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The inverted cone pictured below is initially filled to heig
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13 May 2020, 10:35
13 May 2020, 10:35
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50% (02:06) correct
49% (02:01) wrong based on 145 sessions
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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 =
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.
#greprepclub The inverted cone pictured below is initially filled.jpg [ 29.65 KiB | Viewed 6220 times ]
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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#greprepclub The inverted cone pictured below is initially filled.jpg [ 29.65 KiB | Viewed 6220 times ]
Re: The inverted cone pictured below is initially filled to heig
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13 May 2020, 11:26
13 May 2020, 11:26
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Mpz51ex.png [ 19.92 KiB | Viewed 4968 times ]
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.
General Discussion
Re: The inverted cone pictured below is initially filled to heig
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13 May 2020, 10:36
13 May 2020, 10:36
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Question From Our New Project: GRE Quant Challenge Questions Daily - NEW EDITION!
Question From Our New Project: GRE Quant Challenge Questions Daily - NEW EDITION!
