Probashi Kallyan Bank Ltd. || Officer (General) (10-12-2021) || 2021

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

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

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

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

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

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

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

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

$$\begin{gathered} Here, clearly the sum of all three groups is 50. We know, sum of three odd numbers can not be 50. \\ Therefore, we need one even prime integer, Here, the only prime even integer is 2. \\ Therefore, one group has 2 people Now, we have to allot remaining 48 people in two odd number group. \\ Available combination be \\ \left(i\right) 2, 5, 43 \\ \left(ii\right) 2, 7, 41 \\ \left(iii\right) 2, 11, 37 \\ \left(iv\right) 2, 17, 31 \\ \left(v\right) 2, 19, 29 \\ \therefore There are five ways we can divide 50 people into three groups, so that each group contains members equal to a prime number. \\ answer: 5 ways. \end{gathered}$$

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

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