datavorous/entrance-exam-dataset huggingface dataset:
If speed of light in vacuum $\left(3 \times 10^8 \mathrm{~ms}^{-1}\right)$, acceleration due to gravity $\left(10 \mathrm{~ms}^{-2}\right)$ and mass of electron $\left(9.1 \times 10^{-31} \mathrm{~kg}\right)$ are taken as fundamental physical quantities, then the unit of time in this system is

Options:
A $3 \times 10^3 \mathrm{~s}$
B $5 \times 10^{-19} \mathrm{~s}$
C $3 \times 10^{19} \mathrm{~s}$
D $3 \times 10^7 \mathrm{~s}$

The correct answer is:
$3 \times 10^7 \mathrm{~s}$

Solution:
Speed of light, $\left[\mathrm{L} \mathrm{T}^{-1}\right]=3 \times 10^8 \mathrm{~m} / \mathrm{s}$ ... (1) acceleration due to gravity, $\left[\mathrm{L} \mathrm{T}^{-2}\right]=10 \mathrm{~m} / \mathrm{s}^{-2}$ ... (2)
Divide equation (1) by (2), we have
$\frac{\left[\mathrm{L} \mathrm{T}^{-1}\right]}{\left[\mathrm{LT}^{-2}\right]}=\frac{3 \times 10^8}{10} \Rightarrow[\mathrm{T}]=3 \times 10^7 \mathrm{~s}$

Topic: Physics, Units and Dimensions, AP EAMCET, AP EAMCET 2023 (16 May Shift 2)
If speed of sound in air in $330 \mathrm{~m} \mathrm{~s}^{-1}$, then find the number of overtones present in an open organ pipe of length $1 \mathrm{~m}$ whose frequency $f \leq 1000$

Options:
A $2$
B $4$
C $8$
D $6$

The correct answer is:
$6$

Solution:
Fundamental frequency,
$\begin{aligned} & v=\frac{v}{2 \ell}=\frac{330}{2 \times 1}=165 \mathrm{~Hz} \\ & \therefore \text { Number of overtones allowed }=\frac{1000}{165}=6\end{aligned}$

Topic: Physics, Waves and Sound, AIIMS, AIIMS 2019 (25 May)
If standard enthalpy of formation $\left(\Delta_{\mathrm{f}} \mathrm{H}^{\circ}\right)$ of $\mathrm{CO}_2, \mathrm{H}_2 \mathrm{O}$ and $\mathrm{CH}_4$ are $-393,-286$ and $-74.0 \mathrm{~kJ} \mathrm{~mol}^{-1}$ respectively, the standard enthalpy of combustion of methane in $\mathrm{kJ} \mathrm{mol}^{-1}$ is

Options:
A $-753$
B $-105$
C $-605$
D $-891$

The correct answer is:
$-891$

Solution:
No solution. Refer to answer key.

Topic: Chemistry, Thermodynamics (C), AP EAMCET, AP EAMCET 2017 (25 Apr Shift 1)
If standard molar enthalpy change and standard molar internal energy change measured in bomb calorimeter are equal, which one of the following statements is correct?

Options:
A $\Delta \mathrm{n}>0$, with increase in pressure
B $\Delta \mathrm{n}>0$, with decrease in pressure
C $\Delta \mathrm{n} < 0$, with increase in pressure
D $\Delta \mathrm{n}=0$, at constant pressure

The correct answer is:
$\Delta \mathrm{n}=0$, at constant pressure

Solution:
$\Delta \mathrm{H}^{\circ}=\Delta \mathrm{U}^{\circ}$ (given)
$$
\begin{aligned}
& \Delta \mathrm{H}^{\circ}=\Delta \mathrm{U}^{\circ}+\mathrm{P} \Delta \mathrm{V} \\
& =\Delta \mathrm{U}^{\circ}+\Delta \mathrm{nRT} \\
& =\Delta \mathrm{U}^{\circ}
\end{aligned}
$$
$\Rightarrow \Delta \mathrm{nRT}=0$ which means $\Delta \mathrm{n}=0$ with constant pressure.

Topic: Chemistry, Thermodynamics (C), AP EAMCET, AP EAMCET 2023 (19 May Shift 1)
If sum of squares of the roots of the equation $x^{2}+k x-b=0$ is $2 b$, what is $k$ equal to?

Options:
A 1
B b
C $-b$
D 0

The correct answer is:
0

Solution:
Let the roots of the equation $x^{2}+k x-b=0$ be $\alpha$ and $\beta$. Sum : $\alpha+\beta=-k$ and Product : $\alpha \beta=-b$
According to the question, we have
$\quad \alpha^{2}+\beta^{2}=2 b$
$\Rightarrow(\alpha+\beta)^{2}-2 \alpha \beta=2 b$
$\Rightarrow k^{2}+2 b=2 b \Rightarrow k=0$

Topic: Mathematics, Quadratic Equation, NDA, NDA 2011 (Phase 1)
If sum of the roots of $3 x^{2}+(3 p+1) x-(p+5)=0$ is equal to their product, then what is the value of $p ?$

Options:
A 2
B 3
C 4
D 9

The correct answer is:
2

Solution:
Let $\alpha$ and $\beta$ are the roots of the given equation.
$\therefore$
$\alpha+\beta=\frac{-(3 p+1)}{3}$
and $\quad \alpha \beta=\frac{-(p+5)}{3}$
Now, $\frac{-(3 p+1)}{3}=\frac{-(p+5)}{3}$
$\Rightarrow 3 p+1=p+5$
$\therefore \quad 2 p=4$
$\therefore \quad p=2$

Topic: Mathematics, Quadratic Equation, NDA, NDA 2008 (Phase 2)
If $\mathrm{t}_{\mathrm{r}}$ is the rth term in the expansion of $(1+\mathrm{x})^{101}$, then what is
the ratio $\frac{\mathrm{t}_{20}}{\mathrm{t}_{19}}$ equal to ?

Options:
A $\frac{20 \mathrm{x}}{19}$
B $83 \mathrm{x}$
C $19 \mathrm{x}$
D $\frac{83 x}{19}$

The correct answer is:
$\frac{83 x}{19}$

Solution:
We find $r_{n}$ term:
$\mathrm{t}_{\mathrm{r}}$ is the rth term in the expansion of $(1+\mathrm{x})^{101}$. $\mathrm{t}_{\mathrm{r}}=101 \mathrm{C}_{\mathrm{r}-1} \cdot(\mathrm{x})^{(\mathrm{r}-1)}$
$\therefore \frac{\mathrm{t}_{20}}{\mathrm{t}_{19}}=\frac{101}{{ }^{01} \mathrm{C}_{19}} \cdot \frac{\mathrm{x}^{19}}{\mathrm{x}^{18}}=\frac{101}{{ }^{101} \mathrm{C}_{19} \mathrm{x}}{\mathrm{C}_{18}}=\frac{\frac{101 !}{19 ! 82 !}}{\frac{101 !}{18 ! 83 !}} \mathrm{x}=\frac{83 \mathrm{x}}{19}$

Topic: Mathematics, Binomial Theorem, NDA, NDA 2008 (Phase 1)
If $\tan ^{-1} 2, \tan ^{-1} 3$ are two angles of a triangle, then what is the third angle?

Options:
A $\tan ^{-1} 2$
B $\tan ^{-1} 4$
C $\pi / 4$
D $\pi / 3$

The correct answer is:
$\pi / 4$

Solution:
Let $\mathrm{A}=\tan ^{-1} 2, \mathrm{~B}=\tan ^{-1} 3$ and $\mathrm{C}$ be the angles of a
triangle.
By angle sum property, we have
$\tan ^{-1} 2+\tan ^{-1} 3+\mathrm{C}=180^{\circ}$
$\Rightarrow \tan ^{-1}\left(\frac{5}{-5}\right)=180^{\circ}-\mathrm{C}$
$\Rightarrow \tan ^{-1}(-1)=180^{\circ}-\mathrm{C}$
$\Rightarrow \frac{3 \pi}{4}=\pi-\mathrm{C} \Rightarrow \mathrm{C}=\pi-\frac{3 \pi}{4}=\frac{\pi}{4}$
Hence, third angle is $\frac{\pi}{4}$.

Topic: Mathematics, Inverse Trigonometric Functions, NDA, NDA 2012 (Phase 1)
If $\tan \theta=\frac{1}{2}$ and $\tan \varphi=\frac{1}{3}$, then what is the value of $?

Options:
A 0
B $\frac{\pi}{6}$
C $\frac{\pi}{4}$
D $\frac{\pi}{2}$

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

Solution:
\tan \theta=\frac{1}{2} \\ & \tan \phi=\frac{1}{3} \end{aligned}$
$\tan (\theta+\phi)=\frac{\tan \theta+\tan \phi}{1-\tan \theta \tan \phi}$
$=\frac{\frac{1}{2}+\frac{1}{3}}{1-\left(\frac{1}{2}\right)\left(\frac{1}{3}\right)}=\frac{\frac{5}{6}}{\frac{5}{6}}=1$
$\therefore \theta+\phi=\tan ^{-1}(1)=\frac{\pi}{4}$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2019 (Phase 1)
If $\tan ^{2} \theta=2 \tan ^{2} \phi+1$, then which one of the following is correct?

Options:
A $\cos (2 \theta)=\cos (2 \phi)-1$
B $\cos (2 \theta)=\cos (2 \phi)+1$
C $\cos (2 \theta)=[\cos (2 \phi)-1] / 2$
D $\cos (2 \theta)=[\cos (2 \phi)+1] / 2$

The correct answer is:
$\cos (2 \theta)=[\cos (2 \phi)-1] / 2$

Solution:
Work with option,
$\cos (2 \phi)-1=\frac{1-\tan ^{2} \phi}{1+\tan ^{2} \phi}-1$
$=-\frac{2 \tan ^{2} \phi}{1+\tan ^{2} \phi}=\frac{-\left(\tan ^{2} \theta-1\right)}{1+\frac{\tan ^{2} \theta-1}{2}}$
$=\frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta} \times 2=\cos (2 \theta) 2$
Thus, $\cos 2 \theta=\frac{\cos (2 \phi)-1}{2}$

Topic: Mathematics, Trigonometric Equations, NDA, NDA 2009 (Phase 1)
If tan $(\alpha+\beta)=2$ and $\tan (\alpha-\beta)=1$, then $\tan (2 \alpha)$ is equal
to

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

The correct answer is:
$-3$

Solution:
$\tan (\alpha+\beta)=2$
$\tan (\alpha-\beta)=1$
$\tan 2 \alpha=\tan [(\alpha+\beta)+(\alpha-\beta)]$
$=\frac{\tan (\alpha+\beta)+\tan (\alpha-\beta)}{1-\tan (\alpha+\beta) \cdot \tan (\alpha-\beta)}$
$=\frac{2+1}{1-2 \cdot 1}=\frac{3}{-1}=-3$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2017 (Phase 1)
If $\tan ^{2} \mathrm{~B}=\frac{1-\sin \mathrm{A}}{1+\sin \mathrm{A}}$ then what is the value of $\mathrm{A}+2 \mathrm{~B}$ ?

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

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

Solution:
Given equation is $\tan ^{2} \mathrm{~B}=\frac{1-\sin \mathrm{A}}{1+\sin \mathrm{A}}$
$\Rightarrow$ Applying componendo and dividendo $\frac{1+\tan ^{2} \mathrm{~B}}{1-\tan ^{2} \mathrm{~B}}=\frac{2}{2 \sin \mathrm{A}}$
$\Rightarrow \sin \mathrm{A}=\frac{1-\tan ^{2} \mathrm{~B}}{1+\tan ^{2} \mathrm{~B}} \Rightarrow \sin \mathrm{A}=\cos 2 \mathrm{~B}$
$\Rightarrow \sin A=\sin \left(\frac{\pi}{2}-2 \mathrm{~B}\right)$
$\Rightarrow \mathrm{A}=\frac{\pi}{2}-2 \mathrm{~B} \Rightarrow \mathrm{A}+2 \mathrm{~B}=\frac{\pi}{2}$

Topic: Mathematics, Trigonometric Equations, NDA, NDA 2006 (Phase 2)
If $\tan \mathrm{A}=\frac{1}{2}$ and $\tan \mathrm{B}=\frac{1}{3}$, then what is the value of $(\mathrm{A}+\mathrm{B})$ ?

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

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

Solution:
Let tan $A=\frac{1}{2}$ and $\tan B=\frac{1}{3}$
We know, $\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A \tan B}$
$=\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{2} \cdot \frac{1}{3}}=\frac{\frac{5}{6}}{\frac{5}{6}}=1=\tan \pi / 4$
$\Rightarrow A+B=\pi / 4$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2010 (Phase 1)
If $\tan A=1 / 2$ and $\tan B=1 / 3$, then what is the value of $4 A+4 B ?$

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

The correct answer is:
$\pi$

Solution:
Let tan $A=\frac{1}{2}, \tan B=\frac{1}{3}$
We know,
$\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A \tan B}=\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{2} \times \frac{1}{3}}=\frac{5}{6} \times \frac{6}{5}$
$\tan (A+B)=1$
$\Rightarrow A+B=\tan ^{-1}(1)=\frac{\pi}{4}$
Multiply by 4 on both side,
$4(A+B)=\frac{\pi}{4} \times 4 \Rightarrow 4 A+4 B=\pi$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2011 (Phase 2)
If $\tan A=3 / 4$ and $\tan B=-12 / 5$, then how many values can cot $(A-B)$ have depending on the actual values of $A$ and $B ?$

Options:
A 1
B $2
C 3
D 4

The correct answer is:
4

Solution:
$\tan \mathrm{A}=\frac{3}{4}$ and $\tan \mathrm{B}=-\frac{12}{5}$
$\therefore \quad \cot (\mathrm{A}-\mathrm{B})=\frac{1}{\tan (\mathrm{A}-\mathrm{B})}=\frac{1+\tan \mathrm{A} \tan \mathrm{B}}{\tan \mathrm{A}-\tan \mathrm{B}}$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2010 (Phase 2)
If $\tan A-\tan B=x$ and $\cot B-\cot A=y$, then what is cot
$(A-B)$ equal to?

Options:
A $\frac{1}{y}-\frac{1}{x}$
B $\frac{1}{x}-\frac{1}{y}$
C $\frac{1}{x}+\frac{1}{y}$
D $-\frac{1}{x}-\frac{1}{y}$

The correct answer is:
$\frac{1}{x}+\frac{1}{y}$

Solution:
Let $\tan A-\tan B=x$ and $\cot B-\cot A=y$.
$\Rightarrow \frac{1}{\tan B}-\frac{1}{\tan A}=y$
$\Rightarrow \frac{\tan A-\tan B}{\tan A \tan B}=y \Rightarrow \frac{x}{\tan A \tan B}=y$
Consider $\cot (A-B)=\left(\frac{1}{\tan (A-B)}\right)$
$=\frac{1+\tan A \tan B}{\tan A-\tan B}=\frac{1+\frac{x}{y}}{x}=\frac{y+x}{x y}=\frac{1}{x}+\frac{1}{y}$.

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2011 (Phase 2)
If $\tan \mathrm{A}=\mathrm{x}+1$ and $\tan \mathrm{B}=\mathrm{x}-1$, then $\mathrm{x}^{2} \tan (\mathrm{A}-\mathrm{B})$ has the
value:

Options:
A 1
B $\mathrm{x}$
C 0
D 2

The correct answer is:
2

Solution:
$x^{2} \tan (A-B)$
$=x^{2}\left(\frac{\tan A-\tan B}{1+\tan A \tan B}\right)=x^{2}\left(\frac{(x+1)-(x-1)}{1+\left(x^{2}-1\right)}\right)$
$=x^{2}\left(\frac{2}{x^{2}}\right)=2$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2013 (Phase 1)
If $\tan \theta=\sqrt{m}$, where $\mathrm{m}$ is non-square natural
number, then $\sec 2 \theta$ is

Options:
A a negative number
B a transcendental number
C an irrational number
D a rational number

The correct answer is:
a negative number

Solution:
Let $\tan \theta=\sqrt{m}$, where $\mathrm{m}$ is a non-square natural num-
ber.
$\Rightarrow \sin \theta=\sqrt{m} \cos \theta$
Consider, $\sec 2 \theta=\frac{1}{\cos 2 \theta}=\frac{1}{\cos ^{2} \theta-\sin ^{2} \theta}$
$=\frac{1}{\cos ^{2} \theta-m \cos ^{2} \theta}=\frac{1}{\cos ^{2} \theta(1-m)}$
$=\frac{\sec ^{2} \theta}{1-m}=\frac{1+\tan ^{2} \theta}{1-m}=\frac{1+m}{1-m}$
$=\frac{(1+m)(1-m)}{(1-m)(1-m)}=\frac{\left(1-m^{2}\right)}{(1-m)^{2}}$
Numerator will always be negative and denominator will always be positive
Hence, $\sec 2 \theta=\frac{1-m^{2}}{(1-m)^{2}}$ is a negative number

Topic: Mathematics, Sets and Relations, NDA, NDA 2011 (Phase 2)
If $\tan \theta+\sec \theta=4$, then what is the value of $\sin \theta ?

Options:
A $8 / 17$
B $8 / 15$
C $15 / 17$
D $23 / 32$

The correct answer is:
$15 / 17$

Solution:
$\begin{aligned} & \tan \theta+\sec \theta=4 \\ & \Rightarrow \frac{\sin \theta}{\cos \theta}+\frac{1}{\cos \theta}=4 \\ & \Rightarrow 1+\sin \theta=4 \cos \theta \end{aligned}$
Squaring on both side, $(1+\sin \theta)^{2}=16 \cos ^{2} \theta=16\left(1-\sin ^{2} \theta\right)$
$(1+\sin \theta)^{2}=16(1-\sin \theta)(1+\sin \theta)$
$1+\sin \theta=16-16 \sin \theta$
$17 \sin \theta=16-1$
$\sin \theta=\frac{15}{17}$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2012 (Phase 1)
If the 10 th term of a GP is 9 and 4 th term is 4 , then what is its 7th term?

Options:
A 6
B 14
C $27 / 14$
D $56 / 15$

The correct answer is:
6

Solution:
General term of G.P= ar $^{\text {1-1 }}$. Given : $\mathrm{a}_{10}=9$ and $\mathrm{a}_{4}=4$
$\Rightarrow \mathrm{ar}^{9}=9$ and $\mathrm{ar}^{3}=4$
On dividing we get
$\begin{aligned} \frac{a r^{9}}{a r^{3}}=& \frac{9}{4} \Rightarrow r^{6}=\frac{9}{4} \\ a r^{3}=4 & \Rightarrow\left(a r^{3}\right)^{2}=16 \\ & \Rightarrow a^{2} r^{6}=16 \\ & \Rightarrow a^{2} \times \frac{9}{4}=16 \Rightarrow a^{2}=\frac{64}{9} \Rightarrow a=\frac{8}{3} \\ \text { Thus, } a_{7}=a r^{6}=\frac{8}{3} \times \frac{9}{4}=6 \end{aligned}$

Topic: Mathematics, Sequences and Series, NDA, NDA 2011 (Phase 2)
If the AM and GM between two number are in the ratio $m: n$, then what is the ratio between the two numbers?

Options:
A $\frac{m+\sqrt{m^{2}-n^{2}}}{m-\sqrt{m^{2}-n^{2}}}$
B $\frac{m+n}{m-n}$
C $\frac{m^{2}-n^{2}}{m^{2}+n^{2}}$
D $\frac{m^{2}+n^{2}-m n}{m^{2}+n^{2}+m n}$

The correct answer is:
$\frac{m+\sqrt{m^{2}-n^{2}}}{m-\sqrt{m^{2}-n^{2}}}$

Solution:
Let 'a' and 'b' two numbers.
A.M $=\frac{a+b}{2}$ and $G \cdot M=\sqrt{a b}$
According to the question, $A: G=m: n$
$\Rightarrow \frac{a+b}{2 \sqrt{a b}}=\frac{m}{n} \Rightarrow \frac{(a+b)^{2}}{4 a b}=\frac{m^{2}}{n^{2}}$ ...(i)
and $\frac{(a+b)^{2}-4 a b}{4 a b}=\frac{m^{2}-n^{2}}{n^{2}}$
$\Rightarrow \frac{(a-b)^{2}}{4 a b}=\frac{m^{2}-n^{2}}{n^{2}}$ ...(ii)
Since, on dividing Equation (i) and (ii), we get
$\frac{(a+b)^{2}}{(a-b)^{2}}=\frac{m^{2}}{m^{2}-n^{2}} \Rightarrow \frac{(a+b)}{(a-b)}=\frac{m}{\sqrt{m^{2}-n^{2}}}$
$\Rightarrow \frac{(a+b)+(a-b)}{(a+b)-(a-b)}=\frac{m+\sqrt{m^{2}-n^{2}}}{m-\sqrt{m^{2}-n^{2}}}$
(Using componendo dividendo rule)
$\Rightarrow \quad \frac{2 a}{2 b}=\frac{a}{b}=\frac{m+\sqrt{m^{2}-n^{2}}}{m-\sqrt{m^{2}-n^{2}}}$

Topic: Mathematics, Sequences and Series, NDA, NDA 2010 (Phase 2)
If the AM and GM of two numbers are 5 and 4 respectively, then what is the HM of those numbers?

Options:
A $\frac{5}{4}$
B $\frac{16}{5}$
C $\frac{9}{2}$
D 9

The correct answer is:
$\frac{16}{5}$

Solution:
We know, $\mathrm{HM}=\frac{(\mathrm{GM})^{2}}{\mathrm{AM}}$
$\therefore \mathrm{HM}=\frac{16}{5}$

Topic: Mathematics, Sequences and Series, NDA, NDA 2008 (Phase 2)
If the AM and HM of two numbers are 27 and 12 respectively, then what is their GM equal to? $\quad[2010-I]$

Options:
A 12
B 18
C 24
D 27

The correct answer is:
18

Solution:
Given $\mathrm{AM}=27$ and $\mathrm{HM}=12$ and we know that $(\mathrm{GM})^{2}=(\mathrm{AM})(\mathrm{HM})=27 \times 12$
$\Rightarrow G M=\sqrt{27 \times 12}=\sqrt{3 \times 3 \times 3 \times 3 \times 2 \times 2}$
$\Rightarrow \quad \mathrm{GM}=3 \times 3 \times 2=18$

Topic: Mathematics, Sequences and Series, NDA, NDA 2010 (Phase 1)
If the angle between the lines whose direction ratios are
$(2,-1,2)$ and $(x, 3,5)$ is $\frac{\pi}{4}$, then the smaller value of $x$ is

Options:
A 52
B 2
C 4
D 1

The correct answer is:
2

Solution:
Given direction ratios are $(2,-1,2)$ and $(x, 3,5)$ We know that the angle between the lines whose direction ratios are $\left(a_{1}, b_{1}, c_{1}\right)$ and $\left(a_{2}, b_{2}, c_{2}\right)$ is
$\cos \theta=\frac{a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}} \sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}$
$\Rightarrow \cos \frac{\pi}{4}=\frac{2 x-3+10}{\sqrt{4+1+4} \sqrt{x^{2}+9+25}}=\frac{2 x+7}{\sqrt{9} \cdot \sqrt{x^{2}+34}}$
$\Rightarrow \frac{1}{\sqrt{2}}=\frac{2 x+7}{3 \sqrt{x^{2}+34}} \Rightarrow 2 x+7=3 \sqrt{\frac{x^{2}+34}{2}}$
$\Rightarrow 4 x^{2}+49+28 x=\frac{9\left(x^{2}+34\right)}{2}$ (Squaring on both
sides)
$\Rightarrow 2\left(4 x^{2}+49+28 x\right)=9 x^{2}+306$
$\Rightarrow 8 x^{2}+98+56 x=9 x^{2}+306$
$\Rightarrow x^{2}-56 x+208=0$
$\therefore \mathrm{x}=\frac{56 \pm \sqrt{3136-812}}{2}=\frac{56 \pm 48}{2}=28 \pm 24$
$=4,52 .$
Smaller value $=4$.

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2017 (Phase 2)
If the angle between the vectors $\overrightarrow{\mathbf{a}}$ and $\overrightarrow{\mathbf{b}}$ is $\frac{\pi}{3}$, what is the
angle between $-5 \overrightarrow{\mathbf{a}}$ and $6 \overrightarrow{\mathbf{b}}$ ?

Options:
A $\frac{\pi}{6}$
B $\frac{2 \pi}{3}$
C $\frac{2 \pi}{5}$
D $\frac{3 \pi}{7}$

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

Solution:
From the figure it is clear that the angle between
$6 \overrightarrow{\mathbf{b}}$ and $-5 \overrightarrow{\mathbf{a}}$ is $120^{\circ}$ or $\frac{2 \pi}{3}$
ALTERNATE SOLUTION:
$\theta=\pi-\frac{\pi}{3}=\frac{2 \pi}{3}$

Topic: Mathematics, Vector Algebra, NDA, NDA 2010 (Phase 1)
If the angle between the vectors $\hat{i}-m \hat{j}$ and $\hat{j}+\hat{k}$ is $\frac{\pi}{3}$, then what is the value of $m ?$

Options:
A 0
B 2
C $-2$
D None of these

The correct answer is:
None of these

Solution:
$\cos \frac{\pi}{3}=\frac{(\hat{\mathrm{i}}-\mathrm{m} \hat{\mathrm{j}}) \cdot(\hat{\mathrm{j}}+\hat{\mathrm{k}})}{\left|\sqrt{1+\mathrm{m}^{2}}\right|\left|\sqrt{1^{2}+1^{2}}\right|}$
$\frac{1}{2}=\frac{-\mathrm{m}}{\sqrt{1+\mathrm{m}^{2}} \cdot \sqrt{2}}$

Topic: Mathematics, Vector Algebra, NDA, NDA 2013 (Phase 2)
If the angles of a triangle $\mathrm{ABC}$ are in the ratio $1: 2: 3$. then the corresponding sides are in the ratio

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

The correct answer is:
$1: \sqrt{3}: 2$

Solution:
Given, angles of triangle are in ratio $1: 2: 3$ Consider, $\mathrm{A}=30^{\circ}, \mathrm{B}=60^{\circ}$ and $\mathrm{C}=90^{\circ}$
We know, $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$
$\Rightarrow \frac{a}{\sin 30^{\circ}}=\frac{b}{\sin 60^{\circ}}=\frac{c}{\sin 90^{\circ}}$
$\Rightarrow \frac{a}{\frac{1}{2}}=\frac{b}{\frac{\sqrt{3}}{2}}=\frac{c}{1}$
$\Rightarrow a: b: c=\frac{1}{2}: \frac{\sqrt{3}}{2}: 1=1: \sqrt{3}: 2$

Topic: Mathematics, Properties of Triangles, NDA, NDA 2019 (Phase 1)
If the angles of a triangle are $30^{\circ}$ and $45^{\circ}$ and the included
side is $(\sqrt{3}+1)$, then what is the area of the tringle?

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

The correct answer is:
$\frac{\sqrt{3}+1}{2}$

Solution:
From $\Delta \mathrm{ADB}, \mathrm{AD}=\mathrm{BD}=\mathrm{x}$
In $\Delta \mathrm{ADC}$
$\tan 30^{\circ}=\frac{\mathrm{x}}{\sqrt{3}+1-\mathrm{x}}$
$\Rightarrow \frac{1}{\sqrt{3}}=\frac{\mathrm{x}}{\sqrt{3}+1-\mathrm{x}} \Rightarrow \sqrt{3} \mathrm{x}=\sqrt{3}+1-\mathrm{x}$
$\Rightarrow(\sqrt{3}+1) \mathrm{x}=\sqrt{3}+1$
$x=\frac{\sqrt{3}+1}{\sqrt{3}+1}=1$
Area of $\triangle \mathrm{ABC}=\frac{1}{2} \times(\sqrt{3}+1) \times 1=\frac{\sqrt{3}+1}{2}$

Topic: Mathematics, Properties of Triangles, NDA, NDA 2013 (Phase 2)
If the area of a triangle with vertices $(-3,0),(3,0)$ and $(0, k)$ is 9 sq unit, then what is the value of $k ?$

Options:
A 3
B 6
C 9
D 12

The correct answer is:
3

Solution:
Let the vertices of the $\Delta \mathrm{ABC}$ be $\mathrm{A}(-3,0), \mathrm{B}(3,0)$ and $\mathrm{C}(0, \mathrm{k})$
$\therefore \quad$ Area of $\Delta \mathrm{ABC}=\frac{1}{2}\left|\begin{array}{ccc}-3 & 0 & 1 \\ 3 & 0 & 1 \\ 0 & \mathrm{k} & 1\end{array}\right|$
Given, area is 9
$\Rightarrow \quad 9=\frac{1}{2}\{-3(-\mathrm{k})+1(3 \mathrm{k})\}$
$\Rightarrow \quad 18=3 \mathrm{k}+3 \mathrm{k}$
$\Rightarrow \quad \mathrm{k}=\frac{18}{6}=3$

Topic: Mathematics, Straight Lines, NDA, NDA 2010 (Phase 2)
If the arithmetic and geometric means of two numbers are 10 , 8 respectively, then one number exceeds the other number by

Options:
A 8
B 10
C 12
D 16

The correct answer is:
12

Solution:
Let a and $b$ be two numbers such that
A.M. $=\frac{a+b}{2}=10$
$\Rightarrow \mathrm{a}+\mathrm{b}=20$ ...(i)
and G.M. $=\sqrt{a b}=8 \Rightarrow \mathrm{ab}=64 \quad$...(2)
From (1) and (2), we have $a^{2}-20 a+64=0 \Rightarrow(a-4)(a-16)=0$
$\Rightarrow \mathrm{a}=4,16$
Thus, when $\mathrm{a}=4, \mathrm{~b}=16$ and when $\mathrm{a}=16, \mathrm{~b}=4$
Hence, one number exceeds the other number by $12 .$

Topic: Mathematics, Sequences and Series, NDA, NDA 2011 (Phase 2)
If the carbon atoms fixed by producers already have passed through three species, the trophic level of the last species would be

Options:
A
scavenger

B
tertiary producer

C
tertiary consumer

D
secondary consumer

The correct answer is:
tertiary consumer

Solution:
Producer $\rightarrow$ Ist Trophic level $\rightarrow$
(Primary consumer)
IInd Trophic level $\rightarrow$ IIIrd Trophic level (Secondary consumer) (Tertiary consumer)
Scavengers are the detrivores that feeds on the tissue of dead animals (e.g., vultures) and plays no role in carbon fixation. Producers fix the carbon dioxide which is passed through the different trophic level.

Topic: Biology, Ecosystem
If the cardinality of a set $A$ is 4 and that of a set $B$ is 3 , then what is the cardinality of the set $A \Delta B$ ?

Options:
A 1
B 5
C 7
D Cannot be determined as the sets $A$ and $B$ are not given

The correct answer is:
Cannot be determined as the sets $A$ and $B$ are not given

Solution:
Given, $n(A)=4, n(B)=3$ Since, the sets A and B are not known, then cardinality of the set $\mathrm{A} \Delta \mathrm{B}$ cannot be determined.

Topic: Mathematics, Sets and Relations, NDA, NDA 2010 (Phase 2)
If the centre of the circle passing through the origin is $(3,4)$, then the intercepts cut off by the circle on $\mathrm{x}$ -axis and $\mathrm{y}$ -axis respectively are $\quad[2015-I I]$

Options:
A 3 unit and 4 unit
B 6 unit and 4 unit
C 3 unit and 8 unit
D 6 unit and 8 unit

The correct answer is:
6 unit and 8 unit

Solution:
Equation of circle having radius $\mathrm{r}$ and centre $(3,4)$ is $=(x-3)^{2}+(y-4)^{2}=r^{2}$
if it is passing through $(0,0)$ $\therefore(0-3)^{2}+(0-4)^{2}=\mathrm{r}^{2}$
$\Rightarrow \mathrm{r}^{2}=25$
equation of circle is
$$
(x-3)^{2}+(y-4)^{2}=25
$$
putting $\mathrm{y}=0$ $\therefore \mathrm{x}=6$ unit $=$ interception $\mathrm{x}$ -axis
intercept on y axis (putting $\mathrm{x}=0$ ) is $y-8$ unil

Topic: Mathematics, Circle, NDA, NDA 2015 (Phase 2)
If the centre of the sphere $a x^{2}+b y^{2}+c z^{2}-2 x+4 y+2 z-3=0$ is $(1 / 2,-1,-1 / 2)$, what
is the value of $\mathrm{b}$ ?

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

The correct answer is:
2

Solution:
The given equation of sphere $a x^{2}+b y^{2}+c z^{2}-2 x+4 y+2 z-3=0$
This equation represents a equation of sphere, if coefficient of $x^{2}, y^{2}$ and $z^{2}$ is same.
i.e., $a=b=c$
Equation of sphere can be re-written as $b x^{2}+b y^{2}+b z^{2}-2 x+4 y+2 z-3=0$
$\Rightarrow x^{2}+y^{2}+z^{2}-\frac{2 x}{b}+\frac{4 y}{b}+\frac{2 z}{b}-\frac{3}{b}=0$
The centre of this sphere is $\left(\frac{1}{b}, \frac{-2}{b}, \frac{-1}{b}\right)$
Given that the centre of sphere is $\left(\frac{1}{2},-1,-\frac{1}{2}\right)$
$\frac{1}{b}=\frac{1}{2} \Rightarrow b=2$

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2006 (Phase 2)
If the centroid of a triangle formed by $(7, x),(y,-6)$ and (9,10) is $(6,3)$, then the values of $x$ and $y$ are respectively

Options:
A 5,2
B 2,5
C 1,0
D 0,0

The correct answer is:
5,2

Solution:
Centroid of given triangle $=\left(\frac{7+\mathrm{y}+9}{3}, \frac{\mathrm{x}-6+10}{3}\right)$
$\Rightarrow\left(\frac{16+\mathrm{y}}{3}, \frac{\mathrm{x}+4}{3}\right)=(6,3)$
$\Rightarrow 16+\mathrm{y}=18 ; \mathrm{x}+4=9$
$\Rightarrow \mathrm{y}=2 ; \mathrm{x}=5$

Topic: Mathematics, Straight Lines, NDA, NDA 2017 (Phase 1)