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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Inequalities sign change
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topspin360
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Brent@GMATPrepNow
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Target question: Is x - z > y - z ?topspin360 wrote:Hello,
Is it true that x - z > y - z ?
(1) xz > yz
(2) 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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Mathsbuddy
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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.
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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theCodeToGMAT
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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
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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Mathsbuddy
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True, but what if x = z and y < 0 and Modulus y < z? Would that require further tests?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
