Correct Answer: 90°

Solution :
Given:
Since AD is the median of △ABC,
BD = CD = $\frac{1}{2}$BC
$\therefore$ AD = BD = CD
Thus A, B, and C are in equal distance from point D.
That means D is the circumcentre of △ ABC.
Also, in a right triangle,

Correct Answer: Meghadutam 

Solution : Correct Answer is Meghadutam

The genre of classical Sanskrit poetry is diverse and contains a wide range of poetic styles. The two epics that Indians hold in the utmost awe and respect, the Ramayana and Mahabharata, are the most well-known examples of epic poetry. The

Correct Answer: $4:1$

Solution : Given: $(x^{3}-y^{3}):( x^{2}+xy+y^{2})=5:1$ and $( x^{2}-y^{2}):(x-y)=7:1$
Now, $\frac{x^{3}-y^{3}}{ x^{2}+xy+y^{2}}=\frac{5}{1}$
Using the algebraic identity, $a^3-b^3=(a-b)(a^2+ab+b^2)$
⇒ $\frac{(x-y)(x^{2}+xy+y^{2})}{ x^{2}+xy+y^{2}}=\frac{5}{1}$
⇒ $x-y=5$ ......................... (1)
Again, $\frac{x^{2}-y^{2}}{ x-y}=\frac{7}{1}$
Using the algebraic identity $a^2-b^2=(a-b)(a+b)$
⇒$\frac{(x-y)(x+y)}{ x-y}=\frac{7}{1}$
⇒ $x+y=7$ ........................ (2)
Solving both equations, we get:
⇒ $x= 6, y=1$
Thus,

Correct Answer: 145$^\circ$

Solution :
Given, a cyclic quadrilateral ABCD and diameter AD
And, $\angle$ DAC = 55$^\circ$
In $\triangle$ ACD,
$\angle$ DAC = 55$^\circ$
$\angle$ ACD = 90$^\circ$ (angle in a semicircle)
$\angle$ ADC = 180$^\circ$ – 55$^\circ$ – 90$^\circ$ = 35$^\circ$ (angle sum property)
We know that opposite

Correct Answer: 52

Solution : Given: The lengths of the diagonals of a rhombus are 24 cm and 10 cm.
So, the length of the sides
= $\sqrt{(\frac{24}{2})^2+(\frac{10}{2})^2}$
= $\sqrt{12^2+5^2}$
= $\sqrt{169}$
= $13$
$\therefore$ The perimeter of the rhombus = (4 × 13) = 52 cm
Hence, the correct

Correct Answer: 7 years

Solution : Let the present age of Ram and Ramesh be $3x$ and $5x$.
According to the question,
After 7 years,
$\frac{3x+7}{5x+7}=\frac{4}{5}$
⇒ $15x + 35 = 20x + 28$
⇒ $35 - 28 = 20x - 15x$
⇒ $5x = 7$
⇒ $x = \frac{7}{5}$

Correct Answer: $\text{None of these}$

Solution : Marked price = Rs. 150
Discount = 20% of marked price = 20% of 150 = $\frac{20}{100}$ × 150 = Rs. 30
Price after initial discount = Rs. 150 – Rs. 30 = Rs. 120
Final price = Rs. 108
Additional discount to

Correct Answer: Greater than 1

Solution : $\sin\theta+\cos\theta$ where $0°<\theta<90°$
Putting the value of $\theta=45°$, we have,
$\sin 45°+\cos45°=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}$
⇒ $\sin 45°+\cos45° =\frac{2}{\sqrt{2}}=\sqrt{2}$
Here, $\sqrt{2}=1.414>1$
Hence, the correct answer is greater than 1.

Correct Answer: $4 \sqrt{5} \mathrm{~cm}$

Solution :

We have,
The $ABC$ is a triangle and $D$ is a point on the side $BC$. If $BC = 16 \mathrm{~cm}$, $BD = 11 \mathrm{~cm}$, $\angle ADC = \angle BAC$ (given), $\angle C$ is common angle and $AC$ is common side.
So, $\triangle

Correct Answer: 1

Solution : Taking dice positions of Figures 1 and 2 with 5 as the common face –

Moving clockwise from the common face with 5 –

Pairs of opposite faces –

So, the opposite face to the face containing 3 is 1. Hence, the second option is