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M06-06

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Bunuel
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M06-06 [#permalink] New post   16 Sep 2014, 00:27
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If \(y(u - c) = 0\) and \(j(u - k) = 0\), which of the following must be true, given \(c < k\)?

A. \(yj \lt 0\)
B. \(yj \gt 0\)
C. \(yj = 0\)
D. \(j = 0\)
E. \(y = 0\)
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C
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Bunuel
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Re M06-06 [#permalink] New post   16 Sep 2014, 00:27
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Official Solution:

If \(y(u - c) = 0\) and \(j(u - k) = 0\), which of the following must be true, given \(c < k\)?

A. \(yj \lt 0\)
B. \(yj \gt 0\)
C. \(yj = 0\)
D. \(j = 0\)
E. \(y = 0\)


From \(y(u - c) = 0\), it follows that \(u=c\) or \(y=0\) (or both).

From \(j(u - k) = 0\), it follows that \(u=k\) or \(j=0\) (or both).

Note that \(u=c\) and \(u=k\) cannot simultaneously be correct because if they were, it would imply that \(u=c=k\), but we have been given that \(c < k\). Thus, only one can be correct, which further implies that either \(y=0\) or \(j=0\), making \(yj = 0\) always true.


Answer: C
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pinklemonhat
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Re: M06-06 [#permalink] New post   21 Jun 2020, 07:59
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[y(u-c)]*j = 0*j
[j (u-k)]*y = 0*y

jy (u-c) = 0
jy (u-k) = 0

ADD THEM

jy(u-c) + jy(u-k) = 0
jy (u-c-[u-k]) = 0
jy (u-c-u+k) = 0
jy (k-c) = 0
--> k > c
jy (+ve number) = 0
jy = 0

QED
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kungfury42
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Re: M06-06 [#permalink] New post   26 Aug 2022, 07:54
This is a fairly trippy one given the large number of variables involved. Great question and great explanation by Bunuel.

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Bunuel
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Re: M06-06 [#permalink] New post   12 Jun 2023, 00:28
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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M06-06 [#permalink] New post   02 Apr 2024, 01:56
I find it tough to understand why option d. y = 0 and option e. j = 0 are not always true. Is nt it true that from the 1st equation y = 0 must be true and from the second equation j = 0  must be true since c<k. If not, can someone please explain and elucidate the cases, when y = 0 and j = 0 are not must be true possibles?
Bunuel­
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Bunuel
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Re: M06-06 [#permalink] New post   02 Apr 2024, 02:07
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Shahul^ wrote:
Bunuel wrote:
Official Solution:

If \(y(u - c) = 0\) and \(j(u - k) = 0\), which of the following must be true, given \(c < k\)?

A. \(yj \lt 0\)
B. \(yj \gt 0\)
C. \(yj = 0\)
D. \(j = 0\)
E. \(y = 0\)


From \(y(u - c) = 0\), it follows that \(u=c\) or \(y=0\) (or both).

From \(j(u - k) = 0\), it follows that \(u=k\) or \(j=0\) (or both).

Note that \(u=c\) and \(u=k\) cannot simultaneously be correct because if they were, it would imply that \(u=c=k\), but we have been given that \(c < k\). Thus, only one can be correct, which further implies that either \(y=0\) or \(j=0\), making \(yj = 0\) always true.


Answer: C

­

I find it tough to understand that why option d. y = 0 and option e. j = 0 are not always true. Is nt it true that from the 1st equation y must be true and from the second equation j must be true? If not, can someone please explain and elucidate the cases, when y = 0 and j = 0 are not must be true possibles?
Bunuel

­
D, \(j = 0\), is not necessarily true because­ we can have a case when \(y=0\) and ­\(u=k\) (observer, that both \(y(u - c) = 0\) and \(j(u - k) = 0\) hold true for these values). In this case j can be any number, not necessarily 0.

Similarly, E, \(y = 0\), is not necessarily true because­ we can have a case when \(j=0\) and\(u=c\) (observer, that both \(y(u - c) = 0\) and \(j(u - k) = 0\) hold true for these values). In this case j can be any number, not necessarily 0.
 
Hope it helps.
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Re: M06-06 [#permalink]
02 Apr 2024, 02:07
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