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An hourglass is formed from two identical cones, one kept on

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An hourglass is formed from two identical cones, one kept on [#permalink]

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New post 10 Aug 2008, 01:48
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An hourglass is formed from two identical cones, one kept on top of the other. When the upper cone is full of sand and the lower one is empty, it takes an hour for the sand to flow, at a constant rate, from the upper cone to lower cone. How long does it take for the depth of sand in the lower cone to be one third of the depth of sand in the upper cone?

a. 380/9 min
b. 160/9 min
c. 40 min
d. 20 min
e. 15 mins

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New post 10 Aug 2008, 02:27
Lets say total height = K
Height of sand in lower cone = H
Height of sand in uppercone = H'

H+H' = K
H = 1/3H' (Given in question the depth of sand in the lower cone to be one third of the depth of sand in the upper cone)
=> 1/3H'+H' = 4/3H' = K

Also given in the question, it takes 60 minutes to transfer entire 4/3H' from one cone to the other.

So 4/3H' gets transferred in = 60 mins.
Then 1/3H' will transfer in = 60*1/3H'/4/3H' = 60/4 = 15 Mins.

Answer E.

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New post 10 Aug 2008, 02:56
Sorry Wrong answer!!

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New post 10 Aug 2008, 03:06
In that case answer has to be 19/27*60. 380/9

But I am pretty sure it is not GMAT type question and picked up from somewhere else. Please share the source of OA if possible.

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New post 10 Aug 2008, 03:47
The answer is actually 380/9

U gotta help me with the explanation!! someone send it across as a doubt on my mail

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New post 10 Aug 2008, 05:27
rahulgoyal1986 wrote:
An hourglass is formed from two identical cones, one kept on top of the other. When the upper cone is full of sand and the lower one is empty, it takes an hour for the sand to flow, at a constant rate, from the upper cone to lower cone. How long does it take for the depth of sand in the lower cone to be one third of the depth of sand in the upper cone?

a. 380/9 min
b. 160/9 min
c. 40 min
d. 20 min
e. 15 mins


the hour glass upper cone empties in 1 hr means that v/60 volume of sand is fallen in lower in 1 min

now we want ,x be vol of sand in lower cone and V-x be vol of sand in upper cone.given that
x=1/3 *(v-x) => v/4=x =>
v/60 ------>1min
v/4-------->15min

it takes 15 min to put 1/3 of upper sand to lower cone.
IMO E
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New post 10 Aug 2008, 05:34
15 is not the answer because its not the volume of the sand in lower part which is one third of the volume of the sand in upper portion. Its the height!!!
Very difficult one, I just wish abhijit_sen explains the reason for 380/9 quickly or i will die...

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New post 10 Aug 2008, 05:36
new2gmat wrote:
15 is not the answer because its not the volume of the sand in lower part which is one third of the volume of the sand in upper portion. Its the height!!!
Very difficult one, I just wish abhijit_sen explains the reason for 380/9 quickly or i will die...

i messed up again :x
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New post 12 Aug 2008, 12:07
abhijit_sen wrote:
In that case answer has to be 19/27*60. 380/9

But I am pretty sure it is not GMAT type question and picked up from somewhere else. Please share the source of OA if possible.

abhijit_sen, please help with the explanation for 380/9...

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New post 12 Aug 2008, 12:59
At the required time the depth of sand in the upper cone is two-thirds its original depth. Since volume varies as the cube of any dimension in regular figures, the time required is 1 - (2/3)^3
= 19/27 of an hour = 380/9

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New post 13 Aug 2008, 04:31
after one third height of sand is gone from upper cone the left sand will also be in shape of cone...
height = 2/3 h
and radius = 2/3 r

h and r are height and radius of original cone

1/3 * pi * r^2 * h = goes in 60 minutes
1/3 * pi * (2r/3)^2 * 2h/3 will go in 8 * 60 / 27 minutes = 160/9 minutes

one third hieght of sand will go in = 60 - 160/9 minutes = 380/9

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Re: Help with this qns!!!!!   [#permalink] 13 Aug 2008, 04:31
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