danzig wrote:
A cylindrical water tower with radius 5 m and height 8 m is 3/4 full at noon. Every minute, .08π m3 is drawn from tank, while .03π m3 is added. Additionally, starting at 1pm and continuing each hour on the hour, there is a periodic drain of 4π 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
I agree with the OA. After 21 hours, 147π m3 will be drained. So, there are 3π m3 that have not been drained. Because the constant drain draws 3π 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!
Volume of the cylinder is 150pi
At 1 pm the volume is 147pi (the net drain is 3pi)
Then onwards, the net drain is 7pi per hour.
Setting AP we get 21 hours.
Therefore, total time taken is 22 hours.