Algebra

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Algebra

by ritula » Sun Sep 21, 2008 9:11 pm
If y(u - c) = 0 and j(u - k) = 0, which of the following must be true, assuming c < k?

yj < 0
yj > 0
yj = 0
j = 0
y = 0
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by event_horizon » Sun Sep 21, 2008 9:36 pm
is it C?
OA?

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OA

by ritula » Sun Sep 21, 2008 9:48 pm
YEs it is . Kindly explain
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by event_horizon » Sun Sep 21, 2008 9:54 pm
ok..very simple concept.
y(u-c)=0 & j(u-k)=0

it directly follows from the above equations that y=0 and j=0
out of all the possibilities,the only one correct it yj=0.only choice 4, y=0 is incomplete,only choice 5 , j=0 is also incomplete.both have to be zero.hence the answer c.choices 1,2 are outta question.

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by ritula » Sun Sep 21, 2008 10:07 pm
But u have neglected the fact that frm eq 1, u=c
& frm eq 2, u=k is another possibility.

From y(u-c)=0 , we can say that either y=0 or u=c.
event_horizon wrote:ok..very simple concept.
y(u-c)=0 & j(u-k)=0

it directly follows from the above equations that y=0 and j=0
out of all the possibilities,the only one correct it yj=0.only choice 4, y=0 is incomplete,only choice 5 , j=0 is also incomplete.both have to be zero.hence the answer c.choices 1,2 are outta question.
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by tendays2go » Mon Sep 22, 2008 7:44 am
y(u-c) = 0 & j(u-k) =0
also,
c<k => c-k <0 .....this is the hint here.

but, the only link in 2 eqns is u, so we need to get a value for it.

therefore, in y(u-c)=0 ,assuming that y!= 0 => u =c

now => in j(u-k) = j(c-k)=0 => j =0 [my original answer :( ]

also, y can as well be 0 (contradicting our assumption above)
thus, y*j will definitely be 0. hence, the answer.