Janata Bank Ltd. || AEO (Tellor) (07-02-2020) || 2020

16 women can do the work in =12 days

$\therefore$1 women can do the work in = (16$\times$12) days

$\therefore$1 women can in 1 days =$\frac{1}{12\times 16}$ part of the work

Again, 12 men can do the work in = 8 days

$\therefore$1 man can do the work in = (8 $\times$ 12) days

$\therefore$I man can in 1 days = $\frac{1}{8\times 12}$ part of the work

$\therefore$16 men's 3 day's work = $\frac{1}{8\times 12}\times 3\times 16=\frac{1}{2}$ part of the work

Remaining work =$1-\frac{1}{2}=\frac{1}{2}$

Now, number of men still working = 16-10=6

$\therefore$6 men's 1 days work = $\frac{1}{12\times 8}\times 6=\frac{1}{16}$part

And 4 men's 1 days work =$\frac{1}{12\times 16}\times 4=\frac{1}{48}$part

$\therefore$They can do in 1 day = $\frac{1}{16}+\frac{1}{48}=\frac{3+1}{48}=\frac{4}{48}=\frac{1}{12}$ part

Now, $\frac{1}{12}$part of the work is done in = 1 day

$\therefore$ $\frac{1}{2}$ part of the work is done in = $\frac{12}{2}$ = 6 days.

Let, first, second and third cars are c₁, c₂ be c₃ respectively.

Distance covered by c₁ : c₃ = (240 - 80) : (240 + 80)

=160 : 320 = 1 : 2

Here, we know speed is directly proportional to the distance covered by these cars as time is constant.

$$\begin{gathered} \therefore Ratio of speeds of c_1 : c_2 = 1 : 2 \\ Again, at first meeting of all three cars as times are in arithmetic progression, \\ speed is in harmonic mean. \\ So, c_2\text{'}s speed = \frac{2\times 1\times 2}{1+2}=\frac{4}{3} \\ \therefore Ratio of speed = 1 : \frac{4}{3} : 2= 3: 4 : 6 \\ Then of time = \frac{1}{3} : \frac{1}{4} \frac{1}{6} = 4 : 3 : 2 \\ \therefore Difference between the speed of c_1 and c_3=\frac{240}{2}-\frac{240}{4}=120-60=60 kmph. \end{gathered}$$

A, B এবং C মিলে একটি ব্যবসায় শুরু করলো। তাদের বিনিয়োগের অনুপাত 0.1250.75 0.25 8 মাস পরে A তার বিনিয়োগ 3 গুণ বাড়ানো এবং C তার বিনিয়োগের অর্ধেক তুলে নিল। বছর শেষে তারা 5,800 টাকা মুনাফা করলো। লাভ হিসেবে B কত পেল?

Given that, A : B : C = 0.125:0.75 : 0.25 = 1:6 : 2 [Dividing by 0.125]

Let, the their investment are x, 6x and 2x taka.

A's capital for 12 months = 8x+ (4 $\times$ 4x) = 8x + 16x=24x taka

$\therefore$B's capital for 12 months = 12 $\times$ 6x = 72x taka

$\therefore$C's capital for 12 months = (8 $\times$ 2x) + (4$\times$x) = 16x+4x=20x taka

$\therefore$Ratio of their investment = 24x : 72 x : 20x = 6 : 18 : 5

$\therefore$B's share of profit = 5,800 $\times \frac{18}{6+18+5}$ =5,800$\times$20=3,600 taka

Profit= $\frac{320}{100}\times 100=320$ Tk. and selling price = 420 Tk.

$\therefore$Now, new cost price = 125% of 100Tk .=125 Tk.

$\therefore$New cost price = 420 Tk

$\therefore$Profit 420-125=295 Tk

$\therefore$Percentage of profit on selling price = $\frac{295}{420}\times 100=70.24\%=70\%$

Given that, let the capacity of three vessels are 3x, 2x and x liter.

Now, taking three vessels separately

Vessel - 1: Mixture taken out = $3x\times \frac{1}{3}$ = x liter

$\therefore$Amount of milk=$\frac{5x}{7}$and amount of water = $\frac{2x}{7}$

Vessel - 2; Mixture taken out = $2x\times \frac{1}{2}$ = x liter

$\therefore$Amount of milk$=\frac{4x}{5}$ and amount of water =$\frac{x}{5}$

Vessel - 3: Mixture taken out = $x\times \frac{1}{7}=\frac{x}{7} Liter$

$\therefore$Amount of milk $\frac{4x}{35}$= and amount of water =$\frac{x}{35}$

Total milk $=\frac{5x}{7}+\frac{4x}{5}+\frac{4x}{35}=\frac{25x+28x+4x}{35}=\frac{57x}{35}$

$\therefore$Total volume of mixture=x+x+ $\frac{x}{7}=\frac{7x+7x+x}{7}=\frac{15x}{7}$

$\therefore$Percentage of milk =$\frac{100\times {\displaystyle \frac{57x}{35}}}{{\displaystyle \frac{15x}{7}}}=\frac{100\times 57x}{35}\times \frac{7}{15x}=\frac{39900}{525}=76\%$

$\therefore$Percentage of water = (100 - 76)% = 24%

Given that breadth = 3 meter

$\therefore$Length will bc = 3 $\times$ 1.44 = 4.32 meter

$\therefore$Area = 3 $\times$ 4.32 = 12.96 m²

New breadth=3$\times \frac{125}{100}=3.75 meter$

New length 4.32 $\times \frac{140}{100}=6.048 meter$

$\therefore$ New area =3.75$\times$6.048=22.68 m²

$\therefore$Total cost increases = 45$\times$9.72=437.4 taka.

$$\begin{gathered} Given that, x^2+\frac{1}{x^2}=1 \\ So, (x^2+\frac{1}{x^2}) (x+\frac{1}{x})=(x+\frac{1}{x}) [Multiplying both sides by (x+\frac{1}{x}\left)\right] \\ =x^3+x+\frac{1}{x^3}=(x+\frac{1}{x}) \\ \therefore x^3+\frac{1}{x^3}=0 \\ Now,x^{102}+x^{96}+x^{90}+x^{84}+x^{78}+x^{72}+5 \\ x^{99}(x^3+\frac{1}{x^3})+x^{87}(x^3+\frac{1}{x^3})+x^{75}(x^3+\frac{1}{x^3})+5 \\ =0+0+5=5 \\ Answer: 5 \end{gathered}$$