Is 1 / (a-b) < (b-a)? (1) a < b (2) 1 < |a-b|
Answer: A
Approach:
Rephrase: Is 1<(a-b)(b-a)?
1) a < b
If a, b are both +ve, Both -ve, or -ve & +ve
1 > (a-b) (b-a)
Statement (1) is sufficient.
2) 1 < |a-b|
case 1: a-b>1 if a-b>0 or a>b
case 2: b-a>1 if a-b<0 or a<b
For case 2 we know that the statement 1 > (a-b) (b-a) is always true
Even for case 1: a>b, (a,b) = (9,7) or (-7,-10) or (7,-2) we get: 1 > (a-b) (b-a) is always true
and therefore statement 2 is also sufficient.
My answer is D but the correct answer is A. Can anyone please help me understand why statement 2 is not sufficient and where is the problem with my approach.
Thanks in advance
My answer is different from the correct answer - Please help
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- GMATGuruNY
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When you multiplied each side of the inequality in the question stem by a-b, you presumed that a-b>0.ggmat007 wrote:Is 1 / (a-b) < (b-a)?
(1) a < b
(2) 1 < |a-b|
Rephrase: Is 1<(a-b)(b-a)?
If a-b<0, then the direction of the inequality symbol in your rephrase must FLIP from < to >.
Since the sign of a-b is unknown, it's safer not to rephrase the question stem.
Statement 1: a < b
Thus:
a-b < 0.
b-a > 0.
Here, we can rephrase 1/(a-b) < b-a as follows:
1/(negative) < positive.
negative < positive.
SUFFICIENT.
Statement 2: 1 < |a-b|
Case 1: a-b = 2, implying that b-a = -2.
If we plug a-b = 2 and b-a = -2 into 1/(a-b) < b-a, we get:
1/2 < -2
NO.
Case 2: a-b = -2, implying that b-a = 2.
If we plug a-b = -2 and b-a = 2 into 1/(a-b) < b-a, we get:
1/(-2) < 2
-1/2 < 2.
YES.
Since in the first case the answer is NO, but in the second case the answer is YES, INSUFFICIENT.
The correct answer is A.
Last edited by GMATGuruNY on Mon Jun 03, 2013 4:31 am, edited 1 time in total.
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- bubbliiiiiiii
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Yes -- edited my reply above.bubbliiiiiiii wrote: Statement 2: |a-b| < 1.
I think this should be reverse?
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