Janata & Rupali Bank Ltd. || Officer (24-01-2020) || 2020

Let, C can complete the work in x days.

Now, A's 1 day's work $\frac{1}{2}$ part of the work

$\therefore B\text{'}s 1 day\text{'}s work\frac{1}{15} part of the work$

$$\begin{gathered} \therefore C\text{'}s 1 day\text{'}s work\frac{1}{x} part of the work \\ (A+B+C)\text{'}s 1 day\text{'}s work\frac{1}{5} part of the work \\ According to question \\ \frac{1}{12}+\frac{1}{15}+\frac{1}{x}=\frac{1}{5} \\ =\frac{1}{x}=\frac{1}{5}-\frac{1}{12}-\frac{1}{15}=\frac{12-5-4}{60}=\frac{3}{60}=\frac{1}{20} \\ \therefore \frac{1}{x}=\frac{1}{20}; That mean\text{'}s c\text{'}s 1 days\text{'}s work=\frac{1}{20} \\ Ratio of their work= \frac{1}{12} : \frac{1}{15} :\frac{1}{20}=5 : 4 : 3 \\ \therefore A\text{'}s Share = 960\times \frac{5}{5+4+3}=\frac{960\times 5}{12}=400Tk. \left(answer\right) \end{gathered}$$

Total bottles=6+3+4=13

Here, number of ways of taking 3 drink bottles out of 13=${13}_{c_3}$= $\frac{13\times 12\times 11}{1\times 2\times 3}=286$

Again, here event of taking 3 bottles of the same variety = Event of taking 3 bottles out of 6 or 3

bottles out of 3 or 3 bottles out of 4=$6_{c_3 }+3_{c_3}+4_{c_3}=\frac{6\times 5\times 4}{1\times 2\times 3}+1+\frac{4\times 3\times 2}{1\times 2\times 3}=20+1+4=25$

$\therefore$The probability of taking 3 bottles of the same variety = $\frac{25}{286}$

Then probability of taking 3 bottles are not same variety = 1-$\frac{25}{286}=\frac{286-25}{286}=\frac{261}{286}\left(Answer\right)$

Let, the first pipe can alone fill the tank in x hour

$\therefore$2nd and 3rd pipe will take (x-5) and (x - 9) hours to fill the tank.

According to the question

$$\begin{gathered} \frac{1}{x}+\frac{1}{x-5}=\frac{1}{x-9} \\ =\frac{(x-5)+x}{x(x-5)}=\frac{1}{x-9} \\ =(2x-5) (x-9) =x(x-5) \\ =2x^2-18x-5x+45=x^2-5x \\ =x^2-18x+45=0 \\ =x^2-15x-3x+45=0 \\ =x(x-15)-3(x-15)=0 \\ (x-15) (x-3)=0 \\ Now x-15=0 [x=3 is not acceptable] \\ \therefore x=15 \\ The first pipe will take 15 minutes to fill the \tan k. \end{gathered}$$

$\therefore$The 2nd pipe can fill the tank in = 15 - 5 = 10 hours

And the 3 pipe can fill the tank in = 15 - 9 = 6 hours

2nd and 3rd pipe can fill in 1 hour = $\frac{1}{10}+\frac{1}{6}=\frac{3+5}{30}=\frac{8}{30}=\frac{4}{15}part$

$$\begin{gathered} Now, \frac{4}{15} part of the \tan k is filled by 2nd and 3rd pipe in 1 hour \\ \therefore 1 or complete part of the \tan k is filled by 2nd and 3rd pipe in = \frac{15}{4}=3.75 hours \\ \therefore 2nd and 3rd pipe can fill the \tan k in = 3 hours 45 minutes. \left(answer\right) \end{gathered}$$

$$\begin{gathered} Given that, \frac{x^{24}+1}{x^{12}}=7 \\ =\frac{x^{24}}{x^{12}}+\frac{1}{x^{12}}=7 \\ =x^{12}+\frac{1}{x^{12}}=7 \\ ={(x^{12}+\frac{1}{x^{12}})}^3=7^3 \\ =x^{36}+\frac{1}{x^{36}}+\frac{3\times x^{12}\times 1}{x^{12}}(x^{12}+\frac{1}{x^{12}})=343 \\ =x^{36}+\frac{1}{x^{36}}+(3\times 7)=343 \\ =x^{36}+\frac{1}{x^{36}}+21=343 \\ =x^{36}+\frac{1}{x^{36}}=343-21=322 \\ \therefore \frac{x^{72}+1}{x^{36}}=322 \left(answer\right) \end{gathered}$$

Let us draw a figure according to question

Here, $△$PQT is a right angled-triangle,

So, we can write from Pythagorean Theorem:

$$\begin{gathered} P{T}^2=P{Q}^2+Q{T}^2 \\ =Q{T}^2=P{T}^2-P{Q}^2={15}^2-9^2=225-81=144 \\ \therefore QT=\sqrt{144}=\sqrt{{\left(12\right)}^2}=12 meter \\ Here, △TRS is also a right angled triangle, so we can write from Pythagorean Theorem: \\ {\left(TS\right)}^2={\left(TR\right)}^2+{\left(SR\right)}^2 ={\left(TR\right)}^2={\left(TS\right)}^2-{\left(SR\right)}^2={15}^2-{12}^2=225-144=81 \\ \therefore TR=\sqrt{81}=\sqrt{{\left(9\right)}^2}=9 meter \\ Here, dis\tan ce between houses is QR \\ QR = QT + TR \\ \therefore QR = 12 + 9 = 21 meter \\ \therefore Dis\tan ce 21 meter \left(answer\right) \end{gathered}$$

Let, us draw a diagram according to first condition

Let, x = length; y = breadth

Here, perimeter of the rectangle, 2(x + y) = 28

=$x+y=\frac{28}{2}=14.........\left(i\right)$

Let us draw another diagram as per second condition

Here, length = (x-2) meter and breadth = (y-2) meter.

According to question, (x - 2) (y - 3 )=24............... (ii)

New, we get from equation (i)

x + y = 14

= x = 14 - y ,.....................(iii)

Putting the value of x in equation (ii), we get

(x-2) (y-2)=24

=(14-y-2)(y-2)=24

=(12-y)(y-2)=24

=12y-24-$y^2+2y-24=0$

$$\begin{gathered} =-y^2+14y-487=0 \\ =y^2-14y+48=0 [Multiplying by -1] \\ =y^2-8y-6y+48=0 \\ =y(y-8)-6(y-8)=0 \\ \therefore (y-8) (y-6)=0 \\ Either, y-8=0 \\ \therefore y=8 \\ or, y-6=0 \\ \therefore y=6 \\ Putting y=8 in equation \left(iii\right), \\ We get, x=14-8=6 \\ \therefore x=6 \\ Putting y=6 in equation \left(iii\right), we get \\ x=14-6=8 \\ \therefore x=8 \\ \therefore The dimensions are (8, 6) answer. \end{gathered}$$

Given that, Chandan has 4C; Dewan has 4D

After first round Chandan has = 4C + 2D

Dawan has = 4D - 2D = 2D

After second round Chandan has 4C+ 2D-2C-D=2C+D

Dawan has = 2D+2C+D=3D+2C

Answer: 3D + 2C