot \vec{c}=\bar{c} \vec{b}$ and $\bar{c} \cdot \vec{a}=\bar{a} \cdot \vec{c})$
$\Rightarrow|a|^{2}+|b|^{2}+|c|^{2}=-2[a \cdot b+b . c+c . a]$
$\Rightarrow(3)^{2}+(5)^{2}+(7)^{2}=-2[a \cdot b+b \cdot c+c . a]$
$\Rightarrow a \cdot b+b . c+c . a=\frac{9+25+49}{-2}=-\frac{83}{2}$
Now $a+b+c=0 \quad$ [using eq. (i)]
$\Rightarrow \mathrm{a}+\mathrm{b}=-\mathrm{c}$
On squaring both sides, we get $\Rightarrow \mathrm{a}^{2}+\mathrm{b}^{2}+2 \mathrm{a} \cdot \mathrm{b}=\mathrm{c}^{2}$
$\Rightarrow(3)^{2}+(5)^{2}+2 \mathrm{ab}=(7)^{2}$
$\Rightarrow \overline{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}=\frac{15}{2}$
$\Rightarrow|\overline{\mathrm{a}}||\overrightarrow{\mathrm{b}}| \cos \theta=\frac{15}{2} \Rightarrow 3.5 \cos \theta=\frac{15}{2}$
$\Rightarrow \cos \theta=\frac{1}{2}=\cos \frac{\pi}{3}$
$\theta=\frac{\pi}{3}$
From eq. (i), $\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}=-\overline{\mathrm{a}}$
$\Rightarrow b^{2}+c^{2}+2 b \cdot c=a^{2}$
$\Rightarrow 2 \mathrm{~b} \cdot \mathrm{c}=\mathrm{a}^{2}-\mathrm{b}^{2}-\mathrm{c}^{2}=9-25-49=-65$
$\Rightarrow \mathrm{b.c}=-\frac{65}{2} \Rightarrow|\mathrm{b}||\mathrm{c}| \cos \theta=-\frac{65}{2}$
$\Rightarrow \cos \theta=\frac{65}{2} \times \frac{1}{5} \times \frac{1}{7}=-\frac{13}{14}$
Also, $|\vec{a}+\vec{b}|=|-\vec{c}|=|\vec{c}|=7$

Topic: Mathematics, Vector Algebra, NDA, NDA 2014 (Phase 2)
$\mathrm{A}, \mathrm{B}, \mathrm{C}$ and $\mathrm{D}$ are four sets such that $\mathrm{A} \cap \mathrm{B}=\mathrm{C} \cap \mathrm{D}=\phi$.
Consider the following:
1.$\mathrm{A} \cup \mathrm{C}$ and $\mathrm{B} \cup \mathrm{D}$ are always disjoint.
2.$\mathrm{A} \cap \mathrm{C}$ and $\mathrm{B} \cap \mathrm{D}$ are always disjoint Which of the above statements is/are correct?

Options:
A 1 only
B 2 only
C Both 1 and 2
D Neither 1 nor 2

The correct answer is:
2 only

Solution:
$\quad$ Let $\mathrm{A}=\{1,2\}$
$\mathrm{B}=\{3,4,0\}$
$\mathrm{C}=\{5,6,0\}$
$\mathrm{D}=\{7,8\}$
Such that $(\mathrm{A} \cap \mathrm{B})=(\mathrm{C} \cap \mathrm{D})=\phi$
$\Rightarrow(\mathrm{A} \cup \mathrm{C})=\{1,2,5,6,0\}$
$\Rightarrow(\mathrm{B} \cup \mathrm{D})=\{3,4,7,8,0\}$
$\Rightarrow(\mathrm{A} \cup \mathrm{C}) \cap(\mathrm{B} \cup \mathrm{D})=\{0\}$
So $(\mathrm{A} \cup \mathrm{C})$ and $(\mathrm{B} \cup \mathrm{D})$ are not always dispoint $\Rightarrow(\mathrm{A} \cap \mathrm{C})=\phi$ and $(\mathrm{B} \cap \mathrm{D})=\phi$
So $(\mathrm{A} \cap \mathrm{C})$ and $(\mathrm{B} \cap \mathrm{D})$ are always disjoint.

Topic: Mathematics, Sets and Relations, NDA, NDA 2015 (Phase 2)
A, B, C and D are respectively :-

Common Name	Man	Housefly	C	Wheat

Biological Name	A	Musca domestica	Mangifera indica	Triticum aestivum

Genus	Homo	Musca	Mangifera	Triticum

Family	Hominidae	B	Anacardiaceae	D

Options:
A A - Homo,
B - Muscidae,
C - Neem,
D - Poaceae
B A - Homo sapiens,
B - Muscidae,
C - Mango,
D - Poaceae
C A - Primata,
B - Insecta,
C - Mango,
D - Poales
D A - Primata,
B - Muscidae,
C - Mango,
D - Sapindales

The correct answer is:
A - Homo sapiens,
B - Muscidae,
C - Mango,
D - Poaceae

Solution:
CommonName	Man	Housefly	Mango	Wheat

BiologicalName	Homo sapien	Musca domestica	Mangifera indica	Triticum aestivum

Genus	Homo	Musca	Mangifera	Triticum

Family	Hominidae	Muscidae	Anacardiaceae	Poaceae

Topic: Biology, The Living World, NEET
$\mathrm{a}, \mathrm{b}, \mathrm{c}$ are in G.P. with $1 < \mathrm{a} < \mathrm{b} < \mathrm{n}$, and $\mathrm{n}>1$ is an integer. $\log _{\text {an }}$ $\mathrm{n}, \log _{\mathrm{b}} \mathrm{n}, \log _{\mathrm{c}} \mathrm{n}$ form a sequence. This sequence is which
one of the following ?

Options:
A Harmonic progression
B Arithmetic progression
C Geometric progression
D None of these

The correct answer is:
Harmonic progression

Solution:
If a, b, c are in G.P. then, $\mathrm{b}^{2}=\mathrm{ac} \Rightarrow \mathrm{b}=(\mathrm{ac})^{1 / 2} \quad \ldots .(1)$
Taking $\log _{\mathrm{n}}$ on both the sides of eq. (1).
$\log _{\mathrm{n}} \mathrm{b}=\frac{1}{2}\left[\left(\log _{\mathrm{n}}(\mathrm{ac})\right]=\frac{\log _{\mathrm{n}} \mathrm{a}+\log _{\mathrm{n}} \mathrm{c}}{2}\right.$
or, $\frac{\log _{n} a+\log _{n} c}{2}=\log _{n} b$
So, $\log _{\mathrm{n}} \mathrm{a}, \log _{\mathrm{n}} \mathrm{b}$ and $\log _{\mathrm{n}} \mathrm{c}$ are in $\mathrm{AP}$.
Hence, $\frac{1}{\log _{n} a}, \frac{1}{\log _{n} b}, \frac{1}{\log _{n} c}$ are in H.P.
$\log _{\mathrm{a}} \mathrm{n}=\frac{1}{\log _{\mathrm{n}} \mathrm{a}}$
$\log _{\mathrm{b}} \mathrm{n}=\frac{1}{\log _{\mathrm{n}} \mathrm{b}}$
$\log _{\mathrm{c}} \mathrm{n}=\frac{1}{\log _{\mathrm{n}} \mathrm{c}}$
i.e. $\log _{\mathrm{a}} \mathrm{n}, \log _{\mathrm{b}} \mathrm{n}$, and $\log _{\mathrm{c}} \mathrm{n}$ are in $\mathrm{HP}$.

Topic: Mathematics, Sequences and Series, NDA, NDA 2007 (Phase 1)
$A, B, C, D$ and $E$ are coplanar points and three of them lie in a straight line. What is the maximum number of triangles that can be drawn with these points as their vertices?

Options:
A 5
B 9
C 10
D 12

The correct answer is:
9

Solution:
Number of triangles using 5 points out of three are on a straight line $={ }^{5} C_{3}-{ }^{3} C_{3}=\frac{5 !}{3 ! 2 !}-1=\frac{5 \times 4}{2}-1$
$=10-1=9$

Topic: Mathematics, Permutation Combination, NDA, NDA 2011 (Phase 1)
A baby has been born with a small tail. It is a case exhibiting.

Options:
A Retrogresive evolution
B Mutation
C Reversion
D Metamorphosis

The correct answer is:
Reversion

Solution:
Atavism or reversion is the sudden reappearance of some ancestral features. Atavisms is observed in humans, as with the infants that are born with the vestigial tails called a "coccygeal process" or "coccygeal projection", or "caudal appendage".

Topic: Biology, Evolution, NEET
A bag contains 4 white and 2 black balls and another bag contains 3 white and 5 black balls. If one ball is drawn from each bag, then the probability that one ball is white and one ball is black is

Options:
A $\frac{5}{24}$
B $\frac{13}{24}$
C $\frac{1}{4}$
D $\frac{2}{3}$

The correct answer is:
$\frac{13}{24}$

Solution:
$\mathrm{P}$ (one white ball \& one black ball) $=\mathrm{P}\{[$ black from 1 st bag $\&$ white from $2 \mathrm{nd}]$ or [white from 1 st $\&$ black from $2 \mathrm{nd}]\}$ $=\mathrm{P}\left(\left[\mathrm{B}_{1} \cap \mathrm{W}_{2}\right) \cup\left(\mathrm{W}_{1} \cap \mathrm{B}_{2}\right)\right]$
$=\mathrm{P}\left(\mathrm{B}_{1} \cap \mathrm{W}_{2}\right)+\mathrm{P}\left(\mathrm{W}_{1} \cap \mathrm{B}_{2}\right)$
(By addition theorem for mutually
exclusive events) $=\mathrm{P}\left(\mathrm{B}_{1}\right) \mathrm{P}\left(\mathrm{W}_{2}\right)+\mathrm{P}\left(\mathrm{W}_{1}\right) \mathrm{P}\left(\mathrm{B}_{2}\right)$
$\left(\therefore \mathrm{B}_{2} \& \mathrm{~W}_{2} ; \mathrm{B}_{2} \& \mathrm{~W}_{1}\right.$ are pairs of
independent events) $=\left[\frac{2}{6} \times \frac{3}{8}\right]+\left[\frac{4}{6} \times \frac{5}{8}\right]$
$=\frac{13}{24}$

Topic: Mathematics, Probability, NDA, NDA 2015 (Phase 2)
A bag contains 5 black and 3 white balls. Two balls are drawn at random one after the other without replacement. What is the probability that both are white?

Options:
A $\frac{1}{28}$
B $\frac{1}{14}$
C $\frac{3}{28}$
D None of these above

The correct answer is:
$\frac{3}{28}$

Solution:
Total number of balls $=5+3=8$. Prob (both ball are white)
$=\frac{3}{8} \times \frac{2}{7}(\because$ No. of white ball $=3)$
$=\frac{3}{28}$

Topic: Mathematics, Probability, NDA, NDA 2013 (Phase 1)
A bag contains balls of two colours, 3 black and 3 white. What is the smallest number of balls which must be drawn from the bag, without looking, so that among these three are two of the same colour?

Options:
A 2
B 3
C 4
D 5

The correct answer is:
4

Solution:
Minimum 4 balls have to be drawn so that among these there are two of the same colour

Topic: Mathematics, Probability, NDA, NDA 2013 (Phase 1)
A balloon is directly above one end of a bridge. The angle of depression of the other end of the bridge from the balloon is $48^{\circ} .$ If the height of the balloon above the bridge is $122 \mathrm{~m}$, then what is the length of the bridge?

Options:
A $122 \sin 48^{\circ} \mathrm{m}$
B $122 \tan 42^{\circ} \mathrm{m}$
C $122 \cos 48^{\circ} \mathrm{m}$
D $122 \tan 48^{\circ} \mathrm{m}$

The correct answer is:
$122 \sin 48^{\circ} \mathrm{m}$

Solution:
Let the bridge is $\mathrm{BC}$ and height above the bridge $\mathrm{AB}=122 \mathrm{~m}$
From DABC,
$\begin{aligned} \cot 48^{\circ}=\frac{\mathrm{BC}}{\mathrm{AB}}=\frac{\mathrm{BC}}{122} \\ \mathrm{BC}=122 \cdot \cot \left(48^{\circ}\right) \\=122 \cdot \cot \left(90^{\circ}-42^{\circ}\right) \\=& 122 \cdot \tan 42^{\circ} \end{aligned}$

Topic: Mathematics, Heights and Distances, NDA, NDA 2018 (Phase 2)
A balloon is pumped at the rate of $4 \mathrm{~cm}^{3}$ per second. What is the rate at which its surface area increases and radius is $4 \mathrm{~cm} ?$

Options:
A $1 \mathrm{~cm}^{2} / \mathrm{s}$ ?
B $2 \mathrm{~cm}^{2} / \mathrm{s}$
C $3 \mathrm{~cm}^{2} / \mathrm{s}$
D $4 \mathrm{~cm}^{2} / \mathrm{s}$

The correct answer is:
$2 \mathrm{~cm}^{2} / \mathrm{s}$

Solution:
Let $r$ be the radius of balloon.
Balloon is like a sphere and volume of sphere $=\frac{4}{3} \pi r^{3}$
$\therefore \quad V=\frac{4}{3} \pi r^{3}$
Differentiate both side w.r.t ' $\mathrm{t}^{\prime}$
$\Rightarrow \frac{d V}{d t}=\frac{4}{3} \pi \cdot 3 r^{2} \frac{d r}{d t}$
$\Rightarrow 4=\frac{4}{3} \pi .3(4)^{2} \frac{d r}{d t} \quad\left(\because \frac{d V}{d t}=4 \mathrm{~cm}^{3} / \mathrm{s}\right)$
$\Rightarrow \frac{d r}{d t}=\frac{1}{16 \pi}$
Now, surface area of balloon $=S=4 \pi r^{2}$
$$
\begin{aligned}
\frac{d S}{d t} &=4 \pi \cdot 2 r \frac{d r}{d t} \\
&=4 \pi \cdot 2 \times 4 \frac{1}{16 \pi}(\text { from }(1))=2 \mathrm{~cm}^{2} / \mathrm{s}
\end{aligned}
$$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2010 (Phase 1)
A balloon of mass \(M\) descends with an acceleration \(a( < g)\). What mass need to be removed from the balloon, so that it starts ascending with acceleration, \(a\) ?

Options:
A \(\frac{2 M}{(a+g)}\)
B \(\frac{2 M a}{(a+g)}\)
C \(\frac{2 M a}{(a-g)}\)
D \(\frac{2 M a}{(g-a)}\)

The correct answer is:
\(\frac{2 M a}{(a+g)}\)

Solution:
The given situation is shown in the figure
In above figures, \(M\) is the mass of balloon and \(M^{\prime}\) is the mass remove from the ballon.
According to Newton's law of motion in first case,
\(M g-F=M a \Rightarrow F=M g-M a\) ...(i)
According to Newton's second law of motion, for second case,
\(\begin{aligned}
& F-\left(M-M^{\prime}\right) g=\left(M-M^{\prime}\right) a \\
& \Rightarrow F-M g+M^{\prime} g=M a-M^{\prime} a \\
& \Rightarrow M g-M a-M g+M^{\prime} g=M a-M^{\prime} a \text { [from Eq. (i)] } \\
& \Rightarrow M^{\prime} g+M^{\prime} a=M a+M a \\
& \Rightarrow M^{\prime}(g+a)=2 M a \\
& \Rightarrow M^{\prime}=\frac{2 M a}{g+a} \\
\end{aligned}\)

Topic: Physics, Laws of Motion, AP EAMCET, AP EAMCET 2020 (18 Sep Shift 2)
A battery of emf $10 \mathrm{~V}$ is connected to a uniform wire $A B$ of $1 \mathrm{~m}$ length and having a resistance of $10 \Omega$ in series with a $10 \Omega$ resistor as shown in the figure. Two cells of emf $2 \mathrm{~V}$ and $3 \mathrm{~V}$ having internal resistance $2 \Omega$ and $3 \Omega$, respectively are connected as shown in the figure. If the galvanometer shows null deflection at point $J$ on the wire, then the distance of point $J$ from the point $B$ is.

Options:
A 48 cm
B 50 cm
C 52 cm
D 54 cm

The correct answer is:
52 cm

Solution:
Given $r=10 \Omega, R=10 \Omega, E=10 \mathrm{~V}$ and $L=100 \mathrm{~cm}$.
Now, voltage drop on the wire, $E^{\prime}=\frac{E}{(r+R)} \times R$
$$
\Rightarrow \quad E^{\prime}=\frac{10}{20} \times 10=5 R
$$
So, potential gradient on wire,
Effective emf of combination in secondary circuit,
$$
\begin{array}{rlrl}
& & \frac{V}{r_{\text {eff }}} & =\frac{\varepsilon_1}{r_1}+\frac{\varepsilon_2}{r_2} \\
\Rightarrow \text { Here, } \quad & \frac{1}{r_{\text {eff }}} & =\frac{1}{2}+\frac{1}{3}=\frac{5}{6} \\
\Rightarrow \quad & V & =\frac{6}{5}(1+1)=\frac{12}{5}
\end{array}
$$
Now, at balancing point (from point $A$ ) is
$$
\begin{aligned}
& l=V / x=\frac{12}{5} \times \frac{100}{5} \\
& l=48 \mathrm{~cm}
\end{aligned}
$$
So, length from point $B$ is
$$
100-48=52 \mathrm{~cm}
$$

Topic: Physics, Current Electricity, AP EAMCET, AP EAMCET 2018 (22 Apr Shift 1)
$A B C$ is a triangle in which $A B=6 \mathrm{~cm}, B C=8 \mathrm{~cm}$ and $C A=10 \mathrm{~cm} .$ What is the value of $\cot (A / 4) ?$

Options:
A $\sqrt{5}-2$
B $\sqrt{5}+2$
C $\sqrt{3}-1$
D $\sqrt{3}+1$

The correct answer is:
$\sqrt{5}+2$

Solution:
Here, $\mathrm{AB}=6 \mathrm{~cm}, \mathrm{BC}=8 \mathrm{~cm}$ and $\mathrm{CA}=10 \mathrm{~cm}$
So, $c=6 \mathrm{~cm}, \mathrm{a}=8 \mathrm{~cm}, \mathrm{~b}=10 \mathrm{~cm}$
$\mathrm{S}=\frac{a+b+c}{2}=\frac{24}{2}=12$
$\therefore \tan \frac{A}{2}=\sqrt{\frac{(12-10)(12-6)}{12(12-8)}}$
$\left(\because \tan \frac{\mathrm{A}}{2}=\sqrt{\frac{(s-b)(s-c)}{s(s-a)}}\right)$
$=\sqrt{\frac{1}{4}}=\frac{1}{2}$
$\therefore \cot \frac{A}{2}=2$
Now, $\cot \left(\frac{A}{4}+\frac{A}{4}\right)=\frac{\cot ^{2} \frac{A}{4}-1}{2 \cot \frac{A}{4}}$
$\cot \left(\frac{A}{2}\right)=\frac{\cot ^{2} \frac{A}{4}-1}{2 \cot \frac{A}{4}}$
Let $\cot \left(\frac{A}{4}\right)=x$
$\therefore \quad 2=\frac{x^{2}-1}{2 x}$
$\Rightarrow x^{2}-4 x-1=0$
$\Rightarrow x=\frac{4 \pm \sqrt{16+4}}{2}$
$\Rightarrow x=\frac{4 \pm 2 \sqrt{5}}{2}=2 \pm \sqrt{5}$
So, $\cot \left(\frac{A}{4}\right)=\sqrt{5}+2$ or $2-\sqrt{5}$

Topic: Mathematics, Properties of Triangles, NDA, NDA 2008 (Phase 2)