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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]

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New post 27 Jun 2009, 22:29
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Question Stats:

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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
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Re: water level  [#permalink]

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New post 20 Feb 2010, 13:00
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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]

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New post 22 Jul 2009, 01:17
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]

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New post 20 Feb 2010, 12:08
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]

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New post 20 Feb 2010, 23:39
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]

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New post 13 Mar 2015, 07:48
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]

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New post 13 Mar 2015, 09:04
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]

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New post 14 Mar 2015, 01:59
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
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Re: Water is leaking out from a cylinder container at the rate  [#permalink]

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New post 14 Mar 2015, 05:00
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smartyguy wrote:
.
_____________________________________________________________________________________________________
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]

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New post 14 Mar 2015, 07:27
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]

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New post 14 Mar 2015, 07:28
chetan2u

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

because radius will remain same.
Thanks again chetan

Regards
SG
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Re: Water is leaking out from a cylinder container at the rate  [#permalink]

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New post 18 Jan 2018, 08:51
vcbabu wrote:
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


Since the rate of leaking is 0.31 m^3 per minute, in 10 minutes, 10 x 0.31 = 3.1 m^3 of water has been leaked. Let’s let r = radius of the cylinder. Recall that the formula for the volume of a right circular cylinder is V = πr^2h. Using the volume of water leaked and the height associated with that volume, we can create the following equation:

πr^2 x 0.25 = 3.1

πr^2 / 4 = 3.1

r^2 = 3.1 x 4 / π

Notice that π ≈ 3.1, thus we have:

r^2 = 4

r = 2

Answer: D
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Re: Water is leaking out from a cylinder container at the rate  [#permalink]

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New post 15 Feb 2019, 19:37
vcbabu wrote:
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


Sometimes i tell you,WHY :roll:

You can get the change in volume as 3.1 \(m^3\), 0.31 m^3 per minute, in 10 mins just multiply it with 10

Now level decreases by 0.25, this directly means the new height

3.14 * r^2 * 1/4 = 3.1
r^2 = 4

We can cancel 3.14 from both sides now

For starters, this cannot be 1.5,

root4 = 2
root3 = 1.732
root5 = 2.124

Answer is D
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Re: Water is leaking out from a cylinder container at the rate  [#permalink]

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New post 02 Jun 2019, 09:46
vcbabu wrote:
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


good quesiton
rate of flow 0.31 m^3 per minute
so in 10 mins water flowed out 3.1m^3 and height = .25 mtr
so
pi *r^2*.25=3.1

solve we get r=2
IMO D
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Re: Water is leaking out from a cylinder container at the rate   [#permalink] 02 Jun 2019, 09:46
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