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looking for a good explanation

by finance » Mon Jul 25, 2011 12:32 am
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?

- 20 2/7
- 20 6/7
- 21
- 21 3/7
- 22
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by kenji » Mon Jul 25, 2011 10:37 am
finance wrote: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?

- 20 2/7
- 20 6/7
- 21
- 21 3/7
- 22
Volume of the cyilinder= (25*8)Ï€ m3= 200Ï€ m3
Effective liquid volume= 200Ï€ m3 *3/4= 150Ï€ m3
Quantity of water drawn per hour= ((0.08Ï€*60)-(0.03Ï€*60)+4Ï€)=7Ï€ m3
Time taken to empty the cylinder= (150Ï€ m3) / (7Ï€ m3)= 21.42=21 3/7

ANSWER D
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by finance » Mon Jul 25, 2011 1:20 pm
That was exactly my solution, but that's not the answer:(
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by kenji » Mon Jul 25, 2011 1:37 pm
Volume of the cyilinder= (25*8)Ï€ m3= 200Ï€ m3
Effective liquid volume= 200Ï€ m3 *3/4= 150Ï€ m3
Quantity of water drawn from noon to 1pm((0.08Ï€*60)-(0.03Ï€*60)=3Ï€ m3
Quantity of water drawn per hour= ((0.08Ï€*60)-(0.03Ï€*60)+4Ï€)=7Ï€ m3
Time taken to empty the cylinder= ((150-3)Ï€ m3) / (7Ï€ m3)= 21
+1 previous hour


22
Last edited by kenji on Mon Jul 25, 2011 1:52 pm, edited 2 times in total.
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by dmf5238 » Mon Jul 25, 2011 1:48 pm
22 works. The problem specifies that the 4pi periodic drain starts after one hour, so you need to subtract 3pi for the first hour, THEN divide by the 7pi. This will give you 21 hours, and you need to add back the first hour (in which only 3pi of water drained), 22 hours. :)

I hope this is correct lol!

Edit: I can write out my math if you need me to, but it's essentially the same as kenji posted.
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