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MGMAT - abs again !

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by jayhawk2001 » Sun May 13, 2007 3:13 pm
I spent too much time on this question on the MGMAT test - worse
I got the answer wrong. Any quick ways to solve this ?

OA after a few reply.

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

(1) a < 0

(2) ab >= 0
Last edited by jayhawk2001 on Mon May 14, 2007 6:58 am, edited 1 time in total.
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Source: — Data Sufficiency |
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by f2001290 » Mon May 14, 2007 1:15 am
What is the second statement?
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by jayhawk2001 » Mon May 14, 2007 6:56 pm
f2001290 wrote:What is the second statement?
Oops. cut-n-paste error. I've corrected option 2 above.
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by f2001290 » Wed May 16, 2007 5:51 am
Jay

Is the answer E. Check for (a,b) = (-5,0) and (-5,-3)

These points satisfy both statements but give different answers.

But we should find out systematic approach to solve this.
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by Cybermusings » Wed May 16, 2007 8:23 am
If a and b are integers, and |a| > |b|, is a · |b| < a – b?

(1) a < 0

(2) ab >= 0

Absolute value of a > Absolute value b.....

Statement I: LHS of the statement will always be negative (negative*positive) ; if a =-8;b=-5 a*IbI = -40; and a-b = -8+5 = -3
If a=-8 and b=5; a*IbI = -40 and a-b = -8-5 = -13...Hence A is sufficient

ab>=0

Statement II: One thing is clear a cannot be equal to zero else it's absolute value won't stay more than b...Both can either be positive or both can be negative...If both are positive say a = 6 and b = 5 ; theh a*IbI = 30 and a-b = 1

If a = -8 and b = -5; then a*IbI = -40 and a-b = -8-(-5) = -3...Hence Statement II is insufficient....

I think it should be A...
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by RAGS » Wed May 16, 2007 8:52 am
1 alone
lets try by putting values
let a=-9 b=1 so ans is no
let a=-9 b=-1 so ans is yes hence insufficient

2. alone
let a=9 b=1 so ans is no
let a=9 b=0 so ans is yes hence insufficient

1 & 2 together
let a=-9 b=-1 ans is yes
let a=-9 b=0 ans is no
so insufficient

hence E
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Re: MGMAT - abs again !

by gviren » Thu May 24, 2007 6:53 pm
I think the answer is A

|a| > |b| so a > b and -a < -b
by 1) a < 0 so -a < -b

pick a number a = -5 and b = -3
a.|b| = -15
a-b = -5 + 3 = -2 so it is true and will be true for all the values of -a < -b

By 2) we ab > 0 which is true for both a > b and -a < -b
pick number a= 5 b=3
a.|b| = -15
a-b = 5 - 3 = 2
-15 < 2 true

pick another number a = -5 b=-3
a.|b| = 15
a-b = -5 + 3 = -2
15 < -2 false
so (2) doesn't satisfy

Jay and others - I am kind of weak in mod, please let me know if my approach is correct
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