# GATE2019 CE-2: GA-9

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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. $4.00$

edited
Migrated from GO Civil 1 year ago by Arjun

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: $C$

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1 vote

Let the tank capacity be 20 ltrs.(Any multiple of 4,1,5 like 20,40,60.... will work) .

Since the tank can be filled by X in 5 hrs,X is filling the tank at a rate of 4 ltrs per hour.

Similarly Y is filling the tank at a rate of 5 ltrs per hour.

Similarly drainage hole is emptying the tank at a rate of 1 litre per hour.

If X,Y,Z are open for 1 hr,then at the end of hour,tank is increased by $4+5-1=8$ litres.

Therefore if you start with empty tank in $20/8=2.5hrs$, you can fill the tank completely

1 vote

$Let\ capacity=20L$

$X\rightarrow5H-\ \ 4L/H$

$Y\rightarrow4H-\ \ 5L/H$

$Z\rightarrow 20H-\underline{1L/H}$

$8L/H$

$1H\rightarrow 8L$

$?\rightarrow 20L$

$?=2.5H$

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