Maths

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

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.

$ForMedian=L+\frac{\frac{N}{2}-c,}{f}xh$

$14=12+\frac{\frac{a+b}{2}+13-\left(a+b\right)}{12}*6$              

->a + b = 18                                   -(ii)

Solving (i) & (ii) a = 8 & b = 10

->(a – b)² = 4

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

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

${\left(2x^r+\frac{1}{x^2}\right)}^{10}$

Let the constant term is (k + 1)^th term.

${\therefore }^{10}{C}_k{\left(2x^r\right)}^{10-k}.x^{-2k}$              

${=}^{10}{C}_k{2}^{10-k}.x^{10r-rk-2k}$              

              For constant term 10r – rk – 2k = 0.

$\therefore k=\frac{10r}{r+2}k\in I$              

$\therefore r=3,k=6andr=8,k=8$               

For k = 6

For k = 8  

$\therefore r=8$

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

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

1^st position can be filled in 4 ways as zero cannot appear in 1^st position.

2^nd position can be filled in 4 ways and so on.

Total cases = 4 * 4 * 3 * 2 * 1 = 96

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

 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