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m06 #06

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m06 #06 [#permalink] New post 12 Nov 2008, 13:39
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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

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C

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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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Re: m6 06 [#permalink] New post 12 Nov 2008, 14:28
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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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Re: m6 06 [#permalink] New post 26 Nov 2008, 08:17
Thanks, guys. We'll have to edit the original explanation to be more precise. +1, abhijit_sen.
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Re: m06 #06 [#permalink] New post 17 Feb 2010, 07:12
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

[Reveal] Spoiler: OA
C

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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] New post 23 Feb 2011, 03:49
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] New post 24 Feb 2011, 13:37
How do we arrive at saying: if y(u-c)=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] New post 25 Feb 2012, 12:33
gmatbull, the way I understood/processed the question was as follows:

y(u-c) = 0 so lets assume u-c = 0 or u = c

c < k

therefore u-k < 0

for j(u-k) = 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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Re: m06 #06 [#permalink] New post 28 Feb 2012, 14:15
Yekrut wrote:
gmatbull, the way I understood/processed the question was as follows:

y(u-c) = 0 so lets assume u-c = 0 or u = c

c < k

therefore u-k < 0

for j(u-k) = 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] New post 26 Feb 2013, 06:24
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Expert's post
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] New post 27 Feb 2013, 07:31
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Given y(u-c) = 0 and j(u-k) =0 and c <k.

y(u-c)=0 => either y=0 or u-c=0 => u=c
j(u-k)=0 => either j=0 or u-k=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] New post 21 Apr 2014, 07:52
Since c<k, u-c and u-k cannot be both 0 -> At least one of y and j is 0 -> yj=0
Re: m06 #06   [#permalink] 21 Apr 2014, 07:52
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