Water is leaking out from a cylinder container at the rate

Can you please elaborate that how you did

3.1 * r^2 * .25 = 3.1

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

Cheers!

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..

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

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

chetan2u

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

because radius will remain same.
Thanks again chetan

Regards
SG

chetan2u

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

because radius will remain same.
Thanks again chetan

Regards
SG

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

Sometimes i tell you,WHY

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

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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