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Water is leaking out from a cylinder container at the rate

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Water is leaking out from a cylinder container at the rate [#permalink] New post 27 Jun 2009, 21:29
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Water is leaking out from a cylinder container at the rate of 0.31 m^3 per minute. 10 minutes later, the water level decreases 0.25 meter, what is value of the radius?

(A) 0.5
(B) 1.0
(C) 1.5
(D) 2.0
(E) 2.5
[Reveal] Spoiler: OA
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Re: water level [#permalink] New post 22 Jul 2009, 00:17
Total water loss = 3.1 mtr in 10 min @.31 meter per min
3.14XR^2X.25 = 3.1
r^2 = 4
r=2

(D)
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Re: water level [#permalink] New post 20 Feb 2010, 11:08
irajeevsingh wrote:
Total water loss = 3.1 mtr in 10 min @.31 meter per min
3.14XR^2X.25 = 3.1
r^2 = 4
r=2

(D)


Can you please elaborate that how you did

3.1 * r^2 * .25 = 3.1
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Re: water level [#permalink] New post 20 Feb 2010, 12:00
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Bullet wrote:
irajeevsingh wrote:
Total water loss = 3.1 mtr in 10 min @.31 meter per min
3.14XR^2X.25 = 3.1
r^2 = 4
r=2

(D)


Can you please elaborate that how you did

3.1 * r^2 * .25 = 3.1


In 10 min the volume of the water which leaked out the cylinder would be \(10*0.31=3.1\) m^3.

Volume = \(\pi*r^2*height=3.1\) --> \(\pi*r^2*0.25=3.1\) --> \(\pi=3.14\) --> \(3.14*r^2*0.25=3.1\) --> \(r=2\) (approximately).

Hope it's clear.
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Re: water level [#permalink] New post 20 Feb 2010, 22:39
Bunuel wrote:
Bullet wrote:
irajeevsingh wrote:
Total water loss = 3.1 mtr in 10 min @.31 meter per min
3.14XR^2X.25 = 3.1
r^2 = 4
r=2

(D)


Can you please elaborate that how you did

3.1 * r^2 * .25 = 3.1


In 10 min the volume of the water which leaked out the cylinder would be \(10*0.31=3.1\) m^3.

Volume = \(\pi*r^2*height=3.1\) --> \(\pi*r^2*0.25=3.1\) --> \(\pi=3.14\) --> \(3.14*r^2*0.25=3.1\) --> \(r=2\) (approximately).

Hope it's clear.


Thanks Bunuel, I was confusing 3.14 with the volume where is was pi.

Cheers!
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Re: Water is leaking out from a cylinder container at the rate [#permalink] New post 21 Jan 2015, 12:01
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Re: Water is leaking out from a cylinder container at the rate [#permalink] New post 13 Mar 2015, 06:48
Bunuel wrote:
Bullet wrote:
irajeevsingh wrote:
Total water loss = 3.1 mtr in 10 min @.31 meter per min
3.14XR^2X.25 = 3.1
r^2 = 4
r=2

(D)


Can you please elaborate that how you did

3.1 * r^2 * .25 = 3.1


In 10 min the volume of the water which leaked out the cylinder would be \(10*0.31=3.1\) m^3.

Volume = \(\pi*r^2*height=3.1\) --> \(\pi*r^2*0.25=3.1\) --> \(\pi=3.14\) --> \(3.14*r^2*0.25=3.1\) --> \(r=2\) (approximately).

Hope it's clear.



I have a question, that here it is assumed that the cylinder is standing on the radius and the water is being decreased in the length of the cylinder. Is this a good assumption?
I think this should be stated in the question, as I was trying the problem with the cylinder lying on the length. :(
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Re: Water is leaking out from a cylinder container at the rate [#permalink] New post 13 Mar 2015, 08:04
santorasantu wrote:


I have a question, that here it is assumed that the cylinder is standing on the radius and the water is being decreased in the length of the cylinder. Is this a good assumption?
I think this should be stated in the question, as I was trying the problem with the cylinder lying on the length. :(


hi,
since the question says the water decreased by .25m , it would mean that the height of it .. since if it is lying as you have taken, this .25m does not have same cross section at any place...
Also you have to take a cylinder as on base because that is what is the original shape..
you can take the way you have taken only when it is specified that way..
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Re: Water is leaking out from a cylinder container at the rate [#permalink] New post 14 Mar 2015, 00:59
chetan2u

hi,
since the question says the water decreased by .25m , it would mean that the height of it .. since if it is lying as you have taken, this .25m does not have same cross section at any place...
Also you have to take a cylinder as on base because that is what is the original shape..
you can take the way you have taken only when it is specified that way..
_____________________________________________________________________________________________________
Chetan i am not convinced from the ans i.e 3.14*\(r^2\)*h
why we took height = .25m as it is not the actual total height ??
and total volume =3.1 as it leaked for 10 min only given.

please help what am i missing ?

Regards
SG
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Re: Water is leaking out from a cylinder container at the rate [#permalink] New post 14 Mar 2015, 04:00
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smartyguy wrote:
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_____________________________________________________________________________________________________
Chetan i am not convinced from the ans i.e 3.14*\(r^2\)*h
why we took height = .25m as it is not the actual total height ??
and total volume =3.1 as it leaked for 10 min only given.

please help what am i missing ?

Regards
SG


hi SG,

Water is leaking out from a cylinder container at the rate of 0.31 m^3 per minute. 10 minutes later, the water level decreases 0.25 meter, what is value of the radius?

(A) 0.5
(B) 1.0
(C) 1.5
(D) 2.0
(E) 2.5

in one minute, the volume of water lost is .31\(m^3\)
so in 10 minutes the volume of water lost is 3.1\(m^3\)..(1)

now, we are also said that the height lost is .25m, we are not concerned with the entire height...
what is the volume of water in this .25m...
since it is a cylinder the volume of .25 m= pi*\(r^2\)*h=3.14*\(r^2\)*.25...
3.14*\(r^2\)*.25=3.1...
\(r^2\)=1/.25=4.. so r=2 approx
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Re: Water is leaking out from a cylinder container at the rate [#permalink] New post 14 Mar 2015, 06:27
chetan2u

Hmm it doesn't matter what is the exact total height and volume .

because radius will remain same.
Thanks again chetan

Regards
SG
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Re: Water is leaking out from a cylinder container at the rate [#permalink] New post 14 Mar 2015, 06:28
chetan2u

Hmm it doesn't matter what is the exact total height and volume .

because radius will remain same.
Thanks again chetan

Regards
SG
Re: Water is leaking out from a cylinder container at the rate   [#permalink] 14 Mar 2015, 06:28
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