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If (y+3)(y1)(y2)(y1)=r(y1), what is the value of y? [#permalink]
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Updated on: 13 May 2013, 03:20
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If (y+3)(y1)(y2)(y1)=r(y1), what is the value of y? (1) r^2=25 (2) r=5
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Originally posted by Jasonammex on 24 Aug 2011, 20:04.
Last edited by Bunuel on 13 May 2013, 03:20, edited 1 time in total.
Edited the question.



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Re: Help~ [#permalink]
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24 Aug 2011, 23:23
Jasonammex wrote: if (y+3)(y1)(y2)(y1)=r(y1), what is the value of y?
(1) r^2=25 (2) r=5
I chose A... In the equation above we see that (y1) is the common factor in all terms. So let's take (y1) common out of all the terms. (y+3)(y1)  (y2)(y1)  r(y1) = 0 (y1)[(y+3) (y2)  r] = 0 (y1)(5  r) = 0 Now, the product of these two is 0. This means that at least one of them has to be 0. Either (y1) = 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: Help~ [#permalink]
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26 Aug 2011, 02:49
Correct answer Should be B y1 is the common factor in the question stem Try to break the stem as much as you can So, y1[(y+3)  (y2)] = r(y1) = y+3y+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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Re: Help~ [#permalink]
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26 Aug 2011, 02:50
I am wrong. Sorry guys. Thanks Karishma.
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Re: Help~ [#permalink]
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26 Aug 2011, 06:05
should be E. The questions becomes y1[y+3  y+2)] = r(y1).
r=5. So, can not calculate the value of Y from both 1 and 2.



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Re: if (y+3)(y1)(y2)(y1)=r(y1), what is the value of y? (1) [#permalink]
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07 Jan 2012, 15:31
Rephrasing the stem provides us: (y1)[y+3y+2] = r(y1) (y1)5 = r(y1) (y1)(5r) = 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)(y1)(y2)(y1)=r(y1), what is the value of y? (1) [#permalink]
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13 May 2013, 03:18
Guys I'm bit lost here. (y1)[y+3y+2] = r(y1) (y1)5 = r(y1) (y1) 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)(y1)(y2)(y1)=r(y1), what is the value of y? (1) [#permalink]
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13 May 2013, 03:30
onedayill wrote: Guys I'm bit lost here.
(y1)[y+3y+2] = r(y1) (y1)5 = r(y1)
(y1) 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 (y1)5 = r(y1) by y1 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) (y1)5 = r(y1) by y1, you assume, with no ground for it, that y1 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)(y1)(y2)(y1)=r(y1), what is the value of y?\((y+3)(y1)  (y2)(y1)  r(y1) = 0\) > \((y1)[(y+3) (y2)  r] = 0\) > \((y1)(5  r) = 0\) > 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.
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Re: If (y+3)(y1)(y2)(y1)=r(y1), 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)(y1)(y2)(y1)=r(y1), what is the value of y?
(1) r^2=25 (2) r=5 From the given statement: (y1)[Y+3y+2]=r(y1) Divide both sides by (y1) y+3y+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)(y1)(y2)(y1)=r(y1), 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)(y1)(y2)(y1)=r(y1), what is the value of y?
(1) r^2=25 (2) r=5 From the given statement: (y1)[Y+3y+2]=r(y1) Divide both sides by (y1) y+3y+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: ify3y1y2y1ry1whatisthevalueofy119561.html#p1224280Never 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) (y1)5 = r(y1) by y1, you assume, with no ground for it, that y1 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)(y1)(y2)(y1)=r(y1), what is the value of y? [#permalink]
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15 May 2013, 21:11
Oops by bad, Thanks Bunuel.
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Re: If (y+3)(y1)(y2)(y1)=r(y1), what is the value of y? [#permalink]
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22 May 2016, 11:41
Jasonammex wrote: If (y+3)(y1)(y2)(y1)=r(y1), what is the value of y?
(1) r^2=25 (2) r=5 Given information= (y+3)(y1)(y2)(y1)=r(y1) Question value of y? Inference from given information (y+3)(y1)(y2)(y1)=r(y1) (y+3) (y2)= r (y1) is common at both the sides of the equation. y+3y+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
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