Rajshahi Krishi Unnnayan Bank Ltd Recruitment Test for Office ( Written ) 2017 Examination Held: 15.08.2017 || 2017

Given that, People having passport = 65 

People having both passport and voter ID=30

So, the people having only passport= 65-30 = 35

 According to the question, 

$$\begin{gathered} 30+35+x=115-15 \\ \Rightarrow 65+x=100 \\ \Rightarrow x=100-65 \\ \therefore x= 35 \end{gathered}$$

So, the people having only voter ID = 35 ans.

Rafiq can do in 1 day=$\frac{1}{10}$ portion of work

Rafiq can do in 6 days =$\frac{6}{10}=\frac{3}{5}$ portion of work

Shafiq can do in 1 day= $\frac{1}{20}$ portion of work

Shafiq can do in 4 days = $\frac{4}{20}=\frac{1}{5}$ portion of work

Rafiq & Shafiq completed in first 6 days= $\frac{3}{5}+\frac{1}{5}=\frac{4}{5}$  portion of work

So, the remaining work = $\left(1-\frac{4}{5}\right)=\frac{1}{5}$ portion of work

Arif can do in 1 day =$\frac{1}{10}$ portion of work

Arif can do $\frac{1}{10}$ portion in 1 day

Arif can do $\frac{1}{5}$ portion in = $\frac{10}{5}=2$ days

The required time for completion of the work=6+2=8 days ans.

Let, it will take 'x' months to save 1,06,200 Tk. 

From the savings we get a series.

Such as 1200+ 1300 +1400+...........

Where, a = 1200 Tk & d=1,300 - 1,200 = 100 Tk

We know that, summation = $\frac{n}{2}\left\{2a\left(n-1\right)d\right\}$

According to the question, 

$$\begin{gathered} \frac{n}{2}\left\{2a\left(n-1\right)d\right\} \\ \Rightarrow \frac{n}{2}\left\{2\times 1200+100\left(n-1\right)\right\}=106200 \\ \Rightarrow n\left(1200+50n-50\right)= 106200 \\ \Rightarrow 1200n+50n^2-50n=106200 \\ \Rightarrow 50n^2+1150n=106200 \\ \Rightarrow n^2+23n=2124 \\ \Rightarrow n^2+59n-36n-2124=0 \\ \Rightarrow n\left(n+59\right)-36\left(n+59\right)=0 \\ \Rightarrow \left(n+59\right)\left(n-36\right)=0 \\ \Rightarrow n=36 Or, n=-59 \end{gathered}$$

(Time ie: n cannot be negative)

So, the required time = 36 months = 3 years ans. 

$$\begin{gathered} Let, \frac{x+3}{x-3}=a \\ We get, 2a^2-7a+6=0 \\ \Rightarrow 2a^2-4a-3a+6=0 \\ \Rightarrow 2a\left(a-2\right)-3\left(a-2\right)=0 \\ \Rightarrow \left(a-2\right)\left(2a-3\right)=0 \\ \therefore a=2 or, \frac{3}{2} \\ So, either \frac{x+3}{x-3}=2 \\ \Rightarrow 2x-6=x+3 \\ \therefore x=9 \\ or, \frac{x+3}{x-3}=\frac{3}{2} \\ \Rightarrow 3x-9=2x+6 \\ \therefore x=15 \end{gathered}$$

So, the value of x=9 or 15 ans. 

Let, the width of the rectangle is x meters 

So, the length of the rectangle is = (x+10) Meters

According to the question,

$$\begin{gathered} x\left(x+10\right)=1,200 \\ \Rightarrow x^2+10x-12,00=0 \\ \Rightarrow x^2+40x-30x-1,200=0 \\ \Rightarrow x\left(x+40\right)-30\left(x+40\right)=0 \\ \Rightarrow \left(x+40\right)\left(x-30\right)=0 \\ \Rightarrow x-30 \\ or, x=-40 (But the value of length cannot be a negative) \end{gathered}$$

So, the width of the rectangle is 30 meters, and

The length of the rectangle = 30+10=40 meters.

So, according to the Pythagorean theories.

The length of the diagonal of the rectangle= $\sqrt{{40}^2+{30}^2}=50$ meters. 

Let the length of the other diagonal= d meters

Area of rhombus =$\frac{1}{2}$$(10\times d)=5d$ 

According to the question, 

$$\begin{gathered} 5d-120 \\ d=24 \end{gathered}$$

So, the length of the other diagonal = 24 meters

Half of the one diagonal = $\frac{10}{2}=5$ meters and half of the other diagonal=$\frac{24}{2}=12$ 

One side of the rhombus  =$\sqrt{5^2+{12}^2}=\sqrt{169}=13$

So, the perimeter of the rhombus $4\times 13=52$  meters. ans.

Let, the length of a side of the triangle = a meters

We know, the area of an equilateral triangle= $\frac{\sqrt{3}}{4}{a}^2$

$$\begin{gathered} \frac{\sqrt{3}}{4}{a}^2=3\sqrt{3} \\ \Rightarrow a^2 = 12 \\ \Rightarrow a=\sqrt{12}=2\sqrt{3} \end{gathered}$$ meters.

The radius of the circle =$\frac{{\displaystyle \frac{1}{2}}}{\sqrt{3}}\times 2\sqrt{3}=1$ meter (the length of radius is $\frac{1}{2\sqrt{3}}$ times of a triangle side)

So, the required area $=\mathrm{\pi }\times 1^2=\mathrm{\pi }=\frac{22}{7}=3.1416$ square meters approximately. ans.