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Intern
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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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Director
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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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CIO
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Senior Manager
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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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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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Director
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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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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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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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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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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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