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If y(u  c) = 0 and j(u  k) = 0, which of the following must be true, assuming c < k? (A) yj < 0 (B) yj > 0 (C) yj = 0 (D) j = 0 (E) y = 0 Source: GMAT Club Tests  hardest GMAT questions Expl given  From the first equation of the setup we learn that either y = 0 or u = c. If y = 0, then yj = 0. If u = c then from the second equation j = 0 as c < k. Taking two equations combined, we will arrive to the conclusion that yj = 0. How is colored portion true ?? pls explain.



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12 Nov 2008, 15:28
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Given facts 1) y(u  c) = 0 2) j(u  k) = 0, 3) c < k
Now your question why this statement is true u = c then from the second equation j = 0 as c < k.
Given in 3 that c<k. If u = c, Then u  k cannot be equal to 0 because that also means c  k and as c < k so c  k is not zero.
Now to make statement 2 true. j has to be zero.
Which again means yj = 0



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26 Nov 2008, 09:17



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Re: m06 #06 [#permalink]
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17 Feb 2010, 08:12
mbaobsessed wrote: If y(u  c) = 0 and j(u  k) = 0, which of the following must be true, assuming c < k? (A) yj < 0 (B) yj > 0 (C) yj = 0 (D) j = 0 (E) y = 0 Source: GMAT Club Tests  hardest GMAT questions Expl given  From the first equation of the setup we learn that either y = 0 or u = c. If y = 0, then yj = 0. If u = c then from the second equation j = 0 as c < k. Taking two equations combined, we will arrive to the conclusion that yj = 0. How is colored portion true ?? pls explain. In this from first equation either j=0 or u=k and from second equation either y=0 or u=c now it is given c < k so u=k & u=c cannot be true at same time. Thts why if j=0 then u=c. else if y=0 then u=k.



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Re: m06 #06 [#permalink]
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23 Feb 2011, 04:49
Two cases from 1  y = 0 or c = u Two cases from 2  j = 0 or u = k => c != k so either c = u or k = u but not both if y = 0, then c != u so u = k, but yj = 0 if c = u, then j = 0, and k != u, so j = 0, => yj = 0 so either y = 0 or j = 0 always Hence answer is C
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Re: m06 #06 [#permalink]
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24 Feb 2011, 14:37
How do we arrive at saying: if y(uc)=0, then u = c? it's like saying if y(x) = 0, then x = 0; I don't think so. x (expression inside the braces) can be any value, but the equation (independent variables) reduce to zero. Am i missing something?
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Re: m06 #06 [#permalink]
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25 Feb 2012, 13:33
gmatbull, the way I understood/processed the question was as follows:
y(uc) = 0 so lets assume uc = 0 or u = c
c < k
therefore uk < 0
for j(uk) = 0 to be true j = 0
therefore jy=0
You can also switch the assumption around and you will find that y = 0 and the answer still holds.
Hope it helped a bit.



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28 Feb 2012, 15:15
Yekrut wrote: gmatbull, the way I understood/processed the question was as follows:
y(uc) = 0 so lets assume uc = 0 or u = c
c < k
therefore uk < 0
for j(uk) = 0 to be true j = 0
therefore jy=0
You can also switch the assumption around and you will find that y = 0 and the answer still holds.
Hope it helped a bit. I got the right answer "C" but reading your thought process certainly helped me frame what I was thinking lol



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Re: m06 #06 [#permalink]
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26 Feb 2013, 07:24
If y(u  c) = 0 and j(u  k) = 0, which of the following must be true, assuming c < k?A. yj < 0 B. yj > 0 C. yj = 0 D. j = 0 E. y = 0 \(y(u  c) = 0\) > \(u=c\) or \(y=0\); \(j(u  k) = 0\) > \(u=k\) or \(j=0\); Now, the first option (\(u=c\) and \(u=k\)) cannot be simultaneously correct for both equations because if it is, then it would mean that \(u=c=k\), but we are given that \(c<k\). So, only one can be correct so either \(y=0\) or \(j=0\), which makes \(yj = 0\) is always true. Answer: C.
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Re: m06 #06 [#permalink]
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27 Feb 2013, 08:31
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Given y(uc) = 0 and j(uk) =0 and c <k.
y(uc)=0 => either y=0 or uc=0 => u=c j(uk)=0 => either j=0 or uk=0 => u=k
from above two eqn c=k but c <k so y =0 and j=0
yj=0
Answer : C



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Re: m06 #06 [#permalink]
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21 Apr 2014, 08:52
Since c<k, uc and uk cannot be both 0 > At least one of y and j is 0 > yj=0











