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| OG - QUANT REVIEW PS - #114 |
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givemeanid
Guest
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Is 1/(a-b) < (b-a)?
1) a < b a-b < 0 1/(a-b) < 0 b-a > 0 Sufficient. 2) 1 < |a-b| |a-b| < 1 -1 < a-b < 1 -1 < b-a < 1 So, one can be +ve and the other -ve. Cannot tell which one is which. NOT sufficient. Answer is A. What is OA? |
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givemeanid
Guest
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Typo in the previous post. If this post also has a typo, please beat me back to senses.
Is 1/(a-b) < (b-a)? 1) a < b a-b < 0 1/(a-b) < 0 b-a > 0 Sufficient. 2) 1 < |a-b| |a-b| > 1 a-b < -1 or a-b > 1 b-a > 1 or b-a < -1 When a-b < -1, b-a > 1. When a-b > 1, b-a < -1. Both of those are sufficient. Answer is D. What is OA? |
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givemeanid
Guest
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You know, trying to solve DS at work is a BAD idea. BAD BAD idea.
When a-b < -1, b-a > 1. True. When a-b > 1, b-a < -1. False. Insufficient. Answer should be A. I am done. This is it. Rome is burning, my friends. |
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Stacey Koprince
MGMAT STAFF
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We've all had days like that, givemeanid. :)
You've hit on the issue, though - when multiplying or dividing an inequality by a negative, we have to switch the sign. We're not told whether the quantity (a-b) is pos or neg, so we have to check this both ways. IF a-b is pos, then 1 < (a-b)(b-a) (and you can simplify further from there) IF a-b is neg, then 1 > (a-b)(b-a) (and again you can simplify further from here) But the point is that I have to follow both possibilities through and any particular statement is only sufficient if BOTH equations give me the SAME definitive answer, yes or no. I'll stop there - try this again and come back if you have more questions. |
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| Different approach |
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Guest
Guest
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Is 1/(a-b) < b - a?
Rephrase: 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| This statement mentions |a-b| and nothing about (b-a) (b-a) can be +ve or -ve. Therefore, statement (2) is not sufficient. |
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| OG - QUANT REVIEW PS - #114 |
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