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Bunuel
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I did with and assumption of the tank's capacity:
If tank = 20 liters, tap A fills in 4 min ⇒ rate = 20/4 = 5 liters/min.
Tap B fills in 5 min ⇒ rate = 20/5 = 4 liters/min.
Combined rate (no hole) = 5 + 4 = 9
5+4=9 L/min.

Time to fill without hole = 20/9 min.
With the hole, the two taps together take 30 seconds more or 0.5 min more, so total time = 20/9 + 0.5 = 49/18 mins

If the hole’s outflow rate is h liter/min. Then the filling rate with the hole = (9 − h)

We can write 20 / (9 - h) = 49/18
Solving this we get h = 81/49 Liters/min
So the hole alone will drain out the water in = 20/h
=980/ 81 minutes

And: B.
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