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MBA Applicant 2007/8
 
 

OG - QUANT REVIEW PS - #114

by MBA Applicant 2007/8 Sun Jul 22, 2007 1:28 pm

Is 1/(a-b) < b - a?

1) a < b
2) 1 < |a-b|

Step 1: Rephrase the question:

Is (a - b)^2 < -1?

Step 2
Using,
AD
BCE

Step 3:
(1) - a -b <0 ---> (a - b)^2 is always positive and therefore the answer is NO--- SUFF
(2) if a -b >0 , then a -b >1----> which also means (a-b)^2 >0 and hence NOT less than -1--- NO
if a-b <0, a - b < -1 ------> which means that (a-b)^2 >0 and hence NOT less than -1 --- NO ===SUFF

But the answer is incorrect. Can u please point out the flaw in my reasoning and suggest an improved approach? I didnt like the explanation in OG.

Thank you in advance
givemeanid
 
 

by givemeanid Mon Jul 23, 2007 10:32 am

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?
givemeanid
 
 

by givemeanid Mon Jul 23, 2007 10:38 am

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?
givemeanid
 
 

by givemeanid Mon Jul 23, 2007 11:07 am

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.
StaceyKoprince
ManhattanGMAT Staff
 
Posts: 9360
Joined: Wed Oct 19, 2005 9:05 am
Location: Montreal
 

by StaceyKoprince Mon Jul 23, 2007 10:46 pm

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.
Stacey Koprince
Instructor
Director, Content & Curriculum
ManhattanPrep
Guest
 
 

Different approach

by Guest Tue Jul 24, 2007 12:51 pm

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.