An empty tank can be filled by one pipe in 'x' minutes and emptied by another pipe in (x + 5) minutes. Both the pipes, when opened together, can fill the empty tank in 16.8 minutes. Find the value of x.

Given,

An empty tank can be filled by one pipe in 'x' minutes.

Tank filled by one pipe in a minute = $\dfrac{1}{x}$

Tank is emptied by another pipe in (x + 5) minutes.

Tank emptied by another pipe in a minute = $\dfrac{1}{x + 5}$

When both pipes are working together, tank filled in 1 minute = $\dfrac{1}{x} - \dfrac{1}{x + 5}$

Given,

Both the pipes, when opened together, can fill the empty tank in 16.8 minutes.

So, in one minute, when both the pipes are opened together, tank filled = $\dfrac{1}{16.8}$

$\therefore \dfrac{1}{x} - \dfrac{1}{x + 5} = \dfrac{1}{16.8} \\[1em] \Rightarrow \dfrac{x + 5 - x}{x(x + 5)} = \dfrac{10}{168} \\[1em] \Rightarrow \dfrac{5}{x(x + 5)} = \dfrac{5}{84} \\[1em] \Rightarrow \dfrac{1}{x(x + 5)} = \dfrac{1}{84} \\[1em] \Rightarrow x(x + 5) = 84 \\[1em] \Rightarrow x^2 + 5x = 84 \\[1em] \Rightarrow x^2 + 5x - 84 = 0 \\[1em] \Rightarrow x^2 + 12x - 7x - 84 = 0 \\[1em] \Rightarrow x(x + 12) - 7(x + 12) = 0 \\[1em] \Rightarrow (x - 7)(x + 12) = 0 \\[1em] \Rightarrow (x - 7) = 0 \text{ or } x + 12 = 0 \\[1em] \Rightarrow x = 7 \text{ or } x = -12.$

Since. no. of minutes cannot be negative.

∴ x = 7.

Hence, x = 7.