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In the formula w=p / (v)^1/t integers p and t are positive

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In the formula w=p / (v)^1/t integers p and t are positive [#permalink]

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New post 11 Oct 2012, 09:29
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A
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C
D
E

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0% (00:00) correct 100% (01:45) wrong based on 2 sessions

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In the formula w=p / (v)^1/t integers p and t are positive constants. If w =2 when v = 1 and if w=1/2 when v = 64, then t =
(A) 1
(B) 2
(C) 3
(D) 4
(E) 16
[Reveal] Spoiler: OA

Last edited by abhishek1990p on 14 Oct 2012, 11:08, edited 2 times in total.
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Re: integers [#permalink]

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New post 14 Oct 2012, 09:11
If w =2 when v = 1 and if w=? when v = 64

I think the question is incomplete
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Re: integers [#permalink]

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New post 14 Oct 2012, 09:27
abhishek1990p wrote:
In the formula w=p / (t√v) integers p and t are positive constants. If w =2 when v = 1 and if when v = 64, then t =
(A) 1
(B) 2
(C) 3
(D) 4
(E) 16


hey i dont understand if its (v)^1/t OR t*[(v)^1/2]
can you please be specific...coz i am getting two variable here.
What is the source?
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Kudos [?]: 1 [0], given: 14

Re: In the formula w=p / (v)^1/t integers p and t are positive [#permalink]

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New post 15 Jul 2014, 23:07
Answer is C

Substituting from statement 1:
2=p/(1)^(1/t) -{1}


Substituting from statement 2:
1/2=p/(64)^(1/t) -{2}


{1}/{2} gives:
2/(1/2)=[p/(1)^(1/t)]/[p/(64)^(1/t)]


Simplifying:
4=(64)^1/t
or 4^t=64

answer is 3.

Choice C
Re: In the formula w=p / (v)^1/t integers p and t are positive   [#permalink] 15 Jul 2014, 23:07
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In the formula w=p / (v)^1/t integers p and t are positive

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