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log2bala
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Post subject: Is IaI > IbI  Posted: Tue Nov 10, 2009 1:58 pm |
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Hello Guys
Would anyone kindly explain how to proceed further in this problem?
. Is IaI > IbI ?
a) b < - a
b) a < 0
Answer.
Taking Option B again and eliminating it as it does not have any information on BD is crossed out
Taking Option A now, How do you rephrase this inequality? b < -a
Please clarify on these types how to approach them?
Thank you guys for your help
Bala
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rchitta
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Post subject: Re: Is IaI > IbI Posted: Tue Nov 10, 2009 5:54 pm |
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Assuming IaI & IbI are 3 digit numbers with the digit in the units and hundreds place being the same, aren't they actually asking:
If a < b internally ?
example 121 and 131 should be the same as 2 < 3?
also, 151 and 141 should be same as 5 < 4 ?
1) b < -a
=> a + b < 0 (INSUFFICIENT) not sure where a and b line on the integer line.
i.e, is a +ve, b -ve? or
is a -ve & b +ve? or
is a -ve & b -ve?
So, we cannot determine if a < b.
2. a < 0 (same thing), we have no information about b (INSUFFICIENT)
combining 1 & 2
a + b < 0 and a < 0
this again is INSUFFICIENT, I think. b could be -ve or b could be just +ve making a + b still < 0
So, I would say the answer is E
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rchitta
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Post subject: Re: Is IaI > IbI Posted: Wed Nov 11, 2009 2:01 pm |
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Oops, I think I mis-read the question. I will try to solve it again assuming IaI means absolute value of a and not Integer A Integer. Sorry about the mis-lead.
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rchitta
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Post subject: Re: Is IaI > IbI Posted: Sun Nov 22, 2009 11:37 pm |
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Sorry, I completely forgot about this...
This is what I think...
1. b < -a implies that a + b < 0 (you get this when you add a to either sides of the inequality).
This implies either,
i) both a and b are -ve (cannot say if |a| > |b| here because we are not sure if b is more -ve (far down on the number line from 0) or if a is more -ve. e.g; a = -1 and b = -2, a = -2, b = -1) --> THIS PROVES INSUFFICIENCY.
ii) a is +ve and b -ve: (I'm sure you could come up with some a, b value here)
iii) a is -ve and b is +ve: (I'm sure you could come up with some a, b value here)
2. if a < 0, this doesn't say anything about b and b either could be more -ve or less -ve than a (same thing as above ) INSUFFICIENT.
Combining both: a + b < 0 and a < 0 doesn't tell us much either. either b could be more -ve or less -ve than a. Hence both statements are insufficient together. D
Hope this helps!
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rchitta
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Post subject: Re: Is IaI > IbI Posted: Sun Nov 22, 2009 11:40 pm |
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Sorry again! I meant E is the answer choice: Statements 1 and 2 TOGETHER are NOT sufficient
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esledge
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Post subject: Re: Is IaI > IbI Posted: Tue Feb 02, 2010 1:14 pm |
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Location: St. Louis, MO
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I take pretty much the same pos/neg case approach as rchitta. To save time, make a chart of the scenarios. I actually started with a chart that had 4 possibilities:
a..........b
pos.....pos
pos.....neg
neg.....pos
neg.....neg
Then I add columns for checking which scenarios are allowed by the statement (esp. statement 1: we rule out the pos/pos scenario with that one) and for answering.
At the end, my chart looked something like this:
a...........b......b < -a...........Interpret...............|a| > |b|?
pos......pos.....imposs............(blank)
pos......neg.....OK..............a is closer to 0.............NO
neg.....pos....OK..............b is closer to 0.............YES
neg.....neg....OK..........either a or b closest to ....MAYBE
Bold text indicates what I circled on paper. The chart is for (1), the circle is for (2). Together, we still get a MAYBE result.
_________________
Emily Sledge
Instructor
ManhattanGMAT
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yousuf_azim
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Post subject: Re: Is IaI > IbI Posted: Sun Sep 04, 2011 1:08 am |
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JohnHarris
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Post subject: Re: Is IaI > IbI Posted: Mon Sep 05, 2011 12:05 am |
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Quote: . Is IaI > IbI ?
a) b < - a
b) a < 0
Try a = -1, b = -100. (a) is satisfied, (b) is satisfied, (a) and (b) are satisfied. The inequality is not true. Neither (a), nor (b), nor (a) & (b) are sufficient. What (a) says is that b is to the right of -a on the number line but makes no restrictions on the magnitude of b. Case (b) ignores b and thus makes no restrictions on the magnitude of b. That is, choose any negative a [satisfies (b)] and any b < a. Then b is also negative and, since -a is positive, b < -a [satisfies (a)]. However, -b > -a [multiply previous by minus one], i.e. |b| > |a|. Thus (a) & (b) are not sufficient.
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jnelson0612
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Post subject: Re: Is IaI > IbI Posted: Mon Oct 10, 2011 9:28 pm |
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The answer is E.
_________________
Jamie Nelson
ManhattanGMAT Instructor
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