Data Sufficiency.

[shekhard18]

If ab ≠ 0 and a + b ≠t 0, is 1/a+b < 1/a + 1/b ?

(1) |a| + |b| = a + b

(2) a > b

Is there way that we could solve this using the method of "picking number"? For some reason every time I try doing it. I'm getting stuck.

Thank You.

Regards,
Shekhar.

[RonPurewal]

Picking numbers is a good idea. With the first statement, it's mostly a matter of picking numbers until you notice the kinds of numbers that actually satisfy that statement.

Why don't you tell us a little more about one of your attempts? I.e., what numbers did you try to pick? What did you do with them? Etc.
If I just show you a solution with numbers that I'm picking myself, that will be of little use to you. Remember, this is a test on which you have to come up with such things yourself.

Thanks.

[tomada36]

If ab ≠ 0 and a + b ≠t 0, is 1/a+b < 1/a + 1/b ?

(1) |a| + |b| = a + b

(2) a > b

Is there way that we could solve this using the method of "picking number"? For some reason every time I try doing it. I'm getting stuck.

Thank You.

Regards,
Shekhar.

I chose (A) as the answer, that Statement #1 is sufficient by itself, but Statement #2 is insufficient by itself.

The only way that Statement #1 can be true under all circumstances is if a>0 and b>0. Once this is realized, any selections for 'a' and 'b' will show that 1/a + 1/b > 1/(a+b).
SUFFICIENT

Statement #2 "screams" - to me, at least - "test positive values as well as negative values for 'a' and 'b'."

Looking at the work done for Statement #1, we already know that the answer to the question is yes when a>0 and b>0, and this won't change whether a>b or b>a.

However, when a<0 and b<0 (which is possible without looking at Statement #1), we can suppose that a= -1 and b= -2, which maintains the inequality in Statement #2.

Using these values for 'a' and 'b'...

1/a = 1/(-1) = -1
1/b = 1/(-2) = -1/2
1/a + 1/b = -1 - 1/2 = -3/2

1/(a+b) = 1/(-1-2) = -1/3

For this scenario, 1/a + 1/b is less than 1/(a+b), which yields a no as the answer to the question.

Thus, for Statement #2, the answer to the question depends on whether we select positive or negative values for 'a' and 'b'.
INSUFFICIENT

[RonPurewal]

tomada36,
Yeah, that is pretty much how it works.

What's interesting is that you didn't show your reasoning for the interesting part of the problem, i.e., this part:

The only way that Statement #1 can be true under all circumstances is if a>0 and b>0. Once this is realized, any selections for 'a' and 'b' will show that 1/a + 1/b > 1/(a+b).
SUFFICIENT

To me, the red statement is not immediately obvious. I'd establish it by realizing that either 1/a or 1/b, alone, is already bigger than 1/(a + b), since a + b is bigger than either individual component (and everything is positive). If I add two things that are both individually bigger than something, then the sum will stay bigger than that something.

You did mention "any selections", so maybe you tossed in a bunch of numbers (including weird things like small fractions/decimals, whose behavior is often different from that of other positive numbers) until you were convinced that the result would always be the same. Or maybe you just thought it was so easy that it wasn't worth mentioning. (: