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by karmayogi » Fri Dec 19, 2008 5:52 am
If x, y, and z are nonzero numbers, is xz= 12?
(1) (x^2)yz = 12xy
(2) z/4=3/x
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by cramya » Fri Dec 19, 2008 6:04 am
One approach would be to cancel terms for stmt I and cross multiply in stmt II that will lead to D)

Another approach:

Stmt I

x^2yz-12xy = 0
xy(xz-12) = 0

xy cannot be 0 so xz-12 = 0 xz=12

SUFF
Stmt II

z/4-3/x = 0

xz-12/4x =0

Multiply by 4x

xz-12 = 0
xz=12

SUFF

D)
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by karmayogi » Fri Dec 19, 2008 9:58 am
Thanks everyone...

Actually it was a simple question; I just confused my self with something
:(
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by ronniecoleman » Fri Dec 19, 2008 10:18 am
IMO D

1--> x2yz = 12xy
x( xyz- 12y) = 0

y(xz-12)=0

since x!=0 , y !=0

so xz = 12

2. z/4 = 3/x

cross multiply and get the answer
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