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M21-19

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Math Expert
Joined: 02 Sep 2009
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16 Sep 2014, 01:11
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Difficulty:

95% (hard)

Question Stats:

39% (01:12) correct 61% (14:21) wrong based on 138 sessions

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Is $$p + q \gt \frac{1}{p} + \frac{1}{q}$$ ?

(1) $$p \lt q \lt 1$$

(2) $$pq \lt 1$$
[Reveal] Spoiler: OA

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16 Sep 2014, 01:11
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Official Solution:

Statements (1) and (2) combined are insufficient. Consider $$p = \frac{1}{3}$$, $$q = \frac{1}{2}$$ (the answer is "no") and $$p = -\frac{1}{2}$$, $$q = -\frac{1}{3}$$ (the answer is "yes"). We must remember that $$p$$ and $$q$$ are not necessarily positive.

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21 Jun 2015, 09:26
Hi Bunuel,
Can we not look at this way?

p+q>1/p+1/q can be re-written as (p+q)pq>(p+q). Cancelling (p+q) on both sides, the question can be re-written as is pq>1?

St II says pq<1 and so St II is sufficient.
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21 Jun 2015, 10:57
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dpbe25 wrote:
Hi Bunuel,
Can we not look at this way?

p+q>1/p+1/q can be re-written as (p+q)pq>(p+q). Cancelling (p+q) on both sides, the question can be re-written as is pq>1?

St II says pq<1 and so St II is sufficient.

1. You cannot multiply both sides by pq because we don't know whether it's positive or negative (if it's negative you should flip the sign).
2. You cannot reduce by p+q by the same reason.

Check here: is-p-q-1-p-1-q-154268.html#p1235382
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22 Jun 2015, 08:05
Thanks Bunuel for the explanation.
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Joined: 24 Jun 2015
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03 Jul 2015, 05:42
Bunuel wrote:
dpbe25 wrote:
Hi Bunuel,
Can we not look at this way?

p+q>1/p+1/q can be re-written as (p+q)pq>(p+q). Cancelling (p+q) on both sides, the question can be re-written as is pq>1?

St II says pq<1 and so St II is sufficient.

1. You cannot multiply both sides by pq because we don't know whether it's positive or negative (if it's negative you should flip the sign).
2. You cannot reduce by p+q by the same reason.

Check here: is-p-q-1-p-1-q-154268.html#p1235382

Hi Bunuel,

In this kind of question is it helpfull to try tu reduce the question stem or is better to try numbers directly to the statements?, for instance I try to reduce and the question stem result in (pq)squared - 1 > 0

Thanks a lot.

Regards.

Luis Navarro
Looking for 700
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03 Jul 2015, 06:09
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luisnavarro wrote:
Bunuel wrote:
dpbe25 wrote:
Hi Bunuel,
Can we not look at this way?

p+q>1/p+1/q can be re-written as (p+q)pq>(p+q). Cancelling (p+q) on both sides, the question can be re-written as is pq>1?

St II says pq<1 and so St II is sufficient.

1. You cannot multiply both sides by pq because we don't know whether it's positive or negative (if it's negative you should flip the sign).
2. You cannot reduce by p+q by the same reason.

Check here: is-p-q-1-p-1-q-154268.html#p1235382

Hi Bunuel,

In this kind of question is it helpfull to try tu reduce the question stem or is better to try numbers directly to the statements?, for instance I try to reduce and the question stem result in (pq)squared - 1 > 0

Thanks a lot.

Regards.

Luis Navarro
Looking for 700

First of all if you manipulate with the inequality you don't get (pq)^2 - 1.

Next, it depends on a question and you math skills what approach to choose.

Check this post for inequality DS approaches: time-management-when-testing-cases-in-ds-200020.html#p1538771

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10 Jun 2016, 10:56
Hi bunuel,

I too worked with the same approach. Additionally i also assumed "what if one of the numbers equaled 0" if that was the case then equation in question statement would have an undefined side RHS with 1/0. Based on that i quickly eliminated. is that the right approach considering nothing is mentioned about both numbers being non-zero?
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13 Sep 2017, 14:26
Senthil7 wrote:
Hi bunuel,

I too worked with the same approach. Additionally i also assumed "what if one of the numbers equaled 0" if that was the case then equation in question statement would have an undefined side RHS with 1/0. Based on that i quickly eliminated. is that the right approach considering nothing is mentioned about both numbers being non-zero?

It is safe to assume that both P and Q will not be equal to 0 since that if that were the case it would lead to "undefined" results. I think the assumption is to understand that such scenarios don't exist within the bounds of the GMAT Quantitative Reasoning world.
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29 Nov 2017, 03:04
The answer must be option E. You can have different cases here. Both +ve, both -ve , & one +ve and one -ve. Try out with all the three cases and you will realise that both together are not sufficient.
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Re: M21-19   [#permalink] 29 Nov 2017, 03:04
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