|
Author |
Message |
|
TAGS:
|
|
|
Manager
Joined: 09 Jul 2008
Posts: 111
Location: Dallas, TX
Schools: McCombs 2011
Followers: 1
Kudos [?]:
5
[0], given: 1
|
If a and b are integers, and |a| > |b|, is a |b| < a [#permalink]
 04 Jan 2009, 19:17
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: 37
Followers: 0
Kudos [?]:
0
[0], given: 1
|
Re: Absolutes and Inequality confusion [#permalink]
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: 111
Location: Dallas, TX
Schools: McCombs 2011
Followers: 1
Kudos [?]:
5
[0], given: 1
|
Re: Absolutes and Inequality confusion [#permalink]
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: 37
Followers: 0
Kudos [?]:
0
[0], given: 1
|
Re: Absolutes and Inequality confusion [#permalink]
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: 1592
Followers: 7
Kudos [?]:
132
[0], given: 0
|
Re: Absolutes and Inequality confusion [#permalink]
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: 21
Followers: 0
Kudos [?]:
8
[0], given: 0
|
Re: Absolutes and Inequality confusion [#permalink]
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
|
|
|
|
|
|
|
|
|
Similar topics |
Author |
Replies |
Last post |
|
Similar
Topics:
|
 |
|
|
If a and b are integers, and |a| > |b|, is a |b| < a
|
sujayb |
1 |
22 Nov 2006, 13:25 |
 |
|
|
If a and b are integers, and |a| > |b|, is a |b| < a
|
anindyat |
11 |
26 Dec 2006, 19:36 |
|
|
|
|
If a and b are integers, and |a| > |b|, is a |b| < a
|
ArvGMAT |
2 |
24 Jun 2007, 16:28 |
|
|
|
|
If a and b are integers, and |a| > |b|, is a |b| < a
|
chineseburned |
3 |
06 Apr 2008, 08:13 |
|
|
|
 |
If a and b are integers, and |a| > |b|, is a |b| < a
|
gmatnub |
6 |
29 Jun 2008, 22:09 |
|
|
|
|
|
|
close
