If (y + 3)(y - 1) - (y - 2)(y - 1) = r(y - 1), what is the value of y?

Question

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

KarishmaB · Answer

In the equation above we see that (y-1) is the common factor in all terms. So let's take (y-1) common out of all the terms.

(y+3)(y-1) - (y-2)(y-1) - r(y-1) = 0
(y-1)  = 0
(y-1)(5 - r) = 0

Now, the product of these two is 0. This means that at least one of them has to be 0.
Either (y-1) = 0 or (5 - r) = 0 or both are 0.
So, either y = 1 or r = 5 or both.
Only if we know that r is not 5, then we can say that y must be 1. If r is 5, y may be 1 or may not be 1.

Stmnt 1: r^2 = 25
So r = +- 5
This statement tells us that r can be 5. 
If r = 5, y may or may not be 1.
If r is not 5, y will be 1.
Since we do not know whether r is 5 or not, we cannot say what the value of y is. Not sufficient.

Stmnt 2: r = 5
If r = 5, y may or may not be 1.
Not sufficient.

Both together, r = 5. Again, not sufficient. Answer E

enigma123 · Answer

Correct answer Should be B

y-1 is the common factor in the question stem

Try to break the stem as much as you can

So, y-1  = r(y-1)

= y+3-y+2 = r

r=5. So the question becomes is r=5?

Statement 1 gives us two solution + and _ 5 So insufficient

Statement 2 is sufficient because its same as our new question stem which is r=5.

Hope this helps.

enigma123 · Answer

I am wrong. Sorry guys. Thanks Karishma.

czarczar · Answer

should be E. 
The questions becomes y-1  = r(y-1).

r=5. So, can not calculate the value of Y from both 1 and 2.

shinbhu · Answer

Rephrasing the stem provides us:

(y-1)  = r(y-1)
(y-1)5 = r(y-1)
(y-1)(5-r) = 0
y = 1 or r = 5

1. r = +/- 5, if r = -5, y = 1 and if r = 5, y = anything. Insuff.
2. r = 5, which means y = anything. Insuff.

Combined, only thing common is r = 5, which still provides y = anything. Insuff.

E.

onedayill · Answer

Guys I'm bit lost here.

(y-1)  = r(y-1)
(y-1)5 = r(y-1)

(y-1) is common here

so 5= r

which is what stmt 2 tells us.

I think i should be B.

Bunuel · Answer

Notice that we are asked to find the value of y not r. Also, I guess you reduced (y-1)5 = r(y-1) by y-1 which cannot be done here.

Never reduce an equation by a variable (or expression with a variable), if you are not certain that the variable (or the expression with a variable) doesn't equal to zero. We can not divide by zero.

So, if you divide (reduce) (y-1)5 = r(y-1) by y-1, you assume, with no ground for it, that y-1 does not equal to zero thus exclude a possible solution (notice that both y=1 AND r=5 satisfy the equation).

Hope it's clear.

Complete solution:

If  (y+3)(y-1)-(y-2)(y-1)=r(y-1) , what is the value of y?

(y+3)(y-1) - (y-2)(y-1) - r(y-1) = 0 ;

(y-1)  = 0 ;

(y-1)(5 - r) = 0 ;

y=1  or/and  r=5 .

(1)  r^2=25  -->  r=5  or  r=-5 . Not sufficient.

(2)  r=5  --> y may or may not be 1. Not sufficient.

(1)+(2) The same here: we have that r=5 but y still may or may not be 1. Not sufficient.

Answer: E.

atalpanditgmat · Answer

Hi Buneul, has my logic actually worked here?
 correct me if i am wrong...

From the given statement:
(y-1) =r(y-1)
Divide both sides by (y-1)

y+3-y+2=r
r=5

Here I thought that in the given statement when Y does not exist it is impossible to calculate the value of Y.
Thus both statements are not sufficient.
Ans: E

Bunuel · Answer

Unfortunately that's not correct. Refer to my post above: if-y-3-y-1-y-2-y-1-r-y-1-what-is-the-value-of-y-119561.html#p1224280 Never reduce an equation by a variable (or expression with a variable), if you are not certain that the variable (or the expression with a variable) doesn't equal to zero. We can not divide by zero. So, if you divide (reduce) (y-1)5 = r(y-1) by y-1, you assume, with no ground for it, that y-1 does not equal to zero thus exclude a possible solution (notice that both y=1 AND r=5 satisfy the equation).

atalpanditgmat · Answer

Oops by bad, Thanks Bunuel.

Divyadisha · Answer

Given information= 
(y+3)(y-1)-(y-2)(y-1)=r(y-1)

Question- value of y?

Inference from given information-

(y+3)(y-1)-(y-2)(y-1)=r(y-1) 
(y+3)- (y-2)= r (y-1) is common at both the sides of the equation.
y+3-y+2=r
r=5
(y is cancelling out at both sides of equation, so most likely we need one of the statement to produce value of y )

(1) r^2=25
It does not give us a solution for y

(2) r=5
It does not give us a solution for y

E is the answer

bumpbot · Answer

Automated notice from GMAT Club BumpBot: 

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

 This post was generated automatically.

If (y + 3)(y - 1) - (y - 2)(y - 1) = r(y - 1), what is the value of y?

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards