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finance: Master | Next Rank: 500 Posts; Posts: 150; Joined: Tue Mar 08, 2011 4:38 am; Thanked: 5 times

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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Source: Beat The GMAT — GMAT Strategy | Mon Jul 25, 2011 12:32 am

kenji: Senior | Next Rank: 100 Posts; Posts: 40; Joined: Wed May 20, 2009 1:02 pm; Location: Belgium; Thanked: 6 times; GMAT Score:520

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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finance: Master | Next Rank: 500 Posts; Posts: 150; Joined: Tue Mar 08, 2011 4:38 am; Thanked: 5 times

by finance » Mon Jul 25, 2011 1:20 pm

That was exactly my solution, but that's not the answer:(

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kenji: Senior | Next Rank: 100 Posts; Posts: 40; Joined: Wed May 20, 2009 1:02 pm; Location: Belgium; Thanked: 6 times; GMAT Score:520

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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dmf5238: Junior | Next Rank: 30 Posts; Posts: 29; Joined: Sun Jun 26, 2011 9:19 am; Thanked: 3 times; Followed by:1 members; GMAT Score:710

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.