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Re: from online Manhattan drill [#permalink]
03 Dec 2011, 04:01

What was your answer, and why?

I get E.

From 1: 2a – b < 7 AND |2a – b| < 7 WHERE 2a -b > -7 2a – b < 7 BUT |2a – b| > 7 WHERE 2a -b < -7 INSUFFICIENT

to confirm this you can find values of a & b that satisfy the condition 2a – b < 7 but provide different answers for |2a – b| < 7 ? A: a = 3, b = 0 2a - b = 6 - 0 = 6 6 < 7 and |6| < 7 B: a = -10, b = 1 2a - b = -20 - 1 = -21 -21 < 7 but |-21| > 7

From 2 you can refactor the original question: Is |2(b + 3) - b| < 7? => is |b + 6| < 7? This is true for -13 < b < 1 but false for value outside of this range. Again INSUFFICIENT.

Together, we can further combine in a similar manner as above to see that b + 6 < 7 so we find that b < 1, but we still don't know if b > -13.

So together is INSUFFICIENT.

Again, to see that this is the case, you can substitute numbers that satisfy both 1 & 2, but give different answers to the main question:

A: a = 3, b = 0 2a – b < 7 : 2a - b = 6 - 0 = 6 < 7 a = b + 3 : 3 = 0 + 3 |2a – b| < 7 ? : |6| < 7 TRUE

B: a = -20, b = -17 2a – b < 7 : 2a - b = -40 - (-17) = -23 < 7 a = b + 3 : -20 = -17 + 3 |2a – b| < 7 ? : |-23| < 7 FALSE

See there lies your fallacy.. Both can be possible so it is NOT possible ( SUFFICIENT) to answer conclusively. Had there been some other condition outlining that 2a-b is always greater than 0 , then it would have been SUFFICIENT. Right now as it is stands , it is ambiguous.

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