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by topspin360 » Sun Nov 17, 2013 8:21 am
Hello,

In the following problem, we're not told if the variables are positive or negative. Is statement (1) sufficient?

Is it true that x - z > y - z ?

(1) xz > yz
(2) x + z > y + z
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by Brent@GMATPrepNow » Sun Nov 17, 2013 8:32 am
topspin360 wrote:Hello,

Is it true that x - z > y - z ?

(1) xz > yz
(2) x + z > y + z
Target question: Is x - z > y - z ?

This is a great candidate for rephrasing the target question (more on this important strategy in this free video: https://www.gmatprepnow.com/module/gmat- ... cy?id=1100)

Take x - z > y - z, and add z to both sides to get z > y. So . . .

REPHRASED target question: Is x > y?

Statement 1: xz > yz
Since we don't know whether z is positive or negative, we must consider these two cases:
case a: z is positive. So, if we divide both sides by z, we get x > y
case b: z is negative. So, if we divide both sides by z, we get x < y
Since we can't answer the REPHRASED target question with certainty, statement 1 is NOT SUFFICIENT

Alternatively, we can show that statement 1 is not sufficient by testing possible scenarios that meet the condition that xz > yz .
case a: x = 3, y = 2, and z = 1, in which case x > y
case b: x = 2, y = 3, and z = -1, in which case x < y
Since we can't answer the REPHRASED target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: x + z > y + z
Subtract z from both sides to get x > y
Since we can answer the REPHRASED target question with certainty, statement 2 is SUFFICIENT

Answer = B

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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by Mathsbuddy » Sun Nov 17, 2013 9:03 am
Lets consider 2 separate cases:

Case 1, z=1:

If xz>yz, then x = y + c (where c>0) -EQUATION A

therefore the equation x-z>y-z becomes y+c-1>y-1 which is true.

Case 2, z=-1:

If xz>yz, then -x = -y + c (where c>0) -EQUATION B

therefore the equation x-z>y-z becomes -y+c-1>y--1

This simplifies to the condition that it is only true if c > 2y + 2

Therefore statement 1 is not sufficient (as we don't know if that condition is met).

Statement 2 gives us:

x - 1 > x + c -1

so x > x + c which is not true, and sufficient to say so.
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by theCodeToGMAT » Sun Nov 17, 2013 9:05 am
The important consideration when solving inequalities is that we do not cancel same variable from both sides until we are not sure of sign. The statement #1 is an example of that.
xz > yz
Here, we are tempted to strike-off "z" from both sides; this will be partial solution
We must always consider both negative and positive cases

If "z" is positive, then
x (z) > y (z) ==> x > y

If "z" is negative, then multiply both sides by "-1"
So,
x (-z) > y (-z) ==> y (z) > x (z) ==> y > x

Hence, INSUFFICIENT
R A H U L
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by Mathsbuddy » Sun Nov 17, 2013 10:03 am
theCodeToGMAT wrote:The important consideration when solving inequalities is that we do not cancel same variable from both sides until we are not sure of sign. The statement #1 is an example of that.
xz > yz
Here, we are tempted to strike-off "z" from both sides; this will be partial solution
We must always consider both negative and positive cases

If "z" is positive, then
x (z) > y (z) ==> x > y

If "z" is negative, then multiply both sides by "-1"
So,
x (-z) > y (-z) ==> y (z) > x (z) ==> y > x

Hence, INSUFFICIENT
True, but what if x = z and y < 0 and Modulus y < z? Would that require further tests?
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