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Modulus/Number line

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Modulus/Number line

by bblast » Sun Aug 21, 2011 3:24 am
Is |a|-|b| >= |a-b| ?
(1) b > a
(2) a > 0
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by sumgb » Sun Aug 21, 2011 7:39 am
--------+-------+-------+--------
a = -1; 0; b=1

stmnt 1: b > a

case I :
--------+-------+-------+--------
a = -1; 0; b=1
so |a| - |b| = 0; |a-b| = 2; hence answer NO

case II :

--------+-------+-------+--------
a = -2; b=-1; 0
so |a| - |b| = 1; |a-b| = 1; hence answer YES

2 diff. answers so insuff.
cross off A D

stmnt 2 :

case I:
--------+-------+-------+--------
0; a=1; b=2
so |a| - |b| = -1; |a-b| = 1; hence answer NO

case II :
--------+-------+-------+--------
0; b=1; a=2
so |a| - |b| = 1; |a-b| = 1; hence answer YES

2 diff answers hence insuff
cross off B

Together, we get, b > a and a,b > 0
hence |a| - |b| will always be negative and |a-b| will always be positive so only 1 answer : NO

hence suff
answer C

Hope this helps...

P.S. In the number lines drawn above please read the points at '+' (from left to right, couldnt draw better :-(
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by czarczar » Sun Aug 21, 2011 12:04 pm
bblast wrote:Is |a|-|b| >= |a-b| ?
(1) b > a
(2) a > 0
Case1: take a=3 and b =4 Case2: take a= -4 and b= -3.

from case 1 we will get: -1>=1 ans no.
from case 2 we will get: 1>=1 and yes.
not sufficient.

from equation 2 we know that a >0 we do not know anything about B . So, not sufficient.

combining we know that a >0 , so b should also be >0.

So case 1 remains.

OA:C
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