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# A cylindrical water tower with radius 5 m

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A cylindrical water tower with radius 5 m [#permalink]  20 Nov 2012, 18:00
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37% (03:16) correct 62% (02:25) wrong based on 4 sessions
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!
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Re: A cylindrical water tower with radius 5 m [#permalink]  20 Nov 2012, 21:07
danzig wrote:

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!

Every minute, .05 m^3 is drawn from the tank. In one hour, water drawn from the tank will be 60*.05 = 3 m^3. Since at the end of 21st hr, 3 m^3 water is left in the tank, at the end of the 22nd hour, the entire 3 m^3 water will be drawn and there will be nothing left to drain out in periodic draining.

Further, in 3/7 hrs,
(3/7)*(60)*(.05) = 9/7 m^3 water is drawn

I do not know why they write 3(0.5) = 1.5
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Re: A cylindrical water tower with radius 5 m [#permalink]  26 Nov 2012, 05:58
at 12: 3/4 * 200 = 150 pie m3
at 12 - 1: -.08 + .03 = -.05 * 60 = 3pie m3
at 1 - 147 pie m3 ===> 3 pie m3 + 4 pie m3 = 7 pie m3 ==> so in hour effectively 7 so it takes 21 hours for 147.

thus 21 (1- thereon)+ 1 (12-1) = 22 hours.
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Re: A cylindrical water tower with radius 5 m [#permalink]  17 Dec 2012, 03:24
This one got me, it is so hard to pay attention to every detail and think so much. I initially went with D. So D is wrong why why?

Alright so
initial volume = (3/4)×∏×5²×8 = 150∏
Relative Drain/ min = .08∏ - .03∏ = .05∏ m³/min drain
Relative drain / hour = .05∏×60 = 3∏ m³/ hr

Every one hour starting from 1pm, 4∏ m³ of water is drained. It means that only at the hour the water is drained and NOT “in that 1 hour“

So after 1 hr the relative drain would be 3∏ + 4∏ = 7∏m³ of water drain

What i did initially was formed an equation 150∏ = 7∏×n (n is the number of hrs) so ended up with 21 3/7. This wrong

Look at this way
after 21 hrs the amount of water drain will be 21×7∏ = 147∏ m³
Left over water in the tank after 21 hrs = 3∏ m³
From above we know that it take 1 more hour to drain that 3∏ m³.

So ans is 22hrs
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Re: A cylindrical water tower with radius 5 m [#permalink]  17 Dec 2012, 03:57
Another easy approach to this question is:

Volume to be drained= 150pi
Every minute, the volume that drains is : (0.08pi-0.03pi) metre cube or 0.05pi metre cube.
Every hour 4pi metre cube also drains out, but remember this starts 1 hour later.

We can set up an equation here:
0.08*number[m]ofminutesinyhours+4(y-1)=150[/m]
or
0.08*60*y + 4y=154
or
7y=154
Therefore y=22 hours

I found this approach rather easier.
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Re: A cylindrical water tower with radius 5 m   [#permalink] 17 Dec 2012, 03:57
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