Absolute value

[Bronson]

Is 1/(a-b) < b-a ?

Stat1) a<b
Stat2) 1<!a-b!

[StarDust845]

Is the answer A?

[cris]

I think is C (together are sufficient).

What is the answer?

[jaiswaln]

The answer should be E.

1 <a> (a-b)^2


which is not a true statement. so its not possible to get an answer to the question even by using any of the option or together.

BUMMER....

What is the answer?????

[camitava]

Guys I am having some other opinion at all. The Qs is saying that -
Is 1/(b - a) < (a - b) - But this we can reformat to 1/-(a - b) <a> -1. So we have to show - Is (a - b)^2 > -1?
This we can conclude when option - 1 and option - 2 independently holds true. So IMO D. What's the OA?

[sagar]

the ans is D. both fact statements individually hold good. such problems r best solved by substitution.

[Stuart@KaplanGMAT]

Is 1/(a-b) < b-a ?

Stat1) a<b
Stat2) 1<!a-b!

(1) if a is less than b, then (a-b) is negative. So, 1/(a-b) is also negative.

Also, if a - b is negative, then b-a is positive.

So, 1/(a-b) is negative and b-a is positive.. therefore, 1/(a-b) is definitely less than b-a. Statement (1) is sufficient.


(2) if 1 is less than the abs val of (a-b), then a-b is greater than 1 or a-b is less than -1. However, this doesn't tell us the relationship between a and b, so statement 2 is insufficient.

For example, we could choose the values:

a=10 b=5, since 1 is less than 5. We then ask the original question:

is 1/(10-5) greater than (5-10)? Or is 1/5 greater than -5. Yes it is!

We could also choose the values a= -10 and b = -5 (since 1 is less than 5). We then ask the original question:

Is 1/(-10 --5) greater than (-5 --10)?

In other words, is 1/-5 greater than +5 ? No, it isn't!

Since we can get both a yes and a no, statement (2) is insufficient.

1 is sufficient, 2 isn't.. choice (A) is correct.

ps: I had to edit this post 8 times to finally get it to format correctly - any time I used a greater than or less than sign it messed everythign up.. very frustrating!

[camitava]

Hey Stuart, if I interpret the Qs

Is 1/(a-b) < b-a ?

in the following way -
1/-(b - a) < (b - a)
so -1 < (b - a)^2
So the Qs is coming like -

Is -1 < (b-a)^2 ?

what wrong I am doing? Am I missing something or doing something wrong? Stuart, looking for ur suggestion/ concern...

[gabriel]

Is 1/(a-b) < b-a ?

Stat1) a<b
Stat2) 1<!a-b!

Answer should be A .. a<b means a-b is negative which makes 1/(a-b) negative and also a<b means (b-a) is positive so 1/(a-b)<(b-a).

the second statement could mean a-b is positive or negative so it is insufficient.

[Stuart@KaplanGMAT]

You need to be very careful when manipulating inequalities.

In this question, to get to your final statement, you're multiplying both sides by (b-a). However, if b-1 turns out to be negative, you have to change the direction of the inequality, so you'd actually end up with:

Is -1 greater than (b-a)^2

to which the answer is "no".

So, if b-a is positive, then the answer will be yes. However, if b-a is negative, the answer will be no. That's another way to deal with statement (2) and decide that it's insufficient.

[beatthegmat]

Hi Stuart:

gabriel just let me know about your issues with posting. My apologies for your frustration, I'm investigating right now what the problem is with the greater then, less then symbols...

[beatthegmat]

Quick update. I ran some tests with the symbols and didn't encounter any issues. I'll keep investigating, but this might have been a fluke incident.

Again, my apologies!

[Stuart@KaplanGMAT]

Quick update. I ran some tests with the symbols and didn't encounter any issues. I'll keep investigating, but this might have been a fluke incident.

Again, my apologies!

I disabled HTML, I think that should fix it (since HTML uses < and > for tags).

[jaiswaln]

sorry, i meant the answer is D not E. since -1 > (a-b)^2. So the answer is obvious without the choices.

[beatthegmat]

Quick update. I ran some tests with the symbols and didn't encounter any issues. I'll keep investigating, but this might have been a fluke incident.

Again, my apologies!

I disabled HTML, I think that should fix it (since HTML uses <and> for tags).

That did it. Thanks for the QA work, Stuart.