DS question with p and r

[erin.m.mcelroy]

Can you please link me to mba.com test explanation for the follow DS question:

Is 1/p > r/ (r^2 + 2)?
1) p = r
2) r>0

Thank you!

[RonPurewal]

Can you please link me to mba.com test explanation for the follow DS question:

Is 1/p > r/ (r^2 + 2)?
1) p = r
2) r>0

Thank you!

fyi, for future posts -- you're supposed to give the answer to the problem when you post it.

the key to this question is to realize that you can't "cross-multiply" and you can't multiply by p, because we aren't sure whether p is positive or negative. ("cross multiplying" is really just multiplication by both denominators; you don't have to think about this when you are solving equations, but you definitely do have to think about this when you are solving inequalities!)
therefore, you have to treat the inequality in the prompt the way that it's currently written.

STATEMENT (2) INDIVIDUALLY is pretty clearly insufficient, since p (and therefore 1/p) is completely unrestricted and thus can have pretty much any value you want.

STATEMENT (1) INDIVIDUALLY: this is where most people go wrong with this problem, because "cross multiplying" (which, again, is not allowed here because we don't know signs) will lead you to think that this statement is individually sufficient.
you can't cross-multiply, but you can substitute 'r' for 'p':
the question becomes --> Is 1/r > r/(r^2 + 2)?
let's just try a couple of r's:
if r = 1 --> Is 1 > 1/3? YES
if r = -1 --> Is -1 > -1/3? NO
insufficient.

TOGETHER:
we can still make the same substitution
--> Is 1/r > r/(r^2 + 2)?
this time, though, we can now multiply by the product of the denominators, since the given information is now sufficient to establish that r is positive. (there's no issue with r^2 + 2, as that expression must always be positive regardless of the sign of r.)
if we multiply by both denominators, we get
Is r^2 + 2 > r^2?
the answer to this is yes -- the left-hand side is always greater by exactly 2 -- so, sufficient.

ans (c)

[mba012012]

Thanks RP !

[RonPurewal]

Thanks RP !

sure