datavorous/entrance-exam-dataset huggingface dataset:
If the sum of the first two terms and the sum of the first four terms of a geometric progression with positive common ratio are 8 and 80 respectively, then what is the 6 th term?

Options:
A 88
B 243
C 486
D 1458

The correct answer is:
486

Solution:
Let the Geometric progression be a, ar, ar $^{2}$, ar $^{3}, \ldots$ with common ratio $\mathrm{r}$ and first term 'a '. According to the question, we have
$$
\mathrm{a}+\mathrm{ar}=8 \Rightarrow \mathrm{a}(1+\mathrm{r})=8 ...(i)
$$
and $\mathrm{a}+\mathrm{ar}+\mathrm{ar}^{2}+\mathrm{ar}^{3}=80$
$\Rightarrow \mathrm{a}(1+\mathrm{r})+\operatorname{ar}^{2}(1+\mathrm{r})=80$
$\Rightarrow \mathrm{a}(1+\mathrm{r})\left(1+\mathrm{r}^{2}\right)=80$
$\Rightarrow 8\left(1+r^{2}\right)=80 \quad$ (from (i))
$\Rightarrow 1+r^{2}=\frac{80}{8}=10$
$\Rightarrow \mathrm{r}^{2}=10-1=9$
$\Rightarrow \mathrm{r}=3 \quad(\because \mathrm{r}>0)$
From eq. (i), $a(1+3)=8$ $\Rightarrow \mathrm{a}=2$
Now, $6^{\text {th }}$ term $=$ ar $^{5}=2(3)^{5}=2 \times 243=486$

Topic: Mathematics, Sequences and Series, NDA, NDA 2009 (Phase 1)
If the sum of the matrices $\left[\begin{array}{l}\mathrm{x} \\ \mathrm{x} \\ \mathrm{y}\end{array}\right],\left[\begin{array}{l}\mathrm{y} \\ \mathrm{y} \\ \mathrm{z}\end{array}\right]$ and $\left[\begin{array}{l}\mathrm{z} \\ 0 \\ 0\end{array}\right]$ is the matrix $\left[\begin{array}{c}10 \\ 5 \\ 5\end{array}\right]$, then what is the value of $\mathrm{y} ?$

Options:
A $-5$
B 0
C 5
D 10

The correct answer is:
0

Solution:
We have,
$\left[\begin{array}{l}\mathrm{x} \\ \mathrm{x} \\ \mathrm{y}\end{array}\right]+\left[\begin{array}{l}\mathrm{y} \\ \mathrm{y} \\ \mathrm{z}\end{array}\right]+\left[\begin{array}{l}\mathrm{z} \\ 0 \\ 0\end{array}\right]=\left[\begin{array}{c}10 \\ 5 \\ 5\end{array}\right]$
$\Rightarrow\left[\begin{array}{c}x+y+z \\ x+y \\ y+z\end{array}\right]=\left[\begin{array}{c}10 \\ 5 \\ 5\end{array}\right]$
$\Rightarrow x+y+z=10$ ...(i)
$x+y=5$ ...(ii)
$y+z=5$ ...(iii)
from (ii), $x=5-y$ from (iii), $z=5-y$ $\therefore$ from(i),
$5-y+y+5-y=10$
$\Rightarrow 10-y=10 \Rightarrow y=0$

Topic: Mathematics, Matrices, NDA, NDA 2012 (Phase 2)
If the sum of the roots of a quadratic equation is 3 and the product is 2, then the equation is

Options:
A $\quad 2 x^{2}-x+3=0$
B $x^{2}-3 x+2=0$
C $x^{2}+3 x+2=0$
D $x^{2}-3 x-2=0$

The correct answer is:
$x^{2}-3 x+2=0$

Solution:
Quadratic equation can be given as $x^{2}-($ sum of roots $) x+($ product of roots $)=0$ Hence, Required quadratic equation is $x^{2}-3 x+2=0$

Topic: Mathematics, Quadratic Equation, NDA, NDA 2013 (Phase 1)
If the sum of the rootsof the equation $a x^{2}+b x+c=0$ is equal to the sum of their squares, then

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

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

Solution:
$a \mathrm{x}^{2}+\mathrm{bx}+\mathrm{c}=0$
Let the root be $\alpha$ and $\beta$.
$\alpha+\beta=-\frac{\mathrm{b}}{\mathrm{a}}, \alpha, \beta=\frac{\mathrm{c}}{\mathrm{a}}$
$\Rightarrow \alpha+\beta=\alpha^{2}+\beta^{2} \quad \ldots$ (given)
$\Rightarrow \alpha+\beta=(\alpha+\beta)^{2}-2 \alpha \beta$
$-\frac{\mathrm{b}}{\mathrm{a}}=\left(-\frac{\mathrm{b}}{\mathrm{a}}\right)^{2}-\frac{2 \mathrm{c}}{\mathrm{a}}$
$\Rightarrow \mathrm{b}^{2}+\mathrm{ab}=2 \mathrm{ac}$

Topic: Mathematics, Quadratic Equation, NDA, NDA 2015 (Phase 2)
If the sum of the squares of the distance of the point $(\mathrm{x}, \mathrm{y}, \mathrm{z})$ from the points $(\mathrm{a}, 0,0)$ and $(-\mathrm{a}, 0,0)$ is $2 \mathrm{c}^{2}$, then which one of the following is correct?

Options:
A $x^{2}+a^{2}=2 c^{2}-y^{2}-z^{2}$
B $x^{2}+a^{2}=c^{2}-y^{2}-z^{2}$
C $x^{2}-a^{2}=c^{2}-y^{2}-z^{2}$
D $x^{2}+a^{2}=c^{2}+y^{2}+z^{2}$

The correct answer is:
$x^{2}+a^{2}=c^{2}-y^{2}-z^{2}$

Solution:
Let the point be $\mathrm{P}(\mathrm{x}, \mathrm{y}, \mathrm{z})$ and two points, $(\mathrm{a}, 0,0)$ and $(-a, 0,0)$ be $A$ and $B$
As given in the problem, $\mathrm{PA}^{2}+\mathrm{PB}^{2}=2 \mathrm{c}^{2}$
$\mathrm{so},(\mathrm{x}+\mathrm{a})^{2}+(\mathrm{y}-0)^{2}+(\mathrm{z}-0)^{2}$
$+(\mathrm{x}-\mathrm{a})^{2}+(\mathrm{y}-0)^{2}+(\mathrm{z}-0)^{2}=2 \mathrm{c}^{2}$
or, $(\mathrm{x}+\mathrm{a})^{2}+\mathrm{y}^{2}+\mathrm{z}^{2}+(\mathrm{x}-\mathrm{a})^{2}+\mathrm{y}^{2}+\mathrm{z}^{2}=2 \mathrm{c}^{2}$
$\mathrm{x}^{2}+2 \mathrm{ax}+\mathrm{a}^{2}+\mathrm{y}^{2}+\mathrm{z}^{2}+\mathrm{x}^{2}-2 \mathrm{ax}+\mathrm{a}^{2}+\mathrm{y}^{2}+\mathrm{z}^{2}=2 \mathrm{c}^{2}$
$=2\left(\mathrm{x}^{2}+\mathrm{y}^{2}+\mathrm{z}^{2}+\mathrm{a}^{2}\right)=2 \mathrm{c}^{2}$
$=\mathrm{x}^{2}+\mathrm{y}^{2}+\mathrm{z}^{2}+\mathrm{a}^{2}=\mathrm{c}^{2}$
$=\mathrm{x}^{2}+\mathrm{a}^{2}=\mathrm{c}^{2}-\mathrm{y}^{2}-\mathrm{z}^{2}$

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2007 (Phase 1)
If the sum of the squares of the distances of the point $(x, y)$ from the points $(a, 0)$ and $(-a, 0)$ is $2 b^{2}$, then which one of the following is correct?

Options:
A $x^{2}+a^{2}=b^{2}+y^{2}$
B $x^{2}+a^{2}=2 b^{2}-y^{2}$
C $x^{2}-a^{2}=b^{2}+y^{2}$
D $x^{2}+a^{2}=b^{2}-y^{2}$

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

Solution:
Let $\mathrm{P}(x, y)$ be a point and $\mathrm{A}=(a, 0), \mathrm{B}=(-a, 0)$. Now, $\mathrm{PA}^{2}=(x-a)^{2}+y^{2}$
$\mathrm{PB}^{2}=(x+a)^{2}+y^{2}$
Since the sum of the distances of the point $\mathrm{P}(x, y)$ from the points $\mathrm{A}(a, 0)$ and $\mathrm{B}(-a, 0)$ is $2 b^{2}$. $\therefore \quad \mathrm{PA}^{2}+\mathrm{PB}^{2}=2 b^{2}$
$(x-a)^{2}+(y)^{2}+(x+a)^{2}+(y)^{2}=2 b^{2}$
$\Rightarrow x^{2}+a^{2}-2 a x+y^{2}+x^{2}+a^{2}+2 a x+y^{2}=2 b^{2}$
$\Rightarrow x^{2}+a^{2}+y^{2}=b^{2}$
$\Rightarrow x^{2}+a^{2}=b^{2}-y^{2}$

Topic: Mathematics, Straight Lines, NDA, NDA 2011 (Phase 1)
If the sum of the squares of the roots of $\mathrm{x}^{2}-(\mathrm{p}-2) \mathrm{x}-(\mathrm{p}+1)=0(\mathrm{p} \in \mathrm{R})$ is 5
then what is the value of $\mathrm{p}$ ?

Options:
A 0
B $-1$
C 1
D $\frac{3}{2}$

The correct answer is:
1

Solution:
Let $\alpha$ and $\beta$ be the roots of $x^{2}-(p-2) x-(p+1)=0$
Then $\alpha+\beta=(p-2)$ and $\alpha \beta=-(p+1)$
$\alpha^{2}+\beta^{2}=(\alpha+\beta)^{2}-2 \alpha \beta$
$\Rightarrow(\alpha+\beta)^{2}-2 \alpha \beta=5$
$\Rightarrow(p-2)^{2}+2(p+1)=5$
$\Rightarrow \mathrm{p}^{2}-4 \mathrm{p}+4+2 \mathrm{p}+2=5$
$\Rightarrow \mathrm{p}^{2}-2 \mathrm{p}+1=0$
$\Rightarrow(p-1)^{2}=0$
$\Rightarrow \mathrm{p}=1$

Topic: Mathematics, Quadratic Equation, NDA, NDA 2006 (Phase 1)
If the system of equations $2 x+3 y=7$ and $2 a x+(a+b) y=28$ has infinitely many solutions, then which one of the following is correct?

Options:
A $a=2 b$
B $b=2 a$
C $a=-2 b$
D $b=-2 a$

The correct answer is:
$b=2 a$

Solution:
The given system of equations has infinitely many solution, then
$\frac{2}{2 a}=\frac{3}{a+b}=\frac{7}{28}$
$\Rightarrow \quad a=4$ and $12=a+b$
$\Rightarrow \mathrm{a}=4$ and $b=8 \Rightarrow b=2 a$
$ALTERNATE SOLUTION$ : Given equations are $2 x+3 y=7$
$2 a x+(a+b) y=28$
Matrix form by these equations is $\left[\begin{array}{cc}2 & 3 \\ 2 a & (a+b)\end{array}\right]$
As we know if value of determinant is zero then system of equations have infinitely many solutions.
So, $\left|\begin{array}{cc}2 & 3 \\ 2 a & a+b\end{array}\right|=0$
$\Rightarrow 2 a+2 b-3 \times 2 a=0$
$\Rightarrow 2 a+2 b-6 a=0$
$\Rightarrow \quad 2 b-4 a=0 \Rightarrow b=2 a$

Topic: Mathematics, Determinants, NDA, NDA 2010 (Phase 1)
If the three consecutive vertices of a parallelogram are $(-2,-1),(1,0)$ and $(4,3)$, then what are the coordinates of the fourth vertex?

Options:
A $(1,2)$
B $(1,0)$
C $(0,0)$
D $(1,-1)$

The correct answer is:
$(1,2)$

Solution:
Given vertices of parallelogram are $(-2,-1),(1,0),(4,3)$
Let the fourth vertex be $(\mathrm{x}, \mathrm{y})$ We know, in a parallelogram diagonals bisect each other.
i.e, Midpoint of $\mathrm{AC}=$ Midpoint of $\mathrm{BD}$ $\Rightarrow\left(\frac{-2+4}{2}, \frac{-1+3}{2}\right)=\left(\frac{\mathrm{X}+1}{2}, \frac{\mathrm{Y}+0}{2}\right)$
$\Rightarrow(1,1)=\left(\frac{\mathrm{X}+1}{2}, \frac{\mathrm{Y}+0}{2}\right)$
$\Rightarrow \mathrm{X}+1=2 ; \mathrm{Y}+0=2$
$\Rightarrow \mathrm{X}=1, \mathrm{Y}=2$

Topic: Mathematics, Straight Lines, NDA, NDA 2017 (Phase 1)
If the three observations are $3,-6$ and $-6$, then what is their harmonic mean?

Options:
A 0
B $\infty$
C $-1 / 2$
D $-3$

The correct answer is:
$\infty$

Solution:
Harmonic mean of three number $\mathrm{x} 1, \mathrm{x} 2, \mathrm{x} 3$ is
$\frac{3}{\frac{1}{x_{1}}+\frac{1}{x_{2}}+\frac{1}{x_{3}}}$
$\therefore$ H. $M=\frac{3}{\frac{1}{3}+\left(\frac{-1}{6}\right)+\left(\frac{-1}{6}\right)}=\frac{3}{\frac{1}{3}-\frac{1}{3}}=\frac{3}{0}=\infty$

Topic: Mathematics, Sequences and Series, NDA, NDA 2008 (Phase 2)
If the three vertices of the parallelogram $A B C D$ are $A(1, a), B$ $(3, a), C(2, b)$, then $D$ is equal to

Options:
A $(3, b)$
B $(6, b)$
C $(4, b)$
D $(5, b)$

The correct answer is:
$(4, b)$

Solution:
$\mathrm{ABCD}$ is a parallelogram.
$\mathrm{AB} \| \mathrm{DC}$, then slope of line $\mathrm{AB}=$ slope of line $\mathrm{DC} \frac{\mathrm{a}-\mathrm{a}}{3-1}$
$=\frac{b-y}{2-x} \Rightarrow y=b$
Also, $\mathrm{AD} \| \mathrm{BC}, \frac{\mathrm{a}-\mathrm{y}}{1-\mathrm{x}}=\frac{\mathrm{a}-\mathrm{b}}{3-2}$
$\frac{a-b}{1-x}=\frac{a-b}{3-2} \Rightarrow \frac{a-b}{1-x} \times \frac{1}{a-b}=1$
$1-x=1$
$\mathrm{x}=0$
Then points $\mathrm{D}$ are $(0, \mathrm{~b})$

Topic: Mathematics, Straight Lines, NDA, NDA 2013 (Phase 2)
If the total number of observations is $20, \Sigma x_{i}=1000$ and
$\Sigma x_{i}^{2}=84000$, then what is the variance of the distribution?

Options:
A 1500
B 1700
C 1600
D 180

The correct answer is:
1600

Solution:
Total no. of observation $(n)=20$
$\Sigma x_{i}=1000$
$\bar{x}=\frac{\Sigma x_{i}}{n}=\frac{1000}{20}=50$
Variance $=s d$
$s d=\sqrt{\frac{1}{n} \Sigma x_{i}^{2}-(\bar{x})^{2}}$
$(s d)^{2}=\frac{1}{n} \Sigma x_{i}^{2}-(\bar{x})^{2}=\frac{84000}{20}-(50)^{2}$
$=4200-2500=1700$.
Variance $=1700$

Topic: Mathematics, Statistics, NDA, NDA 2016 (Phase 1)
If the units of length, mass and force are chosen as fundamental units, the dimensions of time would be

Options:
A $\left[\mathrm{M}^{1 / 2} \mathrm{~L}^{-1 / 2} \mathrm{~F}^{1 / 2}\right]$
B $\left[\mathrm{M}^{1 / 2} \mathrm{~L}^{1 / 2} \mathrm{~F}^{1 / 2}\right]$
C $\left[\mathrm{M}^{1 / 2} \mathrm{~L}^{1 / 2} \mathrm{~F}^{-1 / 2}\right]$
D $\left[\mathrm{M}^{1 / 2} \mathrm{~L}^{-1 / 2} \mathrm{~F}^{-1 / 2}\right]$

The correct answer is:
$\left[\mathrm{M}^{1 / 2} \mathrm{~L}^{1 / 2} \mathrm{~F}^{-1 / 2}\right]$

Solution:
$\begin{aligned} \text { As } F=\left[\mathrm{MLT}^{-2}\right] \text { or } & {\left[\mathrm{T}^2\right]=\frac{[\mathrm{ML}]}{[\mathrm{F}]} } \\ \text { or }[\mathrm{T}] & =\left[\mathrm{M}^{1 / 2} \mathrm{~L}^{1 / 2} \mathrm{~F}^{-1 / 2}\right]\end{aligned}$

Topic: Physics, Units and Dimensions, AIIMS, AIIMS 2014
If the value of the determinant $\left|\begin{array}{lll}a & 1 & 1 \\ 1 & b & 1 \\ 1 & 1 & c\end{array}\right|$ is positive, where $\mathrm{a} \neq \mathrm{b} \neq \mathrm{c}$, then the value of abc

Options:
A cannot be less than 1
B is greater than-8
C is less than $-8$
D must be greater than 8

The correct answer is:
is greater than-8

Solution:
$\left|\begin{array}{lll}\mathrm{a} & 1 & 1 \\ 1 & \mathrm{~b} & 1 \\ 1 & 1 & \mathrm{c}\end{array}\right|>0$
$\Rightarrow a(b c-1)-1(c-1)+1(1-b)>0$
$\Rightarrow a b c-a-c+1+1-b>0$
$\Rightarrow a b c+2-(a+b+c)>0$
$\Rightarrow a b c>(a+b+c)-2$
Let; $\mathrm{a}=-1 ; \mathrm{b}=0 \& \mathrm{c}=1$
Then; $0>-2$ [which is correct $]$ Hence, $\mathrm{abc}=0$
$\therefore$ After considering all the option; (b) is correct option.s

Topic: Mathematics, Determinants, NDA, NDA 2015 (Phase 2)
If the values of a set are measured in $\mathrm{cm}$, what will be the $\begin{array}{ll}\text { unit of variance? } & {}\end{array}$

Options:
A $\mathrm{cm}$
B $\mathrm{cm}^{2}$
C $\mathrm{cm}^{3}$
D No unit

The correct answer is:
No unit

Solution:
If the values of a set are measured in $\mathrm{cm}$, then there will not be unit of variance

Topic: Mathematics, Statistics, NDA, NDA 2011 (Phase 1)
If the variance of $6,7,8,9,10,11$ is $\sigma^2$, then the variance of $12,14,16,18,20,22$ is

Options:
A $2 \sigma^2$
B $4 \sigma^2$
C $100+2 \sigma^2$
D $100+4 \sigma^2$

The correct answer is:
$4 \sigma^2$

Solution:
Its a fundamental concept if each data is doubled then variance is 4 times.
So, new variance $=(2)^2 \sigma^2=4 \sigma^2$

Topic: Mathematics, Statistics, AP EAMCET, AP EAMCET 2018 (24 Apr Shift 1)
If the variance of a data is 121 , then the standard deviation of the data is

Options:

None

Solution:
11

Topic: Mathematics, Statistics
If the variance of the data $2,4,5,6,17$ is $\mathrm{v}$, then what is the variance of the data $4,8,10,12,34 ?$

Options:
A $v_{2}$
B $4 v$
C $v^{2}$
D $2 v$

The correct answer is:
$2 v$

Solution:
Given, variance of the data $2,4,5,6,17$ is $v$ Variance of the data $4,8,10,12,34$ is $2 v$. because when each observation is multiplied by 2, ther variance is also multiplied by 2

Topic: Mathematics, Statistics, NDA, NDA 2011 (Phase 2)
If the variance of the distribution
\begin{array}{cccccccc}
\hline \boldsymbol{x}_{\boldsymbol{i}} & 4 & 8 & 11 & 17 & 20 & 24 & 32 \\
\hline \boldsymbol{f}_{\boldsymbol{i}} & 3 & 5 & 9 & 5 & 4 & 3 & 1
\end{array}
is 45.8 , then the variance of the distribution.
\begin{array}{llllllll}
\hline \boldsymbol{x}_{\boldsymbol{i}} & 10 & 18 & 24 & 36 & 42 & 50 & 66 \\
\hline \boldsymbol{f}_{\boldsymbol{i}} & 3 & 5 & 9 & 5 & 4 & 3 & 1 \\
\hline
\end{array}

Options:
A 93.6
B $\sqrt{93.9}$
C 183.2
D $\sqrt{183.2}$

The correct answer is:
183.2

Solution:
Mean of given observation $\left(x_i\right)$ is
$$
\begin{array}{r}
\frac{(4 \times 3)+(8 \times 5)+(11 \times 9)+(17 \times 5)+(20 \times 4)}{+(24 \times 3)+(32 \times 1)} \\
=\frac{420}{30}=14
\end{array}
$$
So,
$$
\begin{aligned}
\mathrm{ce} & =k \frac{10^2+6^2+3^2+3^2+6^2+10^2+18^2}{30} \\
& =45.8 \text { (given) }
\end{aligned}
$$
Now, mean of given observation $\left(y_i\right)$ is
$$
\begin{array}{r}
\begin{array}{r}
(10 \times 3)+(18 \times 5)+(24 \times 9)+(36 \times 5)+(42 \times 4) \\
+(50 \times 3)+(66 \times 1)
\end{array} \\
\begin{aligned}
30 & \frac{900}{30}=30
\end{aligned}
\end{array}
$$
So, variance
$$
\begin{aligned}
& =k \frac{20^2+12^2+6^2+6^2+12^2+20^2+36^2}{30} \\
& =4 \times k\left(\frac{10^2+6^2+3^2+3^2+6^2+10^2+18^2}{30}\right) \\
& =4 \times 45.8=183.2
\end{aligned}
$$

Topic: Mathematics, Statistics, AP EAMCET, AP EAMCET 2018 (23 Apr Shift 2)
If the vectors $\hat{\mathrm{i}}-2 \mathrm{x} \hat{\mathrm{j}}-3 \mathrm{y} \hat{\mathrm{k}}$ and $\hat{\mathrm{i}}+3 \mathrm{x} \hat{\mathrm{j}}+2 \mathrm{y} \hat{\mathrm{k}}$ are
orthogonal to each other, then what is the locus of the point $(\mathrm{x}, \mathrm{y}) ?$

Options:
A A circle
B An ellipse
C A parabola
D A hyperbola

The correct answer is:
A circle

Solution:
Let the given vectors $\hat{\mathrm{i}}-2 \mathrm{x} \hat{\mathrm{j}}-3 \mathrm{y} \hat{\mathrm{k}}$ and $\hat{\mathrm{i}}+3 \mathrm{x} \hat{\mathrm{j}}+2 \mathrm{y} \widehat{\mathrm{k}}$.be $\overrightarrow{\mathrm{A}}$ and $\overrightarrow{\mathrm{B}}$
respectively, and $\theta$ be the angle between
them, so, $\cos \theta=\frac{\overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{B}}}{|\overrightarrow{\mathrm{A}} \| \overrightarrow{\mathrm{B}}|}$
These are orthogonal to each other, $\theta=\pi / 2$ so, $\overrightarrow{\mathrm{A} \cdot \overrightarrow{\mathrm{B}}}=0$
$\Rightarrow 1-(2 x)(3 x)-(3 y)(2 y)=0$
$\Rightarrow 1-6 x^{2}-6 y^{2}=0$
$\Rightarrow x^{2}+y^{2}=\frac{1}{6}$
This equation represents an equation of a circle which is the locus of the point $(\mathrm{x}, \mathrm{y})$

Topic: Mathematics, Vector Algebra, NDA, NDA 2006 (Phase 2)
If the vectors $-\hat{i}-2 x \hat{j}-3 y \hat{k}$ and $\hat{i}-3 x \hat{j}-2 y \hat{k}$ are
orthogonal to each other, then what is the locus of the point $(x, y) ?$

Options:
A a straight line
B an ellipse
C a parabola
D a circle

The correct answer is:
a circle

Solution:
Given vectors $-\hat{i}-2 x \hat{j}-3 y \hat{k}$ and $\hat{i}-3 x \hat{j}-2 y \hat{k}$
are orthogonal to each other.
$\therefore(-\hat{i}-2 x \hat{j}-3 y \hat{k}) \cdot(\hat{i}-3 x \hat{j}-2 y \hat{k})=0$
$\Rightarrow(-1)(1)+(-2 x)(-3 x)+(-3 y)(-2 y)=0$
$\Rightarrow-1+6 x^{2}+6 y^{2}=0$
$\Rightarrow 6 x^{2}+6 y^{2}=1$
$\Rightarrow x^{2}+y^{2}=\left(\frac{1}{\sqrt{6}}\right)^{2}$
Hence, locus of $(x, y)$ is a circle.

Topic: Mathematics, Vector Algebra, NDA, NDA 2011 (Phase 2)
If the vectors $\hat{\mathbf{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}, \hat{\mathrm{i}}+\hat{\mathrm{b}}+\hat{\mathrm{k}}$ and $\hat{\mathrm{i}}=\hat{\mathrm{j}}=\mathrm{c} \hat{\mathrm{k}}(\mathrm{a}, \mathrm{b}, \mathrm{c} \neq 1)$
are coplanar, then the value of $\frac{1}{1-\mathrm{a}}+\frac{1}{1-\mathrm{b}}+\frac{1}{1-\mathrm{c}}$ is equal to

Options:
A 0
B 1
C $a+b+c$
D abc

The correct answer is:
1

Solution:
$a \hat{i}+\hat{j}+\hat{k}, \hat{i}+b \hat{j}+\hat{k}$ and $\hat{i}+\hat{j}+c \hat{k}$ are coplanar.
i.e., $\left|\begin{array}{lll}\mathrm{a} & 1 & 1 \\ 1 & \mathrm{~b} & 1 \\ 1 & 1 & \mathrm{c}\end{array}\right|=0$
$c_{2} \rightarrow c_{2}-c_{3} ; c_{3} \rightarrow c_{3}-c_{1}$
$\left|\begin{array}{ccc}a & 1-a & 1-a \\ 1 & b-1 & 0 \\ 1 & 0 & c-1\end{array}\right|=0$
$\Rightarrow \mathrm{a}(\mathrm{b}-1)(\mathrm{c}-1)-(1-\mathrm{a})(\mathrm{c}-1)-(1-\mathrm{a})(\mathrm{b}-1)=0$
Divide both sides by $(1-a)(1-b)(1-c)$
$\begin{aligned} \Rightarrow & \frac{\mathrm{a}(\mathrm{b}-1)(\mathrm{c}-1)}{(1-\mathrm{a})(1-\mathrm{b})(1-\mathrm{c})}-\frac{(1-\mathrm{a})(\mathrm{c}-1)}{(1-\mathrm{a})(1-\mathrm{b})(1-\mathrm{c})} \\ & \quad-\frac{(1-\mathrm{a})(\mathrm{b}-1)}{(1-\mathrm{a})(1-\mathrm{b})(1-\mathrm{c})}=\frac{0}{(1-\mathrm{a})(1-\mathrm{b})(1-\mathrm{c})} \\ \Rightarrow & \frac{\mathrm{a}}{1-\mathrm{a}}+\frac{1}{1-\mathrm{b}}+\frac{1}{1-\mathrm{c}}=0 \\ \Rightarrow & \frac{1}{1-\mathrm{b}}+\frac{1}{1-\mathrm{c}}=-\frac{\mathrm{a}}{1-\mathrm{a}} \end{aligned}$
Add $\frac{1}{1-\mathrm{a}}$ on both sides.
$\Rightarrow \frac{1}{1-\mathrm{a}}+\frac{1}{1-\mathrm{b}}+\frac{1}{1-\mathrm{c}}=\frac{1}{1-\mathrm{a}}-\frac{\mathrm{a}}{1-\mathrm{a}}=\frac{1-\mathrm{a}}{1-\mathrm{a}}=1 .$

Topic: Mathematics, Vector Algebra, NDA, NDA 2017 (Phase 2)
If the vectors $\alpha \hat{\mathrm{i}}+\alpha \hat{\mathrm{j}}+\gamma \hat{\mathrm{k}}, \hat{\mathrm{i}}+\hat{\mathrm{k}}$ and $\gamma \hat{\mathrm{i}}+\hat{\mathrm{j}}+\beta \hat{\mathrm{k}}$ lie on a
plane, where $\alpha, \beta$ and $\gamma$ are distinct non-negative numbers, then $\gamma$ is

Options:
A Arithmetic mean of $\alpha$ and $\beta$
B Geometric mean of $\alpha$ and $\beta$
C Harmonic mean of $\alpha$ and $\beta$
D None of the above

The correct answer is:
Geometric mean of $\alpha$ and $\beta$

Solution:
If three vectors are co-planar.
$\Rightarrow\left|\begin{array}{lll}\alpha & \alpha & \gamma \\ 1 & 0 & 1 \\ \gamma & \gamma & \beta\end{array}\right|=0$
$\Rightarrow \quad \alpha[0-\gamma]-\alpha[\beta-\gamma]+\gamma[\gamma-0]=0$
$\Rightarrow-\alpha \gamma-\alpha \beta+\alpha \gamma+\gamma^{2}=0$
$\Rightarrow \quad \gamma^{2}=\alpha \beta$
$\Rightarrow \quad$ So $\alpha, \gamma, \beta$ are in G.P.

Topic: Mathematics, Vector Algebra, NDA, NDA 2015 (Phase 2)
If the vectors $\hat{\mathrm{i}}-\mathrm{x} \hat{\mathrm{j}}-\mathrm{y} \hat{\mathrm{k}}$ and $\hat{\mathrm{i}}+\mathrm{x} \hat{\mathrm{j}}+\mathrm{y} \hat{\mathrm{k}}$ are orthogonal to
each other, then what is the locus of the point $(\mathrm{x}, \mathrm{y})$ ?

Options:
A a parabola
B an ellipse
C a circle
D a straight line

The correct answer is:
a circle

Solution:
Since both vectors are orthogonal $\therefore$ their dot product is zero.
$\therefore 1(1)+(-x)(x)+(-y)(y)=0$
$\Rightarrow 1-x^{2}-y^{2}=0$
$\Rightarrow x^{2}+y^{2}=1$
Which is a circle.

Topic: Mathematics, Vector Algebra, NDA, NDA 2012 (Phase 1)
If the vectors $\overrightarrow{\mathrm{k}}$ and $\overrightarrow{\mathrm{A}}$ are parallel to each other, then what
is $\overrightarrow{k k} \times \vec{A}$ equal to?

Options:
A $\mathrm{k}^{2} \overrightarrow{\mathrm{A}}$
B $\overrightarrow{0}$
C $-\mathrm{k}^{2} \overrightarrow{\mathrm{A}}$
D $\overrightarrow{\mathrm{A}}$

The correct answer is:
$\overrightarrow{0}$

Solution:
Given, $\overrightarrow{\mathrm{k}}$ and $\overrightarrow{\mathrm{A}}$ are parallel vectors.
we know, cross product of parallel vectors is $\overrightarrow{\mathrm{O}}$.

Topic: Mathematics, Vector Algebra, NDA, NDA 2018 (Phase 1)
If the velocity of light (c) , gravitational constant (G) and Planck's constant (h) are chosen as fundamental units, then the dimensions of mass in this new system is

Options:
A c 1/2 G 1/2 h 1/2
B c 1/2 G 1/2 h −1/2
C c 1/2 G −1/2 h 1/2
D c −1/2 G 1/2 h 1/2

The correct answer is:
c 1/2 G −1/2 h 1/2

Solution:
Let mâ C x G y h z

By substituting the following dimensions : [C]=L T â1 ; [G]=[ M â1 L 3 T â2 ] and [h]=[M L 2 T â1 ]

Now, comparing both sides, we get

x=1/2; y=â1/2, z=+1/2

So, mâ c 1/2 G â1/2 h 1/2

Topic: Physics, Units and Dimensions, NEET
If the velocity of light $C$, the gravitational constant $G$ and Planck's constant $h$ are chosen as the fundamental units, the dimension of density in the new system is

Options:
A $C^3 G^{-2} h^1$
B $C^5 G^{-2} h^{-1}$
C $C^{-3 / 2} G^{-1 / 2} h^{1 / 2}$
D $C^{9 / 2} G^{-1 / 2} h^{-1 / 2}$

The correct answer is:
$C^5 G^{-2} h^{-1}$

Solution:
Let $\rho=\mathrm{kC}^{\mathrm{a}} \mathrm{G}^{\mathrm{b}} \mathrm{h}^{\mathrm{c}}$
$\begin{aligned}
& \Rightarrow \quad[\rho]=[\mathrm{C}]^{\mathrm{a}}[\mathrm{G}]^{\mathrm{b}}[\mathrm{h}]^{\mathrm{c}} \\
& \Rightarrow \quad \mathrm{ML}^{-3}=\mathrm{L}^{\mathrm{a}} \mathrm{T}^{-\mathrm{a}} \mathrm{M}^{-\mathrm{b}} \mathrm{L}^{3 \mathrm{~b}} \mathrm{~T}^{-2 \mathrm{~b}} \mathrm{M}^{\mathrm{c}} \mathrm{L}^{2 \mathrm{c}} \mathrm{T}^{-\mathrm{c}} \\
& \Rightarrow \quad 1=-\mathrm{b}+\mathrm{c},-3=\mathrm{a}+3 \mathrm{~b}+2 \mathrm{c}, 0=-\mathrm{a}-2 \mathrm{~b}-\mathrm{c} \\
& \Rightarrow \quad \mathrm{a}=5, \mathrm{~b}=-2 \text { and } \mathrm{c}=-1
\end{aligned}$

Topic: Physics, Units and Dimensions, TS EAMCET, TS EAMCET 2022 (20 Jul Shift 1)
If the voltage in the primary coil of a step-up transformer with a transformation ratio 3:2 is 30 V, then the voltage in the secondary coil is

Options:
A 45 V

B 15 V

C 90 V

D 300 V

The correct answer is:
45 V

Solution:
Transformation ratio k=Ns/Np,

since Vs/Vp=Ns/Npâ´Vs=Ns/NpÃVp

Topic: Physics, Alternating Current, NEET
If three events $A, B, C$ are mutually exclusive, then which one of the following is correct ?

Options:
A $\quad P(A \cup B \cup C)=0$
B $\quad P(A \cup B \cup C)=1$
C $\quad P(A \cap B \cap C)=0$
D $\quad P(A \cap B \cap C)=1$

The correct answer is:
$\quad P(A \cap B \cap C)=0$

Solution:
A, B and C are mutually exclusive $A \cap B \cap C=0$
$\mathrm{P}(\mathrm{A} \cap \mathrm{B} \cap \mathrm{C})=0$

Topic: Mathematics, Probability, NDA, NDA 2013 (Phase 2)
If two circles A, B of equal radii pass through the centres of each other, then what is the ratio of the length of the smaller arc to the circumference of the circle A cut off by the circle $\mathrm{B} ?$

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

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

Solution:
When two circles A and B of equal radii pass through the centres of each other, the angle made by arc of $\mathrm{B}$ at
the centre of $\mathrm{B}$ is $90^{\circ}$.
So, length of small $\operatorname{arc}$ of $\mathrm{B}=\frac{2 \pi \mathrm{r} 90^{\circ}}{360^{\circ}}=\frac{\pi \mathrm{r}}{2}$
Hence, circumference of A cut off by the circle $\mathrm{B}$
$=2 \pi r-\frac{\pi r}{2}=\frac{3 \pi r}{2}$
$\therefore \quad$ Required ratio $=\frac{\pi \mathrm{r} / 2}{3 \pi \mathrm{r} / 2}=\frac{1}{3}$

Topic: Mathematics, Circle, NDA, NDA 2006 (Phase 2)