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

(1) p = r
(2) r > 0


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countdown-beginshas-ended-85483-40.html#p649902

Great wayof approaching!Super clear..thanks!
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countdown-beginshas-ended-85483-40.html#p649902

The question can be rephrased as:

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.

Hence the answer is C.

Hope this helps.

I'm happy to help with this one.

Is (1/x) > y/(y^2+3)

(1) y > 0
(2) x = y

Statement #1: y > 0

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.

Mike
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Mike McGarry
Magoosh Test Prep

C. BTW Excellent explanation and approach to problem bunuel

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.

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

(1) p = r
(2) r > 0

r^2 +2 is +ve thus the question becomes

is (r^2 + 2) / p > r?

from one
question becomes is (r^2 + 2) / r > r ..... is r + 2/r > r or ..... is 2/r > 0 .....insuff

from 2

obviously insuff as there is no info about p

both

suff...C