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# If a # -b , is (a -b)/(b+a) < 1 ? 1) b^^2

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Manager
Joined: 17 Dec 2008
Posts: 169
If a # -b , is (a -b)/(b+a) < 1 ? 1) b^^2 [#permalink]

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01 Mar 2009, 17:18
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If $$a # -b$$, is$$(a -b)/(b+a) < 1$$?

1) $$b^^2 > a^^2$$
2) $$a - b > 1$$
Senior Manager
Joined: 08 Jan 2009
Posts: 318

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01 Mar 2009, 18:57
If a # -b, is(a -b)/(b+a) < 1?

i just solved the conditions for to know (a-b)/(a+b) < 1

on solving the above we will get 2b / a+b > 0
for this inequality to hold good either
a. both b>0 and a+b >0 (a>-b) b. both b <0 and a+b<0 (a<-b)

so let move to the stmt :

stmt 1:

b^2 -a^2 >0 therefor (b-a)(b+a) >0.
from the soln set is b>a or b>-a .
Insufficient.

Stmt 2 :
a-b > 1. Does not tell anything.Insufficient.

Combined togther stiill cannot find anything.

Is the Ans E. correct me if i am wrong.
Current Student
Joined: 28 Dec 2004
Posts: 3317
Location: New York City
Schools: Wharton'11 HBS'12

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01 Mar 2009, 19:04
If a # -b, is(a -b)/(b+a) < 1?

1) says |b|>|a| well so if i look at is it is not possible for this equation to be >1 therefore sufficient!

2) a-b>1
oK so if a is much bigger than b and b>0..then this becomes a/a>1 no but if B<0 then yes..insuff

i will go with A
Director
Joined: 25 Oct 2006
Posts: 610

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02 Mar 2009, 03:39
I too got A by plugin values in two stmts. But the process is lengthy.

1) for stmt 1 plugged value of a,b as (-5,-7), (5,7), (5,-8), (-5,7). everytime I got negative.
...suffice
2) plugged in 5,-2 ; 5,2 and -5,-7 and the stmt is insuffice.

My process is bit lenghty so I am looking to find some good method to work this kind of problem.
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If You're Not Living On The Edge, You're Taking Up Too Much Space

Director
Joined: 25 Oct 2006
Posts: 610

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02 Mar 2009, 03:56
FN wrote:
If a # -b, is(a -b)/(b+a) < 1?

1) says |b|>|a| well so if i look at is it is not possible for this equation to be >1 therefore sufficient!

2) a-b>1
oK so if a is much bigger than b and b>0..then this becomes a/a>1 no but if B<0 then yes..insuff

i will go with A

Nice method.
In stmt 2, are you assuming that a is positive? So, a is positive and greater than B. Now for both cases (b=positive/ negative) you are checking the validity.
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If You're Not Living On The Edge, You're Taking Up Too Much Space

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Joined: 17 Jun 2008
Posts: 1507

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02 Mar 2009, 08:28
1
KUDOS
I will also go with A.

From stmt1: a^2 - b^2 < 0
or, (a-b)(a+b) < 0

This means, either (a-b) > 0 and (a+b) < 0
or, (a-b) < 0 and (a+b) > 0

In either case, (a-b)/(a+b) < 0 and hence (a-b)/(a+b) < 1....sufficient.

From stmt2:
(a-b) > 1

This does not guarantee the sign of (a+b).

For example, a = 1, b = -5
or, a = 5 and b = 1.

Hence, insufficient.
Re: DS: Number properties   [#permalink] 02 Mar 2009, 08:28
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