DS - Number Systems

I am Getting the answer as B?

Is it right?

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

The question must be wrong.

.... (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

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

Oops, my ignorance. Thanks for pointing that out.

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.

[quote="gviren"]
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
....
[/quote]

We still cannot determine from the i statement if r =5 or -5. I think the answer is E.

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