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When the oil from a circular cylinder leaked, the formula for the speed of the leak is V = kh^2, where h was the height of the remaining oil and k was constant. If the height of the cylinder is 2, the speed of the leak is V’, When the height of the oil remaining is 4, what was the speed of the leak, in terms of V’?

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Re: Math Revolution and GMAT Club Contest! When the oil from a circular [#permalink]

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29 Nov 2015, 09:40

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The answer is B.

We know speed of the leak is V=4kh^2. (k is constant and h is height of oil remaining). Given - when height of the cylinder is 2 speed of leak is V'. Assuming height of oil in the cylinder is 2 = height of oil remaining, V' =K*2^2. or K=V'/4

Find V when height of oil remaining is 4. V=(V'/4)*4^2 = V'*4 or option B.
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Re: Math Revolution and GMAT Club Contest! When the oil from a circular [#permalink]

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29 Nov 2015, 10:08

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For a cylinder of height 2, that just starts leaking, h = 2. (Appears to be a typo in the question. 2 should be height of remaining oil) \(V` = 2^2k V` = 4k\)

For a cylinder with remaining oil 4, h=4 \(V = 4^2k = 16k\)

Re: Math Revolution and GMAT Club Contest! When the oil from a circular [#permalink]

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29 Nov 2015, 10:08

For a cylinder of height 2, that just starts leaking, h = 2. (Appears to be a typo in the question. 2 should be height of remaining oil) \(V` = 2^2k V` = 4k\)

For a cylinder with remaining oil 4, h=4 \(V = 4^2k = 16k\)

Re: Math Revolution and GMAT Club Contest! When the oil from a circular [#permalink]

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29 Nov 2015, 10:22

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V=Kh^2

at h=2; V'=K(2)^2=4K => K=V'/4 .... (I)

at h=4; V"=K(4)^2=16K .... (II)

substituting .. (I) into (II) , we have V" = 16 (V'/4) = 4V'

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Re: Math Revolution and GMAT Club Contest! When the oil from a circular [#permalink]

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29 Nov 2015, 11:08

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When the oil from a circular cylinder leaked, the formula for the speed of the leak is V = kh^2, where h was the height of the remaining oil and k was constant. If the height of the cylinder is 2, the speed of the leak is V’, When the height of the oil remaining is 4, what was the speed of the leak, in terms of V’?

ok, so we are given the formula: V=kh^2 when h=2, then the speed of leak is V' V' = k*2^2 or 4k

when the height is 4, then V =k*4^2 or 16k

to express this in V', divide 16k by 4k and get 4. V when the height is 4 will be 4 times when the height is 2. in short: 4V'.

Re: Math Revolution and GMAT Club Contest! When the oil from a circular [#permalink]

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30 Nov 2015, 06:19

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Answer can be B

The question is framed poorly. It uses the height of the cylinder to give V'. But it asks for the amount of oil remaining using height of the oil left. So , is the height of the cylinder greater than the height of oil ? If yes, then this is a ridiculously easy question. If not, it is framed incorrectly

Re: Math Revolution and GMAT Club Contest! When the oil from a circular [#permalink]

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01 Dec 2015, 23:06

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When the oil from a circular cylinder leaked, the formula for the speed of the leak is V = kh^2, where h was the height of the remaining oil and k was constant. If the height of the cylinder is 2, the speed of the leak is V’, When the height of the oil remaining is 4, what was the speed of the leak, in terms of V’?

Re: Math Revolution and GMAT Club Contest! When the oil from a circular [#permalink]

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03 Dec 2015, 12:38

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When the oil from a circular cylinder leaked, the formula for the speed of the leak is V = kh^2, where h was the height of the remaining oil and k was constant. If the height of the cylinder is 2, the speed of the leak is V’, When the height of the oil remaining is 4, what was the speed of the leak, in terms of V’?

A. 2V’ B. 4V’ C. V’/2 D. V’/4 E. V’

Soln: \(V = k * h^2\). Given, when \(h = 2, V = V'\). So, \(V' = k * 2^2 = 4k.\) .....\((i)\)

Now, we need to find speed in terms of \(V'\), when \(h = 4.\)

We know speed \(V = k*h^2 = k * 4^2 = 16*k = 4*4k = 4*V' = 4V'.\) (using eqn\((i)\))