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This question was posted in another thread just an hour ago, here is my reply to that post.

Is \frac{1}{p}>\frac{r}{r^2 + 2}?

(1) p=r --> \frac{1}{r}>\frac{r}{r^2+2}? --> as r^2+2 is always positive, multiplying inequality by this expression we'll get: \frac{r^2+2}{r}>r? --> r+\frac{2}{r}>r? --> \frac{2}{r}>0?. This inequality is true when r>0 and not true when r<0. Not sufficient.

(2) r>0. Not sufficient by itself.

(1)+(2) r>0, \frac{2}{r}>0. Sufficient.

Answer: C.

tejal777 again when you simplifying \frac{1}{p} >\frac{r}{r^2 +2} to r^2 +2 > pr, you are making a mistake. You can multiply inequality by r^2 +2 as this expression is always positive, BUT you can not multiply inequality by p, as we don't know the sigh of p.

p < \frac{r^2+2}{r}, by taking the reciprocals on each side (The inequality sign switches here, note that!)

Statement 1: p = r

This means the RHS becomes: \frac{p^2 + 2}{p} = p + \frac{2}{p}. Note that p + \frac{2}{p} will be greater than p (you're adding the term 2/p to it) as long as p > 0. If p < 0, then ultimately, you're subtracting a number from p, and hence it'll be lesser. Since there are two cases here, this is insufficient by itself.

So we are down to B, C and E.

Statement 2: r > 0

Does this say anything at all about the relationship between p and r? No. Insufficient.

So, if you combine the statements, then p = r and r > 0, which means we've answered the only condition we had while solving the first statement and hence the statements together are sufficient.

I think that the answer is B, but the answer is C. Even if we consider that y <0 in the 2nd case, after substituting x=y, still y^2 is going to be positive. I am confused. Please explain

That tells us no information about x, so it's not sufficient by itself.

Statement #2: x = y

Let's say x = y = 1. Then the left is 1, the right side is 1/(1^2 + 3) = 1/4, and the left side is bigger.

BUT, if x = y = -1, then the left side is -1, and the right side is -1/4, and --- here's one of the really tricky things about negatives and inequalities --- the "less negative" number -1/4 is greater than -1, so the right side is bigger. It may be less confusing to think about that in terms of whole numbers ---- for example, 10 > 5, but -5 > -10: it's better to have $10 in your pocket rather than $5 in your pocket, but it's better to be $5 in debt than $10 in debt. Does that make sense?

You are perfectly right --- y^2 is positive whether y is positive or negative, and therefore the denominator (y^2 + 3) is the same whether y is positive or negative, but what's different are whether the fractions themselves are negative, and that's what can reverse the order of the inequality.

Without knowing whether x & y are positive and negative, we cannot determine the direction of the inequality. Statement #2 by itself is not sufficient

Combined: y > 0 AND x = y

Now, we are guaranteed that the fractions are both positive, so multiplying by x or y will not reverse the order of the inequality. Because x = y, we have (1/y) > y/(y^2+3). Cross-multiplying, we get y^2 + 3 > y^2, which is always true. Together, the statements are sufficient. Answer choice = C.

Does that make sense? Please let me know if you have any further questions.

1. y>0. No info on x, insuff. 2. x=y. Substituting 1/y > y^2/(y^2 + 3). However, you can only multiply both sides and not change the inequality if x=y=positive. If negative, inequality changes. Insuff.

Together, it is given that y is positive. So, x=y=positive. So, multiplying both sides with no change in inequality is possible. This implies 1 > y^2/(y^2+3). LHS is obviously < 1, so this inequality is true. Suff. C
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DS - If negative answer only, still sufficient. No need to find exact solution. PS - Always look at the answers first CR - Read the question stem first, hunt for conclusion SC - Meaning first, Grammar second RC - Mentally connect paragraphs as you proceed. Short = 2min, Long = 3-4 min

C. I dont see why you need (2) to solve the problem. Using (1) alone, you know that r^2+2 must be positive so the inequality sign does not change when we simplify to (r^2+2)/r > r.

If r>0, than r^2+2 > r^2 which can be simplified further to 2>0, this is TRUE.

If r<0 than r^2+2 < r^2 which can be simplified further to 2 < 0 which impossible so r must be greater than 0

What am i doing wrong?

if P=0 then R=0. Zero divided by anything is not defined so we cannot solve the equation when p=0.

Re: Is 1/p > r/(r^2+2) ? (1) p = r (2) r > 0 [#permalink]
03 May 2013, 16:17

Kind of incomplete question. It should be given that p cannot be equal to zero. The answer is C. If you know that you can cross multiple with a number only if you know the sign of that number than you can get to the answer C.

Re: Is 1/p > r/(r^2+2) ? (1) p = r (2) r > 0 [#permalink]
04 May 2013, 03:16

Expert's post

Bluelagoon wrote:

Kind of incomplete question. It should be given that p cannot be equal to zero. The answer is C. If you know that you can cross multiple with a number only if you know the sign of that number than you can get to the answer C.

We are given that p = r and r > 0, thus p does not equal 0.

Re: Is 1/p > r/(r^2 + 2)? [#permalink]
21 May 2013, 05:08

Expert's post

mohnish104 wrote:

mr bunuel how does r^2+2/r >r => r^2>0. shouldnt it be r^2+ 2 >r^2 instead

\frac{1}{r}>\frac{r}{r^2+2}? --> as r^2+2 is always positive, multiplying inequality by this expression we'll get: \frac{r^2+2}{r}>r? --> r+\frac{2}{r}>r? --> \frac{2}{r}>0?.

As for your solution: we cannot cross-multiply \frac{1}{r}>\frac{r}{r^2+2} since we don't know whether r is positive or negative.