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If a and b are integers, and |a| > |b|, is a |b| < a

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Manager
Joined: 01 Feb 2006
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If a and b are integers, and |a| > |b|, is a |b| < a [#permalink]

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22 Nov 2006, 13:25
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If a and b are integers, and |a| > |b|, is a Â· |b| < a â€“ b?

(1) a < 0

(2) ab >= 0

Ans : E
VP
Joined: 28 Mar 2006
Posts: 1372
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Re: MGMAT: DS : Inequalities [#permalink]

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22 Nov 2006, 17:20
sujayb wrote:
If a and b are integers, and |a| > |b|, is a Â· |b| < a â€“ 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 Â· |b| < a â€“ 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 Â· |b| < a â€“ 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 Â· |b| < a â€“ 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 Â· |b| < a â€“ 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
Re: MGMAT: DS : Inequalities   [#permalink] 22 Nov 2006, 17:20
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