datavorous/entrance-exam-dataset huggingface dataset:
A pipe open at one end has length $0.8 \mathrm{~m}$. At the open end of the tube a string $0.5 \mathrm{~m}$ long is vibrating in its $1^{\text {st }}$ overtone and resonates with fundamental frequency of pipe. If tension in the string is $50 \mathrm{~N}$, the mass of string is (speed of sound $=320 \mathrm{~m} / \mathrm{s}$ )

Options:
A 25 gram
B 15 gram
C 20 gram
D 10 gram

The correct answer is:
10 gram

Solution:
$2 \times\left[\frac{1}{2 \ell_{1}} \sqrt{\frac{T}{m}}\right]=\frac{v}{4 \ell_{2}}$
$\frac{1}{0.5} \sqrt{\frac{50}{m}}=\frac{320}{4 \times 0.8}$
$\therefore \quad$ Total mass of the string $=0.02 \times 0.5 \mathrm{~kg}=10 \mathrm{gm}$
$\therefore \quad 02 \mathrm{~kg} / \mathrm{m}$

Topic: Physics, Waves and Sound, MHT CET, MHT CET 2020 (12 Oct Shift 1)
A plane P passes through the line of intersection of the planes $2 x-y+3 z=2, x+y-z=1$ and the point $(1,0,1)$. If the plane $\mathrm{P}$ touches the sphere $\mathrm{x}^{2}+\mathrm{y}^{2}+\mathrm{z}^{2}=\mathrm{r}^{2}$, then what is $r$ equal to?

Options:
A $\frac{2}{\sqrt{29}}$
B $\frac{4}{\sqrt{29}}$
C $\frac{5}{\sqrt{29}}$
D 1

The correct answer is:
$\frac{5}{\sqrt{29}}$

Solution:
Plane $\mathrm{P}$ touches the sphere $\mathrm{x}^{2}+\mathrm{y}^{2}+\mathrm{z}^{2}=\mathrm{r}^{2}$ then
$\mathrm{r}=$ Distane between centre of sphere $(0,0,0)$ to plance $\mathrm{P}$.
$\Rightarrow \mathrm{r}=\left|\frac{5(0)+2(0)-5}{\sqrt{5^{2}+2^{2}+(0)^{2}}}\right|$
$\frac{5}{\sqrt{25+4}}$

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2016 (Phase 1)
A plane P passes through the line of intersection of the planes $2 x-y+3 z=2, x+y-z=1$ and the point $(1,0,1)$. What are the direction ratios of the line of intersection of the given planes?

Options:
A $\langle 2,-5,-3\rangle$
B $\langle 1,-5,-3\rangle$
C $\langle 2,5,3\rangle$
D $\langle 1,3,5\rangle$

The correct answer is:
$\langle 2,-5,-3\rangle$

Solution:
Tofind, intersection point first put $z=0$ $2 x-y=2$
$\frac{x+y=1}{3 x=3}$
$x+y=$
$y=1-$
$\mathrm{At} x=1, v=0$
$(x, y, z)=(1,0,0)$
Putting $x=0$
$y-z=1$
$\frac{-y+3 z=2}{2 z=3}$
$z=\frac{3}{2}$
$y=1+z=1+\frac{3}{2}=\frac{5}{2}$
$(x, y, z) \equiv\left(0, \frac{5}{2}, \frac{3}{2}\right)$
Point of intersection $\left(x_{1}, y_{1}, z_{1}\right)=(1,0,0)$
$\left(x_{2}, y_{2}, z_{2}\right)=\left(0, \frac{5}{2}, \frac{3}{2}\right)$
Hence direction ratios of the line of intersection of given plane $ < 2,-5,-3>$

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2016 (Phase 1)
A plane P passes through the line of intersection of the planes $2 x-y+3 z=2, x+y-z=1$ and the point $(1,0,1)$. What is the equation of the plane P? $\quad$

Options:
A $\quad 2 x+5 y-2=0$
B $5 x+2 y-5=0$
C $x+z-2=0$
D $2 x-y-2 z=0$

The correct answer is:
$5 x+2 y-5=0$

Solution:
Eq. of plane through two given planes is:
$(2 x-y+3 z-2)+\lambda(x+y-z-1)=0$
It passes through $(1,0,1)$ $\therefore 3-\lambda=0 \Rightarrow \lambda=3$
Eq. of plane is:
$5 x+2 y-5=0$

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2016 (Phase 1)
A plane passes through the point (3,5,7). If the direction ratios of its normal are equal to the intercepts made by the plane x+3y+2z=9 with the coordinate axes, then the equation of that plane is

Options:
A x+y+z=5
B 6x+2y+3z=105
C 12x+4y+6z=49
D 6x+2y+3z=49

The correct answer is:
6x+2y+3z=49

Solution:
Given:

x+3y+2z=9

⇒x9+y3+z92=1

Comparing with the standard form xa+yb+zc=1, we get

a=9→x-intercept

b=3→y-intercept

c=92→z-intercept

So, the normal vector for the required plane is n→=9i^+3j^+92k^

a→=3i^+5j^+7k^

Equation of required plane,

⇒r→·n→=a→·n→

⇒r→·9i^+3j^+92k^=3i^+5j^+7k^·9i^+3j^+92k^

⇒r→·9i^+3j^+92k^=27+15+632

⇒xi^+yj^+zk^·9i^+3j^+92k^=42+632

⇒9x+3y+92z=1472

⇒3x+y+32z=492

⇒6x+2y+3z=49.

Topic: Mathematics, Three Dimensional Geometry, AP EAMCET, AP EAMCET 2018 (25 Apr Shift 1)
A plane wave front of wavelength $\lambda$ is incident on a single slit of width $b$. What is the angular width for secondary maximum?

Options:
A $\frac{\lambda}{2 b}$
B $\frac{\lambda}{b}$
C $\frac{2 \lambda}{b}$
D $\frac{b}{\lambda}$

The correct answer is:
$\frac{\lambda}{b}$

Solution:
$\beta=\frac{\lambda D}{d}$ and $\theta=\frac{\beta}{D}$
$\therefore$
$$
\theta=\frac{\lambda}{d}
$$
Width of single slit $=b$
So,
$$
\theta=\frac{\lambda}{b}
$$

Topic: Physics, Wave Optics, MHT CET, MHT CET 2011
A point is chosen at random inside a circle. What is the probability that the point is closer to the centre of the circle than to its boundary?

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

The correct answer is:
$\frac{1}{4}$

Solution:
Probability $=\frac{\pi\left(\frac{\mathrm{r}}{2}\right)^{2}}{\pi \mathrm{r}^{2}}$
$=\frac{\pi\left(\frac{r^{2}}{4}\right)}{A r^{2}}$
$=\frac{r^{2}}{4 r^{2}}=\frac{1}{4}$

Topic: Mathematics, Probability, NDA, NDA 2017 (Phase 1)
A point is chosen at random inside a rectangle measuring 6 inches by 5 inches. What is the probability that the randomly selected point is at least one inch from the edge of the rectangle?

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

The correct answer is:
$\frac{2}{5}$

Solution:
Probability that the randomly selected point is at least one inch from the edge of the rectangle $=\frac{4 \times 3}{6 \times 5}=\frac{12}{30}=\frac{2}{5}$

Topic: Mathematics, Probability, NDA, NDA 2016 (Phase 1)
A point $P(1,2,3)$ is one vertex of a cuboid formed bythe coordinate planes and the planes passing through $P$ and parallel to the coordinate planes. What is the equation of the plane passing through $P(1,2,3)$ and parallel to $x y$ -plane?

Options:
A $x+y=3$
B $x-y=-1$
C $z=3$
D $x+2 y+3 z=14$

The correct answer is:
$z=3$

Solution:
Equation of plane passing through $(1,2,3)$ and parallel to $x y$ -plane is $z=3$.
$\therefore \quad$ Option $(\mathrm{c})$ is correct.

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2015 (Phase 1)
A point $P(1,2,3)$ is one vertex of a cuboid formed bythe coordinate planes and the planes passing through $P$ and parallel to the coordinate planes. What is the length of one of the diagonals of the cuboid?

Options:
A $\sqrt{10}$ units
B $\sqrt{14}$ units
C 4 units
D 5 units

The correct answer is:
$\sqrt{14}$ units

Solution:
Length of one of the diagonal of cube
$=\sqrt{(1)^{2}+(2)^{2}+(3)^{2}}$
$=\sqrt{1+4+9}=\sqrt{14}$ units
$\therefore \quad$ Option (b) is correct.

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2015 (Phase 1)
A point $P$ moves such that its distances from $(1,2)$ and $(-2,3)$ are equal. Then the locus of $P$ is

Options:
A straight line
B Parabola
C ellipse
D hyperbola

The correct answer is:
straight line

Solution:
Let moving point be $\mathrm{P}(\mathrm{x}, \mathrm{y})$
$\sqrt{(\mathrm{y}-2)^{2}+(\mathrm{x}-1)^{2}}=\sqrt{(\mathrm{y}-3)^{2}+(\mathrm{x}+2)^{2}}$
$\Rightarrow(\mathrm{y}+2)^{2}+(\mathrm{x}-1)^{2}=(\mathrm{y}-3)^{2}+(\mathrm{x}+2)^{2}$
$\Rightarrow y^{2}+4+4 y+x^{2}+1-2 x=y^{2}+9-6 y+x^{2}+4 x+4$
$\Rightarrow 10 \mathrm{y}-6 \mathrm{x}-8=0$
$\therefore$ Locus of $\mathrm{P}$ is a straight line.

Topic: Mathematics, Straight Lines, NDA, NDA 2013 (Phase 2)
A point $P$ moves such that the difference of its distances from two given points $(c, 0)$ and $(-c, 0)$ is constant. What is the locus of the point $P$ ?

Options:
A Circle
B Ellipse
C Hyperbola
D Parabola

The correct answer is:
Hyperbola

Solution:
When a point $P$ mowes such that the difference of its distances from two given points $(c, 0)$ and $(-c, 0)$ is constant, then the locus of the point $\mathrm{P}$ is hyperbola. It is the definition of hyper bola also.

Topic: Mathematics, Hyperbola, NDA, NDA 2010 (Phase 2)
A pole is floating in a liquid with 80 cm of its length immersed. It is pushed down a certain distance and then released. The time period of vertical oscillation is

Options:
A 4π7 s
B 3π7 s
C 2π7 s
D Ï7 s

The correct answer is:
4π7 s

Solution:
Using the standard result of the time period of floating bodies.
⇒T=2πl⁡immersedg s
⇒T=2π0.809.8=4π7 s

Topic: Physics, Oscillations, NEET
A polygon has 44 diagonals. The number of its sides is

Options:
A 11
B 10
C 8
D 7

The correct answer is:
11

Solution:
No. of diagonals in a polygen $={ }^{\mathrm{n}} \mathrm{C}_{2}-\mathrm{n}$
$\Rightarrow 44={ }^{\mathrm{n}} \mathrm{C}_{2}-\mathrm{n}$
$\Rightarrow 44=\frac{\mathrm{n} !}{2 !(\mathrm{n}-2) !}-\mathrm{n}$
$\Rightarrow \quad 44=\frac{\mathrm{n}(\mathrm{n}-1)}{2}-\mathrm{n}$
$\Rightarrow \quad 44=\frac{\mathrm{n}(\mathrm{n}-3)}{2}$
$\Rightarrow \mathrm{n}^{2}-3 \mathrm{n}-88=0$
$\Rightarrow(\mathrm{n}-11)(\mathrm{n}+8)=0$
$\mathrm{n} \neq-8$
$\mathrm{n}=11$

Topic: Mathematics, Permutation Combination, NDA, NDA 2015 (Phase 2)
A population of Paramecium caudatum was growth in a culture medium. After 5 days the culture medium became overcrowded with Paramecium and had depleted nutrients. What will happen to the population and what type of growth will the population attain? Draw the growth curve.

Options:

None

Solution:
After the lag phase initially, the population will grow in an exponential manner as the nutrients and space will be abundant.
When depletion of the food sources, the population density starts decreasing and ends in an asymptote phase, after which the population density reaches the carrying
capacity (maximum number of individuals of a population or species that a given environment can sustain indefinitely).
The population shows a logistic of pattern growth giving an S-shaped curve.
Where $K=$ carrying capacity
$N=$ population density at time ' $\mathrm{t}$ '
$r=$ Intrinsic rate of natural increase

Topic: Biology, Organisms and Populations
A positive acute angle is divided into two parts whose
tangents are $\frac{1}{8}$ and $\frac{7}{9}$. What is the value of this angle?

Options:
A $\frac{\pi}{3}$
B $\frac{\pi}{4}$
C $\frac{\pi}{6}$
D $\frac{\pi}{12}$

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

Solution:
Let two parts of an angle $\theta$ are $\phi$ and $\psi$. So, $\theta=\phi+\psi$ So, $\tan \theta=\tan (\phi+\psi)$
$=\frac{\tan \phi+\tan \psi}{1-\tan \phi \tan \psi}=\frac{\frac{1}{8}+\frac{7}{9}}{1-\frac{1}{8} \cdot \frac{7}{9}}=\frac{\frac{9+56}{72}}{\frac{72-7}{72}}=\frac{\frac{65}{72}}{\frac{65}{72}}=1$
$=\tan \frac{\pi}{4} \Rightarrow \theta=\frac{\pi}{4}$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2006 (Phase 1)
A potentiometer balances at $44 \mathrm{~cm}$ when a cell of internal resistance $1 \Omega$ is in the secondary circuit. To obtain the balancing point at $40 \mathrm{~cm}$, the resistance to be connected parallel to cell is

Options:
A $20 \Omega$
B $10 \Omega$
C $30 \Omega$
D $5 \Omega$

The correct answer is:
$10 \Omega$

Solution:
(b) Internal Resistance, $\mathrm{r}=1 \Omega$
Balancing Point, $\ell_1=44 \mathrm{~cm}$.
Obtain Balancing Point $\ell_2=40 \mathrm{~cm}$.
Internal Resistance is given by
$$
\mathrm{r}=\mathrm{R}\left[\frac{\ell_1-\ell_2}{\ell_2}\right]
$$
$\begin{aligned} & \mathrm{R}=\frac{\mathrm{r} \ell_2}{\ell_1-\ell_2}=\frac{1 \times 40}{44-40} \\ & \Rightarrow \mathrm{R}=\frac{40}{4}=10 \\ & \Rightarrow \mathrm{R}=10 \Omega\end{aligned}$

Topic: Physics, Current Electricity, TS EAMCET, TS EAMCET 2023 (12 May Shift 2)
A potentiometer has a uniform wire of length $5 \mathrm{~m}$. A battery of emf $10 \mathrm{~V}$ and negligible internal resistance is connected between its ends. A secondary cell connected to the circuit gives balancing length at $200 \mathrm{~cm}$. The emf of the secondary cell is

Options:
A $4 \mathrm{~V}$
B $6 \mathrm{~V}$
C $2 \mathrm{~V}$
D $8 \mathrm{~V}$

The correct answer is:
$4 \mathrm{~V}$

Solution:
Given, length of potentiometer wire
$l=5 \mathrm{~m}$
emf of battery, $E=10 \mathrm{~V}$
Potential gradient across the potentiometer wire,
$K=\frac{E}{l}=\frac{10}{5}=2 \mathrm{Vm}^{-1}$
When secondary cell is connected to the circuit of potentiometer, then balancing length,
$l_{1}=200 \mathrm{~cm}=2 \mathrm{~m}$
$\therefore$ emf of the secondary cell, $E_{s}=K l_{1}=2 \times 2=4 \mathrm{~V}$

Topic: Physics, Current Electricity, KCET, KCET 2020
A power transmission line feeds input power at $2300 \mathrm{~V}$ to a step down transformer with its primary windings having $\mathbf{4 0 0 0}$ turns. What should be the number of turns in the secondary in order to get output power at $230 \mathrm{~V}$ ?

Options:

None

Solution:
Here $E_p=2300 \mathrm{~V}, N_p=400$ turns, $\mathrm{E}_{\mathrm{s}}=230 \mathrm{~V}, \mathrm{~N}_{\mathrm{s}}=$ ?
We know in a transformer, $\frac{E_S}{E_p}=\frac{N_S}{N_p}$
$$
N_s=\frac{E_s N_p}{E_p}=\frac{230 \times 4000}{2300}=400 \mathrm{turns}
$$

Topic: Physics, Alternating Current
A primary amine, $\mathrm{RNH}_2$ can be reacted with $\mathrm{CH}_3-\mathrm{X}$ to get secondary amine, $\mathrm{R}-\mathrm{NHCH}_3$ but the only disadvantage is that $3^{\circ}$ amine and quaternary ammonium salts are al so obtained as side products. Can you suggest a method where $\mathrm{RNH}_2$ forms only $2^{\circ}$ amine?

Options:

None

Solution:
Carbylamine reaction can be used for conversion of primary amine into secondary amine.

Topic: Chemistry, Amines
A primary amine, RNH2can be reacted with CH3—X to get secondary amine, R—NHCH3but the only disadvantage is that 3° amine and quaternary ammonium salts are also obtained as side products. Is any method possible where RNH2forms only 2° amine?

Options:
A Perform the reaction under high pressure to favor the formation of secondary amines.
B Use a large excess of CH3—X to drive the reaction toward the formation of secondary amines.
C Carbylamine reaction is shown by 1° amine only which results in the replacement of two hydrogen atoms attached to the nitrogen atom of the NH2group by one carbon atom. On catalytic reduction the isocyanide will give a secondary amine with one methyl group.
D Conduct the reaction at a low temperature to favor the formation of secondary amines.

The correct answer is:
Carbylamine reaction is shown by 1° amine only which results in the replacement of two hydrogen atoms attached to the nitrogen atom of the NH2group by one carbon atom. On catalytic reduction the isocyanide will give a secondary amine with one methyl group.

Solution:
Correct Option is : (C)
Carbylamine reaction is shown by 1° amine only which results in the replacement of two hydrogen atoms attached to the nitrogen atom of the NH2group by one carbon atom. On catalytic reduction the isocyanide will give a secondary amine with one methyl group.

Topic: Chemistry, Amines, NEET
A primary coil and secondary coil are placed close to each other. A current, which changes at the rate of $25 \mathrm{~A}$ in a millisecond, is present in the primary coil. If the mutual inductance is $92 \times 10^{-6} \mathrm{H}$, then the value of induced emf in the secondary coil is

Options:
A $4.6 \mathrm{~V}$
B $2.3 \mathrm{~V}$
C $0.368 \mathrm{mV}$
D $0.23 \mathrm{mV}$

The correct answer is:
$2.3 \mathrm{~V}$

Solution:
The induced emf in secondary coil.
$e=M \frac{d i}{d t}=92 \times 10^{-6} \times \frac{25}{1 \times 10^{-3}}=2.3 \mathrm{~V}$

Topic: Physics, Electromagnetic Induction, AP EAMCET, AP EAMCET 2012
A primary coil and secondary coil are placed close to each other. A current, which changes at the rate of $25 \mathrm{~A}$ in a millisecond, is present in the primary coil. If the mutual inductance is $92 \times 10^{-6} \mathrm{H}$, then the value of induced emf in the secondary coil is

Options:
A $4.6 \mathrm{~V}$
B $2.3 \mathrm{~V}$
C $0.368 \mathrm{mV}$
D $0.23 \mathrm{mV}$

The correct answer is:
$2.3 \mathrm{~V}$

Solution:
The induced emf in secondary coil.
$e=M \frac{d i}{d t}=92 \times 10^{-6} \times \frac{25}{1 \times 10^{-3}}=2.3 \mathrm{~V}$

Topic: Physics, Electromagnetic Induction, TS EAMCET, TS EAMCET 2012
A problem in statistics is given to three students $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ whose chances of solving it independently are $\frac{1}{2}, \frac{1}{3}$ and $\frac{1}{4}$ respectively. The probability that the problem will be solved is

Options:
A $\frac{1}{12}$
B $\frac{11}{12}$
C $\frac{1}{2}$
D $\frac{3}{4}$

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

Solution:
$\quad$ Let $\mathrm{P}(\mathrm{A})=\frac{1}{2}, \mathrm{P}(\mathrm{B})=\frac{1}{3} \& \mathrm{P}(\mathrm{C})=\frac{1}{4}$
$\mathrm{P}(\mathrm{A} \cup \mathrm{B} \cup \mathrm{C})=1-\mathrm{P}(\overline{\mathrm{A}}) \mathrm{P}(\overline{\mathrm{B}}) \mathrm{P}(\overline{\mathrm{C}})$
$=1-\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)$
$=1-\left[\frac{1}{2} \times \frac{2}{3} \times \frac{3}{4}\right]=1-\frac{1}{4}=\frac{3}{4}$

Topic: Mathematics, Probability, NDA, NDA 2015 (Phase 2)
A problem is given to three students $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ whose
probabilities of solving the problem are $\frac{1}{2}, \frac{3}{4}$ and $\frac{1}{4}$ respectively. What is the probability that the problem will be solved if they all solve the problem independently?

Options:
A $\frac{29}{32}$
B $\frac{27}{32}$
C $\frac{25}{32}$
D $\frac{23}{32}$

The correct answer is:
$\frac{29}{32}$

Solution:
Given, $P(A)=\frac{1}{2}, P(B)=\frac{3}{4}, P(C)=\frac{1}{4}$
$P(\bar{A})=1-\frac{1}{2}=\frac{1}{2}$
$P(\bar{B})=1-\frac{3}{4}=\frac{1}{4}$
$P(\bar{C})=1-\frac{1}{4}=\frac{3}{4}$
$\therefore$ Probability that problem will be solved if they solve independently is
$1-(P(\bar{A}) \cdot P(\bar{B}) \cdot P(\bar{C})$
$=1-\left(\frac{1}{2}\right)\left(\frac{1}{4}\right)\left(\frac{3}{4}\right)=1-\frac{3}{32}=\frac{29}{32}$

Topic: Mathematics, Probability, NDA, NDA 2019 (Phase 1)
A quadratic polynomial with two distinct roots has one real root. Then, the other root is

Options:
A not necessarily real, if the coefficients are real
B always imaginary
C always rea
D real, if the coefficients are real

The correct answer is:
always rea

Solution:
We know that, if a quadratic polynomial with two distinct root has one real root, then
$b^{2}-4 a c>0$
Then other root will always real.

Topic: Mathematics, Quadratic Equation, NDA, NDA 2008 (Phase 2)
A question is given to three students A, B and C whose
chances of solving it are $\frac{1}{2}, \frac{1}{3}$ and $\frac{1}{4}$ respectively. What is the probability that the question will be solved?

Options:
A $\frac{1}{24}$
B $\frac{1}{4}$
C $\frac{3}{4}$
D $\frac{23}{24}$

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

Solution:
Required probability $=1-\left(\mathrm{P}^{\prime}(\mathrm{A}) . \mathrm{P}^{\prime}(\mathrm{B}) . \mathrm{P}^{\prime}(\mathrm{C})\right)$
$=1-\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)$
$=\left(1-\frac{1}{2} \times \frac{2}{3} \times \frac{3}{4}\right)$
$=1-\frac{1}{4}=\frac{3}{4}$

Topic: Mathematics, Probability, NDA, NDA 2017 (Phase 1)
A radioactive element disintegrates at a rate proportional to the quantity of substance $\mathrm{Q}$ present at any time $\mathrm{t}$. What is the differential equation of the disintegration?

Options:
A $\frac{\mathrm{dQ}}{\mathrm{dt}}=-\mathrm{Q}$
B $\frac{\mathrm{dQ}}{\mathrm{dt}}=-\mathrm{kQ}, \mathrm{k} < 0$
C $\frac{d Q}{d t}=-k Q, k>0$
D $\frac{\mathrm{dQ}}{\mathrm{dt}}=\mathrm{Q} \quad$

The correct answer is:
$\frac{d Q}{d t}=-k Q, k>0$

Solution:
A radioactive element disintegrates at a rate proportional to the quantity of substance $Q$ present at any time $\mathrm{t}$.
$\Rightarrow \quad \frac{\mathrm{dQ}}{\mathrm{dt}} \propto-\mathrm{Q}$
$\Rightarrow \frac{\mathrm{dQ}}{\mathrm{dt}}=-\mathrm{kQ}, \mathrm{k}>0$ is a constant.
This is required differential equation.

Topic: Mathematics, Differential Equations, NDA, NDA 2007 (Phase 2)
A random sample of 20 people is classified in the followin, 1
\begin{array}{|c|c|}
\hline Age & Frequency \\
\hline 15-25 & 2 \\
\hline 25-35 & 4 \\
\hline 35-45 & 6 \\
\hline 45-55 & 5 \\
\hline 55-65 & 3 \\
\hline
\end{array} the mean age of this group of people?

Options:
A $41.0$
B 415
C 420
D $42.5$

The correct answer is:
415

Solution:
\begin{array}{|c|c|c|c|}
\hline Age & Mid value \mathbf{x}_{\mathbf{i}} & Frequency \mathbf{f}_{\mathbf{i}} & \mathbf{f}_{\mathbf{i}} \mathbf{x}_{\mathbf{i}} \\
\hline 15-25 & 20 & 2 & 40 \\
\hline 25-35 & 30 & 4 & 120 \\
\hline 35-45 & 40 & 6 & 240 \\
\hline 45-55 & 50 & 5 & 250 \\
\hline 55-65 & 60 & 3 & 180 \\
\hline & & \mathbf{\Sigma} \mathbf{f}_{\mathrm{i}}=20 & \mathbf{\Sigma x}_{\mathbf{i}} \mathbf{f}_{\mathbf{i}}=830 \\
\hline
\end{array}
$\Rightarrow$ Mean age $=\frac{\Sigma x_{i} f_{i}}{\Sigma f_{i}}=\frac{830}{20}=41.5$

Topic: Mathematics, Statistics, NDA, NDA 2016 (Phase 2)
A random variable $X$ follows the binomial distribution and $X \sim B(n, 0.3)$. If the mean of $X$ is three times as large as the standard deviation of $X$, then $n=$

Options:
A $9$
B $21$
C $27$
D $3$

The correct answer is:
$21$

Solution:
Given, $p=0.3, q=1-p=0.7$
Also, Mean $(\mu)=3 \times$ Standard deviation $(\sigma)$.
$\Rightarrow \quad \mu=3(\sigma)$...(i)
$\begin{aligned} & \text { Now, } \mu=n p=0.3 n \\ & \text { and } \sigma=\sqrt{n p q}=\sqrt{n(0 \cdot 3)(0 \cdot 7)}=\sqrt{0.21 n} \\ & \text { From Eq. (i), }(0.3) n=3 \sqrt{(0.21) n}\end{aligned}$
$\begin{array}{ll}\Rightarrow & (0.1) n=\sqrt{(0.21) n} \\ \Rightarrow & \left(\frac{n}{10}\right)^2=\left(\frac{21 n}{100}\right) \\ \Rightarrow & n^2-21 n=0 \Rightarrow n(n-21)=0 \\ \Rightarrow & n=0,21\end{array}$
But $n \neq 0$
Thus, $n=21$.

Topic: Mathematics, Statistics, AP EAMCET, AP EAMCET 2022 (08 Jul Shift 2)
A random variable $X$ has the following distribution
\begin{array}{lllllllll}
\hline \begin{array}{l}
Values of \\
X(x)
\end{array} & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\
\hlineP(X=x) \quad 0 & k & 2 k & 2 k & 3 k & k^2 & 2 k^2 & 7 k^2+k \\
\hline
\end{array}

Options:
A $\frac{9}{10}$
B $\left(\frac{9}{10}\right)^2$
C $\left(\frac{3}{10}\right)$
D $\frac{1}{10}$

The correct answer is:
$\left(\frac{9}{10}\right)^2$

Solution:
$$
\begin{aligned}
& \text { Since, } \Sigma P\left(X=x_i\right)=1 \\
& \Rightarrow 0+K+2 K+2 K+3 K+K^2+2 K^2 \\
& +7 K^2+K=1 \\
& \Rightarrow \quad 9 K+10 K^2=1 \\
& \Rightarrow \quad K=\frac{1}{10}, \quad[\because K>0] \\
&
\end{aligned}
$$
So,
$$
\begin{aligned}
& P(0 < K < 6)=K+2 K+2 K+3 K+K^2 \\
& =8 K+K^2=\frac{8}{10}+\frac{1}{100}=\frac{81}{100}=\left(\frac{9}{10}\right)^2
\end{aligned}
$$

Topic: Mathematics, Probability, AP EAMCET, AP EAMCET 2018 (23 Apr Shift 2)
A random variable $X$ has the probability distribution
\begin{array}{l|c|c|c|c|c|c|c|c}
\hlinex & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\
\hlineP(x) & 0.15 & 0.23 & 0.12 & 0.10 & 0.20 & 0.08 & 0.07 & 0.05 \\
\hline
\end{array}
For the events $E=\{x$ is prime number $\}$ and $F=\{x < 4\}$ the probability of $P(E \cup F)$ is

Options:
A $0.50$
B $0.77$
C $0.35$
D $0.87$

The correct answer is:
$0.77$

Solution:
Given, $E=\{x$ is a prime number $\}$
$$
\begin{aligned}
P(E) &=P(2)+P(3)+P(5)+P(7) \\
&=0.23+0.12+0.20+0.07=0.62 \\
\text { and } \quad F &=\{x < 4\} \\
P(F) &=P(1)+P(2)+P(3) \\
&=0.15+0.23+0.12=0.50 \\
\text { and } \quad P(E \cap F) &=P(2)+P(3) \\
=& 0.23+0.12=0.35 \\
\therefore \quad P(E \cup F) &=P(E)+P(F)-P(E \cap F) \\
&=0.62+0.50-0.35 \\
&=0.77
\end{aligned}
$$

Topic: Mathematics, Probability, MHT CET, MHT CET 2008
A random variable \(X\) has the probability distribution
\(\begin{array}{lllllll} \hline X=x_i & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline P\left(X=x_i\right) & 0.2 & 0.3 & 0.12 & 0.1 & 0.2 & 0.08 \\ \hline \end{array}\)
If \(A=\left\{x_i / x_i\right.\) is a prime number}, \(B=\left\{x_i / x_i < 4\right\}\) are two events, then \(P(A \cup B)=\)

Options:
A 0.31
B 0.62
C 0.82
D 0.41

The correct answer is:
0.82

Solution:
The given probability distribution for random variable \(x\)
\(\begin{array}{lllllll}
\hline x=x_i & 1 & 2 & 3 & 4 & 5 & 6 \\
\hline P\left(X=x_i\right) & 0.2 & 0.3 & 0.12 & 0.1 & 0.2 & 0.08 \\
\hline
\end{array}\)
and two events \(A=\left\{x_i \mid x_i\right.\) is a prime number \(\}\) \(=\{2,3,5\}\)
and \(B=\left\{x_i \mid x_i < 4\right\}=\{1,2,3\}\)
So,
\(\begin{aligned}
& P(A \cup B)=P(X=1)+P(X=2)+P(X=3)+P(X=5) \\
& =0.2+0.3+0.12+0.2=0.82
\end{aligned}\)
Hence, option (3) is correct.

Topic: Mathematics, Probability, AP EAMCET, AP EAMCET 2019 (20 Apr Shift 1)