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

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Is 1/p > r/(r^2 + 2)? [#permalink]

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01 Nov 2009, 16:06
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Is 1/p > r/(r^2 + 2) ?

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

[Reveal] Spoiler:
1/p > r /r^2 +2
r^2 +2 > pr ??

From stmt 1 we know:
r^2 +2 > r^2 ..so why s the answer not A
[Reveal] Spoiler: OA

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Last edited by Bunuel on 04 May 2013, 04:11, edited 2 times in total.
Edited the question and added the OA
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01 Nov 2009, 16:33
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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.

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.

Hope it's clear.
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01 Nov 2009, 19:11
Great wayof approaching!Super clear..thanks!
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04 Nov 2009, 01:17
Thanks Bunuel , your approach to the problem was very good. !

thanks
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26 Oct 2010, 17:59
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.

Hope this helps.
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08 Jan 2012, 14:17
Is (1/x) > y/(y^2+3)

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

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

Thanks
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08 Jan 2012, 15:55
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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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08 Jan 2012, 17:16
Good one.

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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20 Jan 2012, 06:42
C. BTW Excellent explanation and approach to problem bunuel
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Is 1/p > r/(r^2 + 2) ? [#permalink]

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03 May 2013, 09:41
robertrdzak wrote:
Is 1/p > r/(r^2+2) ?

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

[Reveal] Spoiler:
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.
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Re: Is 1/p > r/(r^2+2) ? (1) p = r (2) r > 0 [#permalink]

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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.
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Re: Is 1/p > r/(r^2+2) ? (1) p = r (2) r > 0 [#permalink]

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04 May 2013, 04:16
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.

Check here: is-1-p-r-r-86165.html#p645975
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Re: Is 1/p > r/(r^2 + 2)? [#permalink]

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04 May 2013, 10:40
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
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Re: Is 1/p > r/(r^2 + 2)? [#permalink]

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10 May 2013, 10:11
Is 1/p > r/(r^2 + 2) ?

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

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

from 1 the numerator becomes 2 and the sign of the denominator depends on r.... insuff

from 2........insuff

both
r = p >0 .... suff ...c
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Re: Is 1/p > r/(r^2 + 2)? [#permalink]

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21 May 2013, 04:06
mr bunuel how does r^2+2/r >r => r^2>0. shouldnt it be r^2+ 2 >r^2 instead
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Re: Is 1/p > r/(r^2 + 2)? [#permalink]

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21 May 2013, 06:08
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.

Hope it's clear.
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Re: Is 1/p > r/(r^2 + 2)? [#permalink]

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21 May 2013, 07:27
hey thanks a lot Mr Bunuel for the explanation.
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Re: Is 1/p > r/(r^2 + 2)? [#permalink]

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01 Jan 2014, 06:36
tejal777 wrote:
Is 1/p > r/(r^2 + 2) ?

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

[Reveal] Spoiler:
1/p > r /r^2 +2
r^2 +2 > pr ??

From stmt 1 we know:
r^2 +2 > r^2 ..so why s the answer not A

Good question +1

r^2 + 2 /p > r?

(r^2 + 2 - pr)/p > 0?

Statement 1

r = p so numerator must be 2, but we don't know whether p is positive or negative

Statement 2

R is positive, but we don't know anything about P

(1) + (2) We know that p = r and r>0, Hence p>0

Sufficient

Hope it helps
Cheers!

J
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Re: Is 1/p > r/(r^2 + 2)? [#permalink]

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Is 1/p > r/(r^2 + 2)? [#permalink]

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27 Mar 2015, 03:45
1/p > r/r^2+2
=> r^2 + 2 > pr
=> r^2 - pr > -2
=> r(r - p) > -2
now if p=r than r-p becomes 0
so => 0>-2 true

Why not statement A is suffiecient
Is 1/p > r/(r^2 + 2)?   [#permalink] 27 Mar 2015, 03:45

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