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A cylindrical water tower with radius 5 m and height 8 m is

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A cylindrical water tower with radius 5 m and height 8 m is [#permalink] New post 20 Nov 2012, 17:00
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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] New post 20 Nov 2012, 20:07
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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] New post 26 Nov 2012, 04:58
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] New post 17 Dec 2012, 02:24
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] New post 17 Dec 2012, 02:57
Expert's post
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:
\(0.08*number[m]\)of\(\)minutes\(\)in\(\)y\(\)hours\(+\)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 and height 8 m is [#permalink] New post 07 Aug 2014, 00:26
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Re: A cylindrical water tower with radius 5 m and height 8 m is [#permalink] New post 07 Aug 2014, 08:58
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12 noon: 150π m3

For every minute there is a net drain of 0.08π-0.03π=0.05π m3. So for every hour=0.05*60=3π m3/hour
This means at 1pm there is 150π-3π=147π m3

And from 1pm there is an additional drain of 4π m3 so net drain= 3π+4π=7π m3/hour
So the time taken to drain 147π m3 is 147π/7π=21 hours. and lets not forget the 1 hour between 12 and 1pm.

Therefore, total time taken is 21+1=22 hours
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Re: A cylindrical water tower with radius 5 m and height 8 m is [#permalink] New post 21 Aug 2014, 10:58
3/4 of 200pi = 150pi.

0.08pi removed & 0.03pi added for each minute. for hours 4.8pi removed & 1.8pi added. consolidated will be 3.0pi removed for each hour. Note: This starts at noon.

147pi.

starting from 1pm for each hour 4 pi is removed from the total. (along with the regular 3.0pi) , it will be 7pi removed from 1 pm .

147pi/7pi = 21 hours.

Adding the hour between 12 to 1 pm - 21+1 = 22 hours.
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Re: A cylindrical water tower with radius 5 m and height 8 m is [#permalink] New post 22 Apr 2015, 00:41
Marcab wrote:
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:
\(0.08*number[m]\)of\(\)minutes\(\)in\(\)y\(\)hours\(+\)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.



hi,
I have not understood why we have used .08 instead of .05(net drainage)
please help
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Re: A cylindrical water tower with radius 5 m and height 8 m is [#permalink] New post 22 Apr 2015, 02:56
Expert's post
ssriva2 wrote:
Marcab wrote:
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:
\(0.08*number[m]\)of\(\)minutes\(\)in\(\)y\(\)hours\(+\)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.



hi,
I have not understood why we have used .08 instead of .05(net drainage)
please help


Hi ssriva2,

You are right, it has to be 0.05, which is net drainage and not 0.08. The equation writes 0.08 but solves using 0.05. Assuming 0.08 in the equation will not give the right answer.

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Re: A cylindrical water tower with radius 5 m and height 8 m is [#permalink] New post 22 Jun 2015, 07:03
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.
Re: A cylindrical water tower with radius 5 m and height 8 m is   [#permalink] 22 Jun 2015, 07:03
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