Correct Answer: (32.5, 37.0, 43.5)

Solution : Given:
(10.5, 15.0, 21.5) 

Here, add 4.5 to the first number to obtain the second number, and then add 6.5 to the second number to obtain the third number –
10.5 + 4.5 = 15.0; 15.0 + 6.5 = 21.5
Let's check the

Correct Answer: E

Solution : Given:
(i) E is not at an end, and D is second to the left of F. C is the neighbor of E and is sitting diagonally opposite to D.

 (ii) B is the neighbor of F.

Correct Answer: Only conclusion I follows

Solution : Let's analyse the conclusions –
Conclusion (I): The Government understands that the condition of poor farmers needs immediate attention:
→This conclusion follows as the PM announced the annual pension for farmers.
Conclusion (II): No benefits are announced for other sections of society:

Correct Answer: $7 \frac{1}{2}$

Solution : One day work of A = $\frac{1}{8}$
One day work of B = $\frac{1}{7}$
Working alternatively,
Two day work of A + B = $\frac{1}{8}+\frac{1}{7}$
= $\frac{15}{56}$
Six days work of (A + B) = $3\times\frac{15}{56}$
= $\frac{45}{56}$
On the seventh day, A works

Correct Answer: Arteries

Solution : The correct option is Arteries.

The blood vessels that carry blood from the heart to various parts of the body are called arteries. Arteries are responsible for transporting oxygenated blood (except for the pulmonary artery, which carries deoxygenated blood to the lungs for oxygenation) and

Correct Answer: 65°

Solution :
Given: PA and PB are tangents
$\angle OAP = 90^\circ $
$\angle OBP = 90^\circ $
As, OAPB is a quadrilateral
$\angle OAP + \angle APB + \angle PBO + \angle BOA = 360^\circ $
⇒ $90^\circ + 130^\circ + 90^\circ + \angle BOA =

Correct Answer: $\left(2-{e}^2\right) ^\frac{3}{2}$

Solution : Given,
$\cot^2θ = 1 - e^2$
Consider,
$\operatorname{cosec} θ + \cot^3θ \sec θ$
$=\frac{1}{\sinθ} + \frac{\cos^3θ}{\sin^3θ}\frac1{\cosθ}$
$=\frac{\sin^2θ+\cos^2θ}{\sin^3θ}$
$=\frac{1}{\sin^3θ}$ [As $\sin^2θ+\cos^2θ=1$]
$=\operatorname{cosec^3}θ$
Also, we know that,
$\operatorname{cosec^2}θ=1+\cot^2θ$
⇒ $\operatorname{cosec^2}θ=1+1-e^2$
⇒ $\operatorname{cosec^2}θ=2-e^2$
⇒ $\operatorname{cosec}θ=(2-e^2)^{\frac12}$
⇒ $\operatorname{cosec^3}θ=(2-e^2)^{\frac32}$
Hence, the correct answer is $(2-e^2)^{\frac32}$.

Correct Answer: School

Solution : Given:
A scientist is related to the laboratory.

The first term is a professional and the second term is the workplace –
A scientist is a professional and a laboratory is a workplace or environment.
Similarly, a teacher is a professional and a school is

Correct Answer: 7 days

Solution : Given: 42 persons can consume 144 kg of wheat in 15 days.
Let in $x$ days 30 persons consume 48 kg of wheat.
We know that $\frac{M1D1}{W1}=\frac{M2D2}{W2}$
According to the question,
$\frac{42×15}{144}=\frac{30×x}{48}$
⇒ $x=7$
Hence, the correct answer is 7 days.

Correct Answer: $\frac{8}{3}$

Solution : $\operatorname{cosec}(90^{\circ} - \theta) = \sec \theta$
$\tan\theta = \frac{1}{\cot\theta}$
So, $4\left(\operatorname{cosec}^2 57^{\circ}-\tan ^2 33^{\circ}\right)-\cos 90^{\circ}-y \tan ^2 66^{\circ} \tan ^2 24^{\circ}=\frac{y}{2}$
⇒ $(\operatorname{cosec}^{2}(90-33)^{\circ} - \tan^{2}33^{\circ}) - 0- y×\tan^{2}66^{\circ} × \tan^{2}(90-66)^{\circ}$ = $\frac{y}{2}$ 
⇒ $4(\sec^{2}57^{\circ}- \tan^{2}33^{\circ}) -  y × \tan^{2}66^{\circ} × \cot^{2}66^{\circ} = \frac{y}{2}$
⇒ $4