Two pipes can fill a tank in 76 minutes if both are turned on.If only one is used,it would take 27 minutes longer for the smaller pipe to fill

Two pipes can fill a tank in 76 minutes if both are turned on. If only one is used, it would take 27 minutes longer for the smaller pipe to fill the tank than the larger pipe. How long will it take for the smaller pipe to fill the tank? (Round your answer to the nearest tenth)

Detailed Solution:

Let us say that the smaller pipe takes X minutes to fill the tank alone and the larger pipe takes.

Therefore in one minute, the smaller tap fills up $\dfrac {1}{x}$ of the tank

The larger pipe fills the tank faster. The larger pipe takes 27 minutes less time than the smaller pipe to fill the tank. Therefore the time taken by the larger pipe is X-27 minutes.

Therefore in one minute, the Larger tap fills up $\dfrac {1}{x-27}$ of the tank.

So when both taps are open, in 1 minute $\dfrac {1}{x}+\dfrac {1}{x-27}$ will be filled.

we are given in the problem that, when both taps are open, they can fill the tank in 76 minutes.

So in one minute, both tanks can fill $\dfrac {1}{76}$ th part of the tank.

Here both the expressions we arrived at are equal since both give the part of the tank filled in 1 minute.

$\dfrac {1}{x}+\dfrac {1}{x-27}=\dfrac {1}{76}$

LCM of x and x -27 is x(x-27) .Multiplying both sides with x(x-27)to get rid of the fractions.

$\begin{aligned}\dfrac {1}{x}+\dfrac {1}{x-27}=\dfrac {1}{76}\\ \Rightarrow x\left( x-27\right) \left[ \dfrac {1}{x}+\dfrac {1}{x-27}\right] =\dfrac {x\left( x-27\right) }{76}\\ \Rightarrow x-27+x=\dfrac {x^{2}-27x}{76}\\ \Rightarrow 76\left( 2x-27\right) =x^{2}-27x\\ \Rightarrow x^{2}-179x+2052=0\end{aligned}$

Solving this quadratic equation using the quadratic formula the value of would be x=166.69 and 12.31

we need to ignore the x as 12.31 because the x-27 leads to a negative time

Hence the x=166.69 minutes rounded to nearest tenth decimal.

Therefore it would take 166.7 minutes for the smaller tap to fill the tank alone.

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