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An oil tank can be filled by pipe $X$ in $5$ hours and pipe $Y$ in $4$ hours, each pump working on its own. When the oil tank is full and the drainage hole is open, the oil is drained in $20$ hours. If initially the tank was empty and someone started the two pups together but left the drainage hole open, how many hours will it take for the tank to be filled? (Assume that the rate of drainage is independent of the Head)

  1. $1.50$
  2. $2.00$
  3. $2.50$
  4. $4.00$
in Numerical Ability by Veteran (421k points)
edited by | 40 views
Migrated from GO Civil 4 months ago by Arjun

1 Answer

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Best answer
Part filled by pipe $X$ in $1$ hour $= \frac{1}{5}$
Part filled by pipe $Y$ in $1$ hour $= \frac{1}{4}$
Part emptied by drainage hole in $1$ hour $= \frac{1}{20}$

Net part filled by pipes $X$ & $Y$ and when drainage hole is left open in $1$ hour $= \frac{1}{5} +\frac{1}{4} -\frac{1}{20} = \frac{2}{5}$

So, the oil tank can be filled in $\dfrac{1}{\frac{2}{5}} = 2.5 \;\text{hours}$

Correct Answer: $C$
by Boss (15k points)
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