datavorous/entrance-exam-dataset huggingface dataset:
The angle subtended at the centre of a circle of radius $3 \mathrm{~cm}$ by an arc of length $1 \mathrm{~cm}$ is :

Options:
A $\frac{30^{\circ}}{\pi}$
B $\frac{60^{\circ}}{\pi}$
C $60^{\circ}$
D None of the above

The correct answer is:
$\frac{60^{\circ}}{\pi}$

Solution:
Let $\theta$ be the required angle
$\therefore \quad \theta=\frac{\operatorname{arc}}{\text { radius }}=\frac{1}{3}$ radians
Now, 1 radian $=\frac{180^{\circ}}{\pi}$
$\therefore \quad \frac{1}{3}$ radian $=\frac{180^{\circ}}{\pi} \times \frac{1}{3}=\frac{60^{\circ}}{\pi}$
Hence, Required angle $=\theta=\frac{60^{\circ}}{\pi}$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2012 (Phase 2)
The angles $\mathrm{A}, \mathrm{B}, \mathrm{C}$ of a triangle are in the ratio $2: 5: 5$. What is the value of $\tan \mathrm{B} \tan \mathrm{C} ?$

Options:
A $4+\sqrt{3}$
B $4+2 \sqrt{3}$
C $7+4 \sqrt{3}$
D $3+3 \sqrt{3}$

The correct answer is:
$7+4 \sqrt{3}$

Solution:
Let the angles $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ of a triangle are $2 \mathrm{x}, 5 \mathrm{x}$ and $5 \mathrm{x}$, respectively So, $2 x+5 x+5 x=180^{\circ}$
$\Rightarrow \mathrm{x}=\frac{180^{\circ}}{12}=15^{\circ}$
Angles are $30^{\circ}, 75^{\circ}, 75^{\circ}$
$\angle \mathrm{B}=75^{\circ}$ and $\angle \mathrm{C}=75^{\circ}$
$\tan B \tan C=\left(\tan 75^{\circ}\right)^{2}=\left(\tan \left(45^{\circ}+30^{\circ}\right)\right)^{2}$
$=\left(\frac{\tan 45^{\circ}+\tan 30^{\circ}}{1-\tan 45^{\circ} \tan 30^{\circ}}\right)^{2}=\left(\frac{1+\frac{1}{\sqrt{3}}}{1-\frac{1}{\sqrt{3}}}\right)^{2}=\left(\frac{\sqrt{3}+1}{\sqrt{3}-1}\right)^{2}$
$=\left(\frac{(\sqrt{3}+1)^{2}}{3-1}\right)^{2}=\frac{1}{4}[3+1+2 \sqrt{3}]^{2}$
$=\frac{1}{4}[4+2 \sqrt{3}]^{2}=\frac{1}{4}[16+12+16 \sqrt{3}]$
$=\frac{1}{4}[28+16 \sqrt{3}]=7+4 \sqrt{3}$

Topic: Mathematics, Properties of Triangles, NDA, NDA 2006 (Phase 2)
The angles of a triangle are in AP and the least angle is $30^{\circ}$. What is the greatest angle (in radian)?

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

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

Solution:
We have, $\mathrm{a}=30^{\circ}, \mathrm{n}=3$,
$\mathrm{S}_{3}=180^{\circ}$
$\mathrm{S}_{3}=\frac{3}{2}\left[2 \times 30^{\circ}+(3-1) \mathrm{d}\right]$
$180=\frac{3}{2}[60+2 \mathrm{~d}]$
$180=3[30+\mathrm{d}]$
$30+d=60$
$\mathrm{d}=30$
Now, largest angle $=\mathrm{a}+2 \mathrm{~d}=30+60=90^{\circ}=\frac{\pi}{2}$

Topic: Mathematics, Sequences and Series, NDA, NDA 2012 (Phase 2)
The angles of elevation of the top of a tower from the top and foot of a pole are respectively $30^{\circ}$ and $45^{\circ} .$ If $\mathrm{h}_{\mathrm{T}}$ is the height of the tower and $\mathrm{h}_{\mathrm{p}}$ is the height of the pole, then which of the following are correct?
1.$\quad \frac{2 \mathrm{~h}_{\mathrm{P}} \mathrm{h}_{\mathrm{T}}}{3+\sqrt{3}}=\mathrm{h}_{\mathrm{P}}^{2} \quad$
2.$\quad \frac{\mathrm{h}_{\mathrm{T}}-\mathrm{h}_{\mathrm{P}}}{\sqrt{3}+1}=\frac{\mathrm{h}_{\mathrm{P}}}{2}$
3.$\quad \frac{2\left(\mathrm{hp}+\mathrm{h}_{\mathrm{T}}\right)}{\mathrm{h}_{\mathrm{P}}}=4+\sqrt{3}$
Select the correct answer using the code given below.

Options:
A 1 and 3 only
B 2 and 3 only
C 1 and 2 only
D 1,2 and 3

The correct answer is:
1 and 2 only

Solution:
Let the distance between tower and pole be ' $\mathrm{x}$.
$\tan 45^{\circ}=\frac{\mathrm{h}_{\mathrm{t}}}{\mathrm{x}} \Rightarrow \frac{\mathrm{h}_{\mathrm{t}}}{\mathrm{x}}=1$
$\Rightarrow \mathrm{h}_{\mathrm{t}}=\mathrm{x}$
Now, $\tan 30^{\circ}=\frac{\mathrm{h}_{\mathrm{t}}-\mathrm{h}_{\mathrm{p}}}{\mathrm{x}} \Rightarrow \frac{1}{\sqrt{3}}=\frac{\mathrm{h}_{\mathrm{t}}-\mathrm{h}_{\mathrm{p}}}{\mathrm{x}}$
$\Rightarrow \frac{1}{\sqrt{3}}=\frac{\mathrm{h}_{\mathrm{t}}-\mathrm{h}_{\mathrm{p}}}{\mathrm{h}_{\mathrm{t}}}$
$\Rightarrow \frac{1}{\sqrt{3}-1}=\frac{h_{t}-h_{p}}{h_{t}-\left(h_{t}-h_{p}\right)}$
$\Rightarrow \frac{\sqrt{3}+1}{2}=\frac{\mathrm{h}_{\mathrm{t}}-\mathrm{h}_{\mathrm{p}}}{\mathrm{h}_{\mathrm{p}}} \Rightarrow \frac{\mathrm{h}_{\mathrm{t}}-\mathrm{h}_{\mathrm{p}}}{\sqrt{3}+1}=\frac{\mathrm{h}_{\mathrm{p}}}{2}$
So, statement ' 2 ' is correct.
Now, $\frac{\mathrm{h}_{\mathrm{t}}-\mathrm{h}_{\mathrm{p}}}{\mathrm{h}_{\mathrm{p}}}=\frac{\sqrt{3}+1}{2} \Rightarrow \frac{\mathrm{h}_{\mathrm{t}}-\mathrm{h}_{\mathrm{p}}+2 \mathrm{~h}_{\mathrm{p}}}{\mathrm{h}_{\mathrm{p}}}=\frac{\sqrt{3}+1+4}{2}$
$\Rightarrow \frac{\mathrm{h}_{\mathrm{t}}+\mathrm{h}_{\mathrm{p}}}{\mathrm{h}_{\mathrm{p}}}=\frac{5+\sqrt{3}}{2}$
So, statement ' 3 ' is incorrect.

Topic: Mathematics, Heights and Distances, NDA, NDA 2017 (Phase 2)
The angles of elevation of the top of a tower standing on a horizontal plane from two points on a line passing through the foot of the tower at distances $49 \mathrm{~m}$ and $36 \mathrm{~m}$ are $43^{\circ}$ and $47^{\circ}$ respectively. What is the height of the tower?

Options:
A $40 \mathrm{~m}$
B $42 \mathrm{~m}$
C $45 \mathrm{~m}$
D $47 \mathrm{~m}$

The correct answer is:
$42 \mathrm{~m}$

Solution:
$\mathrm{AB}=\mathrm{h}$ (height of the tower $\mathrm{BD}=36 \mathrm{~m}$
$\mathrm{BC}=49 \mathrm{~m}$
$\angle \mathrm{D}=47^{\circ}$
$\angle \mathrm{C}=43^{\circ}$
Now, in $\triangle \mathrm{ABD}$,
$\tan 47^{\circ}=\frac{\mathrm{h}}{36 \mathrm{~m}}$
and in $\triangle \mathrm{ABC}$,
$\tan 43^{\circ}=\frac{\mathrm{h}}{49 \mathrm{~m}}$
$\tan \left(90^{\circ}-47^{\circ}\right)=\frac{\mathrm{h}}{49}$
$\therefore \quad \cot 47^{\circ}=\frac{\mathrm{h}}{49}$
Multiplying equations (i) and (ii)
$\tan 47^{\circ} \cdot \cot 47^{\circ}=\frac{\mathrm{h}}{36} \times \frac{\mathrm{h}}{49}=1=\frac{\mathrm{h}^{2}}{36 \times 49}$
$\mathrm{h}=6 \times 7=42 \mathrm{~m}$
$\therefore \quad$ Option (b) is correct

Topic: Mathematics, Heights and Distances, NDA, NDA 2015 (Phase 1)
The antrum is

Options:
A Fluid filled cavity of Graffian follicle
B An inner theca interna
C The secondary follicle
D Fluid filled cavity of primary follicle

The correct answer is:
Fluid filled cavity of Graffian follicle

Solution:
The antrum is the fluid-filled space found in antral or Graafian follicles; also called the follicular cavity.

Topic: Biology, Human Reproduction, NEET
The area bounded by the coordinate axes and the curve $\sqrt{\mathrm{x}}+\sqrt{\mathrm{y}}=1$, is

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

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

Solution:
$\begin{aligned} \text { Area }=\int y d x \\=& \int_{0}^{1}(1-\sqrt{x})^{2} d x \end{aligned}$
$[\because$ curve makes the intercept of 1 on both axes] $=\int_{0}^{1}(1+x-2 \sqrt{x}) d x$
$=[\mathrm{x}]_{0}^{1}+\frac{1}{2}\left[\mathrm{x}^{2}\right]_{0}^{1}-\frac{4}{3}\left[\mathrm{x}^{\frac{3}{2}}\right]_{0}^{1}$
$=1+\frac{1}{2}-\frac{4}{3}=\frac{3}{2}-\frac{4}{3}=\frac{1}{6} \mathrm{sq}$ unit

Topic: Mathematics, Area Under Curves, NDA, NDA 2015 (Phase 2)
The area bounded by the curve $x=\mathrm{f}[\mathrm{y}]$, the $\mathrm{y}$ -axis and the two lines $\mathrm{y}=\mathrm{a}$ and $\mathrm{y}=\mathrm{b}$ is equal to: $\quad\mathrm{}$

Options:
A $\int_{\mathrm{a}}^{\mathrm{b}} \mathrm{y} \mathrm{dx}$
B $\int_{\mathrm{a}}^{\mathrm{b}} \mathrm{y}^{2} \mathrm{dx}$
C $\int_{\mathrm{a}}^{\mathrm{b}} \mathrm{x} \mathrm{dy}$
D None of the above

The correct answer is:
$\int_{\mathrm{a}}^{\mathrm{b}} \mathrm{x} \mathrm{dy}$

Solution:
Required Area $=\int_{\mathrm{y}=\mathrm{a}}^{\mathrm{b}} \mathrm{f}(\mathrm{y}) \mathrm{d} \mathrm{y}=\int_{\mathrm{a}}^{\mathrm{b}} \mathrm{x} \mathrm{dy}$

Topic: Mathematics, Area Under Curves, NDA, NDA 2012 (Phase 1)
The area bounded by the curve $|\mathrm{x}|+|\mathrm{y}|=1$ is

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

The correct answer is:
2 square units

Solution:
$|\mathrm{x}|+|\mathrm{y}|=1$
We know, $|x|=\left\{\begin{array}{l}x, x \geq 0 \\ -x, x < 0\end{array}\right.$
$\therefore|x|+|y|=1$ is
$\left\{\begin{array}{cc}x+y=1 & \text { for } x>0, y>0 \\ -x+y=1 & \text { for } x < 0, y>0 \\ x-y=1 & \text { for } x>0, y < 0 \\ -x-y=1 & \text { for } x < 0, y < 0\end{array}\right.$
If we plot graphs of these equations, we get
The curve is symmetrical about $\mathrm{x}$ any $\mathrm{y}$ -axis. $\therefore$ Area $=4 \times$ Area of $A O D$
$=4 \times \int_{0}^{1} \mathrm{ydx}=4 \times \int_{0}^{1}(1-\mathrm{x}) \mathrm{d} \mathrm{x}=4\left[\mathrm{x}-\frac{\mathrm{x}^{2}}{2}\right]_{0}^{1}$
$=4\left[1-\frac{1^{2}}{2}-\left(0-\frac{0^{2}}{2}\right)\right]$
$=4\left(1-\frac{1}{2}\right)=4 \times \frac{1}{2}=2$

Topic: Mathematics, Area Under Curves, NDA, NDA 2017 (Phase 2)
The area bounded by the curve, $y=-x^2$, x-axis, $x=1$ and $x=4$, is

Options:
A 21 sq. units
B 10 sq. units
C 20 sq. units
D $\frac{21}{2}$ sq. units

The correct answer is:
21 sq. units

Solution:
$-\int_1^4\left(-x^2\right) \mathrm{d} x=\left[\frac{x^3}{3}\right]_1^4=\frac{64}{3}-\frac{1}{3}=21$ sq. units

Topic: Mathematics, Area Under Curves, MHT CET, MHT CET 2022 (08 Aug Shift 1)
The area of a rectangle whose length is five more than twice its width is 75 square unit. The length is

Options:
A 5 unit
B 10 unit
C 15 unit
D 20 unit

The correct answer is:
15 unit

Solution:
Letwidth of the rectangle $=\mathrm{x}$ so, length $=2 \mathrm{x}+5$
Given, Area of rectangle $=75$ $\Rightarrow x(2 x+5)=75 \Rightarrow 2 x^{2}+5 x-75=0$
$\Rightarrow 2 x^{2}+15 x-10 x-75=0$
$\Rightarrow(x-5)(2 x+15)=0$
$\Rightarrow \quad x=5,-\frac{15}{2}$
Since, width can not be negative.
$\mathrm{x}=5$
length $=2 \mathrm{x}+5=15$ unit

Topic: Mathematics, Quadratic Equation, NDA, NDA 2013 (Phase 1)
The area of a triangle, whose vertices are $(3,4),(5,2)$ and the point of intersection of the lines $x=a$ and $y=5$, is 3 square units. What is the value of $a$ ?

Options:
A 2
B 3
C 5
D 4

The correct answer is:
4

Solution:
Area of $\Delta \mathrm{ABC}=3 \mathrm{sq}$. unit
$\therefore \quad \frac{1}{2}\left|\begin{array}{lll}3 & 4 & 1 \\ 5 & 2 & 1 \\ \mathrm{a} & 5 & 1\end{array}\right|=3$
$\therefore\left|\begin{array}{lll}3 & 4 & 1 \\ 5 & 2 & 1 \\ \mathrm{a} & 5 & 1\end{array}\right|=6$
$\begin{array}{ll}\therefore & 3(2-5)-4(5-a)+1(25-2 a)=6 \\ & -9-20+4 a+25-2 a=6 \\ & 2 a=10 \\ & a=5\end{array}$
Option (d) is correct

Topic: Mathematics, Area Under Curves, NDA, NDA 2015 (Phase 1)
The area of the figure formed by the lines $a x+b y+c=0$, $a x-b y+c=0, a x+b y-c=0$ and $a x-b y-c=0$ is

Options:
A $\frac{\mathrm{c}^{2}}{\mathrm{ab}}$
B $\frac{2 \mathrm{c}^{2}}{\mathrm{ab}}$
C $\frac{c^{2}}{2 a b}$
D $\frac{c^{2}}{4 a b}$

The correct answer is:
$\frac{2 \mathrm{c}^{2}}{\mathrm{ab}}$

Solution:
Area of triangle
$\Delta \mathrm{AOB}=\frac{1}{2} \times \frac{\mathrm{c}}{\mathrm{b}} \times \frac{\mathrm{c}}{\mathrm{a}}=\frac{\mathrm{c}^{2}}{2 \mathrm{ab}}$
Total area $=4 \times$ area $\triangle \mathrm{AOB}$
$=4 \times \frac{\mathrm{c}^{2}}{2 \mathrm{ab}}$
$=\frac{2 c^{2}}{a b}$

Topic: Mathematics, Area Under Curves, NDA, NDA 2015 (Phase 2)
The area of the square, one of whose diagonals is $3 \hat{i}+4 \hat{j}$ is

Options:
A 12 square unit
B $12.5$ square unit
C 25 square unit
D $156.25$ square unit

The correct answer is:
$12.5$ square unit

Solution:
Length of diagonal
$=\mathrm{D}=\sqrt{3^{2}+4^{2}}$
$\Rightarrow \mathrm{D}=5$
$\therefore$ Area $=\frac{1}{2}(\mathrm{D})^{2}=\frac{25}{2}$
$=12.5$ units

Topic: Mathematics, Vector Algebra, NDA, NDA 2015 (Phase 2)
The arithmetic mean and standard deviation of a data of nine numbers are 13 and 5 respectively. If 3 is included as the 10th item of the data, then the variance of the data of ten number is

Options:
A 23.5
B 21.5
C 31.5
D 27

The correct answer is:
31.5

Solution:
$\because \frac{\sum_{i=1}^9 x_i}{9}=13 \Rightarrow \sum_{i=1}^9 x_i=117$.
$\begin{array}{rlrl} & & & \\ \text { and } & & \sigma^2=25=\frac{\sum_{i=1}^9 x_i^2}{9}-(13)^2 \\ \Rightarrow & \quad \sum_{i=1}^9 x_i^2=9[25+169]\end{array}$
Now, after including $10^{\text {th }}$ item as ' 3 '
$$
\begin{gathered}
\text { New mean } \bar{y}=\frac{\sum_{i=1}^9 x_i+3}{10}=\frac{117+3}{10}=12 \\
\text { and new variance }=\frac{\left(\sum_{i=1}^9 x_i^2+9\right)}{10}-(12)^2 \\
=\frac{(169+25+1) 9}{10}-144=\frac{1755-1440}{10}=31.5
\end{gathered}
$$

Topic: Mathematics, Statistics, AP EAMCET, AP EAMCET 2018 (22 Apr Shift 2)
The arithmetic mean of $1,8,27,64, \ldots . .$ up ton terms is given
by

Options:
A $\frac{\mathrm{n}(\mathrm{n}+1)}{2}$
B $\frac{n(n+1)^{2}}{2}$
C $\frac{n(n+1)^{2}}{4}$
D $\frac{\mathrm{n}^{2}(\mathrm{n}+1)^{2}}{4}$

The correct answer is:
$\frac{n(n+1)^{2}}{4}$

Solution:
Sum of the given terms $=1+8+27+64 \ldots \ldots .+$ upton terms
Sum $=1^{3}+2^{3}+3^{3}+4^{3}+\ldots \ldots .+$ upto $n$ terms
Sum $=\left(\frac{n(n+1)}{2}\right)^{2}=\frac{n^{2}(n+1)^{2}}{4}$
$\mathrm{AM}=\frac{\mathrm{n}^{2}(\mathrm{n}+1)^{2}}{4} \times \frac{1}{\mathrm{n}}=\frac{\mathrm{n}(\mathrm{n}+1)^{2}}{4}$

Topic: Mathematics, Statistics, NDA, NDA 2015 (Phase 2)
The arithmetic mean of 4 numbers is 15 . The arithmetic mean of another 6 numbers is $12 .$ What is the arithmetic mean of the combined 10 numbers?

Options:
A $12.2$
B $12.8$
C 132
D $13.8$

The correct answer is:
132

Solution:
Arithmetic mean of 10 numbers $=\frac{4 \times 15+6 \times 12}{10}=\frac{60+72}{10}=13.2$

Topic: Mathematics, Statistics, NDA, NDA 2008 (Phase 2)
The arithmetic mean of marks in Mathematics for four divisions A, B, C and D were 80, 75, 70 and 72 respectively. Their standard deviations were $12,6,8$ and and 10 respectively. Then, division
has more uniformity.

Options:
A D
B B
C C
D A

The correct answer is:
B

Solution:
$$
\begin{aligned}
& \text { We know that C.V. }=\frac{\text { Standard Deviation }}{\text { Mean }} \\
& \therefore(\text { C.V. })_A=\frac{12}{18}=0.15 \text { and }(\text { C.V. })_B=\frac{6}{75}=0.08 \\
& (\text { C.V. })_C=\frac{8}{70}=0.11 \text { and }(\text { C.V. })_D=\frac{10}{72}=0.14
\end{aligned}
$$
C.V. is the least for division B.

Topic: Mathematics, Statistics, MHT CET, MHT CET 2021 (23 Sep Shift 1)
The arithmetic mean of numbers $\mathrm{a}, \mathrm{b}, \mathrm{c}, \mathrm{d}, \mathrm{e}$ is $\mathrm{M}$. What is the value of $(a-M)+(b-M)+(c-M)+(d-M)+(e-M) ?$

Options:
A $\mathrm{M}$
B $a+b+c+d+e$
C 0
D $5 \mathrm{M}$

The correct answer is:
0

Solution:
Given $M=\frac{a+b+c+d+e}{5}$ $\Rightarrow a+b+c+d+c=5 M$
$\rightarrow a+b+c+d+c-5 m-0$
$\Rightarrow(\mathrm{a}-\mathrm{M})+(\mathrm{b}-\mathrm{M})+(\mathrm{c}-\mathrm{M})+(\mathrm{d}-\mathrm{M})+(\mathrm{c}-\mathrm{M})$
$=0$
Hence, Required value $=0$

Topic: Mathematics, Statistics, NDA, NDA 2012 (Phase 2)
The arithmetic mean of the squares of the first $n$ natura numbers is

Options:
A $\frac{n(n+1)(2 n+}{6}$
B $\frac{n(n+1)(2 n}{2}$
C $\frac{1[(2 n+1)}{3}$
D $\frac{(n+1)(2 n}{3}$

The correct answer is:
$\frac{1[(2 n+1)}{3}$

Solution:
Sum of squares of first 'n' natural numbers
$=\frac{n(n+1)(2 n+1)}{6}$
Mean of the squares of first 'n' natural numbers
$=\frac{(\mathrm{n}+1)(2 \mathrm{n}+1) \times \mathrm{n}}{6 \times \mathrm{n}}=\frac{(\mathrm{n}+1)(2 \mathrm{n}+1)}{6}$

Topic: Mathematics, Statistics, NDA, NDA 2013 (Phase 2)
The arithmetic mean of two numbers exceeds their geometric mean by 2 and the geometric mean exceeds their harmonic mean by $1.6$. What are the two numbers?

Options:
A 16,4
B 81,9
C 256,16
D 625,25

The correct answer is:
16,4

Solution:
Let $\mathrm{A}, \mathrm{G}$ and $\mathrm{H}$ be the arithmetic mean, geometric mean and Harmonic mean of two numbers a and $b$ respectively. According to the Question $\mathrm{G}=\mathrm{H}+1.6$
and $\mathrm{A}=\mathrm{H}+1.6+2=\mathrm{H}+3 \cdot 6$
We have $\mathrm{AH}=\mathrm{G}^{2}$
$(\mathrm{H}+3.6) \mathrm{H}=(\mathrm{H}+1.6)^{2}$
$\Rightarrow \mathrm{H}^{2}+3.6 \mathrm{H}=\mathrm{H}^{2}+2.56+3.2 \mathrm{H}$
$\Rightarrow \quad H=\frac{2.56}{0.4}=6.4$
$\therefore \quad A=6.4+3.6=10$
and $\mathrm{G}=6.4+1.6=8$
Now, $\mathrm{A}=\frac{a+b}{2} \Rightarrow \mathrm{a}+\mathrm{b}=2 \mathrm{~A}$
$\Rightarrow \quad a+b=20$ ...(i)
and $a b=G^{2}=64$ ...(ii)
We know that, $(a-b)^{2}=(a+b)^{2}-4 a b$ $=400-256=144$
$\Rightarrow \quad a-b=12$ ...(iii)
On solving Eqs. (i) and (iii), we get $a=16$ and $b=4$

Topic: Mathematics, Sequences and Series, NDA, NDA 2010 (Phase 2)
The available biomass for the consumption to herbivores and decomposers in an ecosystem is

Options:
A Gross primary productivity
B Net primary productivity
C Secondary productivity
D Tertiary productivity

The correct answer is:
Net primary productivity

Solution:
Gross primary productivity of an ecosystem is the rate of production of organic matter during photosynthesis. A considerable amount of GPP is utilised by plants in respiration. Gross primary productivity minus respiration losses (R), is the net primary productivity (NPP). It is the available biomass for the consumption to heterotrophs (herbivores and decomposers)

Topic: Biology, Ecosystem, NEET
The average age of 20 students in a class is $15 \mathrm{yr}$. If the teacher's age is included, the average increases by one. What is the teacher's age? $\quad$

Options:
A $30 \mathrm{yr}$
B $21 \mathrm{yr}$
C $42 \mathrm{yr}$
D $36 \mathrm{yr}$

The correct answer is:
$36 \mathrm{yr}$

Solution:
Let the teacher's age is $x$ years According to question
$15+1=\frac{20 \times 15+x}{21}$
$16 \times 21=300+x$
$\Rightarrow x=336-300=36$ years

Topic: Mathematics, Statistics, NDA, NDA 2008 (Phase 2)
The average age of a combined group of men and women is 25 years. If the average age of the group of men is 26 years and the of the group of women is 21 years, then the percentage of men and women in the group is respectively

Options:
A 20,80
B 40,60
C 60,40
D 80,20

The correct answer is:
80,20

Solution:
Let No. of Man $=\mathrm{M}$ Let No. of Women $=\mathrm{W}$ $26 \mathrm{M}+21 \mathrm{~W}=25(\mathrm{M}+\mathrm{W})$
$\mathrm{M}=4 \mathrm{~W}$
$\mathrm{M}: \mathrm{W}=4: 1$
$\therefore$ Percent of Men and Women $=80 \%, 20 \%$

Topic: Mathematics, Statistics, NDA, NDA 2018 (Phase 2)
The average marks obtained by the students in a class are 43 . If the average marks obtained by 25 boys are 40 and the average marks obtained by the girl students are 48 , then what is the number of girl students in the class?

Options:
A 15
B 17
C 18
D 20

The correct answer is:
15

Solution:
Let Number of girls student be $x$
Sum of marks $=25 \times 40+\mathrm{x} \times 48$
Total students $=25+\underline{x}$
$43=\frac{25 \times 40+x \times 48}{x+25}$
$\Rightarrow 43 \mathrm{x}+43 \times 25=25 \times 40+\mathrm{x} \times 48$
$\Rightarrow 5 \mathrm{x}=3 \times 25$
$\Rightarrow x=15$

Topic: Mathematics, Statistics, NDA, NDA 2013 (Phase 1)
The average sales and standard deviation of sales for four months for a company are as follows:
\begin{array}{lcccc}
\hline & Month 1 & Month 2 & Month 3 & Month 4 \\
\hline Average sales & 30 & 57 & 82 & 28 \\
Standard & 2 & 3 & 4 & 2 \\
deviation of sales & & & & \\
\hline During which month are the sales most consistent?
\end{array}

Options:
A Month 1
B Month 2
C Month 3
D Month 4

The correct answer is:
Month 3

Solution:
From visual observation of given table we can say that during month 3 , the sales are most consistent

Topic: Mathematics, Statistics, NDA, NDA 2009 (Phase 1)
The axis of the parabola $y^{2}+2 x=0$ is

Options:
A $x=0$
B $y=0$
C $x=2$
D $y=2$

The correct answer is:
$y=0$

Solution:
$y^{2}+2 x=0 \Rightarrow y^{2}=-2 x$, which is in the form $y^{2}=-4 a x .$ Therefore axis of parabola is $x$ -axis and its equation is $\mathrm{y}=0$.

Topic: Mathematics, Parabola, NDA, NDA 2013 (Phase 2)
The bark of a tree includes

Options:
A both secondary and primary phloem.
B secondary cortex, cork cambium and cork.
C periderm and primary/secondary phloem.
D all of these

The correct answer is:
all of these

Solution:
Bark is the feature of woody plants which is a result of secondary growth and includes both living and non-living tissues. Secondary and primary phloem, secondary cortex, cork, cork cambium and periderm all are present within a bark.

Topic: Biology, Anatomy of Flowering Plants, NEET
The below curve shows the graph of $\mathrm{a}^{\mathrm{x}}$ under which one of the following conditions?

Options:
A $a \geq 1$
B $\mathrm{a}>1$
C $0 < \mathrm{a} \leq 1$
D $\quad 0 < \mathrm{a} < 1$

The correct answer is:
$\quad 0 < \mathrm{a} < 1$

Solution:
The given curve shows the graph of a $^{x}$ which is decreasing when $\mathrm{x}$ is increasing. This happens when $0 < \mathrm{a} < 1$.

Topic: Mathematics, Functions, NDA, NDA 2006 (Phase 2)
The below image represents a type of animal which belongs to the genus

Options:
A Ophiura
B Palaeomon
C Ascidia
D Limulus

The correct answer is:
Palaeomon

Solution:
The given diagram is of a common prawn belonging to the genus Palaemon. This genus is under the class of crustaceans having a chitin based exoskeleton and joint appendages. They are part of the largest animal phylum Arthropoda. Ascidia is a type of genus living in marine habitats coming under the subphylum Urochordata. Ophiura is an example of Echinoderms. Limulus is called as king Crab also belongs to the phylum Arthropoda.

Topic: Biology, Animal Kingdom, NEET
The bhagirathi and the Alaknanda merge at Deoprayag toform the Ganga with their speeds in the ratio $1: 1.5$. The cross-sectional areas of the Bhagirathi, the Alaknanda and the Ganga are in the ratio $1: 2: 3$. Assuming stremline flow, the ratio of the speed of Ganga to that of the Alaknands is

Options:
A $7: 9$
B $4: 3$
C $8: 9$
D $5: 3$

The correct answer is:
$8: 9$

Solution:
By equation of continuity
Area of Bhagirathi $\Rightarrow \mathrm{A}$
Area of Alaknanda $\Rightarrow 2 \mathrm{~A}$
Area of Ganga $\Rightarrow 3 \mathrm{~A}$
$\mathrm{V}_{\mathrm{B}}: \mathrm{V}_{\mathrm{AL}}: \mathrm{V}_{\mathrm{G}}: \Rightarrow \mathrm{V}: \frac{3}{2} \mathrm{~V}: \mathrm{V}_{1}$
By equation of continuity
$\begin{array}{l}
\mathrm{AV}+\frac{3}{2} \mathrm{~A} \cdot 2 . \mathrm{V} \Rightarrow 3 \mathrm{~A} \cdot \mathrm{V}_{1} \\
\mathrm{~V}_{\text {ganga }}=\frac{4}{3} \mathrm{~V} \\
\frac{\mathrm{V}_{\text {Alaknanda }}}{\mathrm{V}_{\text {ganga }}} \Rightarrow \frac{\frac{3}{2} \mathrm{~V}}{\frac{4}{3} \mathrm{~V}}=\frac{9}{8} \\
\mathrm{~V}_{\text {ganga }}: \mathrm{V}_{\text {Alaknanda }}=8: 9
\end{array}$

Topic: Physics, Mechanical Properties of Fluids, KVPY, KVPY 2017 (19 Nov SA)
The binary number $0.111111$.... (where thedigit 1 is recurring) is equivalent in decimal system to which one of the following?

Options:
A $\frac{1}{10}$
B $\frac{11}{10}$
C 1
D $\frac{10}{11}$

The correct answer is:
1

Solution:
Let binary number $0.1111111 \ldots=\mathrm{x}$ $\Rightarrow \mathrm{x}=2^{-1}+2^{-2}+2^{-3}+\ldots . \infty=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\ldots . . \infty$
This is an infinite G.P. series with first term $=\frac{1}{2}$ and
common ratio $=\frac{1}{2}$
$\Rightarrow \mathrm{x}=\frac{1 / 2}{1-\frac{1}{2}}=\frac{1 / 2}{1 / 2}=1$
$=128+8+4=140$ (decimal numbers)
and $1101101=1 \times 2^{6}+1 \times 2^{5}+1 \times 2^{3}+1 \times 2^{2}+1 \times 2^{0}$
$=64+32+8+4+1=109$
their difference $=140-109=31$

Topic: Mathematics, Sets and Relations, NDA, NDA 2008 (Phase 1)
The binary number expression of the decimal number 31 is

Options:
A 1111
B 10111
C 11011
D 11111

The correct answer is:
11111

Solution:
Decimal number $=31$
\begin{array}{l|c|l}
2 & 31 & 1 \\
2 & 15 & 1 \\
2 & 7 & 1 \\
2 & 3 & 1 \\
2 & 1 & 1 \\
& 0 &
\end{array}
So, binary form of 31 is 11111 .

Topic: Mathematics, Sets and Relations, NDA, NDA 2018 (Phase 1)
The binary representation of the decimal number 45 is

Options:
A 110011
B 101010
C 1101101
D 101101

The correct answer is:
101101

Solution:
Binary form of $45=101101$
1 \begin{array}{l|l|l}
2 & 45 & 1 \\
\hline 2 & 22 & 0 \\
\hline 2 & 11 & 1 \\
\hline & 5 & 1 \\
\hline 2 & 2 & 0 \\
\hline 2 & 1 &
\end{array}
$\therefore$ Binary form of $45=101101$

Topic: Mathematics, Sets and Relations, NDA, NDA 2013 (Phase 1)
The binomial distribution has:

Options:
A only one parameter
B two parameters
C three parameters
D four parameters

The correct answer is:
two parameters

Solution:
Binomial distribution has two parameters $\mathrm{n}$ and $\mathrm{p}$, where $\mathrm{n}$ is number of trials and $\mathrm{p}$ is probability of success.

Topic: Mathematics, Probability, NDA, NDA 2013 (Phase 1)