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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]
13 May 2013, 02:30

1

This post received KUDOS

Expert's post

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?

Re: if (y+3)(y-1)-(y-2)(y-1)=r(y-1), what is the value of y? (1) [#permalink]
07 Jan 2012, 14:31

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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DS - If negative answer only, still sufficient. No need to find exact solution. PS - Always look at the answers first CR - Read the question stem first, hunt for conclusion SC - Meaning first, Grammar second RC - Mentally connect paragraphs as you proceed. Short = 2min, Long = 3-4 min

Re: If (y+3)(y-1)-(y-2)(y-1)=r(y-1), what is the value of y? [#permalink]
14 May 2013, 20:21

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]
15 May 2013, 00:53

Expert's post

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

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).
_________________