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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 _________________
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. _________________
I am the master of my fate. I am the captain of my soul. Please consider giving +1 Kudos if deserved!
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? (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? [#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 _________________
Do not forget to hit the Kudos button on your left if you find my post helpful.
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). _________________
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