# A tank can be filled by one pipe in $x$ minutes and emptied by another in $\left( {x + 5} \right)$ minutes. Both the pipes opened together can fill an empty tank in $16.8$minutes. What is the value of $x$?

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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.

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.