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Math Revolution and GMAT Club Contest! When the oil from a circular

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

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New post 04 Dec 2015, 05:06
1
answer is B : 4V

Explanation: V=kh^2
when h=2, then V=k2^2=4k, therefore k=V/4.
when h=4, then speed of leak=kh^2=(V/4)x4^4=V/4 x 16=4V.
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Re: Math Revolution and GMAT Club Contest! When the oil from a circular  [#permalink]

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New post 04 Dec 2015, 06:46
1
V'=2^2 k

let new speed be s

s= k 4^2

so, s=4V'

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

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New post 04 Dec 2015, 23:19
V=k *2^2 = 4k
V'= k * 4^2 = 16k
V' = 4V . V= V'/4
Speed of leak in terms of V ' is V'/4. Answer choice D.
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New post 06 Dec 2015, 03:39
1
When h=2: V' = k*2^2 = 4k
When h = 4, V = k*4^2 = 16k
=> V = 4V'
Hence B.
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Re: Math Revolution and GMAT Club Contest! When the oil from a circular  [#permalink]

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New post 06 Dec 2015, 08:26
What to find? Speed of leak when height is 4.

Given
1. Formula for speed V=Kh^2
2. Speed= V at height 2

From the given information calculate the value of constant K by using given information.

V=K*2^2

Therefore K= V/4

Use the value of constant K to find out speed at height 4.

Speed of leak when the height of the oil remaining 4 = Kh^2 =V/4*4^4 =4V.
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Re: Math Revolution and GMAT Club Contest! When the oil from a circular  [#permalink]

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New post 14 Dec 2015, 08:03
Bunuel wrote:

Math Revolution and GMAT Club Contest Starts!



QUESTION #4:

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’


Check conditions below:



Math Revolution and GMAT Club Contest

The Contest Starts November 28th in Quant Forum


We are happy to announce a Math Revolution and GMAT Club Contest

For the following four (!) weekends we'll be publishing 4 FRESH math questions per weekend (2 on Saturday and 2 on Sunday).

To participate, you will have to reply with your best answer/solution to the new questions that will be posted on Saturday and Sunday at 9 AM Pacific.
Then a week later, the forum moderator will be selecting 2 winners who provided most correct answers to the questions, along with best solutions. Those winners will get 6-months access to GMAT Club Tests.

PLUS! Based on the answers and solutions for all the questions published during the project ONE user will be awarded with ONE Grand prize:

PS + DS course with 502 videos that is worth $299!



All announcements and winnings are final and no whining :-) GMAT Club reserves the rights to modify the terms of this offer at any time.


NOTE: Test Prep Experts and Tutors are asked not to participate. We would like to have the members maximize their learning and problem solving process.

Thank you!



MATH REVOLUTION OFFICIAL SOLUTION:

This type of question appears within a score range of 45 to 48. It is important to first find the value of a constant like this speed question. According to the question, when the height of the cylinder is 2, the speed of the leak is V’. Therefore, the first step would be: V’=k(2)^2=k4, k=V’/4. Because the question asks for the speed of the leak when the height is 4, the calculation would be: V=k(4)^2=(V’/4)(16)=V’/4. So, B is the correct answer.
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Re: Math Revolution and GMAT Club Contest! When the oil from a circular  [#permalink]

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New post 05 Jan 2016, 04:22
V=kh^2
when h=2, V'=4k===>k=V'/4
when h=4, V=16k===>16*V'/4===>4V'

Ans: B
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Re: Math Revolution and GMAT Club Contest! When the oil from a circular   [#permalink] 10 May 2019, 03:00

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