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# Engineer X, Y, Z each assemble computers at their respecitve

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Engineer X, Y, Z each assemble computers at their respecitve [#permalink]  04 Feb 2006, 16:59
Engineer X, Y, Z each assemble computers at their respecitve constant rates. If Rx is the ratio of Engineer X's constant rate to Engineer Z's constant rate and Ry is the ratio of Engineer Y's constant rate to Engineer Z's constant rate, is Engineer Z's constant rate the greatest of the three?

1) Rx < Ry

2) Ry < 1
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It looks like C.
from 1 => Vx<Vy=> not suf
from 2=>Vy<Vz => not suf
Together Vx<Vy<Vz suf.
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Re: Engineers work [#permalink]  05 Feb 2006, 03:33
C it is.

Rx = X/Z
Ry = Y/Z

1) X/Z < Y/Z, Only X < Y can be concluded. No info about Z. INSUFF

2) Y/Z < 1. No indo about X. INSUFF

Combining:

X/Z < Y/Z < Z/Z (=1)
=> X < Y < Z
Hence C!
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Agree with (C).
Nice explanation giddi
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Giddi , excellant explanation... i got C too ... same approach!
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late but C. happy that i actually got it !!

Rx = X/Z

Ry = Y/Z

1) Rx<Ry. Since Z is the same in both cases, this must mean that X<Y, but doesnt tell us anything about Z. INSUFF, so eliminate A and D

2) Ry<1. This tells us that Y must be less than Z, but still no comparison with X. INSUFF, so eliminate B.

Putting together, we know that X<Y, and Y<Z, therefore it must follow that X<Y<Z, so yes, Z is the greatest rate.

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X/Z = Rx

Y/Z = Ry

1) Insufficient.
If X/Z = Rx = 1, then X = 2. If Y/Z = 2, then Y = 2Z. In this case, Y is the fastest.

But if X/Z = Rx = 1/4 = 4X = Z. If Y/Z = 1/2 then 2Y = Z. In this case, Z is the fastest.

2) Ry is less than 1, but information about Rx. Insufficient.

Using 1) and 2), Z is the fastest.

Ans C
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