DS - Number Systems

This topic has expert replies
User avatar
Master | Next Rank: 500 Posts
Posts: 400
Joined: Sat Mar 10, 2007 4:04 am
Thanked: 1 times
Followed by:1 members

DS - Number Systems

by f2001290 » Fri Jun 15, 2007 4:41 am
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

Junior | Next Rank: 30 Posts
Posts: 22
Joined: Mon Sep 18, 2006 10:17 am

by simplythebest » Fri Jun 15, 2007 5:32 am
I am Getting the answer as B?

Is it right?

Senior | Next Rank: 100 Posts
Posts: 54
Joined: Mon Aug 06, 2007 12:04 pm

Re: DS - Number Systems

by bingojohn » Fri Aug 10, 2007 11:42 am
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.

Senior | Next Rank: 100 Posts
Posts: 44
Joined: Fri Mar 30, 2007 3:11 pm

Re: DS - Number Systems

by gviren » Mon Aug 13, 2007 9:16 am
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

Senior | Next Rank: 100 Posts
Posts: 54
Joined: Mon Aug 06, 2007 12:04 pm

Re: DS - Number Systems

by bingojohn » Mon Aug 13, 2007 12:53 pm
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.

User avatar
Senior | Next Rank: 100 Posts
Posts: 77
Joined: Mon Jul 09, 2007 6:54 pm
Location: US of A

Re: DS - Number Systems

by Auzbee » Sun Aug 19, 2007 4:23 pm
[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.

Junior | Next Rank: 30 Posts
Posts: 22
Joined: Fri May 04, 2007 3:51 am

by krishnamurthyu » Mon Aug 20, 2007 8:09 am
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