If a # -b , is (a -b)/(b+a)

Question

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

tkarthi4u · Answer

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.

FN · Answer

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

priyankurml · Answer

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.

priyankurml · Answer

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.

scthakur · Answer

I will also go with A.

From stmt1: a^2 - b^2 0 and (a+b) 0

In either case, (a-b)/(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.

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

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