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Re: The inverted cone pictured below is initially filled to heig [#permalink]
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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]
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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]
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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]
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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]
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