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# If a and b are integers, and |a| > |b|, is a |b| < a

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Manager
Joined: 09 Jul 2008
Posts: 112
Location: Dallas, TX
Schools: McCombs 2011
If a and b are integers, and |a| > |b|, is a |b| < a [#permalink]

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04 Jan 2009, 19:17
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

If a and b are integers, and |a| > |b|, is a · |b| < a – b?

(1) a < 0

(2) ab >= 0

Interested in seeing efficient approach for this. My approach of trying every combination takes way too long. Thanks!
Intern
Joined: 30 Sep 2008
Posts: 36
Re: Absolutes and Inequality confusion [#permalink]

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04 Jan 2009, 21:02
Is it the OA is E ?

I tried the no. also but when see the problem like this
I usaully assume that the no. in absolute is zero
Manager
Joined: 09 Jul 2008
Posts: 112
Location: Dallas, TX
Schools: McCombs 2011
Re: Absolutes and Inequality confusion [#permalink]

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04 Jan 2009, 21:31
GMATpp wrote:
Is it the OA is E ?

I tried the no. also but when see the problem like this
I usaully assume that the no. in absolute is zero

Not sure if I understand your method.. Can you elaborate. Do you assume a and b = 0 ??
Intern
Joined: 30 Sep 2008
Posts: 36
Re: Absolutes and Inequality confusion [#permalink]

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04 Jan 2009, 22:34
I assume B = 0 according to l a l > l b l
so it would be easy to think about value of a.

then try the numbers
SVP
Joined: 17 Jun 2008
Posts: 1548
Re: Absolutes and Inequality confusion [#permalink]

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04 Jan 2009, 23:06
I also got E with the following approach:

From the question |a| > |b| or, a^2 > b^2
or, (a-b)(a+b) > 0 and this means, either a > b and a > -b or, a < b and a < -b

Now, a > b and a > -b is possible only if a and b are both positive
and a < b and a < -b is possible only if a < 0.

Now, if a > 0 then a.|b| > 0 and a-b > 0 but, a.|b| < a-b may not be true.
Similarly, if a < 0 then both a.|b| and a-b will be < 0 but again, inequality may not be true.

Now stmt1 does not give any extra information. Insufficient.
Stmt2 also does not give any extra information. Insufficient.
Intern
Joined: 19 Jun 2008
Posts: 20
Re: Absolutes and Inequality confusion [#permalink]

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07 Jan 2009, 11:36
scthakur wrote:
I also got E with the following approach:

From the question |a| > |b| or, a^2 > b^2
or, (a-b)(a+b) > 0 and this means, either a > b and a > -b or, a < b and a < -b

Now, a > b and a > -b is possible only if a and b are both positive
and a < b and a < -b is possible only if a < 0.

Now, if a > 0 then a.|b| > 0 and a-b > 0 but, a.|b| < a-b may not be true.
Similarly, if a < 0 then both a.|b| and a-b will be < 0 but again, inequality may not be true.

Now stmt1 does not give any extra information. Insufficient.
Stmt2 also does not give any extra information. Insufficient.

When you say that a > b and a > -b is possible only if a and b are both positive, does that mean in all cases or only in some? If a =2 and b =-1 it holds true as well.
Re: Absolutes and Inequality confusion   [#permalink] 07 Jan 2009, 11:36
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# If a and b are integers, and |a| > |b|, is a |b| < a

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