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M03-25

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M03-25 [#permalink]

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New post 16 Sep 2014, 00:21
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A
B
C
D
E

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48% (00:30) correct 52% (00:19) wrong based on 203 sessions

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New post 16 Sep 2014, 00:21
Official Solution:


(1) \(p = q\). Not sufficient, since no info about \(r\).

(2) \(q^2 = r^2\). Since given that \(p\) and \(q\) are positive numbers then \(q=r\). Not sufficient since no info about \(p\).

(1)+(2) As \(p = q\) and \(q=r\) then \(p=q=r\). Sufficient.


Answer: C
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Re: M03-25 [#permalink]

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New post 11 Feb 2015, 12:01
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Hi Bunuel,

Thanks for this question.

I agree with the solution provided. Concerning the old debate whether 0^2=1 or 0^2=0, could it be assumed that for the purposes of the GMAT 0^2=0. With regards to this question this could change the answer substantially, as if not, it could mean that in (2) p and q can b both 0 and 1, and so the sanswer would be (E).

Thank you!
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New post 12 Feb 2015, 06:22
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bgpower wrote:
Hi Bunuel,

Thanks for this question.

I agree with the solution provided. Concerning the old debate whether 0^2=1 or 0^2=0, could it be assumed that for the purposes of the GMAT 0^2=0. With regards to this question this could change the answer substantially, as if not, it could mean that in (2) p and q can b both 0 and 1, and so the sanswer would be (E).

Thank you!


There is no debate whatsoever:

0^(positive) = 0, for example, 0^2 = 0.

(anything but 0)^0 = 1, for example, 3^0 = 1.

0^0 is undefined and not tested on the GMAT.

For the question at hand none of the variables can be 0, because we are told that p, q, and r are positive numbers, while 0 is neither positive nor negative.

Hope it helps.
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Re: M03-25 [#permalink]

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New post 12 Feb 2015, 06:28
Got it! Thanks so much! Seems I was wrong based on some other old discussions.

0^(any positive) = 0
3^0 = 1.
0^0 is undefined and not tested on the GMAT :)
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M03-25 [#permalink]

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New post 16 Nov 2015, 14:05
What if p = 2 q= 2 and r=\(\sqrt{2}\), then these values are not equal?

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New post 16 Nov 2015, 22:13

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Re: M03-25 [#permalink]

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New post 21 Apr 2016, 14:05
I observed in another problem whose equation is as follows,
q^2 - r^2 = 0

In that problem, it was said q cannot be equal to r, since the equation can be further reduced to |q| - |r| = 0.

Does it apply here as well?

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New post 21 Apr 2016, 14:12
atturhari wrote:
I observed in another problem whose equation is as follows,
q^2 - r^2 = 0

In that problem, it was said q cannot be equal to r, since the equation can be further reduced to |q| - |r| = 0.

Does it apply here as well?


Not sure I fully understand what you mean there but q^2 - r^2 = 0 can be written as |q| - |r| = 0:

q^2 - r^2 = 0
q^2 = r^2
|q| = |r| (by taking the square root from both sides, which we can safely do since both sides are non-negative).

If we knew that \(q \neq r\), then we could further simply it to q = -r.
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: M03-25 [#permalink]

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New post 13 Oct 2016, 11:40
THIS IS CRAZY . I CANNOT BELIEVE I DID NOT READ P,Q,R ARE POSITIVE. DAMN **** :(

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Re: M03-25 [#permalink]

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New post 06 Nov 2017, 15:14
q2=r2 if
q = - 2 and
r = 2 than both are 4, but are not the same, or am I wrong?

Than the answer must be E

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New post 06 Nov 2017, 20:42

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Re: M03-25   [#permalink] 06 Nov 2017, 20:42
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