Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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 _________________

I am the master of my fate. I am the captain of my soul. Please consider giving +1 Kudos if deserved!

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]

Show Tags

03 May 2013, 17:17

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]

Show Tags

04 May 2013, 04:16

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.

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.

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

Part 2 of the GMAT: How I tackled the GMAT and improved a disappointing score Apologies for the month gap. I went on vacation and had to finish up a...

So the last couple of weeks have seen a flurry of discussion in our MBA class Whatsapp group around Brexit, the referendum and currency exchange. Most of us believed...

This highly influential bestseller was first published over 25 years ago. I had wanted to read this book for a long time and I finally got around to it...