Class 12th

Let fare for business class and economy class be Rs. 3x and Rs. 2x respectively. Let the passenger in business class and economy class be 12y and 25y respectively.

Total fare = 3x * 12y + 2x * 25y

=86xy = 430000

xy = 5000

Absolute difference = (50 – 36) * 5000

= Rs.70, 000

A camera lens focuses the light onto a light-sensitive surface like film or a digital sensor for capturing an image. The camera lens gathers as much light as possible for forming a clear and bright image. This lens focuses the collected light on a specific point like camera's film or its digital sensor. The focus is achieved due to the curvature of lens and the arrangement of its optical elements. This lens refracts light rays so that they converge at focal plane where film/sensor is located.

Let X = Total Amount of Gold

& Y = Total Amount of Platinum

$\frac{}{}$ $\frac{X}{Y}=\frac{2}{5}$  . (1)

Given U1 + P1 = U2 + P2. (2)

U1 + U1 = X & P1 + P2 = Y. (3)

Also, given U1 = 3U2

From (2) P2 − P1 = 2U2

Also, X = 4U2 & 2P1 = Y − 2U2

$\frac{X}{Y}=\frac{2}{5}=\frac{4U_2}{2P_1+2U_2}$

4P1 + 4U2 = 20U2

4P1 = 16U2 $=\frac{16}{3}{U}_1$

$\frac{U_1}{P_1}=\frac{4*3}{16}=\frac{3}{4}$ $\frac{}{}{}_{}$

U1 : P1 = 3 : 4

A telescope functions on the working principle of gathering light form distant source and then, forming image which can be magnified for observation. Based on their design, there are two types of telescopes including refracting telescopes and reflecting telescopes. A refracting telescope has a front lens called as objective lens. This front lens is a convex lens that collects light from distant objects and then bend it to the focal point. In a reflecting telescope, a concave mirror is the primary mirror that collects light and then, reflects it to a focal point.

Let Gopal's share be G, Abhishek's be A, Sadiq's be S and Pawan's be P.

Given that $G+3=A+\frac{A}{3}=\frac{80S}{100}=P-4$ $\frac{}{}\frac{}{}$

You get

G = P – 7. (1)

$$ $A=\frac{3}{4}(P-4)$  . (2)

$$ $S=\frac{5}{4}(P-4)$  …. (3)

Also given that G + A + P + S = 80…. (4)

Substituting the values from (1), (2) and (3) in (4), we get

$\frac{}{}\frac{}{}$  $\left[P+\frac{3P}{4}+\frac{5P}{4}+P\right]-7-3-5=80$ or P = 23.75

For n = 14, 39 (1 + 33 + 36 + 35)

= 39 (1 + 27 + 729 + 243)

= 39 * 103.

Volume of tube $=\frac{2}{3}\pi a^3+\pi a^{2}b$ $\frac{}{}{}^{}{}^{}$

When it is half-full $=\frac{1}{3}\pi a^3+\frac{1}{2}\pi a^{2}b$ $\frac{}{}{}^{}\frac{}{}{}^{}$

Volume of Cylinder

$=\frac{1}{3}\pi a^3+\frac{1}{2}\pi a^{2}b-\frac{2}{3}\pi a^3=\frac{1}{2}\pi a^{2}b-\frac{1}{3}\pi a^3$

Length in the cylinder

$=\left(\frac{1}{2}\pi a^{2}b-\frac{1}{3}\pi a^3\right)÷\pi a^2=\frac{1}{2}b-\frac{1}{3}a$

The height above the lowest point

$\frac{1}{2}b-\frac{1}{3}a+a=\frac{2}{3}a+\frac{1}{2}b$

When it is $\frac{1}{4}$ $\frac{}{}$ full, volume $=\frac{1}{6}\pi a^3+\frac{1}{4}{\pi }^{a}b$ $\frac{}{}{}^{}\frac{}{}{}^{}$

Volume of cylinder

$=\frac{1}{6}\pi a^3+\frac{1}{4}\pi a^{2}b-\frac{2}{3}\pi a^3=\frac{1}{4}\pi a^{2}b-\frac{1}{2}\pi a^3$

Length in the cylinder

$=\left(\frac{1}{4}\pi a^{2}b-\frac{1}{2}\pi a^3\right)÷\pi a^2=\frac{1}{4}b-\frac{1}{2}a$

The height above the lowest point

$=\frac{1}{4}b-\frac{1}{2}a+a=\frac{1}{2}a+\frac{1}{4}b$

When it is full, volume 

$=\frac{3}{4}\left(\frac{2}{3}\pi a^3+\pi a^{2}b\right)=\frac{1}{2}\pi a^3+\frac{3}{4}\pi a^{2}b$

Volume in cylinder

$=\frac{1}{2}\pi a^2+\frac{3}{4}\pi a^{2}b-\frac{2}{3}\pi a^3=\frac{3}{4}\pi a^{2}b-\frac{1}{6}\pi a^3$

Length in the cylinder

$=\left(\frac{3}{4}\pi a^{2}b-\frac{1}{6}\pi a^3\right)÷\pi a^2=\frac{3}{4}b-\frac{1}{6}a$

The height above the lowest point

$=\frac{3}{4}b-\frac{1}{6}a+a=\frac{5}{6}a+\frac{3}{4}b$

The average age of n men is X year. Total age is given by N years

$\begin{array}{l}Newaverageage =\frac{NX+X-1+X+1+X+2}{N+3}\\ =\frac{NX+3X+2}{N+3}\\ =\frac{X[N+3]}{N+3}+\frac{2}{N+3}\\ \Rightarrow X+\frac{2}{N+3}\end{array}$