Abs: is a*|b| < a-b ?

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Abs: is a*|b| < a-b ?

by euro » Tue Nov 02, 2010 1:34 am
If 'a' and 'b' are integers, and |a| > |b|, is a*|b| < a-b ?

(1) a < b

(2) a*b >= 0

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by vaibhavtripathi » Tue Nov 02, 2010 3:15 am
euro wrote:If 'a' and 'b' are integers, and |a| > |b|, is a*|b| < a-b ?

(1) a < b
(2) a*b >= 0
'|a|>|b|' also means that '|a|>0', since minimum value for |b| can be 0 and |a| has to be greater than that.

(1). '|a|>|b|' and 'a<b' (where '|a|>0') means 'a' is negetive; 'b' however could be negetive, zero, or positive. Not Sufficient
(2). 'a*b>=0' tells us that 'a' and 'b' are either both positive or both negetive, or 'b' is 0. Not Sufficient

Taking both statemenst now:
From (1) We know that 'a' is negetive. Putting this knowledge in (2), 'b' must either be negetive or 0.
Lets plug a=-2, and b=-1 in a*|b| < a-b
LHS = -2*|-1| = -2
RHS = -2-(-1) = -1
So LHS<RHS

Lets plug a=-2, and b=0 in a*|b| < a-b
LHS = -2*0 = 0
RHS = -2-0 = -2
So LHS>RHS

Both together not sufficient. E
Thanks,
Vaibhav

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by goyalsau » Tue Nov 02, 2010 10:41 am
Even i did the same way as vaibhav did,
But that was a very long and confusing way,
Can't we do it any other way,
Saurabh Goyal
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EveryBody Wants to Win But Nobody wants to prepare for Win.

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by diebeatsthegmat » Wed Nov 03, 2010 5:42 am
euro wrote:If 'a' and 'b' are integers, and |a| > |b|, is a*|b| < a-b ?

(1) a < b

(2) a*b >= 0
i think the answer is E


|a|>|b| and a<b
this could only happen when a<0 and b>=o
if a= -3 b= 2 so a|b|=-6<-3-2=-5
if a=-3 and b=0 so 0 >-3
insufficient

+ |a|>|b| and a*b>=0 this could happen only when a<0 and b<0 and a must <b for sure
for example: a=-3 nd b=-2 so |-3|>|-2| ( yes) and -3*-2>0
so -3*|-2|=-6<-5 ( as A)
and that conditions only happen when a>0 and b>0 and a>b
for example a=3 and b=2 so |3|>|2| and 3*2>0
and 3*2=6>3-2=1
when b=0 and a<0 or a>0 the condition |a|>|b| with all number of a and b=0 and a*b>=0
if a=3 b=0 so 3*|0|=0< 3-0 ( the same as A)
but if a =-3 so -3*0=0>-3 (the same as A)
insufficient
1+2 / is not sufficient
E

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by fskilnik@GMATH » Thu Nov 04, 2010 9:58 am
euro wrote:If 'a' and 'b' are integers, and |a| > |b|, is a*|b| < a-b ?

(1) a < b

(2) a*b >= 0
The answer is really "E"... let´s look into a more "analytical/technical approach" taking into account one of my 'golden rules' for Data Sufficiency: "simplify the question stem and/or the statements whenever possible". This is particularly useful here, because inequalities and modulus are (in general) difficult to deal with, especially when he have little info on the signs of each individual parameter/variable!!
vaibhavtripathi wrote: '|a|>|b|' also means that '|a|>0', since minimum value for |b| can be 0 and |a| has to be greater than that.
Excellent, therefore we know, for sure, that a is non-zero!

Now let us simplify the question stem dividing in two cases:

(*) If b is null, the question is equivalent to: is "a positive?" (verify that)!

(**) If b is NOT null, the question is equivalent to (verify that!):

> "Is b less than a.(1-b) ?" If b is positive ;

> "Is b less than a.(1+b) ?" If b is negative ;


That´s all we need to work pretty quick from here... have a look:

(1) a< b

BIFURCATES, therefore insufficient:

> Take b null , then a is certainly negative (why?), therefore from equivalence (*) we answer in the NEGATIVE ;

Practical example: b = 0 and a = -1.

> Take b negative, then (from |a|> |b|) we are sure (think geometrically on the number line) that a < b < 0, therefore the equivalence (**) is certainly satisfied (answering in the POSITIVE), because the LHS is negative and the RHS is positive or null (null only when b is equal to -1, do not forget that a and b are integers!).

Practical example: b = -1 and a = -2.


(2) ab is non-negative

BIFURCATES with exactly the same cases explored in statement (1)!

(1+2) BIFURCATES immediately (why?), therefore "E".

Best Regards,
Fabio.
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
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