I am Getting the answer as B?
Is it right?
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DS - Number Systems
Source: Beat The GMAT — Data Sufficiency |
The question must be wrong.f2001290 wrote: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
.... (y+3)(y-1) – (y-2)(y-1) = r(y-1)
=> (y-1) {(y+3)-(y-2)} = r(y-1)
=> y+3-y+2 = r
=> 5 = r
the value of y cannot be determined. Answer [E].
The title/subject of this topic is meaningless. This problem has nothing to do with number systems.
bingojohn - On the step 2, you cannot divide the eqn with (y-1) as it is not stated that y!=1
I think the answer is A
=> (y-1) {(y+3)-(y-2)} = r(y-1)
=> 5 (y-1) = r(y-1)
=> r=25 => r=5 or -5
=> With r=5, this equn is insovable
=> When r=-5, 5(y-1) = -5(y-1)
=> y-1=-y+1
=> y=1
Let me know if anyone has different opinion
I think the answer is A
=> (y-1) {(y+3)-(y-2)} = r(y-1)
=> 5 (y-1) = r(y-1)
=> r=25 => r=5 or -5
=> With r=5, this equn is insovable
=> When r=-5, 5(y-1) = -5(y-1)
=> y-1=-y+1
=> y=1
Let me know if anyone has different opinion
Oops, my ignorance. Thanks for pointing that out.gviren wrote:bingojohn - On the step 2, you cannot divide the eqn with (y-1) as it is not stated that y!=1
I think the answer is A
=> (y-1) {(y+3)-(y-2)} = r(y-1)
=> 5 (y-1) = r(y-1)
=> r=25 => r=5 or -5
=> With r=5, this equn is insovable
=> When r=-5, 5(y-1) = -5(y-1)
=> y-1=-y+1
=> y=1
Let me know if anyone has different opinion
However, statement (1) is not giving us any more information than is already presented in the question itself... how is the answer [A]?
It should still be [E], because we don't know for sure if r = -5.
Given :
(y+3)(y-1) – (y-2)(y-1) = r(y-1)
(y+3)(y-1) – (y-2)(y-1) - r (y-1 ) = 0;
(y-1) [ (y+3)- (y-2) - r ] =0
(y-1) [ y+3 - y + 2 - r ] =0
(y-1) (5-r) =0
i.e y=1 or r =5
y = ?
(1) r^2 = 25
r = (5,-5)
r = 5 : Equation holds good , y=1 maybe may not be true
r =-5 ; Equation doesnt hold good ; y=1 ;
Not Sufficient.
2)r = 5
r = 5 : Equation holds good , y=1 maybe may not be true.
Not Sufficient.
1+2
r = 5 : Equation holds good , y=1 maybe may not be true.
Not Sufficient.
ANS:E
(y+3)(y-1) – (y-2)(y-1) = r(y-1)
(y+3)(y-1) – (y-2)(y-1) - r (y-1 ) = 0;
(y-1) [ (y+3)- (y-2) - r ] =0
(y-1) [ y+3 - y + 2 - r ] =0
(y-1) (5-r) =0
i.e y=1 or r =5
y = ?
(1) r^2 = 25
r = (5,-5)
r = 5 : Equation holds good , y=1 maybe may not be true
r =-5 ; Equation doesnt hold good ; y=1 ;
Not Sufficient.
2)r = 5
r = 5 : Equation holds good , y=1 maybe may not be true.
Not Sufficient.
1+2
r = 5 : Equation holds good , y=1 maybe may not be true.
Not Sufficient.
ANS:E
