Kaplan DS

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Kaplan DS

by davy420 » Mon Sep 15, 2008 11:36 pm
Can somebody please explain this problem. Thanks

If x+y=3 what is the value x-y/y-z

A. y+z=4
B. x+z=5

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by gdrea3 » Tue Sep 16, 2008 4:45 am
I'm sure I'm doing this the long way, but here it goes:

Statment 1 is insufficient b/c the equation does not give you a value for x.
Statement 2 is insufficient for the same reason--you are not given a value for y.

Statements 1 and 2 together--write out equations and subsitute values:
x+y=5-->x=5-z
y+z=4-->y=4-z

So in the equation given in the question, x+y=3 subsitute the x for 5-z:
5-z+y=3
Subsitute the y for 4-z:
5-z+4-z=3-->subtract 5 from both sides-->-2z+4=-2-->subtract 4 from both sides-->-z=-6-->Z=6

Substitute 6 in for z for one of the equations:y+z=4-->y+6=4-->y=-2

Substitute -2 in for y and from here you shouldn't do anymore work--you have enough info to solve for x-y/y-z, so the answer is C
:D

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by davy420 » Tue Sep 16, 2008 11:55 am
Im curious but how did you get x+y=5-->x=5-z

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by gdrea3 » Wed Sep 17, 2008 5:13 am
Oh, sorry, my bad, I was writing out too many of the equations at once. Here's a clearer approach:

Statements 1 and 2 ALONE are both insufficient b/c they are missing a variable (statement 1 is missing X and statement 2 is missing Y)

Line all the equations up and convert 2 of them:
X+Y=3-->X=3-Y
Y+Z=4-->Z=4-Y

Substitue the X in this equation for 3-Y
X+Z=5-->3-Y+Z=5
Subsitute the Z for 4-Y-->3-Y+4-Y=5-->Y=-2

Put in -2 for Y in any equation: Y+Z=4-->-2+Z=4-->Z=6

Therefore you can solve for all variables and thus answer the original question. Sorry for the mix up earlier!

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by bluementor » Fri Sep 19, 2008 3:27 am
This approach could be (only slightly) faster:

I hope its clear that both statements are insufficient on their own.

Both statements (A and B) together:

B - A = 1 = x-y
x + y - B = -2 = y-z (x+y = 3 is given in the question stem)

Therefore, both statments are sufficient (C).

BlueMentor