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Algebra

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Source: — Problem Solving |

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
Philosophers have interpreted world in various ways, the point is to change it!
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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.
Philosophers have interpreted world in various ways, the point is to change it!
Join the discussion

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.
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