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

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

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New post Updated on: 04 Apr 2019, 23:08
3
35
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A
B
C
D
E

Difficulty:

  55% (hard)

Question Stats:

64% (01:51) correct 36% (01:55) wrong based on 497 sessions

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


(1) p = r

(2) r > 0


1/p > r /r^2 +2
the question is asking:
r^2 +2 > pr ??

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

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Originally posted by tejal777 on 01 Nov 2009, 16:06.
Last edited by Bunuel on 04 Apr 2019, 23:08, edited 3 times in total.
Edited the question and added the OA
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Is 1/p > r/(r^2 + 2) ? (1) p = r (2) r > 0  [#permalink]

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New post 01 Nov 2009, 16:33
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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) As from (2) \(r>0\), then \(\frac{2}{r}>0\) is true. 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.

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

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New post 01 Nov 2009, 19:11
Great wayof approaching!Super clear..thanks!
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Re: Is 1/p > r/(r^2 + 2) ? (1) p = r (2) r > 0  [#permalink]

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New post 04 Nov 2009, 01:17
Thanks Bunuel , your approach to the problem was very good. !

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

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New post 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.

Hence the answer is C.

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

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

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New post 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) ? (1) p = r (2) r > 0  [#permalink]

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New post 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) ? (1) p = r (2) r > 0  [#permalink]

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

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New post 27 Mar 2015, 05:15
architbansal wrote:
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


We cannot cross-multiply in this case.

We can multiply 1/p > r/(r^2+2) by r^2+2 because r^2+2=non-negative+positive=positive but we cannot multiply 1/p > r/(r^2+2) by p because we don't know its sign:
If p is positive, then we'd get 1 > p*r/(r^2+2): keep the sign when multiplying by positive value;
If p is negative, then we'd get 1 < p*r/(r^2+2): flip the sign of the inequality when multiplying by negative value.

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

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New post 15 Sep 2018, 09:31
Hi,

here are my two cents.

From Stmt 1:
p=r
Then
\(\frac{1}{r} > \frac{r}{r^2+2}\)

then

\(\frac{1}{r} - \frac{r}{r^2+2}\)>0

\(\frac{r^2+2 - r^2}{r(r^2+2)}\) > 0

\(\frac{2}{r(r^2+2)}\) > 0

So if we know that r > 0 then we can say that the exp is >0

But we don't know if r>0

Stmt 2:

r>0

That means \(\frac{r}{r^2+2}\)is positive or >0

Now question is IS \(\frac{1}{p}\) >0 or p>0
Since we have no information about p hence insufficient.

Now combining Stmt 1 and Stm2 we have
p=r and r>o so p>0

Hence sufficient.

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

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New post 09 Oct 2018, 06:58
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tejal777 wrote:
Is 1/p > r/(r² + 2) ?

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


Target question: Is 1/p > r/(r² + 2) ?

Statement 1: p = r
Take the target question and replace p with r to get: Is 1/r > r/(r² + 2) ?
Since r² + 2 is always POSITIVE, we can safely multiply both side of the inequality by r² + 2.
When we do this, we get: Is (r² + 2)/r > r ?

Now we can apply the following fraction property: (a + b)/c = a/c + b/c
We get: Is r²/r + 2/r > r ?
Simplify to get: Is r + 2/r > r ?
Subtract r from both sides to get: Is 2/r > 0 ?
At this point, we can see that statement 1 is not sufficient.
When r is POSITIVE, the answer to our revised target question is YES, it IS the case that 2/r > 0
When r is NEGATIVE, the answer to our revised target question is NO, it is NOT the case that 2/r > 0
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: r > 0
Since we have no information about p, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined
From statement 1 we learned that . . .
When r is POSITIVE, the answer to our revised target question is YES, it IS the case that 2/r > 0
When r is NEGATIVE, the answer to our revised target question is NO, it is NOT the case that 2/r > 0

Statement 2 tells us that r is POSITIVE
So, we can be certain that the answer to our revised target question is YES, it IS the case that 2/r > 0
Since we can answer the target question with certainty, the combined statements are SUFFICIENT

Answer: C

Cheers,
Brent
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Re: Is 1/p > r/(r^2 + 2) ? (1) p = r (2) r > 0   [#permalink] 09 Oct 2018, 06:58
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