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The inverted cone pictured below is initially filled to heig

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The inverted cone pictured below is initially filled to heig [#permalink] New post   13 May 2020, 10:35
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The inverted cone pictured below is initially filled to height \(h_0\) with a volume of \(1,000 \pi\) water. 60 percent of the water is drained from the cone and the water level falls to height \(h_1\). For a cone with radius r and height h, volume =


Image


Quantity A
Quantity B
\(\frac{h_1}{h_0}\)
\(0.4\)



A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.



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A
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rishabhrbs96
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Re: The inverted cone pictured below is initially filled to heig [#permalink] New post   13 May 2020, 11:26
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Lets assume,
when the cone is filled till h0, the radius of circular cross section of cone till that height = r0 (which is AB)
when the cone is filled till h1, the radius of circular cross section of cone till that height = r1 (which is CD)

From, similarity of triangels,
AB/CD = OA/OC
=> r0/r1 = h0/h1 ........... (1)

Now,
60% of water was drained
=> initial volume / final volume = 100 / 40
=> [ (1/3) * pi * r0 * r0 * h0] / [ (1/3) * pi * r1 * r1 * h1] = 1 / 0.4

from (1) we can replace the value of r0/r1
=> (h0*h0*h0)/(h1*h1*h1) = 1/0.4
=> h1/h0 = cube root of 0.4 = 0.7368 > 0.4

So, the answer is A.
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Re: The inverted cone pictured below is initially filled to heig [#permalink] New post   13 May 2020, 10:36
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Post A Detailed Correct Solution For The Above Questions And Get A Kudos.
Question From Our New Project: GRE Quant Challenge Questions Daily - NEW EDITION!
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Re: The inverted cone pictured below is initially filled to heig [#permalink] New post   13 May 2020, 12:14
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No need to use external hosts, even though there is nothing wrong with them

Just use upload attachment under the windows when you post a reply or create a post.

So the image is directly showed up into the post.

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Re: The inverted cone pictured below is initially filled to heig [#permalink] New post   24 Jul 2020, 18:22
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Another solution:

1) Define r_1 is the radius that corresponds to h_1, and r_0 is the radius that corresponds to h_0
2) We can set (1 / 3) * pi * r_1^2 * h_1/(1 / 3) * pi * r_0 ^2 * h_0 = 4 / 10
3) From step 2, it follows that h_1 / h_0 = (4 / 10) * (r_0 / r_1) ^2
4) Since (r_0 / r_1) > 1, h_1 / h_0 > 4 / 10, so the answer is A
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Re: The inverted cone pictured below is initially filled to heig [#permalink] New post   24 Jul 2020, 20:18
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The inverted cone pictured below is initially filled to height h0 with a volume of 1,000π water. 60 percent of the water is drained from the cone and the water level falls to height h1. For a cone with radius r and height h, volume =

60% of 1000 = 600
h1/h0 = 600/1000 = 0.6 > 0.4
Quantity A

Did I miss something here?
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Re: The inverted cone pictured below is initially filled to heig [#permalink] New post   27 Jul 2020, 00:30
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1000pi=(1/3)pi r0^2 h0
r0^2 h0 = 3000
400pi (as it's 40 per cent of the water) = (1/3)pi r1^2 h1
r1^2 h1 = 1200

(h1 r1^2)/(r0^2 h0) = 1200/3000 = 0.4
but! r1 < r0 - that is why h1/h0 > 0.4
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Re: The inverted cone pictured below is initially filled to heig [#permalink] New post   07 Apr 2024, 10:30
Hello from the GRE Prep Club BumpBot!

Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: The inverted cone pictured below is initially filled to heig [#permalink]
07 Apr 2024, 10:30
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