datavorous/entrance-exam-dataset huggingface dataset:
If the roots of the equation $3 x^{2}-5 x+q=0$ are equal, then what is the value of $q$ ?

Options:
A 2
B $5 \sqrt{1}$.
C $12 / 25$
D $25 / 12$

The correct answer is:
$25 / 12$

Solution:
Givenquadratic equation is $3 x^{2}-5 x+q=0$
Since, roots of this equation are equal therefo $b^{2}-4 a c=0$
$\Rightarrow(-5)^{2}-4(3)(q)=0$
$\Rightarrow 25-12 q=0$
$\Rightarrow q=\frac{25}{12}$

Topic: Mathematics, Quadratic Equation, NDA, NDA 2011 (Phase 2)
If the roots of the equation $4 \beta^{2}+\lambda \beta-2=0$ are of the from
$\frac{\mathrm{k}}{\mathrm{k}+1}$ and $\frac{\mathrm{k}+1}{\mathrm{k}+2}$, then what is the value of $\lambda$ ?

Options:
A $2 \mathrm{k}$
B 7
C 2
D $\mathrm{k}+1$

The correct answer is:
7

Solution:
Let $\frac{\mathrm{k}}{\mathrm{k}+1}$ and $\frac{\mathrm{k}+1}{\mathrm{k}+2}$ are the roots of the equation $4 \beta^{2}+\lambda, \beta-2=0$, then
Sum of the roots $=\frac{k}{k+1}+\frac{k+1}{k+2}=-\frac{\lambda}{4}$...(i)
and product of the roots, $\frac{k}{k+1} \times \frac{k+1}{k+2}=-\frac{2}{4}$
$\Rightarrow \frac{\mathrm{k}}{\mathrm{k}+2}=-\frac{1}{2} \Rightarrow 2 \mathrm{k}=-\mathrm{k}-2 \Rightarrow \mathrm{k}=-\frac{2}{3}$
Putting the value of $\mathrm{k}$ in(i), we get $\frac{-\frac{2}{3}}{-\frac{2}{3}+1}+\frac{-\frac{2}{3}+1}{-\frac{2}{3}+2}=-\frac{\lambda}{4}$
$\Rightarrow \frac{-\frac{2}{3}}{\frac{1}{3}}+\frac{\frac{1}{3}}{\frac{4}{3}}=-\frac{\lambda}{4} \Rightarrow-2+\frac{1}{4}=-\frac{\lambda}{4}$
$\Rightarrow \lambda=7$

Topic: Mathematics, Quadratic Equation, NDA, NDA 2006 (Phase 1)
If the roots of the equation $\left(a^{2}+b^{2}\right) x^{2}-2 b(a+c) x+\left(b^{2}+c^{2}\right)=0$ are equal, then
which one of the following is correct?

Options:
A $2 b=a+c$
B $b^{2}=a c$
C $b+c=2 a$
D $\bar{b}=a c$

The correct answer is:
$b^{2}=a c$

Solution:
Since, the roots of the equation $\left(a^{2}+b^{2}\right) x^{2}-2 b(a+c) x+\left(b^{2}+c^{2}\right)=0$ are equal.
therefore, discriminant $=0$
$\Rightarrow[2 b(a+c)]^{2}-4\left(a^{2}+b^{2}\right)\left(b^{2}+c^{2}\right)=0$
$4 b^{2}(a+c)^{2}=4\left(a^{2}+b^{2}\right)\left(b^{2}+c^{2}\right)$
$\Rightarrow b^{2}\left(a^{2}+c^{2}+2 a c\right)=a^{2} b^{2}+b^{2} c^{2}+a^{2} c^{2}+b^{4}$
$\Rightarrow a^{2} b^{2}+b^{2} c^{2}+2 a c b^{2}=a^{2} b^{2}+b^{2} c^{2}+a^{2} c^{2}+b^{4}$
$\Rightarrow b^{4}+a^{2} c^{2}-2 a c b^{2}=0$
$\Rightarrow\left(b^{2}-a c\right)^{2}=0 \Rightarrow b^{2}=a c$

Topic: Mathematics, Quadratic Equation, NDA, NDA 2010 (Phase 2)
If the roots of the equation $x^{2}-4 x-\log _{3} N=0$ are real, then what is the minimum value of $N ?$

Options:
A $1 / 256$
B $1 / 27$
C $1 / 81$
D $1 / 6$

The correct answer is:
$1 / 6$

Solution:
Given equation is $x^{2}-4 x-\log _{3} N=0$
Since, roots are real $\therefore b^{2}-4 a c=0 \Rightarrow(4)^{2}-4\left(-\log _{3} N\right) \geq 0$
$\Rightarrow 16 \geq-4 \log _{3} N$
$\Rightarrow 4 \geq-\log _{3} N$
$\Rightarrow 4 \geq \log _{3} N^{-1}$
$\Rightarrow N^{-1} \geq 3^{4} \geq 81$
$\Rightarrow N \geq \frac{1}{81}$
Hence, minimum val ue of $\mathrm{N}$ is $\frac{1}{81}$.

Topic: Mathematics, Quadratic Equation, NDA, NDA 2011 (Phase 2)
If the roots of the equation $x^{2}-n x+m=0$ differ by 1, then

Options:
A $\mathrm{n}^{2}-4 \mathrm{~m}-1=0$
B $\mathrm{n}^{2}+4 \mathrm{~m}-1=0$
C $\quad \mathrm{m}^{2}+4 \mathrm{n}+1=0$
D $\mathrm{m}^{2}-4 \mathrm{n}-1=0$

The correct answer is:
$\mathrm{n}^{2}-4 \mathrm{~m}-1=0$

Solution:
Let the root be $\alpha$ and $\beta$ $x^{2}-n x+m=0$
$\Rightarrow \alpha+\beta=n ; \alpha . \beta=\mathrm{m}$
$\alpha-\beta=$
$\Rightarrow(\alpha+\beta)^{2}=(\alpha-\beta)^{2}+4 \alpha \beta$
$\Rightarrow \mathrm{n}^{2}=1+4 \mathrm{~m}$
- $n^{2}-4 m-1=0$

Topic: Mathematics, Quadratic Equation, NDA, NDA 2015 (Phase 2)
If the roots of the equation $x^{2}+p x+q=0$ are in the same ratio as those of the equation $x^{2}+b x+m=0$, then which one of the following is correct?

Options:
A $\mathrm{p}^{2} \mathrm{~m}=R^{2} \mathrm{q}$
B $\mathrm{m}^{2} \mathrm{p}=\hat{\mathrm{p}}$
C $m^{2} p=q^{2} y$
D $m^{2} p^{2}=R^{2} q$

The correct answer is:
$\mathrm{p}^{2} \mathrm{~m}=R^{2} \mathrm{q}$

Solution:
Let $\mathrm{a}, \mathrm{b}$ be the roots of $x^{2}+p / x^{2}+q=0$
$\mathrm{So}, \mathrm{a}+\mathrm{b}=-\mathrm{p} / \mathrm{x}^{2}, \mathrm{ab}=\mathrm{q}$...(i)
Let $c, d$ be the roots of $x^{2}+b x+m=0$
So, $\mathrm{c}+\mathrm{d}=-1, \mathrm{~cd}=\mathrm{m}$...(ii)
Given that roots of both the equations are in the
Same ratio
So, $\frac{a}{b}=\frac{c}{d} \quad$...(iii)
$\Rightarrow \frac{\mathrm{b}}{\mathrm{a}}=\frac{\mathrm{d}}{\mathrm{c}}\ldots(\mathrm{iv})$
$(\mathrm{iii})+(\mathrm{iv}) \Rightarrow \frac{\mathrm{a}}{\mathrm{b}}+\frac{\mathrm{b}}{\mathrm{a}}=\frac{\mathrm{c}}{\mathrm{d}}+\frac{\mathrm{d}}{\mathrm{c}}$
$\Rightarrow \frac{\mathrm{a}^{2}+\mathrm{b}^{2}}{\mathrm{ab}}=\frac{\mathrm{c}^{2}+\mathrm{d}^{2}}{\mathrm{ed}} \Rightarrow \frac{\mathrm{a}^{2}+\mathrm{b}^{2}}{\mathrm{ab}}+2=\frac{\mathrm{c}^{2}+\mathrm{d}^{2}}{\mathrm{~cd}}+2$
$\Rightarrow \frac{\mathrm{a}^{2}+\mathrm{b}^{2}+2 \mathrm{ab}}{\mathrm{ab}}=\frac{\mathrm{c}^{2}+\mathrm{d}^{2}+2 \mathrm{~cd}}{\mathrm{~cd}}$
$\Rightarrow \frac{(\mathrm{a}+\mathrm{b})^{2}}{\mathrm{ab}}=\frac{(\mathrm{c}+\mathrm{d})^{2}}{\mathrm{ed}}$
$\Rightarrow \frac{(-\mathrm{P})^{2}}{\mathrm{q}}=\frac{(-1)^{2}}{\mathrm{~m}} \quad($ from (i) and(ii))
$\Rightarrow \mathrm{P}^{2} \mathrm{~m}=1^{2} \mathrm{q}$

Topic: Mathematics, Quadratic Equation, NDA, NDA 2017 (Phase 1)
If the roots of the equation $x^{2}+p x+q=0$ are $\tan 19^{\circ}$ and $\tan 26^{\circ}$, then which one of the following is correct?

Options:
A $q-p=1$
B $\mathrm{p}-\mathrm{q}=1$
C $\mathrm{p}+\mathrm{q}=2$
D $\mathrm{p}+\mathrm{q}=3$

The correct answer is:
$q-p=1$

Solution:
Given, $\tan 19^{\circ}$ and $\tan 26^{\circ}$ are roots of $x^{2}+p x+q=0$
$\therefore \tan 19^{\circ}+\tan 26^{\circ}=\frac{-p}{1}=-p$
$\left(\tan 19^{\circ}\right)\left(\tan 26^{\circ}\right)=\frac{q}{1}=q$
$\tan \left(19^{\circ}+26^{\circ}\right)=\frac{\tan 19^{\circ}+\tan 26^{\circ}}{1-\tan 19^{\circ} \tan 26^{\circ}}$
$\tan 45^{\circ}=\frac{-p}{1-q} \Rightarrow 1=\frac{-p}{1-q}$
$\Rightarrow 1-q=-p$
$\Rightarrow q-p=1$

Topic: Mathematics, Quadratic Equation, NDA, NDA 2019 (Phase 1)
If the roots of the equations $x^{2}-(a-1) x+(a+b)=0$ and $a x^{2}-2 x+b=0$ are identical, then what are the values of a and $\mathrm{b}$ ?

Options:
A $a=2, b=4$
B $a=2, b=-4$
C $\mathrm{a}=1, \mathrm{~b}=\frac{1}{2}$
D $\mathrm{a}=-1, \mathrm{~b}=-\frac{1}{2}$

The correct answer is:
$a=2, b=-4$

Solution:
Let $\alpha$ and $\beta$ be the roots of both the equations $x^{2}-(a-1) x+(a+b)=0$
$\Rightarrow \alpha+\beta=(a-1)$ and $\alpha \beta=(a+b)$
and $a x^{2}-2 x+b=0$
$\Rightarrow \alpha+\beta=\frac{2}{a}$ and $\alpha \beta=\frac{b}{a}$
Equating the sums of roots
$a^{2}-a-2=0 \Rightarrow a=-1,2$
Equating the products of roots and
and if $a=2, b=$ $1=$
From the given option, $=-4$ matches.

Topic: Mathematics, Quadratic Equation, NDA, NDA 2007 (Phase 1)
If the roots of $x^{2}-2 m x+m^{2}-1=0$ lie between $-2$ and 4 , then which one of the following is correct?

Options:
A $-1 \leq \mathrm{m} \leq 3$
B $-3 \leq \mathrm{m} \leq 3$
C $-3 \leq \mathrm{m} \leq 5$
D $-1 \leq \mathrm{m} \leq 5$

The correct answer is:
$-3 \leq \mathrm{m} \leq 3$

Solution:
Since, the roots of $x^{2}-2 m x+m^{2}-1=0$ lie between
$-2$ and 4 i.e., $\mathrm{b} 2-4 \mathrm{ac} \geq 0$ and $-2 < \frac{-\mathrm{b}}{2 \mathrm{a}} < 4$
$(2 m)^{2}-4\left(m^{2}-1\right) \geq 0$...(1)
and $-2 < \frac{2 \mathrm{~m}}{2} < 4$
$\Rightarrow-2 < m < 4$
From (i) $4 m^{2}-4 m^{2}+4 \geq 0$
$\Rightarrow m \in R$
and $\mathrm{f}(-2)>0$, also $\mathrm{f}(4)>0$
$4+4 m+m^{2}-1>0,16-8 m+m^{2}-1>0$
$\Rightarrow \mathrm{m}^{2}+4 \mathrm{~m}+3>0, \mathrm{~m}^{2}-8 \mathrm{~m}+15>0$
$\Rightarrow(m+1)(m+3)>0,(m-3)(m-5)>0$
$\Rightarrow-3 < \mathrm{m} < -1$ and $3 < \mathrm{m} < 5$
Thus, the inteval in which it lies is $-1 \leq \mathrm{m} \leq 5$

Topic: Mathematics, Quadratic Equation, NDA, NDA 2006 (Phase 1)
If the second term of a GP is 2 and the sum of its infinite term is 8, then the GP is

Options:
A $8,2, \frac{1}{2}, \frac{1}{8}, \ldots . .$
B $10,2, \frac{2}{5}, \frac{2}{25}, \ldots .$
C $4,2,1, \frac{1}{2}, \frac{1}{2^{2}}, \ldots . .$
D $6,3, \frac{3}{2}, \frac{3}{4}, \ldots . .$

The correct answer is:
$4,2,1, \frac{1}{2}, \frac{1}{2^{2}}, \ldots . .$

Solution:
Let the first term and common ratio of the G.P. is a and
r respectively.
$\begin{aligned} \text { Then, } a r=2, \frac{a}{1-r}=& 8 \\ \Rightarrow \quad \frac{a r}{r-r^{2}}=8 \quad \Rightarrow \quad 2=8 r-8 r^{2} \\ \Rightarrow \quad 4 r^{2}-4 r+1=0 \quad \Rightarrow \quad r=\frac{1}{2} \\ \therefore \text { GP. is } 4,2,1, \frac{1}{2}, \ldots . \end{aligned}$

Topic: Mathematics, Sequences and Series, NDA, NDA 2018 (Phase 2)
If the sequence $\left\{\mathrm{S}_{\mathrm{n}}\right\}$ is a geometric progression and $\mathrm{S}_{2} \mathrm{~S}_{11}=\mathrm{S}_{\mathrm{p}} \mathrm{S}_{8}$, then what is the value of $\mathrm{p}$ ?

Options:
A 1
B 3
C 5
D cannot be determined

The correct answer is:
5

Solution:
Given $\mathrm{S}_{2} \mathrm{~S}_{11}=\mathrm{S}_{\mathrm{p}} \mathrm{S}_{8}$
$\Rightarrow($ ar $)\left(\mathrm{ar}^{10}\right)=\mathrm{ar}^{\mathrm{P}-1}\left(\mathrm{ar}^{7}\right)$
where 'a' is the first term and ' $\mathrm{r}$ ' is the common ratio of GP. $\Rightarrow r^{11}=r^{7+p-1} \Rightarrow r^{11}=r^{6+p}$
$\Rightarrow 11=6+\mathrm{p} \Rightarrow \mathrm{p}=5$

Topic: Mathematics, Sequences and Series, NDA, NDA 2012 (Phase 1)
If the sides of a triangle are $6 \mathrm{~cm}, 10 \mathrm{~cm}$ and $14 \mathrm{~cm}$, then what is the largest angle included by the sides?

Options:
A $90^{\circ}$
B $120^{\circ}$
C $135^{\circ}$
D $150^{\circ}$

The correct answer is:
$120^{\circ}$

Solution:
We know that largest side has greatest angle opposite it. $\therefore \mathrm{a}=14 \mathrm{~cm}, \mathrm{~b}=10 \mathrm{~cm}$ and $\mathrm{c}=6 \mathrm{~cm}$
$\therefore \cos A=\frac{c^{2}+b^{2}-a^{2}}{2 b c}$
$=\frac{36+100-196}{2 \times 6 \times 10}=-\frac{1}{2}=\cos 120^{\circ}$
$\Rightarrow \angle \mathrm{A}=120^{\circ}$

Topic: Mathematics, Properties of Triangles, NDA, NDA 2009 (Phase 1)
If the sides of a triangle are in the ratio $2: \sqrt{6}: 1+\sqrt{3}$, then what is the smallest angle of the triangle?

Options:
A $75^{\circ}$
B $60^{\circ}$
C $45^{\circ}$
D $30^{\circ}$

The correct answer is:
$45^{\circ}$

Solution:
Let $\mathrm{ABC}$ be a triangle with sides $\mathrm{a}=1+\sqrt{3}, \mathrm{~b}=2$ and
$c=\sqrt{6}$
So, $\cos B=\frac{a^{2}+c^{2}-b^{2}}{2 a c}=\frac{(1+\sqrt{3})^{2}+(\sqrt{6})^{2}-4}{2(1+\sqrt{3})(\sqrt{6})}$
$=\frac{2 \sqrt{3}+6}{2 \sqrt{6}+\sqrt{18}}=\frac{3+\sqrt{3}}{\sqrt{6}+3 \sqrt{2}}$
$=\frac{\sqrt{3} \sqrt{3}+\sqrt{3}}{(\sqrt{3}+\sqrt{3} \sqrt{3}) \sqrt{2}}=\frac{1}{\sqrt{2}}$
$\Rightarrow B=45^{\circ}$ is the smallest angle.
$(\because$ smallest side is $b=2$ )

Topic: Mathematics, Properties of Triangles, NDA, NDA 2011 (Phase 2)
If the slopes of the line of regression of $Y$ and $X$ and of $X$ and $\mathrm{Y}$ are $30^{\circ}$ and $60^{\circ}$ respectively, then $\mathrm{r}(\mathrm{X}, \mathrm{Y})$ is:

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

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

Solution:
Slope of line of regression of $\mathrm{Y}$ and $\mathrm{X}$, is $30^{\circ}$. So $\mathrm{b}_{\mathrm{yx}}=\tan 30^{\circ}=\frac{1}{\sqrt{3}}$ and for $\mathrm{X}$ and $\mathrm{Y}$ it is $60^{\circ}$
Hence, $\frac{1}{\mathrm{~b}_{\mathrm{xy}}}=\tan 60^{\circ}=\sqrt{3}$
$b_{y x}=\frac{1}{\sqrt{3}}$ and $b_{x y}=\frac{1}{\sqrt{3}}$
$\mathrm{r}(\mathrm{x}, \mathrm{y})=\mathrm{r}^{2}=\mathrm{b}_{\mathrm{yx}} \cdot \mathrm{b}_{\mathrm{xy}}=\frac{1}{3}$
so, $\mathrm{r}=\pm \frac{1}{\sqrt{3}}$
Since, $b_{y x}$ and $b_{x y}$ are both positive, $r=+\frac{1}{\sqrt{3}}$

Topic: Mathematics, Statistics, NDA, NDA 2007 (Phase 1)
If the solution of the differential equation $\quad$
$\frac{d y}{d x}=\frac{a x+3}{2 y+f}$
represents a circle, then what is the value of $a$ ?

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

The correct answer is:
$-2$

Solution:
Given differential equation is
$$
\frac{d y}{d x}=\frac{a x+3}{2 y+f}
$$
By separating the variable we get $(2 y+f) d y=(a x+3) d x$
Integrate on both side,
$$
\begin{array}{l}
\int(2 y+f) d y=\int(a x+3) d x \\
\Rightarrow \quad y^{2}+f y=\frac{a x^{2}}{2}+3 x
\end{array}
$$
This equation represents a circle, if
$$
-1=\frac{a}{2} \Rightarrow a=-2
$$

Topic: Mathematics, Differential Equations, NDA, NDA 2011 (Phase 1)
If the speed of light (c), acceleration due to gravity (g) and pressure (p) are taken as the fundamental quantities, then the dimension of gravitational constant is

Options:
A c 2 g 0 p −2
B c 0 g 2 p −1
C c g 3 p −2
D c −1 g 0 p −1

The correct answer is:
c 0 g 2 p −1

Solution:
Let [G]∝ c x g y p z

by substituting the following dimensions :

[G]=[M−1L3T−2],[c]=[LT−1],[g]=[LT−2]

[p]=[ML−1T−2]

and by comparing the powers of both sides we can get x=0, y=2, z=−1
∴ [G]∝ c 0 g 2 p −1

Topic: Physics, Units and Dimensions, NEET
If the standard deviation of data is 12 and mean is 72 , then coefficient of variation is

Options:
A $15.67 \%$
B $14.67 \%$
C $13.67 \%$
D $16.67 \%$

The correct answer is:
$16.67 \%$

Solution:
$\begin{aligned} & \text { Coefficient of variation }=\frac{\text { Standard Deviation }}{\text { Mean }} \times 100 \\ & \frac{12}{72} \times 100 \%=16.67 \%\end{aligned}$

Topic: Mathematics, Statistics, MHT CET, MHT CET 2021 (24 Sep Shift 1)
If the standard deviation of first $n$ natural numbers is 2 , then the value of $n$ is

Options:
A 7
B 5
C 4
D 6

The correct answer is:
7

Solution:
$\begin{aligned} & \text { S.D. }=\sqrt{\frac{\sum x_i^2}{n}-(\bar{x})^2} \\ & \Rightarrow 2=\sqrt{\frac{\sum n^2}{n}-\left(\frac{\sum n}{n}\right)^2} \\ & \Rightarrow 4=\frac{n(n+1)(2 n+1)}{6 n}-\left(\frac{n(n+1)}{2 n}\right)^2 \\ & \Rightarrow 4=\frac{n+1}{2}\left\{\frac{2 n+1}{3}-\frac{n+1}{2}\right\} \\ & \Rightarrow 4=\frac{n+1}{2}\left(\frac{n-1}{6}\right) \\ & \Rightarrow n^2=49 \Rightarrow n=7\end{aligned}$

Topic: Mathematics, Statistics, MHT CET, MHT CET 2022 (07 Aug Shift 1)
If the standard deviation of the numbers -1 , $0,1, k$ is $\sqrt{5}$ where $k>0$, then $k$ is equal to

Options:
A $4 \sqrt{\frac{5}{3}}$
B $\sqrt{6}$
C $2 \sqrt{\frac{10}{3}}$
D $2 \sqrt{6}$

The correct answer is:
$2 \sqrt{6}$

Solution:
Given, numbers are $-1,0,1, k$.
Standard deviation, $\sigma=\sqrt{5}$
$$
\sigma^2=\frac{\Sigma x_i^2}{n}-\left(\frac{\Sigma x_i}{n}\right)^2
$$
$$
\begin{aligned}
& \Rightarrow \quad 5=\frac{1+0+1+k^2}{4}-\left(\frac{-1+0+1+k}{4}\right)^2 \\
& \Rightarrow \quad 5=\frac{2+k^2}{4}-\frac{k^2}{16} \Rightarrow 80=8+4 k^2-k^2 \\
& \Rightarrow \quad 72=3 k^2 \Rightarrow k^2=24 \Rightarrow k=2 \sqrt{6} .
\end{aligned}
$$

Topic: Mathematics, Statistics, KCET, KCET 2022
If the standard deviation of the numbers 2,3 , \(2 x\) and 11 is 3.5. Find the possible values of \(x\).

Options:
A \(2 \cdot \frac{7}{2}\)
B \(3 \cdot \frac{5}{3}\)
C \(2 \cdot \frac{5}{2}\)
D \(3 \cdot \frac{7}{3}\)

The correct answer is:
\(3 \cdot \frac{7}{3}\)

Solution:
\(\sigma=3.5\)
\(\begin{gathered}
\text {Mean }(\bar{x})=\frac{2+3+2 x+11}{4}=\left(4+\frac{x}{2}\right) \\
\sigma^2=\frac{1}{n} \sum\left(x_i-\bar{x}\right)^2 \\
\Rightarrow \quad(3 \cdot 5)^2=\frac{1}{4}\left[\left(-2-\frac{x}{2}\right)^2+\left(-1-\frac{x}{2}\right)^2+\left(\frac{3 x}{2}-4\right)^2+\left(7-\frac{x}{2}\right)^2\right] \\
\Rightarrow 3 x^2-16 x+21=0 \Rightarrow x=\frac{7}{3}, 3
\end{gathered}\)

Topic: Mathematics, Statistics, AP EAMCET, AP EAMCET 2020 (17 Sep Shift 1)
If the standard deviation of the random variable X is 3pq and mean is 3p then Ex2=….

Options:
A 3pq+3q2
B 3p1+2p
C 3pq+3p2
D 3q1+2q ]

The correct answer is:
3p1+2p

Solution:
Key Idea: Use p +q =1 and VarX=EX2-EX2
We have standard deviation of X = 3pq
⇒VarX=3pq
and mean, EX=3p
Now, 3pq=Ex2-3p2
âµVarX=EX2-EX2
⇒EX2=3pq+9p2
=3p1-p+9p2âµp+q=1
=3p-3p2+9p2
⇒3p+6p2⇒3p1+2p

Topic: Mathematics, Probability, MHT CET, MHT CET 2019 (Shift 2)
If the standard enthalpy change $\left(\Delta_{\mathrm{r}} \mathrm{H}^\theta\right)$ for a certain reaction at $298 \mathrm{~K}$ and constant pressure is $-1860 \mathrm{~kJ}$ $\mathrm{mol}^{-1}$, the standard entropy change $\left(\Delta_{\text {sys }} \mathrm{s}^\theta\right)$ of the same reaction is $-550 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$, which one of the following statements is correct?

Options:
A $\left(\Delta_{\text {sys }} S^\theta\right)+\Delta_{\text {surr }} S^\theta=-7692 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$, the reaction is spontaneous
B $\left(\Delta_{\text {sys }} \mathrm{S}^\theta\right)+\Delta_{\text {surr }} \mathrm{S}^\theta=-5692 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$, the reaction is non-spontaneous
C $\left(\Delta_{\text {sys }} \mathrm{S}^\theta\right)+\Delta_{\text {surr }} \mathrm{S}^\theta=+5692 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$, the reaction is spontaneous
D $\left(\Delta_{\text {sys }} S^\theta\right)+\Delta_{\text {surr }} S^\theta=+7692 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$, the reaction is non-spontaneous

The correct answer is:
$\left(\Delta_{\text {sys }} \mathrm{S}^\theta\right)+\Delta_{\text {surr }} \mathrm{S}^\theta=+5692 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$, the reaction is spontaneous

Solution:
$$
\begin{aligned}
& \text { } \Delta \mathrm{G}^{\circ}=\Delta \mathrm{H}^{\circ}-\mathrm{T} \Delta \mathrm{s}^{\circ} \\
& =\left(-1860 \times 10^3\right)-298(-550) \\
& =-1696.1 \mathrm{~kJ} \mathrm{Mol}^{-1}
\end{aligned}
$$
Thus, the reaction is Spontaneous as $\Delta \mathrm{G}^{\circ}$ is negative. Now, Since the system loses the heat, the surrounding must gain it and therefore,
$$
\begin{aligned}
& \Delta \mathrm{H}^{\circ}{ }_{\text {Surrounding }}=+1860 \times 10^3 \mathrm{~J} \mathrm{~mol}^{-1} . \\
& \Rightarrow \Delta \mathrm{S}_{\text {surrounding }}^{\circ}=\frac{\Delta \mathrm{H}_{\text {surrounding }}^{\circ}}{\mathrm{T}}
\end{aligned}
$$
$\begin{aligned} & =\frac{+1860 \times 10^3}{298} \\ & =6241.6 \mathrm{~J} \\ & \Rightarrow \Delta S_{\text {total }}^{\circ}=\Delta S_{\text {system }}^{\circ}+\Delta S_{\text {surrounding }}^{\circ} \\ & =(-550)+(6241.6) \\ & =+5692 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1} .\end{aligned}$

Topic: Chemistry, Thermodynamics (C), AP EAMCET, AP EAMCET 2023 (15 May Shift 1)
If the standard enthalpy change $\left(\Delta_{\mathrm{r}} \mathrm{H}^\theta\right)$ for reaction $\mathrm{H}_2(\mathrm{~g})+\mathrm{Br}_2(\mathrm{l}) \rightarrow 2 \mathrm{HBr}(\mathrm{g})$ is $-72.8 \mathrm{~kJ}$, the standard enthalpy of formation $\left(\Delta_{\mathrm{f}} \mathrm{H}^\theta\right)$ of $\mathrm{HBr}(\mathrm{g})$ (in $\left.\mathrm{kJ} \mathrm{mol}^{-1}\right)$ is

Options:
A -36.4
B +36.4
C -18.2
D -18.2

The correct answer is:
-36.4

Solution:
$\begin{aligned} & \text { } \Delta \mathrm{H}_{\mathrm{f}}^{\circ}\left(\mathrm{H}_2 \text { gas }\right)=0.0 \mathrm{~kJ} \mathrm{~mol}^{-1}, \Delta \mathrm{H}_{\mathrm{f}}^{\circ}\left(\mathrm{Br}_2 \text { liquid) }\right. \\ & =0.0 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ & \Delta \mathrm{H}_{\mathrm{T}}^{\circ}=-72.8 \mathrm{~kJ} \text { for } 2 \text { moles of } \mathrm{HBr} \text {. } \\ & \Rightarrow \Delta \mathrm{H}_{\mathrm{r}}^{\circ}=\left[2 \Delta \mathrm{H}_{\mathrm{f}}^{\circ} \text { ( } \mathrm{H} \mathrm{Br} \text { gas }\right)-\left(\Delta \mathrm{H}_{\mathrm{f}}^{\circ}\left(\mathrm{H}_2 \text { gas }\right)\right) \\ & \left.+\left(\Delta \mathrm{H}_{\mathrm{f}}{ }^0\left(\mathrm{Br}_2 \text { liq. }\right)\right)\right] \\ & -72.8=2 \Delta \mathrm{H}_{\mathrm{f}}^{\circ}(\mathrm{HBr} \text { gas })-(0.0+0.0) \\ & =2 \Delta \mathrm{H}_{\mathrm{f}}^{\circ}(\mathrm{HBr} \text { gas }) \\ & \Rightarrow \Delta \mathrm{H}_{\mathrm{f}}^{\circ}(\mathrm{HBr} \text { gas })=\frac{-72.8}{2}=,-36.4 \mathrm{~kJ} \mathrm{~mol}^{-1} \text {. } \\ & \end{aligned}$

Topic: Chemistry, Thermodynamics (C), AP EAMCET, AP EAMCET 2023 (15 May Shift 1)
If the standard enthalpy of sublimation $\left(\Delta_{\mathrm{gos}} \mathrm{H}^{\circ}\right)$ of solid $\mathrm{CO}_2$, naphthalene, $\mathrm{Li}$ and $\mathrm{Na}$ are $25.2,73.0,162,108$ $\mathrm{kJ}$ mol respectively, the order of sublimation of these substances with temperature is

Options:
A $\mathrm{Na}>\mathrm{Li}>$ Naphthalene $>$ Solid $\mathrm{CO}_2$
B Solid $\mathrm{CO}_2>\mathrm{Na}>$ Naphthalene $>\mathrm{Li}$
C Solid $\mathrm{CO}_2>$ Naphthalene $>\mathrm{Na}>\mathrm{Li}$
D $\mathrm{Li}>\mathrm{Na}>$ Naphthalene $>$ Solid $\mathrm{CO}_2$

The correct answer is:
Solid $\mathrm{CO}_2>$ Naphthalene $>\mathrm{Na}>\mathrm{Li}$

Solution:
The lower the sublimation enthalpy, the lower is the heat required to sublime it and easier for that substance to get sublimed.
Thus, the order of ease of sublimation will be:solid $\mathrm{CO}_2>\mathrm{Na}>\mathrm{Na}>\mathrm{Li}$

Topic: Chemistry, Thermodynamics (C), AP EAMCET, AP EAMCET 2023 (16 May Shift 1)
If the straight linc $\frac{x-x_{0}}{\ell}=\frac{y-y_{0}}{m}=\frac{z-z_{0}}{n}$ is parallcl to the plane $a x+b y+c z+d=0$ then which one of the following is correct?

Options:
A $\ell+\mathrm{m}+\mathrm{n}=0$
B $a+b+c=0$
C $\frac{\mathrm{a}}{\ell}+\frac{\mathrm{b}}{\mathrm{m}}+\frac{\mathrm{c}}{\mathrm{n}}=0$
D $\mathrm{a} \ell+\mathrm{bm}+\mathrm{cn}=0$

The correct answer is:
$\mathrm{a} \ell+\mathrm{bm}+\mathrm{cn}=0$

Solution:
If the line is parallel to the plane then $\mathrm{a} \ell+\mathrm{bm}+\mathrm{cn}=0$

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2013 (Phase 1)
If the straight lines $x-2 y=0$ and $k x+y=1$ intersect at the
point $\left(1, \frac{1}{2}\right)$, then what is the value of $k ?$

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

The correct answer is:
$1 / 2$

Solution:
Since. the straight lines $x-2 y=0$ and $k x+y=1$ intersect at the point $\left(1, \frac{1}{2}\right)$.
$\therefore \quad$ The point $\left(1, \frac{1}{2}\right)$ satisfies the equation $k x+y=1$
$\therefore \quad$ Put $x=1$, and $y=\frac{1}{2}$ in $\mathrm{eq}^{\mathrm{n}} k x+y=1$
we get $k .1+\frac{1}{2}=1 \Rightarrow k=\frac{1}{2}$

Topic: Mathematics, Straight Lines, NDA, NDA 2010 (Phase 2)
If the sum of $\mathrm{m}$ terms of an AP is $\mathrm{n}$ and the sum of $\mathrm{n}$ terms is $\mathrm{m}$, then the sum of $(\mathrm{m}+\mathrm{n})$ terms is

Options:
A $m n$
B $m+n$
C $2(m+n)$
D $-(m+n)$

The correct answer is:
$-(m+n)$

Solution:
Given, $\mathrm{S}_{\mathrm{m}}=\mathrm{n}$ and $\mathrm{S}_{\mathrm{n}}=\mathrm{m}$
This is direct formula $\mathrm{S}_{\mathrm{m}+\mathrm{n}}=-(\mathrm{m}+\mathrm{n})$

Topic: Mathematics, Sequences and Series, NDA, NDA 2017 (Phase 1)
If the sum of ' $n$ ' terms of an arithmetic progression is $n^{2}-2 n$, then what is the $n^{\text {th }}$ term?

Options:
A $3 n-n^{2}$
B $2 n-3$
C $2 n+3$
D $2 n-5$

The correct answer is:
$2 n-3$

Solution:
Given,
$\begin{aligned} & \mathrm{S}_{\mathrm{n}}=\mathrm{n}^{2}-2 \mathrm{n} \\ \therefore & \quad \mathrm{ a}_{\mathrm{n}}=\mathrm{S}_{\mathrm{n}}-\mathrm{S}_{\mathrm{n}-1} \\=& \mathrm{n}^{2}-2 \mathrm{n}-\left[(\mathrm{n}-1)^{2}-2(\mathrm{n}-1)\right] \\=& \mathrm{n}^{2}-2 \mathrm{n}-\left[\mathrm{n}^{2}+1-2 \mathrm{n}-2 \mathrm{n}+2\right]=2 \mathrm{n}-3 \end{aligned}$

Topic: Mathematics, Sequences and Series, NDA, NDA 2008 (Phase 2)