Answer

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**Hint:**First you need to follow the given condition in the question, i.e., in \[16.8\] minutes the tank gets filled when both pipes are operated simultaneously. Initially make equation such that portion of tank filled in 1 min (i.e. $\dfrac{1}{{16.8}}{\text{ }}$) is equal to portion of tank filled in 1 min if both the pipes operate simultaneously & finally solve this equation to get the answer.

**Complete step-by-step answer:**

In 1-minute portion of tank filled be filling pipe $ = \dfrac{1}{x}$

In 1 minute, portion of tank filled in 1 minute if both operate simultaneously =

${

\left[ {\dfrac{1}{x}} \right] - \left[ {\dfrac{1}{{x + 5}}} \right] \\

= \dfrac{{x + 5 - x}}{{x\left( {x + 5} \right)}} \\

= \dfrac{5}{{(x + 5)x}}{\text{ }}...{\text{(1)}} \\

} $

According to the given condition it takes a minute to fill both pipes - \[16.8\].

So, in 1-minute portion of tank filled $ = \dfrac{1}{{16.8}}{\text{ }}...{\text{(2)}}$

From (1) and (2) we have

$\dfrac{5}{{\left( {x + 5} \right)x}} = \dfrac{1}{{16.8}}$

${x^2} + 5x - 84 = 0$

$ \Rightarrow {x^2} + \left( {12 - 7} \right)x - 84 = 0$

$ \Rightarrow {x^2} + 12x - 7x - 84 = 0$ [ Taking $(x + 12)$as common multiple]

$ \Rightarrow x\left( {x + 12} \right) - 7\left( {x + 12} \right) = 0$

$ \Rightarrow \left( {x + 12} \right)\left( {x - 7} \right) = 0$

$\therefore x = - 12 {\text{ & }} x = 7$

As time can never be negative so

$x = 7{\text{ min}}{\text{.}}$ is the correct answer .

Hence a tank can be filled by one pipe in $7{\text{ min}}{\text{.}}$

**Note:**When you get such a problem first of all read the questions carefully & make equations according to given information. You have to make a linear equation and find out the value of an unknown term i.e. $x$ . While solving the equation, the concept of factorisation should also be applied.

Do the calculations carefully to ensure that you get an accurate answer.

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