datavorous/entrance-exam-dataset huggingface dataset:
In a triangle $A B C, B C=\sqrt{39}, A C=5$ and $A B=7$. What is the measure of the angle $A ?$

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

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

Solution:
Let $\mathrm{a}, \mathrm{b}, \mathrm{c}$ be the sides of $\Delta A B C$ and $\angle A=\theta$
$\therefore \quad a=\sqrt{39}, b=5$ and $c=7$
and $\cos A=\frac{b^{2}+c^{2}-a^{2}}{2 b c}=\frac{25+49-39}{2 \times 5 \times 7}$
$=\frac{1}{2}=\cos \frac{\pi}{3}$
$\Rightarrow \quad A=\frac{\pi}{3}$

Topic: Mathematics, Properties of Triangles, NDA, NDA 2010 (Phase 1)
In a triangle $A B C$, if $A=\tan ^{-1} 2$ and $\mathrm{B}=\tan ^{-1} 3$, then $\mathrm{C}$ is equal to

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

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

Solution:
We have $A=\tan ^{-1} 2 \Rightarrow \tan A=2$
and $B=\tan ^{-1} 3 \Rightarrow \tan B=3$.
Since, $A, B, C$ are angles of a triangle $\therefore A+B+C=\pi$
$\Rightarrow C=\pi-(A+B)$ ...(1)
Now, $A+B=\tan ^{-1} 2+\tan ^{-1} 3$
$=\pi+\tan ^{-1}\left(\frac{2+3}{1-2.3}\right)$
$\left[\because \tan ^{-1} x+\tan ^{-1} y=\pi+\tan ^{-1} \frac{x+y}{1-x y}\right]$
$=\pi+\tan ^{-1}(-1)=\pi-\tan ^{-1}(1)$
$=\pi-\frac{\pi}{4}=\frac{3 \pi}{4}$
$\therefore$ from $(1), C=\pi-\frac{3 \pi}{4}=\frac{\pi}{4}$

Topic: Mathematics, Properties of Triangles, NDA, NDA 2011 (Phase 2)
In a triangle $\mathrm{ABC}, \mathrm{a}=2 \mathrm{~b}$ and $\angle \mathrm{A}=3 \angle \mathrm{B}$. Which one of the following is correct?

Options:
A The triangle is isosceles
B The triangle is equilateral
C The triangle is right-angled
D Such triangle does not exist

The correct answer is:
The triangle is right-angled

Solution:
From properties of triangle we know that
$\frac{\mathrm{a}}{\sin \mathrm{A}}=\frac{\mathrm{b}}{\sin \mathrm{B}}=\frac{\mathrm{c}}{\sin \mathrm{C}}$
Given that $\mathrm{a}=2 \mathrm{~b}$ and $\mathrm{A}=3 \mathrm{~B}$.
We get $\frac{2 b}{\sin 3 B}=\frac{b}{\sin B}$
$\Rightarrow \frac{\sin 3 B}{\sin B}=2 \Rightarrow \frac{3 \sin B-4 \sin ^{3} B}{\sin B}=2$
$\Rightarrow 3-4 \sin ^{2} B=2$
$\Rightarrow \sin ^{2} B=\left(\frac{1}{2}\right)^{2} \Rightarrow \sin B=\frac{1}{2}=\sin \frac{\pi}{6}$
$\Rightarrow B=\frac{\pi}{6}$ since $A=3 B$
So, $\angle A=\frac{\pi}{2}$
Thus, triangle is right angled triangle.

Topic: Mathematics, Properties of Triangles, NDA, NDA 2006 (Phase 1)
In a triangle $\mathrm{ABC}, \mathrm{a}-2 \mathrm{~b}+\mathrm{c}=0 .$ The value of $\cot \left(\frac{\mathrm{A}}{2}\right) \cot \left(\frac{\mathrm{C}}{2}\right)$ is

Options:
A $\frac{9}{2}$
B 3
C $\frac{3}{2}$
D 1

The correct answer is:
3

Solution:
Given, $\mathrm{a}-2 \mathrm{~b}+\mathrm{c}=0 \Rightarrow \mathrm{a}+\mathrm{c}=2 \mathrm{~b} \cot \left(\frac{\mathrm{A}}{2}\right) \cot \left(\frac{\mathrm{C}}{2}\right)$
$=\sqrt{\frac{\mathrm{s}(\mathrm{s}-\mathrm{a})}{(\mathrm{s}-\mathrm{b})(\mathrm{s}-\mathrm{c})} \times \frac{\mathrm{s}(\mathrm{s}-\mathrm{c})}{(\mathrm{s}-\mathrm{a})(\mathrm{s}-\mathrm{b})}}=\sqrt{\frac{\mathrm{s}^{2}}{(\mathrm{~s}-\mathrm{b})^{2}}}=\frac{\mathrm{s}}{\mathrm{s}-\mathrm{b}}$
$=\frac{2 \mathrm{~s}}{2 \mathrm{~s}-2 \mathrm{~b}}=\frac{\mathrm{a}+\mathrm{b}+\mathrm{c}}{\mathrm{a}+\mathrm{b}+\mathrm{c}-2 \mathrm{~b}}=\frac{2 \mathrm{~b}+\mathrm{b}}{2 \mathrm{~b}-\mathrm{b}}=\frac{3 \mathrm{~b}}{\mathrm{~b}}=3$

Topic: Mathematics, Properties of Triangles, NDA, NDA 2017 (Phase 2)
In a triangle $\mathrm{ABC}, a=8, b=10$ and $c=12$. What is the angle C equal to?

Options:
A $A / 2$
B $2 \mathrm{~A}$
C $3 \mathrm{~A}$
D $3 \mathrm{~A} / 2$

The correct answer is:
$2 \mathrm{~A}$

Solution:
$\cos C=\frac{a^{2}+b^{2}-c^{2}}{2 a b}=\frac{164-144}{2(8)(10)}=\frac{1}{8}$
$\cos A=\frac{b^{2}+c^{2}-a^{2}}{2 b c}=\frac{3}{4}$
$\operatorname{Sin} C=\frac{3 \sqrt{7}}{8}$ and $\operatorname{Sin} A=\frac{\sqrt{7}}{4}$
$\frac{\cos C}{\cos A}=\frac{1}{6} < 1 \Rightarrow \cos C < \cos A \Rightarrow \mathrm{C}>\mathbf{A}$
$\cos (C-A)=\cos C \cos A+\sin C \sin A$
$=\frac{1}{8} \times \frac{3}{4}+\frac{3 \sqrt{7}}{8} \times \frac{\sqrt{7}}{4}=\frac{3}{4}=\cos A$
$\Rightarrow \mathrm{C}-\mathrm{A}=\mathrm{A} \Rightarrow \mathrm{C}=2 A$

Topic: Mathematics, Properties of Triangles, NDA, NDA 2011 (Phase 2)
In a triangle $\mathrm{ABC}, \mathrm{b}=\sqrt{3} \mathrm{~cm}, \mathrm{c}=1 \mathrm{~cm}, \angle \mathrm{A}=30^{\circ}$, what is
the value of a?

Options:
A $\sqrt{2} \mathrm{~cm}$
B $2 \mathrm{~cm}$
C $1 \mathrm{~cm}$
D $\frac{1}{2} \mathrm{~cm}$

The correct answer is:
$1 \mathrm{~cm}$

Solution:
As given: In a triangle $\mathrm{ABC}, \mathrm{AC}=\mathrm{b}=\sqrt{3} \mathrm{~cm}, \mathrm{AB}$
$=\mathrm{c}=1 \mathrm{~m}$ and $\angle \mathrm{A}=30^{\circ}$
From cosine formulae
$\cos A=\frac{b^{2}+c^{2}-a^{2}}{2 b c}=\frac{(\sqrt{3})^{2}+1^{2}-a^{2}}{2 \sqrt{3} \cdot 1}$
$\Rightarrow \frac{\sqrt{3}}{2}=\frac{3+1-a^{2}}{2 \sqrt{3}} \Rightarrow 3=4-a^{2}$
$\Rightarrow \mathrm{a}^{2}=4-3=1 \Rightarrow \mathrm{a}=1 \mathrm{~cm}$

Topic: Mathematics, Properties of Triangles, NDA, NDA 2008 (Phase 1)
In a triangle $\mathrm{ABC}, \mathrm{c}=2, \mathrm{~A}=45^{\circ}, \mathrm{a}=2 \sqrt{2}$, than what is $\mathrm{C}$
equal to ?

Options:
A $30^{\circ}$
B $15^{\circ}$
C $45^{\circ}$
D None of these

The correct answer is:
$30^{\circ}$

Solution:
According to sine rule,
$\frac{\mathrm{a}}{\sin \mathrm{A}}=\frac{\mathrm{b}}{\sin \mathrm{B}}=\frac{\mathrm{c}}{\sin \mathrm{C}}$
$\therefore \frac{\mathrm{a}}{\sin \mathrm{A}}=\frac{\mathrm{c}}{\sin \mathrm{C}}$
$\Rightarrow \sin C=\frac{\mathrm{c} \cdot \sin \mathrm{A}}{\mathrm{a}}=\frac{2 \cdot \sin 45^{\circ}}{2 \sqrt{2}}$
$=\frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}}=\frac{1}{2}=\sin 30^{\circ}$
$\mathrm{C}=30^{\circ}$

Topic: Mathematics, Properties of Triangles, NDA, NDA 2014 (Phase 2)
In a triangle $\mathrm{ABC}$ if $\mathrm{a}=2, \mathrm{~b}=3$ and $\sin \mathrm{A}=\frac{2}{3}$, then what is angle B equal to?

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

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

Solution:
In $\Delta \mathrm{ABC}, \mathrm{a}=2, \mathrm{~b}=3$ and $\sin \mathrm{A}=\frac{2}{3}$
We know, $\frac{\sin A}{a}=\frac{\sin B}{b}$
$\Rightarrow \frac{\frac{2}{3}}{2}=\frac{\sin B}{3}$
$\Rightarrow \frac{2}{6}=\frac{\sin B}{3} \Rightarrow \sin B=\frac{6}{6}=1$
$\Rightarrow B=\sin ^{-1}(1)$ ...(1)
$=\frac{\pi}{2}$.

Topic: Mathematics, Properties of Triangles, NDA, NDA 2018 (Phase 1)
In a triangle $\mathrm{ABC}$, if $\mathrm{a}=2 \mathrm{~b}$ and $\mathrm{A}=3 \mathrm{~B}$ then which one of the following is correct?

Options:
A The triangle is obtuse-angled
B The triangle is acute-angled but not right-angled
C The triangle is right-angled
D The triangle is isosceles but not obtuse-angled

The correct answer is:
The triangle is right-angled

Solution:
We know from the Sine law that
$\frac{\mathrm{a}}{\sin \mathrm{A}}=\frac{\mathrm{b}}{\sin \mathrm{B}}$
$\Rightarrow \frac{2 b}{\sin 3 B}=\frac{b}{\sin B}$
$\Rightarrow 2 \sin B=\sin 3 B$
$\Rightarrow 2 \sin B=3 \sin B-4 \sin ^{3} B$
$\Rightarrow \sin B-4 \sin ^{3} B=0$
$\Rightarrow \sin B\left(1-4 \sin ^{2} B\right)=0$
$\Rightarrow \sin B=0$ or $1-4 \sin ^{2} B=0 \Rightarrow B=0$ or $B=30^{\circ}$
$\Rightarrow \mathrm{B}=30^{\circ}$ and $\mathrm{A}=3 \times 30^{\circ}=90^{\circ}$
$\Rightarrow \mathrm{B}=0$ is not possible so, $\mathrm{B}=30^{\circ}$ and $\mathrm{A}=3 \times 30^{\circ}=90^{\circ}$
$\Rightarrow$ The triangle is right angled triangle.

Topic: Mathematics, Properties of Triangles, NDA, NDA 2007 (Phase 1)
In a triangle $\mathrm{ABC}$, $\mathrm{if} \cos \mathrm{A}=\cos \mathrm{B} \cos \mathrm{C}$, what is the value of
$\tan A-\tan B-\tan C$ ?

Options:
A 0
B $-1$
C $1+\tan A \tan B \tan C$
D $\tan \mathrm{A} \tan \mathrm{B} \tan \mathrm{C}-1$

The correct answer is:
0

Solution:
As given, $\cos \mathrm{A}=\cos \mathrm{B} \cos \mathrm{C}$
$\tan A-\tan B-\tan C$
$=\frac{\sin A}{\cos A}-\frac{\sin B}{\cos B}-\frac{\sin C}{\cos C}$
$=\frac{\sin A}{\cos A}-\frac{(\sin B \cos C+\cos B \sin C)}{\cos B \cos C}$
$=\frac{\sin \mathrm{A}-\sin (\mathrm{B}+\mathrm{C})}{\cos \mathrm{A}} \quad \ldots[\mathrm{using}(1)]$
$\begin{array}{rr}=\frac{\sin A-\sin (\pi-A)}{\cos A} & {[\text { Since, } A+B+C=\pi} \\ & \text { So, } B+C=\pi-A]\end{array}$
$=\frac{\sin A-\sin A}{\cos A}=0$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2006 (Phase 1)
In a triangle $\mathrm{ABC}$, if taken in order, consider the following statements:
$1.$ $\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}}+\overrightarrow{\mathrm{CA}}=\overrightarrow{0}$
$2.$ $\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}}-\overrightarrow{\mathrm{CA}}=\overrightarrow{0}$
$3.$ $\overrightarrow{\mathrm{AB}}-\overrightarrow{\mathrm{BC}}+\overrightarrow{\mathrm{CA}}=\overrightarrow{0}$
$4.$ $\overrightarrow{\mathrm{BA}}-\overrightarrow{\mathrm{BC}}+\overrightarrow{\mathrm{CA}}=\overrightarrow{0}$
How many of the above statements are correct?

Options:
A One
B Two
C Three
D Four

The correct answer is:
One

Solution:
From triangle law of vector addition
$\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}}+\overrightarrow{\mathrm{CA}}=0$
Only statement $(1)$ is correct.

Topic: Mathematics, Vector Algebra, NDA, NDA 2018 (Phase 2)
In a triangle $\mathrm{ABC}$ if the angles $\mathrm{A}, \mathrm{B}, \mathrm{C}$ are in $\mathrm{AP}$, then which one of the following is correct?

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

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

Solution:
Since $\mathrm{A}, \mathrm{B}, \mathrm{C}$ are in A.P. $\therefore 2 \mathrm{~B}=\mathrm{A}+\mathrm{C}$
Also, $\mathrm{A}+\mathrm{B}+\mathrm{C}=180^{\circ} \Rightarrow 2 \mathrm{~B}+\mathrm{B}=180^{\circ}$
$\Rightarrow 3 \mathrm{~B}=180^{\circ} \Rightarrow \mathrm{B}=60^{\circ}$
Now, we know
$\cos B=\frac{a^{2}+c^{2}-b^{2}}{2 a c}$
$\Rightarrow \cos 60^{\circ}=\frac{\mathrm{a}^{2}+\mathrm{c}^{2}-\mathrm{b}^{2}}{2 \mathrm{ac}}$
$\Rightarrow \frac{1}{2}=\frac{\mathrm{a}^{2}+\mathrm{c}^{2}-\mathrm{b}^{2}}{2 \mathrm{ac}} \Rightarrow \mathrm{ac}=\mathrm{a}^{2}+\mathrm{c}^{2}-\mathrm{b}^{2}$
$\Rightarrow b^{2}=a^{2}+c^{2}-a c$

Topic: Mathematics, Properties of Triangles, NDA, NDA 2012 (Phase 1)
In a triangle $\mathrm{ABC}, \sin \mathrm{A}-\cos \mathrm{B}=\cos \mathrm{C}$, then what is $\mathrm{B}$ equal
to ?

Options:
A $\pi$
B $\pi / 3$
C $\pi / 2$
D $\pi / 4$

The correct answer is:
$\pi / 2$

Solution:
In a $\triangle \mathrm{ABC}$, we have $\sin A-\cos B=\cos C \Rightarrow \sin A=\cos B+\cos C$
$\Rightarrow 2 \sin \frac{A}{2} \cdot \cos \frac{A}{2}=2 \cos \left(\frac{B+C}{2}\right) \cdot \cos \left(\frac{B-C}{2}\right)$
$[\because \sin 2 \mathrm{~A}=2 \sin \mathrm{A} \cdot \cos \mathrm{A}]$
and $\cos B+\cos C=2 \cos \left(\frac{B+C}{2}\right) \cdot \cos \left(\frac{B-C}{2}\right)$
$\Rightarrow 2 \sin \frac{\mathrm{A}}{2} \cdot \cos \frac{\mathrm{A}}{2}=2 \cos \left(90^{\circ}-\frac{\mathrm{A}}{2}\right) \cdot \cos \left(\frac{\mathrm{B}-\mathrm{C}}{2}\right)$
$\left[\because \mathrm{A}+\mathrm{B}+\mathrm{C}=180^{\circ} \Rightarrow\left(\frac{\mathrm{B}+\mathrm{C}}{2}\right)=90^{\circ}-\frac{\mathrm{A}}{2}\right]$
$\Rightarrow 2 \sin \frac{\mathrm{A}}{2} \cdot \cos \frac{\mathrm{A}}{2}=2 \sin \frac{\mathrm{A}}{2} \cdot \cos \left(\frac{\mathrm{B}-\mathrm{C}}{2}\right)$
$\left[\because \cos \left(90^{\circ}-\theta\right)=\sin \theta\right]$
$\Rightarrow \quad \cos \frac{\mathrm{A}}{2}=\cos \left(\frac{\mathrm{B}-\mathrm{C}}{2}\right)$
$\Rightarrow \frac{A}{2}=\frac{B-C}{2}$
$\Rightarrow \quad \mathrm{A}+\mathrm{C}=\mathrm{B}$ ....(i)
Also, $A+C=180^{\circ}-B$ ....(ii)
So, $180^{\circ}-\mathrm{B}=\mathrm{B}$
$\Rightarrow \quad 2 \mathrm{~B}=180^{\circ}$
$\therefore \quad \mathrm{B}=90^{\circ}$

Topic: Mathematics, Properties of Triangles, NDA, NDA 2014 (Phase 2)
In $\Delta \mathrm{ABC}$, if $\angle \mathrm{A}: \angle \mathrm{B}: \angle \mathrm{C}=1: 2: 3$, then what is $\mathrm{BC}: \mathrm{CA}$ :
$\mathrm{AB}$ ?

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

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

Solution:
Ratio of angles is given by $\angle \mathrm{A}: \angle \mathrm{B}: \angle \mathrm{C}=1: 2: 3$
Let $\angle \mathrm{A}=\mathrm{x}, \angle \mathrm{B}=2 \mathrm{x}$ and $\angle \mathrm{C}=3 \mathrm{x}$
We know that in a triangle $\angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}=180^{\circ}$
$\Rightarrow \mathrm{x}+2 \mathrm{x}+3 \mathrm{x}=180^{\circ}$
$\Rightarrow \quad x=\frac{180^{\circ}}{6}=30^{\circ}$
So, $\angle \mathrm{A}=30^{\circ}, \angle \mathrm{B}=60^{\circ}$ and $\angle \mathrm{C}=90^{\circ}$
From sin law
$\frac{\mathrm{BC}}{\sin \mathrm{A}}=\frac{\mathrm{CA}}{\sin \mathrm{B}}=\frac{\mathrm{AB}}{\sin \mathrm{C}}=\mathrm{K}$
$\frac{\mathrm{BC}}{\sin 30^{\circ}}=\frac{\mathrm{CA}}{\sin 60^{\circ}}=\frac{\mathrm{AB}}{\sin 90^{\circ}}=\mathrm{K}$
$\mathrm{BC}=\mathrm{K} \sin 30^{\circ}$
$\mathrm{CA}=\mathrm{K} \sin 60^{\circ}$
$\mathrm{AB}=\mathrm{K} \sin 90^{\circ}$
$\mathrm{BC}: \mathrm{CA}: \mathrm{AB}$
$=\sin 30^{\circ}: \sin 60^{\circ}: \sin 90^{\circ}=\frac{1}{2}: \frac{\sqrt{3}}{2}: 1=1: \sqrt{3}: 2$

Topic: Mathematics, Properties of Triangles, NDA, NDA 2006 (Phase 2)
In an annual ring, the light coloured part is known as:

Options:
A Early wood
B Late wood
C Heartwood
D Sapwood

The correct answer is:
Early wood

Solution:
Annual ring consists of two concentric light and dark colour ring, which are formed due unequal activity of cambium in different season. In spring season the cambium is more active producing more amount of secondary xylem with wider lumen which is light in colour, in autumn season cambium is less active and produces less amount of xylem with narrow lumen and which is dark in colour.

Topic: Biology, Anatomy of Flowering Plants, NEET
In an AP, the $m^{\text {th }}$ term $1 / n$ and $n^{\text {th }}$ term is $1 / m$. What is its $(m n)^{\text {th }}$ term?

Options:
A $1 /(\mathrm{mn})$
B $\mathrm{m} / \mathrm{n}$
C $n / m$
D 1

The correct answer is:
1

Solution:
Let 'a' be the first term and 'd' be the common difference of an A.P
Now, Given $\mathrm{m}^{\text {th }}$ term $=\frac{1}{n}$
and $\mathrm{n}^{\text {th }}$ term $=\frac{1}{m}$
$\Rightarrow \mathrm{a}+(\mathrm{m}-1) \mathrm{d}=\frac{1}{n}$ and
$a+(n-1) d=\frac{1}{m}$
By subtracting the above two eqns, we get
$(\mathrm{m}-1-\mathrm{n}+1) \mathrm{d}=\frac{1}{n}-\frac{1}{m}$
$\Rightarrow(\mathrm{m}-\mathrm{n}) \mathrm{d}=\frac{m-n}{m n}$
$\Rightarrow d=\frac{1}{m n}$
Now, $(\mathrm{mn})^{\text {th }}$ term $=\mathrm{a}+(\mathrm{mn}-1) \mathrm{d}$
$=a+(m n-1) \frac{1}{m n}=a+1-\frac{1}{m n}$
Now, $a=\frac{1}{m}-(n-1) d$
$=\frac{1}{m}-(n-1) \frac{1}{m n}=\frac{1}{m}-\frac{n}{m n}+\frac{1}{m n}=\frac{1}{m n}$
$\therefore(\mathrm{mn})^{\text {th }}$ term $=\frac{1}{m n}+1-\frac{1}{m n}=1$

Topic: Mathematics, Sequences and Series, NDA, NDA 2009 (Phase 2)
In an ecosystem the rate of production of organic matter during photosynthesis is termed as:

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

The correct answer is:
Gross primary productivity

Solution:
Gross primary productivity is the amount of organic matter produced at given length of time during photosynthesis.

Topic: Biology, Ecosystem, NEET, NEET 2015 (Phase 1)
In an ecosystem the rate of production of organic matter during photosynthesis is termed as:

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

The correct answer is:
Gross primary productivity

Solution:
Gross primary productivity is the amount of organic matter produced at given length of time during photosynthesis.

Topic: Biology, Ecosystem, NEET
In an ellipse, if the distance between the foci is 6 units and the length of its minor axis is 8 units, then its eccentricity is

Options:
A 12
B 75
C 15
D 35

The correct answer is:
35

Solution:
Consider the standard equation of the ellipse

x2a2+y2b2=1,

The distance between it's foci is 6 , so, 2c=6 ⇒c=3

and given length of minor axis is 8 units,

hence, 2b=8 ⇒ b=4

So, the value of a=b2+c2=42+32=5

Hence, e=ca=35

Topic: Mathematics, Ellipse, AP EAMCET, AP EAMCET 2021 (19 Aug Shift 2)
In an examination, $40 \%$ of candidates got second class When the data are represented by a pie chart, what is the angle corresponding to second class'

Options:
A 4
B 90
C $144^{5}$
D $320^{\circ}$

The correct answer is:
$144^{5}$

Solution:
First, convert $40 \%$ into fraction
$40 \%=\frac{40}{100}=\frac{2}{5}$
Angle $=\frac{2}{5} \times 360=144^{\circ}$

Topic: Mathematics, Statistics, NDA, NDA 2017 (Phase 1)
In an examination, $70 \%$ students passed in Physics, $80 \%$ students passed in Chemistry, $75 \%$ students passed in Mathematics and $85 \%$ students passed in Biology, and $x \%$ students failed in all the four subjects. What is the minimum value of $\mathrm{x}$ ?

Options:
A 10
B 12
C 15
D None of the above

The correct answer is:
None of the above

Solution:
Here, maximum number of students failed in all the four subjects $=15 \%$ But, minimum number of students failed in all the four subjects varies from 0 to $15 \%$. So, correctoption is (d).

Topic: Mathematics, Sets and Relations, NDA, NDA 2016 (Phase 2)
In an examination out of 100 students, 75 passed in English
60 passed in Mathematics and 45 passed in both English and Mathematics. What is the number of students passed in exactly one of the two subjects?

Options:
A 45
B 60
C 75
D 90

The correct answer is:
45

Solution:
Total no. of students $=100$ Let $E$ denote the students who have passed in English Let $M$ denote the students who have passed in Maths $n(E)=75, n(M)=60$ and $n(E \cap M)=45$
we know $n(E \cup M)=n(E)+n(M)-n(E \cap M)$
$=75+60-45=90$
Required number of students $=90-45=4^{5}$

Topic: Mathematics, Sets and Relations, NDA, NDA 2009 (Phase 1)
In an examination, the probability of a candidate solving a question is $\frac{1}{2}$. Out of given 5 questions in the examination, what is the probability that the candidate was able to solve at least 2 questions?

Options:
A $\frac{1}{64}$
B $\frac{3}{16}$
C $\frac{1}{2}$
D $\frac{13}{16}$

The correct answer is:
$\frac{13}{16}$

Solution:
$\begin{aligned} & P={ }^{5} C_{2}\left(\frac{1}{2}\right)^{2}\left(\frac{1}{2}\right)^{3}+{ }^{5} C_{3}\left(\frac{1}{2}\right)^{3}\left(\frac{1}{2}\right)^{2} \\ &+{ }^{5} C_{4}\left(\frac{1}{2}\right)^{4}\left(\frac{1}{2}\right)^{1}+{ }^{5} C_{5}\left(\frac{1}{2}\right)^{5}\left(\frac{1}{2}\right)^{0} \\ &=\left(\frac{1}{2}\right)^{5}\left[{ }^{5} C_{2}+{ }^{5} C_{3}+{ }^{5} C_{4}+{ }^{5} C_{5}\right] \\ &=\frac{1}{3^{2}}[10+10+5+1] \\ &=\frac{1}{3^{2}} \times 26=\frac{13}{16} \end{aligned}$

Topic: Mathematics, Probability, NDA, NDA 2015 (Phase 1)
In an examination, there are 3 multiple choice questions and each question has 4 choices. If a student randomly selects answer for all the 3 questions, what is the probability that the student will not answer all the 3 questions correctly?

Options:
A $1 / 64$
B $63 / 64$
C $1 / 12$
D $11 / 12$

The correct answer is:
$63 / 64$

Solution:
Since, probability of answering all the three questions correctly $=\frac{1}{4} \times \frac{1}{4} \times \frac{1}{4}=\frac{1}{64}$
$\therefore$ Probability of not answering all the three questions correctly $=1-\frac{1}{64}=\frac{63}{64}$

Topic: Mathematics, Probability, NDA, NDA 2011 (Phase 1)
In an ideal step up transformer, if the input voltage and input power are $\mathrm{V}_1$ and $\mathrm{P}_1$ respectively and the output voltage and output power are $\mathrm{V}_2$ and $\mathrm{P}_2$ respectively, then

Options:
A $\mathrm{V}_1=\mathrm{V}_2 ; \mathrm{P}_1=\mathrm{P}_2$
B $\mathrm{V}_1>\mathrm{V}_2 ; \mathrm{P}_1>\mathrm{P}_2$
C $\mathrm{V}_1 < \mathrm{V}_2 ; \mathrm{P}_1 < \mathrm{P}_2$
D $\mathrm{V}_1 < \mathrm{V}_2 ; \mathrm{P}_1=\mathrm{P}_2$

The correct answer is:
$\mathrm{V}_1 < \mathrm{V}_2 ; \mathrm{P}_1=\mathrm{P}_2$

Solution:
In step up transformer the output voltage is greater than input voltage because the number of turns on the secondary is more than the number of turns in the primary. Therefore $\mathrm{V}_2>\mathrm{V}_1$
Input power is equal to output power.
$$
\therefore \mathrm{P}_1=\mathrm{P}_2
$$
Hence (4) is correct.

Topic: Physics, Alternating Current, TS EAMCET, TS EAMCET 2023 (12 May Shift 2)
In an ideal step-down transformer, out of the following quantities, which quantity increases in the secondary coil?

Options:
A Power
B Voltage
C Current
D Frequency

The correct answer is:
Current

Solution:
For an ideal transformer,
$$
\frac{\mathrm{V}_{\mathrm{s}}}{\mathrm{V}_{\mathrm{p}}}=\frac{\mathrm{I}_{\mathrm{p}}}{\mathrm{I}_{\mathrm{s}}}
$$
Hence, if $\mathrm{V}_{\mathrm{s}} < \mathrm{V}_{\mathrm{p}}$ then $\mathrm{I}_{\mathrm{p}} < \mathrm{I}_{\mathrm{s}}$

Topic: Physics, Alternating Current, MHT CET, MHT CET 2021 (20 Sep Shift 1)
In an old tree trunk of mango, which tissue will be present in the maximum amount?

Options:
A Primary xylem
B Primary phloem
C Secondary xylem
D Secondary cortex

The correct answer is:
Secondary xylem

Solution:
In an old trunk, increase in thickness occurs due to secondary growth. The secondary growth in the stellar region occurs due to activity of vascular cambium, which forms more amount of secondary xylem and this forms the heart of the trunk.

Topic: Biology, Anatomy of Flowering Plants, NEET
In an open beaker containing reactants, what is considered the boundary of the system?

Options:
A The walls of the beaker
B The reactants inside the beaker
C An imaginary surface enclosing the beaker and reactants
D The air surrounding the beaker

The correct answer is:
An imaginary surface enclosing the beaker and reactants

Solution:
In this context, the boundary is defined as an imaginary surface that encloses both the beaker and the reactants inside it.

Topic: Chemistry, Thermodynamics (C), NEET
In Andalusian fowls, the mating of black and white fowls will produce

Options:
A Black fowls
B White fowls
C Blue fowls
D Grey fowls

The correct answer is:
Blue fowls

Solution:
The colour of Andalusian fowl exhibits incomplete dominance. Black fowls are homozygous dominant with the genotype BB. White fowls are homozygous recessive with the genotype bb. When they mate, all fowls are heterozygous having the genotype Bb which is expressed as blue.

Topic: Biology, Principles of Inheritance and Variation, NEET
In Angiosperms, microsporogenesis and megasporogenesis:

Options:
A Form gametes without further divisions
B Involve meiosis
C Occur in ovule
D Occur in anther

The correct answer is:
Involve meiosis

Solution:
The Angiosperms are heterosporous. The stamens of the flower are the male sexual parts and they are highly modified leaves called microsporophylls. The microsporophylls produce anthers or microsporangia. Within the anther, there are special diploid cells called pollen mother cell or microspore mother cells. The microspore mother cells undergo meiosis to produce haploid microspores or pollen grains. The microspore produces a few celled male gametophytes, each male gametophyte has two cells tube cell and a generative cell. This process is called microsporogenesis.

Gynoecium or pistil is the female sexual parts. The gynoecium is made up of highly modified leaves called megasporophylls or carpels. The megasporophylls fused to form ovaries. The function of carpels is to produce ovules or megasporangia by a process called megasporogenesis. Within the ovules, there are specialized cells called the megaspore mother cells (2n). The megaspores mother cells are diploid (2n) which undergo meiosis or reduction division to form haploid megaspores (n). The megaspores produce a multicellular female gametophyte or embryo sac. The mature embryo sac has a group of three antipodal cells, two synergid cells, one female gamete or egg cell. There are two haploid polar nuclei which fuse together to produce a diploid secondary nucleus.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET, NEET 2015 (Phase 2)