Class 12th

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

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

$A={\left(\begin{array}{ccc}\hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right)}_{3*3}$

 B is a matrix of same order with entries from {1,2,3,4,5}. and satisfying AB = BA.

$LetB=\left(\begin{array}{ccc}\hfill a_{11}\hfill & \hfill a_{12}\hfill & \hfill a_{13}\hfill \\ \hfill a_{21}\hfill & \hfill a_{22}\hfill & \hfill a_{23}\hfill \\ \hfill a_{31}\hfill & \hfill a_{32}\hfill & \hfill a_{33}\hfill \end{array}\right)$           

$\therefore \left(\begin{array}{ccc}\hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right)\left(\begin{array}{ccc}\hfill a_{11}\hfill & \hfill a_{12}\hfill & \hfill a_{13}\hfill \\ \hfill a_{21}\hfill & \hfill a_{22}\hfill & \hfill a_{23}\hfill \\ \hfill a_{31}\hfill & \hfill a_{32}\hfill & \hfill a_{33}\hfill \end{array}\right)=\left(\begin{array}{ccc}\hfill a_{11}\hfill & \hfill a_{12}\hfill & \hfill a_{13}\hfill \\ \hfill a_{21}\hfill & \hfill a_{22}\hfill & \hfill a_{23}\hfill \\ \hfill a_{31}\hfill & \hfill a_{32}\hfill & \hfill a_{33}\hfill \end{array}\right)\left(\begin{array}{ccc}\hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right)$

$a_{21}=a_{12},a_{22}=a_{11},a_{23}=a_{13},a_{31}=a_{32}$

$\therefore$ there exist only 5 distinct entries in the matrix B so that possible case = 5⁵ = 3125

Total time = 182s

2 + 4 + 6 + 8 +……………+ n terms = 182

i.e. n=13

Total distance covered = 2* [1² + 2 ²+ 3² +……. …………. + 13²]

= 2 * 13 * 14 * $\frac{27}{6}$  = 9 * 13 * 14m

Average speed = 9 * 13 * $\frac{4}{182}$ = 9m/s

$x^2=-2\left(y-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)$  

y² = -4 (x – 1)

Required area = $2{\displaystyle {\int }_{-1}^0\left[2\sqrt[]{x+1}-\frac{1-x^2}{2}\right]dx}$  

$=2{\left[\frac{4}{3}{\left(x+1\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-\frac{1}{2}\left(x-\frac{x^3}{3}\right)\right]}_{-1}^0$          

= 2