sujayb wrote:
If a and b are integers, and |a| > |b|, is a · |b| < a – b?
(1) a < 0
(2) ab >= 0
From premise |a| > |b| the value of a > value of b (here we are not taking the positive or negetive but only the intrinsic value)
From(1) a is negetive
So lets take a=-2 and b=-1 in a · |b| < a – b
-2*|-1| < -2 -(-1) which is -2<-1 which is true
lets take a = -2 and b=1 we can take this as |-2| > |1|
-2*|1| < -2 -(1) whihc is -2 < -3 which is false
So A ruled out
From (2) ab>=0
means a,b are both same sign either +ve or negetive
take values a=-2 and a=-1 in a · |b| < a – b
-2*|-1| < -2 -(-1) which is -2<-1 which is true
but if a=2 and b=1
2*|1| < 2 -(1) which is 2<1 which is false...
So (2) ruled out
Combining both yields nothing...
E stands
Ans : E
From premise |a| > |b| the value of a > value of b (here we are not taking the positive or negetive but only the intrinsic value)
From(1) a is negetive
So lets take a=-2 and b=-1 in a · |b| < a – b
-2*|-1| < -2 -(-1) which is -2<-1 which is true
lets take a = -2 and b=1 we can take this as |-2| > |1|
-2*|1| < -2 -(1) whihc is -2 < -3 which is false
So A ruled out
From (2) ab>=0
means a,b are both same sign either +ve or negetive
take values a=-2 and a=-1 in a · |b| < a – b
-2*|-1| < -2 -(-1) which is -2<-1 which is true
but if a=2 and b=1
2*|1| < 2 -(1) which is 2<1 which is false...
So (2) ruled out
Combining both yields nothing...
E stands
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