Bangladesh development bank ltd, Recruitment Test for senior officer, Examination held on: 06.01.2018 || 2018

Here, $p=3x^2-35x+50$

Now, if the company makes a profit, then

$$\begin{gathered} p>0 \\ \Rightarrow 3x^2-35x+50 >0 \\ \Rightarrow 3x^2-30x-5x+50 >0 \\ \Rightarrow 3x\left(x-10\right)-5\left(x-10\right) >0 \\ \Rightarrow \left(x-10\right)\left(3x-5\right) >0...............\left(i\right) \end{gathered}$$

As this equation (i) is greater than 0, so the value of the two roots must have different values in different intervals.

Now, from the equation (i), we have 

The value of x is greater than 10 ie: x> 10

Or, the value of x is less than $\frac{5}{3}$ and greater than or equal to 0 i.e. $0\le x<\frac{5}{3}$ 

Because advertising cost can not be negative.

So, if the company makes a profit,

The values of $x=\left\{0\le x\le \frac{5}{3} or, x>10\right\}$ . 

576 Tk. would be saved when the breadth = 3 meter less

1 Tk. would be saved when the breadth=$\frac{3}{576 }$  meter less

7,200 Tk. would be saved when the breadth$=\frac{3\times 7200}{576}=37.5$ meter less

So, the breadth of the room is 37.5 meter 

Since the sum of the 3-digits is II and each digit represents a prime number. 

So the 3-digits may be 2, 2, 7 or 3, 3, 5 because 2+2+7=11 and 3+3+5=11

Now using the digit 2, 2, 7 we have the prime number 227. Because other two numbers i.e. 722 and 272 are by 2 and thus are not prime. 

Again, using the digit 3, 3, 5 we have the number 353 which is a prime number. But 533 which is divisible by 13 and thus 533 is not prime and 335 is divisible by 5 and thus 335 is not prime.

So, the three-digit prime number whose sum of the digits is 11 and each digit representing a prime number is 227 & 353. (Answer)

Let's break down the problem step by step:Cyclic Parallelogram: A cyclic parallelogram is a quadrilateral whose opposite angles are supplementary and all its vertices lie on a single circle.Parallelogram: A parallelogram is a quadrilateral with opposite sides parallel.To prove that a cyclic parallelogram must be a rectangle, we need to show that all its angles are right angles (90 degrees).Here's the proof:In a cyclic quadrilateral, opposite angles are supplementary, meaning the sum of the measures of the opposite angles is 180 degrees.Let's label the angles of the cyclic parallelogram as A, B, C, and D.In a parallelogram, opposite angles are equal. So, let's say angle A is equal to angle C, and angle B is equal to angle D.Now, if we add the measures of angle A and angle C, and the measures of angle B and angle D, the sum should be 180 degrees, because the opposite angles in a cyclic quadrilateral are supplementary.So, ( A + C = 180^\circ ) and ( B + D = 180^\circ ).Since A = C and B = D, we can substitute:[ A + A = 180^\circ ] and [ B + B = 180^\circ ].Thus, ( A = 90^\circ ) and ( B = 90^\circ ), which implies all angles in the parallelogram are right angles.Therefore, the cyclic parallelogram is a rectangle.

$$\begin{gathered} Let, \frac{a}{q-r}=\frac{b}{r-p}=\frac{c}{p-q}=k \\ so, \frac{a}{q-r}=k \\ a=k\left(q-r\right);b=k\left(r-p\right);c=k\left(p-q\right) \\ L.H.S.=a+b+c \\ =k\left(q-r\right)+k\left(r-p\right)+k\left(p-q\right) \\ =k\left(q-r+r-p+p-q\right) \\ =k\times 0 \\ =0 \\ R.H.S = pa+qb+rc \\ =p\left[k\left(q-r\right)\right]+q\left[k\left(r-p\right)\right]+r\left[k\left(p-q\right)\right] \\ =p(kq-kr)+q(kr-kp)+r(kp-kq) \\ =pkq-pkr+qkr-pkq+pkr-qkr \\ =0 \end{gathered}$$

Therefore, a+b+c=pa+qb+ rc [Showed] 

Let,

The original speed = x km/h

ATQ, 

108/x - 108/x + 3 = 3

=> 108x + 324 - 108x/x² + 3x = 3

=> 3x² + 9x - 324 = 0

=> 3(x² + 3x - 108) = 0

=> x² + 3x - 108 = 0

=> x² + 12x - 9x - 108 = 0

=> (x + 12)(x - 9) = 0

=> x = 9

Therefore, original speed 9.

$$\begin{gathered} \frac{x}{2}+\frac{6}{y}=9 \\ \Rightarrow \frac{xy+12}{2y}=9 \\ xy+12=18y...........\left(i\right) \\ \frac{x}{3}+\frac{2}{y}=4 \\ \Rightarrow \frac{xy+6}{3y}=4 \\ xy+6=12y...........\left(ii\right) \\ Now, \left[\right(i)-(ii\left)\right] we have \\ \left(xy+12=18\right)-\left(xy+6=12y\right)=\left(6=6y\right)=\left(y=1 \right) \\ putting the value of \text{'}y\text{'} in \left(ii\right), we have \\ \left(x\times 1\right)+6=12\times 1 \\ \Rightarrow x=12-6 \\ x=6 \\ (x,y)=(6,1) \end{gathered}$$