In how many ways 50 people can be divided into 3 groups, so that cache group contains members equal to a prime number?

$Here, Principal = p; Rate of interest = r; Time = n$

$\therefore Compounded amount = p{(1+\frac{r}{100})}^n$

$$\begin{gathered} Now, according to question, p{(1+\frac{r}{100})}^{15}=2p \\ \therefore {(1+\frac{r}{100})}^{15}=2...............\left(i\right) \\ Let, n be the number of years when principal will become 8 times. \\ \therefore p{(1+\frac{r}{100})}^n=8p ={(1+\frac{r}{100})}^n=8 =(1+\frac{r}{100})=2^3 \\ ={(1+\frac{r}{100})}^n={(1+\frac{r}{100})}^{3\times 15} ={(1+\frac{r}{100})}^n={(1+\frac{r}{100})}^{45} \therefore n=45 \\ \therefore Required times 45 years. \end{gathered}$$

$$\begin{gathered} Let, the radius and height of the cylinder be r and h cm respectively. \\ We know,surface area of cylinder is=2\mathrm{\pi rh} {\mathrm{cm}}^2 \end{gathered}$$

$$\begin{gathered} And volume of the cylinder is={\mathrm{\pi r}}^2\mathrm{h} {\mathrm{cm}}^3 \\ \mathrm{According} \mathrm{to} \mathrm{question} , 2\mathrm{\pi rh} =100.........\left(\mathrm{i}\right) \\ \mathrm{And}, {\mathrm{\pi r}}^2\mathrm{h}=150.......\left(\mathrm{ii}\right) \\ \mathrm{Dividing} \mathrm{equation} \left(\mathrm{ii}\right) \mathrm{by} \mathrm{equation} \left(\mathrm{i}\right), \mathrm{we} \mathrm{get} \end{gathered}$$

$$\begin{gathered} \frac{{\mathrm{\pi r}}^2\mathrm{h}}{2\mathrm{\pi rh}}=\frac{150}{100} \\ =\frac{r}{2}=\frac{3}{2} \\ \therefore r=3cm \\ Putting, r = 3 in equation \left(i\right), we get \\ 2\mathrm{\pi rh}=100 \\ \therefore \mathrm{h}=\frac{100}{2\mathrm{\pi r}}=\frac{100}{2\times 3\times 3.1416}=5.3 \mathrm{cm} \\ \therefore \mathrm{Required} \mathrm{radius} = 3\mathrm{cm} \mathrm{and} \mathrm{height} = 5.3\mathrm{cm}. \end{gathered}$$

$$\begin{gathered} Let us depict a figure according to question \\ Here, ABCD is a 3 by 4 rec\tan gle which is inscribed in a circle. \\ \therefore Diagonal of the rectangle area, AC=D^{\sqrt{A{D}^2+C{D}^2}}=C^{\sqrt{3^2+4^2}}=\sqrt{9+16}=\sqrt{25}=5 unit \end{gathered}$$

$$\begin{gathered} \therefore Radius of the circle =\frac{5}{2} unit \\ \therefore Circumference of the circle = 2\pi r [Here, r = radius] \\ =2\times 3.1416 \times \frac{5}{2}=15.71 unit. \end{gathered}$$

$Given, cot (A + B) = 1$

$$\begin{gathered} = cot (A+B)=cot 45° \left[\right[As, cot 45°-1] \\ \therefore A+B= 45°..............(i) \\ again, cot (A-B)=\sqrt{3} \\ =cot (A-B)=cot 30° [As, cot 30°=\sqrt{3}) \\ \therefore A-B=30°.................(ii) \\ Adding equation (i) with equation (ii) we get \\ \frac{A+B=45° \\ A-B=30°}{2A=75°} \\ \therefore A=\frac{75°}{2}=37.5° \\ Subtracting equation (ii) from (i), we get \\ \frac{A+B=45° \\ A-B=30°}{2B=15°} \\ \therefore B=\frac{15°}{2}=7.5° \\ \therefore Required A =37.5° and B = 7.5° \end{gathered}$$