Is |a|-|b| >= |a-b| ?
(1) b > a
(2) a > 0
Modulus/Number line
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- bblast
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Quant 47-Striving for 50
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- sumgb
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--------+-------+-------+--------
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
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
- czarczar
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Case1: take a=3 and b =4 Case2: take a= -4 and b= -3.bblast wrote:Is |a|-|b| >= |a-b| ?
(1) b > a
(2) a > 0
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