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If xy > 0 does (x-1)(y-1)=1 ? (1) x + y = [#permalink]
 03 Aug 2010, 05:25
Question Stats:
72% (01:52) correct
27% (01:45) wrong based on 2 sessions
If xy > 0 does (x-1)(y-1)=1? In addition to this question, can you confirm my understanding of the question. When a question reads "... does A=B ?" We could also understand it to mean, "under the conditions given does A always = B?" As opposed to "can A=B?"
Last edited by TallJTinChina on 03 Aug 2010, 05:59, edited 1 time in total.
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Re: Quantitative Guide 80 does (x-1)(y-1)=1? [#permalink]
07 Sep 2010, 05:19
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guytree wrote: I am bit sceptic to post this. But I wanted to check if this approach is right.
From the question we know that X and Y both are greater than 0.
In the statement 2 we could use simple plug-ins. If x=y=2 then (x-1)(y-1)=1. However, if x=y=3 then (x-1)(y-1) is not equal to 1.
I would greatly appreciate if you let me understand any loopholes in this approach.
Cheers If xy>0 does (x-1)(y-1)=1? (1) x + y = xy(2) x=yxy>0 means that either both x and y are positive or both are negative (so neither of unknowns equals to zero: x\neq{0} and y\neq{0}). Question: is (x-1)(y-1)=1? --> is xy-x-y+1=1? is x+y=xy? (1) x+y=xy --> directly gives us the answer YES. Sufficient. (2) x=y --> question becomes: is x+x=x^2? --> is x(x-2)=0? --> is x=0 or x=2? --> as given that x\neq{0}, then the question becomes is x=2? We don't know that, hence this statement is not sufficient. Answer: A.
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Re: Quantitative Guide 80 does (x-1)(y-1)=1? [#permalink]
02 Sep 2010, 21:40
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zuperman wrote: From Statement 2 you still get x^2-2x+1=1 or x*(x-2)=0. So x is either 0 or 2. Obviously x cannot be 0 hence x=2 and thence y=2. This is sufficient to find the value of (x-1)(y-1)=1.
Am I still missing something here? You're misreading the question - they're ASKING you if (x-1)(y-1)=1, not TELLING you. You're assuming that's true and solving for x to fit the question. But by the criteria in statement 2, what if x=y=5? Then (x-1)(y-1) = (4)(4) = 16 =/= 1.
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Re: Quantitative Guide 80 does (x-1)(y-1)=1? [#permalink]
03 Aug 2010, 05:39
Hi,
The question is:
(x-1)(y-1)=1 or xy-y-x+1=1 or xy=y+x
Thus first statement is sufficient.
Second statement x=y
x^2=2x which is not sufficient to answer the question.
So the right answer should be A. Are you sure the OA is C?
regards,
Jack
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Re: Quantitative Guide 80 does (x-1)(y-1)=1? [#permalink]
01 Sep 2010, 15:30
How is statement 2 alone insufficient? Substitute y for x in the question stem to get x^2 = 2x. x^2=2x has two solutions, x=0 and x=2. Zero is ruled out by the question stem, leaving x=y=2. Thus, (x-1)(y-1)=1. Any help would be appreciated
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Re: Quantitative Guide 80 does (x-1)(y-1)=1? [#permalink]
01 Sep 2010, 17:01
TallJTinChina wrote: If xy > 0 does (x-1)(y-1)=1? In addition to this question, can you confirm my understanding of the question. When a question reads "... does A=B ?" We could also understand it to mean, "under the conditions given does A always = B?" As opposed to "can A=B?" Does A=B is basically a yes/no type of DS question. Here you have to determine whether the given statements can provide you a unique solution to the question whether A=B is yes or no. Always simplify the question. (x-1)*(y-1)=1 xy -x-y +1 = 1=> xy - x - y = 0is to be answered. Statement 1. apparently its sufficient. Statement 2. x=yxy -x-y when x = ygives x^2 - 2x = x*(x-2)..not sufficient. Hence A
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Re: Quantitative Guide 80 does (x-1)(y-1)=1? [#permalink]
02 Sep 2010, 09:29
How is statement 2 alone insufficient? Substitute y for x in the question stem to get x^2 = 2x. x^2=2x has two solutions, x=0 and x=2. Zero is ruled out by the question stem, leaving x=y=2. Thus, (x-1)(y-1)=1. Any help would be appreciated
Statement 2 is telling you that x=y. Which means that you can write the left side of the equation x^2-2x+1. Now you need to determine if x^2-2x+1 is equal to 1. For different values of x you can get different answers and hence statement 2 is not sufficient.
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Re: Quantitative Guide 80 does (x-1)(y-1)=1? [#permalink]
02 Sep 2010, 10:13
sjgudapa wrote: How is statement 2 alone insufficient? Substitute y for x in the question stem to get x^2 = 2x. x^2=2x has two solutions, x=0 and x=2. Zero is ruled out by the question stem, leaving x=y=2. Thus, (x-1)(y-1)=1. Any help would be appreciated
Statement 2 is telling you that x=y. Which means that you can write the left side of the equation x^2-2x+1. Now you need to determine if x^2-2x+1 is equal to 1. For different values of x you can get different answers and hence statement 2 is not sufficient. Your approach is wrong. You are combining both the statements. If one of the statement is sufficient then answer is either A or D. Answer is D only when B is also sufficient alone. We need to check if (x-1)(y-1) = 1 put x=y we get if (x-1) ^2 = 1. Since we can not solve this equation as we do not have the value of x. This is not sufficient. Hence A
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Re: Quantitative Guide 80 does (x-1)(y-1)=1? [#permalink]
02 Sep 2010, 10:32
gurpreetsingh , I am not combining 2 statements. I was explaining to BeavisMan why Statement 2 is insufficient.
If you look at the question stem it is asking if (x-1) (y-1) = 1?
By just looking at Statement 2 you are told that x=y which means that I can re-write the left side of the equation that I am being questioned about as (x-1) (x-1) = x^2 - 2x +1 . But this still does not tell me where it is equal to 1 or not. If I substitute x as 2 , I will get an answer "YES" , if I substitute other values of x , then I will not get a value equal to 1 and the answer is "NO" . Hence Statement 2 is insufficient.
In your approach you are changing a YES / NO data sufficiency question to finding out the value of x. I do not think that is the right approach.
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Re: Quantitative Guide 80 does (x-1)(y-1)=1? [#permalink]
02 Sep 2010, 10:36
I forgot to mention that I was explaining for the red portion. Pls let me know where I have changed the question from yes/no to finding the value of x?
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Re: Quantitative Guide 80 does (x-1)(y-1)=1? [#permalink]
02 Sep 2010, 10:51
Sorry Gurpreet, i felt you were solving an equation to get x. But I see what you were getting at.
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Re: Quantitative Guide 80 does (x-1)(y-1)=1? [#permalink]
02 Sep 2010, 16:01
From Statement 2 you still get x^2-2x+1=1 or x*(x-2)=0. So x is either 0 or 2. Obviously x cannot be 0 hence x=2 and thence y=2. This is sufficient to find the value of (x-1)(y-1)=1.
Am I still missing something here?
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Re: Quantitative Guide 80 does (x-1)(y-1)=1? [#permalink]
03 Sep 2010, 02:47
TehJay wrote: zuperman wrote: From Statement 2 you still get x^2-2x+1=1 or x*(x-2)=0. So x is either 0 or 2. Obviously x cannot be 0 hence x=2 and thence y=2. This is sufficient to find the value of (x-1)(y-1)=1.
Am I still missing something here? You're misreading the question - they're ASKING you if (x-1)(y-1)=1, not TELLING you. You're assuming that's true and solving for x to fit the question. But by the criteria in statement 2, what if x=y=5? Then (x-1)(y-1) = (4)(4) = 16 =/= 1.  I was over-engineering it I think.Cheers.
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Re: Quantitative Guide 80 does (x-1)(y-1)=1? [#permalink]
07 Sep 2010, 02:51
I am bit sceptic to post this. But I wanted to check if this approach is right.
From the question we know that X and Y both are greater than 0.
In the statement 2 we could use simple plug-ins. If x=y=2 then (x-1)(y-1)=1. However, if x=y=3 then (x-1)(y-1) is not equal to 1.
I would greatly appreciate if you let me understand any loopholes in this approach.
Cheers
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Re: Quantitative Guide 80 does (x-1)(y-1)=1? [#permalink]
07 Sep 2010, 02:58
guytree wrote: I am bit sceptic to post this. But I wanted to check if this approach is right.
From the question we know that X and Y both are greater than 0.
In the statement 2 we could use simple plug-ins. If x=y=2 then (x-1)(y-1)=1. However, if x=y=3 then (x-1)(y-1) is not equal to 1.
I would greatly appreciate if you let me understand any loopholes in this approach.
Cheers That approach works to show that (B) isn't sufficient, but note that you don't know that both X and Y are greater than 0 - they could both be less than 0 as well.
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Re: Quantitative Guide 80 does (x-1)(y-1)=1? [#permalink]
07 Sep 2010, 03:02
hey thanks Tehjay.
I have another question related to this. What will be the primary goal here to prove (B) insufficient or do we need to prove the accuracy of the values of x and y.
I mean even when x and y could be less than 0; the point would be still be that the statement 2 gives a both yes and no as an answer or is it not?
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Re: If xy > 0 does (x-1)(y-1)=1 ? (1) x + y = [#permalink]
14 Jan 2012, 16:45
If xy > 0 does (x-1)(y-1)=1?
Rewrite the question as:
does xy - x - y + 1 = 1?
does xy - x - y = 0?
(1) x + y = xy
Subtract (x + y) from both sides.
xy - x - y = 0
By definition, every value of (x, y) satisfies the rephrased equation.
(2) x = y
While some values for x = y satisfy the prompt equation, other values do not.
If x = y = 2, then xy - x - y = 0, and the answer to the question is yes.
If x = y = 1, then xy - x - y = -1, and the answer to the question is no.
As per the above comments, notice that substituting x for y leads to another question, whose answer may be yes or no depending on actual values of x=y.
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Re: If xy > 0 does (x-1)(y-1)=1 ? (1) x + y =
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14 Jan 2012, 16:45
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