pls help

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pls help

by anjaligeorge1 » Wed Jan 07, 2009 11:16 am
If x > 0, then 1/[v(2x)+vx] =

A. 1/v(3x)
B. 1/[2v(2x)]
C. 1/(xv2)
D. (v2-1)/vx
E. (1+v2)/vx

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by vivek.kapoor83 » Wed Jan 07, 2009 11:17 am
is ans A ?

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by DavoodBeater » Wed Jan 07, 2009 11:58 am
if nothing is not to the power of something else (!) (i mean if i didnt misunderstand)
then:
1/[v(2x)+vx] = 1/[vx(2+1)] = 1/[3vx]

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Re: pls help

by Stuart@KaplanGMAT » Wed Jan 07, 2009 2:20 pm
anjaligeorge1 wrote:If x > 0, then 1/[v(2x)+vx] =

A. 1/v(3x)
B. 1/[2v(2x)]
C. 1/(xv2)
D. (v2-1)/vx
E. (1+v2)/vx
If you see the algebra, it's very quick:

1/[v(2x)+vx] = 1/(2vx + 1vx) = 1/3vx

If you don't see the algebra, then this is a perfect question for picking numbers.

let's let v=3 and x=7 (picking fairly random wacky numbers reduces the chance that more than 1 answer will work out).

Original: 1/(3*2*7 + 3*7) = 1/(42+21) = 1/63

Choices:

a) 1/(3*3*7) = 1/63.. yay!

if we want to be 100% sure:

b) 1/(2*3*2*7) = 1/84
c) 1/(7*3*2) = 1/42
d) way too big, don't bother
e) way too big, don't bother
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by nervesofsteel » Wed Jan 07, 2009 11:57 pm
Yes put X=2
and only A satisfies the questions