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M06-06

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Math Expert
Joined: 02 Sep 2009
Posts: 46264

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16 Sep 2014, 00:27
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15% (low)

Question Stats:

85% (01:10) correct 15% (01:30) wrong based on 106 sessions

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

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

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Math Expert
Joined: 02 Sep 2009
Posts: 46264

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16 Sep 2014, 00:27
Official Solution:

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

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

$$y(u - c) = 0$$. Either $$u=c$$ or $$y=0$$;

$$j(u - k) = 0$$. Either $$u=k$$ 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 $$c \lt k$$. So, only one can be correct so either $$y=0$$ or $$j=0$$, which makes $$yj = 0$$ always true.

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Current Student
Joined: 18 Jun 2015
Posts: 41

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12 Sep 2016, 12:43
Thanks for the explanation. Somehow I missed to guess the logic of u=c and u=k can't happen simultaneously.
And got this wrong. The explanation is crisp and concise.
Intern
Joined: 01 Jan 2016
Posts: 1

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26 Dec 2017, 20:39
Hello Bunuel,
besides the fact that c<k, can't it be "and/or" instead of "or" for each assumption?
y(u−c)=0. Either u=c AND/or y=0? It doesn't change the answer but I would like to know why you didn't include it in your solution. Thank you
Math Expert
Joined: 02 Sep 2009
Posts: 46264

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26 Dec 2017, 21:01
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Guimeister wrote:
Hello Bunuel,
besides the fact that c<k, can't it be "and/or" instead of "or" for each assumption?
y(u−c)=0. Either u=c AND/or y=0? It doesn't change the answer but I would like to know why you didn't include it in your solution. Thank you

Yes it's an inclusive OR, which means that Either $$u=c$$ or $$y=0$$ (or both).
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Re: M06-06   [#permalink] 26 Dec 2017, 21:01
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