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

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

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24 Aug 2011, 20:04
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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
[Reveal] Spoiler: OA

Last edited by Bunuel on 13 May 2013, 03:20, edited 1 time in total.
Edited the question.
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24 Aug 2011, 23:23
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Jasonammex 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

I chose A...

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)[(y+3) -(y-2) - r] = 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
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Re: if (y+3)(y-1)-(y-2)(y-1)=r(y-1), what is the value of y? (1) [#permalink]

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13 May 2013, 03:30
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onedayill wrote:
Guys I'm bit lost here.

(y-1)[y+3-y+2] = 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.

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?

$$(y+3)(y-1) - (y-2)(y-1) - r(y-1) = 0$$ --> $$(y-1)[(y+3) -(y-2) - r] = 0$$ --> $$(y-1)(5 - r) = 0$$ --> 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.

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26 Aug 2011, 02:49

y-1 is the common factor in the question stem

Try to break the stem as much as you can

So, y-1[(y+3) - (y-2)] = 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.
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26 Aug 2011, 02:50
I am wrong. Sorry guys. Thanks Karishma.
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26 Aug 2011, 06:05
should be E.
The questions becomes y-1[y+3 - y+2)] = r(y-1).

r=5. So, can not calculate the value of Y from both 1 and 2.
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Re: if (y+3)(y-1)-(y-2)(y-1)=r(y-1), what is the value of y? (1) [#permalink]

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07 Jan 2012, 15:31
Rephrasing the stem provides us:

(y-1)[y+3-y+2] = 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.
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Re: if (y+3)(y-1)-(y-2)(y-1)=r(y-1), what is the value of y? (1) [#permalink]

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13 May 2013, 03:18
Guys I'm bit lost here.

(y-1)[y+3-y+2] = 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.
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Re: If (y+3)(y-1)-(y-2)(y-1)=r(y-1), what is the value of y? [#permalink]

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14 May 2013, 21:21
Hi Buneul, has my logic actually worked here?
correct me if i am wrong...
Jasonammex 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

From the given statement:
(y-1)[Y+3-y+2]=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
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Re: If (y+3)(y-1)-(y-2)(y-1)=r(y-1), what is the value of y? [#permalink]

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15 May 2013, 01:53
atalpanditgmat wrote:
Hi Buneul, has my logic actually worked here?
correct me if i am wrong...
Jasonammex 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

From the given statement:
(y-1)[Y+3-y+2]=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

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).
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Re: If (y+3)(y-1)-(y-2)(y-1)=r(y-1), what is the value of y? [#permalink]

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15 May 2013, 21:11
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Re: If (y+3)(y-1)-(y-2)(y-1)=r(y-1), what is the value of y? [#permalink]

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Re: If (y+3)(y-1)-(y-2)(y-1)=r(y-1), what is the value of y? [#permalink]

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22 May 2016, 11:41
Jasonammex 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

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

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Re: If (y+3)(y-1)-(y-2)(y-1)=r(y-1), what is the value of y?   [#permalink] 22 May 2016, 11:41
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