Janata Bank Ltd. Recruitment Test AEO ( RC) Written Examination Held On : 22.09.2017 || 2017

Here, n(total) = 400

n(chess ) = 18, n(TT) = 20, n (carom) = 27.

n(chess & TT)=7, n(TT & carom)= 12 and n(Chess & Carom)=?

n(chess & TT & carom)=4

We have, 

n(total) = n(chess) + n(TT) + n (carom)-n(chess & TT) - n(TT & carom)- n(chess & carom) +

n(chess & TT & carom)

$\Rightarrow$ 40=18+20+27-7-12-n(chess & carom) + 4

n(chess & carom) = 50 - 40 = 10

So, total number of students who play both chess & carom is = 10.

But there are 4 students who play all the three games. So, the number of students who play chess and carom but not table tennis = 10-4=6. ans.

According to the question,

5% of total profit goes to workers' provident fund. 

So, the remaining 95% of profit is distributed between Aslam and Babul. 

Let, Aslam's profit share is = 3x 

So, Babul's profit share is = 2x 

So, total profit after workers' provident fund deduction (i.e: 95% of total profit)

=3x+2x=5x

According to the question, 

$$\begin{gathered} 3x=855000 \\ \Rightarrow 5x=855000\times \frac{5}{3}=1425000 \\ So, 95\% of total profit=14,25,000 \\ 100\% of total profit = 14,25,000 \times \frac{100}{95}=15,00,000 \end{gathered}$$ ans.

Let, the amount invested in Bank Y was = Tk. p

So, the amount invested in Bank X was=  Tk. (13,900-p)

According to the question,

$$\begin{gathered} p\times 2\times 11\%+\left(13,900-p \right)\times 2\times 14\%=3508 \\ \Rightarrow \frac{22p}{100}+\frac{28\left(13900-p\right)}{100}=3508 \\ \Rightarrow 22p+389200-28p=350800 \\ \Rightarrow 6p=3,89,200 - 350800 \\ \Rightarrow 6p=38400 \\ \therefore p=\frac{38400}{6}=6400 TK. \end{gathered}$$

So, the amount deposited in Bank Y is 6,400 Tk. ans.

When Mahi starts, Rashed =$120 \times 1$ = 120 km away from him

Relative speed of Mahi over Rashed = (140-120) kmph = 20 kmph 

Required time to catch up Rashed = $\frac{120}{20}=6$ hr. ans 

Let, the width of the table = x feet.

The length of the table = 3x feet 

According to the question

3x-5=x+5

⇒ 2x=10

x=5 

The length of the table is=3 * 5= 15 feet. 

and the width of the table is = 5 feet. 

$$\begin{gathered} Given, a=3+2\sqrt{2} \\ \Rightarrow a=2+1+2\sqrt{2} \\ \Rightarrow a={\sqrt{\left(2\right)}}^2+2\times \sqrt{2}\times 1+1^2 \\ \Rightarrow a={\left(\sqrt{2}+1\right)}^2 \\ \Rightarrow \sqrt{a}=\sqrt{2}+1 \\ and, \frac{1}{\sqrt{a}}=\frac{1}{\sqrt{2}+1}=\frac{\sqrt{2}-1}{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}=\frac{\sqrt{2}-1}{{\left(\sqrt{2}\right)}^2-{\left(1\right)}^2}=\frac{\sqrt{2}-1}{2-1}=\sqrt{2}-1 \\ Now, \left(\sqrt{a}-\frac{1}{\sqrt{a}}\right)=\sqrt{2}+1-\sqrt{2}+1=2\left[[Putting the value of \sqrt{a} and \frac{1}{\sqrt{a}}\right] \end{gathered}$$

Answer: 2 

Let, be the price of one shirt, y be the price of one pair of trousers and z be the price of one hat.

According to the question,

 4x+4y+2z=5,600…………..(i) 

9x+9y+6z=1,2900…………..(ii) 

3x+3y+2z=4,300…………..(iii) 

Now, we have {(i)-(iii)}

 4x+4y+2z=5,600…………..(i) - 3x+3y+2z=4,300…………..(iii) = x+y=1300 ………………(iv) 

Again, from equation (iii)

$$\begin{gathered} 3x+3y+2z=4,300 \\ \Rightarrow 3\left(x+y\right)+2z=4300 \\ \Rightarrow 3\times 1300+2z=4300 \\ z=200 \\ x+y+z=1300+200=1500\mathbf{ }\mathit{a}\mathit{n}\mathit{s}\mathbf{.} \end{gathered}$$