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

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

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29 Nov 2015, 09:57
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Question Stats:

77% (01:07) correct 23% (01:34) wrong based on 292 sessions

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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’

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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, 10:03
2
As Given
V = kh^2

1. h=2 , V can be calculated as

V = Kh^2
V' =4k

2. h=4 ,

V =16k = 4(4k)= 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, 10:40
2

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, 11:05
1
v' = 4k
v at h=4 is 16k, which is equal to 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
3
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
$$V = 4^2k = 16k$$

Thus, $$V = 4V$$

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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
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
$$V = 4^2k = 16k$$

Thus, $$V = 4V$$

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

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Updated on: 29 Nov 2015, 11:52
2
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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Originally posted by Nightfury14 on 29 Nov 2015, 11:22.
Last edited by Nightfury14 on 29 Nov 2015, 11:52, edited 1 time in total.
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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, 11:27
1
$$V = kh^2 = k4^2 = 16k$$

$$V' = kh^2 = k2^2 = 4k$$

V is four times bigger than V' so $$V=4V'$$

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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, 11:36
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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, 12:08
1
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'.
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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, 12:25
1
Solution: V'= k* 2^2 or V' = k* 4 or k= V'/4;
then v" (new speed)= k* 4^2 or v"= 16 *k or 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, 22:32
1
V = kh^2, where h was the height of the remaining oil and k was constant.

Height of the cylinder is 2 --> Speed of the leak is V' = 4k
Height of the oil remaining is 4 --> Speed of the leak = 16k = 4*4k = 4V'

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

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30 Nov 2015, 07:19
1

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

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30 Nov 2015, 09:09
1
B?

Given, V=kh^2
When height of the oil remaining is 4,
V=k4^2=16k

If height of the cylinder=2
Speed V'=k2^2=4k (considering it is fully filled)

Hence V=4V'
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Math Revolution and GMAT Club Contest! When the oil from a circular  [#permalink]

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Updated on: 07 Dec 2015, 09:24
1
when h = 2

V` = k * 2^2 => v#=4k => k = v#/4

when h =4

$$=> k * 4 * 4 => v = \frac{v}{4}*4 * 4 = 4 v$$

Originally posted by appsy01 on 30 Nov 2015, 23:44.
Last edited by appsy01 on 07 Dec 2015, 09:24, edited 1 time in total.
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Re: Math Revolution and GMAT Club Contest! When the oil from a circular  [#permalink]

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01 Dec 2015, 10:17
1
v=K(2x2)=4K

If h=4, new speed will be K(4x4)=16K=4V.

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

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01 Dec 2015, 13:15
1
Formula for the speed of leak is v = k*h^2. Let's write the question in terms of this formula:

"If the height of the cylinder is 2, the speed of the leak is V’" in terms for above formula can be stated as:

v' = k*2^2 (v' is the velocity when the hight is 2 )

v' = 4k

k = v'/4

"When the height of the oil remaining is 4, what was the speed of the leak" in terms of formula can be stated as:

v = k*4^2

v = 16k

v = 16*(v'/4)

v = 4v' (this is the velocity of leak in terms of v', when hight of remaining oil is 4)

Option (B) is correct.
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Re: Math Revolution and GMAT Club Contest! When the oil from a circular  [#permalink]

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02 Dec 2015, 00:06
1
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’

When h=2 v=V'

So V' = k.4
Hence k = V'/4

Now when h=4

v=(V'/4).4^2
v=V'.16/4
v=4V'

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

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03 Dec 2015, 12:08
1
When h = 2,
V' = k(2)^2 = 4k
when h = 4,
V = k(4)^2 = 4.4k = 4V'

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

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03 Dec 2015, 13:38
1
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$$(i)$$)

Re: Math Revolution and GMAT Club Contest! When the oil from a circular &nbs [#permalink] 03 Dec 2015, 13:38

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