If (y+3)(y-1)-(y-2)(y-1)=r(y-1), what is the value of y?
1) r^2=25
2) r=5
I do not understand why we need to find y at all....Here's my approach...
(y+3)(y-1)-(y-2)(y-1)=r(y-1)
or (y-1)[(y+3)-(y-2)]=r(y-1)
or [(y+3)-(y-2)]=r
or y+3-y+2=r
or 5=r
So ultimately, there is no y. So both the statements are irrelevant.
But can we have a DS question where we do not need to find/prove anything? I may have missed something here, i didn't get the point of this question.
Please explain. Thanks.
OA is E [spoiler][/spoiler]
Confusing equation - Set11 Q5
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Hey arocks,
The Q asks "what is the value of y?", and that is what we need to find here, though we can find r without using y.
now
(y+3)(y-1)-(y-2)(y-1)=r(y-1)
stmt 1 :
r^2 =25 so r =-5 or 5
in both cases i.e plugging r=5 or -5 we will end up with a quadratic eqn with y raised to power of 2, hence it will have atleast two solns the two values of r
not suff
stmt 2 :
r =5
again the same case, if we plug in the value of 5 we will get a quadratic eqn of y so it will have two roots so INSUFF
combine: same as stmt 2
so E
The Q asks "what is the value of y?", and that is what we need to find here, though we can find r without using y.
now
(y+3)(y-1)-(y-2)(y-1)=r(y-1)
stmt 1 :
r^2 =25 so r =-5 or 5
in both cases i.e plugging r=5 or -5 we will end up with a quadratic eqn with y raised to power of 2, hence it will have atleast two solns the two values of r
not suff
stmt 2 :
r =5
again the same case, if we plug in the value of 5 we will get a quadratic eqn of y so it will have two roots so INSUFF
combine: same as stmt 2
so E
Regards
Samir
Samir