The posted weight limit for a wooden bridge in Bandarban is 6,000 kg. A delivery truck that is carrying x identical boses each weighting 14 Ag will pass over the bridge. If the combined weight of the empty delivery truck and its driver is 4.500 kg, what is the maximum possible value of x that will keep the combined weight of the truck, driver and boxes below the bridge's weight limit?

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

Given that, total notes = 12
Let, the number of Tk. 10 notes = x 

The number of Tk. 5 notes = 12-x
According to the question 

$$\begin{gathered} \left(10\times x\right)+5\left(12-x\right)<95 \\ \Rightarrow 10x+60-5x<95 \\ \Rightarrow 5x<95-60 \\ \Rightarrow x<\frac{35}{5} \\ \therefore x=7 \end{gathered}$$

So, the number of Tk. 10 notes will be less than 7, i.e 6.

Repeat: BDBL (SO) 2018 

$$\begin{gathered} x^2-12x+27<0 \\ \Rightarrow x^2-9x-3x+27<0 \\ \Rightarrow x\left(x-9\right)-3\left(x-3\right)<0 \\ Here, if we take x<3 \\ Then, \left(x^2-12x+27\right)<0 will not be true and if we take x<9. \\ So, x must lie between 3<x<9, \\ Because for each value between 3<x<9, (x^2-12x+27)<0 will be true. \\ \mathit{a}\mathit{n}\mathit{s}. 3 <x<9. \end{gathered}$$

$$\begin{gathered} x^2-13x+40\ge 0 \\ \Rightarrow x^2-8x-5x+40\ge 0 \\ \Rightarrow x(x-8)-5(x-8)\ge 0 \\ \Rightarrow (x-8) (x-5)\ge 0 \\ Here, the product of (x-8) and (x-5) is grater than zero or equal to zero. \\ So the value of these must be both positive or negative. \\ When both positive, (x-8)\ge 0 \\ \Rightarrow x>8; (x-5)\ge 0 \\ \Rightarrow x\ge 5 [It is not true] \\ When both negative, -(x-8)\ge 0 \\ \Rightarrow x\le 8; [It is not true] \\ -(x-5)\ge 0 \\ \Rightarrow x\le 5; \\ So, the range x\ge 8 Or, x\le 5. \end{gathered}$$