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Re: If xy > 0 does (x1)(y1)=1 ?
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18 May 2015, 03:49
Bunnel,
Option B turns out to be x=0 or x=2.X cannot be equal to zero but we can use x=2 right.In that case we will get (x1)(y1)=1



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Re: If xy > 0 does (x1)(y1)=1 ?
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18 May 2015, 04:29
kirtivardhan wrote: Bunnel,
Option B turns out to be x=0 or x=2.X cannot be equal to zero but we can use x=2 right.In that case we will get (x1)(y1)=1 Dear kirtivardhanYour question tells me that you analyzed St. 2 in one of the following two ways. I'll list them both here and discuss the error in them. Way 1 You did your analysis of St. 2 as follows:"Put x = y in x + y = xy => \(2x = x^2\) Upon solving, x = 0 or x = 2"
The error a student who analyses St. 2 in this way does is that he is not considering the equation x = y alone (which is the only piece of info that St. 2 gives) but instead has mistakenly carried over information from St. 1 (x + y = xy) into his analysis of St. 2. Way 2 You did your analysis of St. 2 as follows:"Given that x = y We need to find if \((x1)^2 = 1\)? That is, if \((x1)^2  1 = 0\) or (x11)(x1+1) = 0 That is, x(x2) = 0 That is, x = 0 or x = 2
We're given that xy > 0. This means, x cannot be EQUAL TO zero. So, x = 2"The error a student who analyses St. 2 in this way does is that he has used the equation \((x1)^2 = 1\) as a FACT, not as something to be verified ____________ The correct analysis of St. 2 would be as under: From the question statement, we know that x and y have same sign and both are not equal to zero
From St. 2, x = y
Using this, we've to determine if (x1)(y1) = 1? That is, if \((x1)^2 = 1\)? That is, if x(x2) = 0
That is, we need to determine if x = 0 or x = 2.
We know that x cannot be equal to 0
So, we need to determine if x = 2. If x = 2, the equation in the question will hold true. For other values of x, the equation in the question will not hold true.
Since we don't know if x = 2 or not, St. 2 is insufficient to arrive at a unique answer for the given question.Hope this discussion helped! Regards, Japinder
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Re: If xy > 0 does (x1)(y1)=1 ?
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18 May 2015, 04:39
Thanks Japinder,
You mean to say that we plugged option b in the question asked and we got x=0 and x=2.If x would have been mentioned to be 2 somehow then option b would have been suff.
I am making sense?
In short ,what we are trying to do is we are plugging an option in to the question asked and checking whether the result is there in the fact
Regards



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Re: If xy > 0 does (x1)(y1)=1 ?
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18 May 2015, 05:05
kirtivardhan wrote: Thanks Japinder,
You mean to say that we plugged option b in the question asked and we got x=0 and x=2.If x would have been mentioned to be 2 somehow then option b would have been suff.
I am making sense?
In short ,what we are trying to do is we are plugging an option in to the question asked and checking whether the result is there in the fact
Regards Dear kirtivardhanYes, you're absolutely right in your first statement. If we were given, either in the question statement or in St. 2 itself that x = 2, then Option B would have been sufficient. To answer your last statement, let me reiterate what we are trying to do here in Analysis of St. 2: 1. Info given in Question statement: xy > 0. This is a FACT 2. Info given in St. 2: x = y. This is also a FACT Using these 2 facts, we need to confirm if (x1)(y1) = 1? This is the QUESTION. By using Facts 1 and 2 to simplify the question, we saw that the answer to this question is YES, if x = 2 and NO, if x has some other value. Hope this clarification helped! Japinder
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Re: If xy > 0 does (x1)(y1)=1 ?
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19 May 2015, 03:57
Bunuel wrote: 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 plugins. If x=y=2 then (x1)(y1)=1. However, if x=y=3 then (x1)(y1) is not equal to 1.
I would greatly appreciate if you let me understand any loopholes in this approach.
Cheers If \(xy>0\) does \((x1)(y1)=1\)? (1) \(x + y = xy\) (2) \(x=y\) \(xy>0\) 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 \((x1)(y1)=1\)? > is \(xyxy+1=1\)? 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 \(x(x2)=0\)? > 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. Hi Bunuel, I cannot find any values that represent xy = x+y Can you provide some explanation?



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Re: If xy > 0 does (x1)(y1)=1 ?
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19 May 2015, 04:12
LaxAvenger wrote: Bunuel wrote: If \(xy>0\) does \((x1)(y1)=1\)? (1) \(x + y = xy\) (2) \(x=y\)
\(xy>0\) 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 \((x1)(y1)=1\)? > is \(xyxy+1=1\)? 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 \(x(x2)=0\)? > 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. Hi Bunuel, I cannot find any values that represent xy = x+y Can you provide some explanation? LaxAvengerxy = x+y So, xy  x = y => x(y1) = y => \(x = \frac{y}{(y1)}\) From this equation, you can find a number of (x,y) pairs. Example, when y = 2, x = 2 When y = 3, x = 3/2 etc. Please note that you're not told that x and y are integers. So, you should not assume it. Hope this helped! Japinder
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Re: If xy > 0 does (x1)(y1)=1 ?
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26 Jun 2016, 05:28
Is it valid to solve 1) as: is xyxy+1=1 ? > since xy=x+y we get: is x+yxy=0 ? same as is 0=0 ? answer: yes it is. sufficient.



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Re: If xy > 0 does (x1)(y1)=1 ?
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19 Sep 2016, 07:17
A is correct. Here's why:
(1) x+y = xy > Plug into main equation
xyxy+1 = 1 xy xy = 0 x+yxy = 0 0 = 0
SUFFICIENT
(2) x=y > Rewrite as (y1)^2 = 1 > (y1) = +/ 1
NOT SUFFICIENT



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Re: If xy > 0 does (x1)(y1)=1 ?
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21 May 2017, 05:17
jakolik wrote: Hi,
The question is: (x1)(y1)=1 or xyyx+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 hey just a question, why can we not divide both sides by x to get x=2 I know its wrong but could someone pls let me know why it is wrong ?
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Re: If xy > 0 does (x1)(y1)=1 ?
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21 May 2017, 05:40
TehJay wrote: zuperman wrote: From Statement 2 you still get x^22x+1=1 or x*(x2)=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 (x1)(y1)=1. Am I still missing something here? You're misreading the question  they're ASKING you if (x1)(y1)=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 (x1)(y1) = (4)(4) = 16 =/= 1. I made exactly the same mistake. I assumed that (x1)(y1) = 1 and then x=y > 2=2
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Re: If xy > 0 does (x1)(y1)=1 ?
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23 May 2017, 06:51
daviddaviddavid wrote: jakolik wrote: Hi,
The question is: (x1)(y1)=1 or xyyx+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 hey just a question, why can we not divide both sides by x to get x=2 I know its wrong but could someone pls let me know why it is wrong ? Hey, If we do not whether the variable x is greater than zero or not, we cannot divide both sides by x and get x = 2. \(x^2 = 2x\) \(x^2 2x = 0\) \(x(x2) = 0\)
Therefore the value of x can be 0 or 2. If you divide \(x^2 = 2x\) by \(x\) on both sides, you are assuming that x is not equal to 0 and thus this division would make sense: \(\frac{x^2}{x} = \frac{2x}{x}\) . If \(x = 0\) then, \(\frac{x^2}{0} = \frac{2x}{0}\) is not valid as such. Hence, if one does not know whether x is 0 or nonzero, we take all the terms to the lefthand side of the equation and then solve for the value of x. Thanks, Saquib Quant Expert eGMAT
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Re: If xy > 0 does (x1)(y1)=1 ?
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