Teasing my brain- need help

[johny01]

Hi,

If a ≠-b, is a-b/b+a <1?
1.b2 >a2
2.a — b > 1
Can you solve this. This question is from http://TestCircle.com I am being used this website for preparing my GMAT?

Thanks!

[eravi11]

Statement 1 yields: Either b > a or b < -a
Now a-b/a+b <1 can be rewritten as [1-(b/a)]/[1+(b/a)]<1 Putting values it can be proven that statement 1 is insuffiicient.

Using Statement 2 we can prove that a-b/a+b <1 or a-b/a+b>1 Hence INSUFFICIENT

Combining 1 and 2 we have b<-a and a-b>1 INSUFFICIENT.

I think it is E..What is OA?

[spacepure]

(a-b)/(b+a) <1
<=> (a-b)/(b+a) -1<0
<=> (-2b)/(a+b)<0
<=> b>0

Thus, it has nothing to do with 1 and 2

[GMATQuantCoach]

Audia8 is absolutely right. The answer is A.

[audia8]

I go with A:
we want to find if (a-b)/(b+a) < 1

1) b^2 > a^2
a^2-b^2<0=>(a+b)(a-b)<0=>(a+b),(a-b) must be one positive(+), one negative(-), therefore (a-b)/(b+a)<0 =>(a-b)/(b+a)<1 sufficient

2) a-b>1 insufficient

[india_chintan]

Bow wow wow yippee yo yippee yay!!!!11 audia. i had a different explanation for the same but then the answer is same. it must be A.

[wakajoola]

Yes, the answer is A. As the first statement is sufficient if we put an imaginary number to it like b=2 and a=1 then we get the desired answer.

While in statement 2, since a-b>1 cannot guarantee <1.

[sonikamadala]

Originally Posted by johny01

Hi,

If a ≠-b, is a-b/b+a <1?
1.b2 >a2
2.a — b > 1
Can you solve this. This question is from http://TestCircle.com I am being used this website for preparing my GMAT?

Thanks!

Hi IMo answer is B.

Here is my explanation..

Ploblem is easy if we substitute samll numbers.

consider FS1:
------------
given b2 >a2 => b>a

if b = 2, a=1 then a-b/b+a <1
if b= 2 , a= -3 then a-b/b+a >1

so FS1 alone is not sufficient.


Consider FS2:
-------------
given a — b > 1 => a > 1+b

for any values of a and b, which satisfy this condition the value of
a-b/b+a is >1

Ex : b= -10, a = -8
b = 1, a= 3
b = -10 a= 8 etc

so FS2 alone is sufficient.

So answer is B.

I hope this is clear.

[rpi_dude09]

I concur with Audia 8. Answer is A.
Cond 1 yields: (a+b)(a-b)<0 Hence one negative and one positive term. Ratio will always be negative i.e. <0 . Hence the ratio will always be <1 (sufficient)
Cond 2: Not sufficient.

[sonikamadala]

Originally Posted by sonikamadala

Hi IMo answer is B.

Here is my explanation..

Ploblem is easy if we substitute samll numbers.

consider FS1:
------------
given b2 >a2 => b>a

if b = 2, a=1 then a-b/b+a <1
if b= 2 , a= -3 then a-b/b+a >1

so FS1 alone is not sufficient.


Consider FS2:
-------------
given a — b > 1 => a > 1+b

for any values of a and b, which satisfy this condition the value of
a-b/b+a is >1

Ex : b= -10, a = -8
b = 1, a= 3
b = -10 a= 8 etc

so FS2 alone is sufficient.

So answer is B.

I hope this is clear.

small correction :
for any values of a and b, which satisfy this condition the value of
a-b/b+a is <1 not a-b/b+a is >1. Mistakenly written.