A cylindrical water tower

[metallicafan]

A cylindrical water tower with radius 5 m and height 8 m is 3/4 full at noon. Every minute, .08Ï€ m3 is drawn from tank, while .03Ï€ m3 is added. Additionally, starting at 1pm and continuing each hour on the hour, there is a periodic drain of 4Ï€ m3. From noon, how many hours will it take to drain the entire tank?

A. 20 2/7
B. 20 6/7
C. 21
D. 21 3/7
E. 22

OA is E.

I agree with the OA. After 21 hours, 147Ï€ m3 will be drained. So, there are 3Ï€ m3 that have not been drained. Because the constant drain draws 3Ï€ m3 per hour, we will have to wait until the 22th hour. Probably, the periodic drain will draw little water in the 22th hour. Please, confirm whether my reasoning is Ok.

In this sense, I don't understand this part of the OE:
"Had we divided 150/7, we'd land on , but we have to consider how the 3/7 remainder actually leaves the tank.

Now we have to deal with remainders.

With 3 m3 remaining, after another 3/7 hours, only 3(0.5) = 1.5 m3 will be drained. So the tank will not actually be empty until 22 hours, when the periodic draw empties the remainder."
I don't understand the OE does this: 3(0.5) = 1.5 m3 :S !

Thanks!

Source: www.grockit.com

[Bill@VeritasPrep]

Your reasoning is fine; there will be little water left for the 22nd periodic draw, so it will remove all of the remaining water.

The OE is showing why 21 3/7 doesn't work. We're not draining a steady 7pi per hour, we're draining a steady 3pi per hour with an additional 4pi one-time draw each hour.