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Re: If x^2 = y^2, is it true that x > 0? [#permalink]
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19 May 2013, 13:46
yezz wrote: ... and thus y is either 1/3 or y = 1
I believe 1/3 is discounted in the question where it states x^2 = y^2. If y = 1/3 then x = 2/3; these values for x and y contradict the question. y = 1/3 is rejected, leaving only y = 1. Why is my position not correct?



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Re: If x^2 = y^2, is it true that x > 0? [#permalink]
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20 May 2013, 04:32
Bunuel wrote: Please refer to the solutions provided. That is what I am doing, and the solutions provided sometimes appear to me to be incorrect. This is one of those times. Here is my working: stormbind wrote: If x^2 = y^2, is it true that x > 0?
1. x = 2y + 1 2. y <= 1
I answered "Statement 1 alone is sufficient"; Can you please show why my working is not correct?
x^2 = y^2 x^2 = (2y + 1)(2y + 1)
If x=1 or x=1 then x^2 = 1.
1 = (2y + 1)(2y + 1)
Using number substitution, only 1 is substitutable for y without breaking the x^2 = y^2 statement in the question:
i.e. (2*1 + 1)(2*1 + 1) = (4  2 2 + 1) = 1 = x^2.
Therefore y = 1
x = 2y + 1; so x = 1 also.
Statement 1 is sufficient. Where did I go wrong?
Thanks.
Official Solution states Statement 1 is insufficient because it renders y = 1/3 or y = 1. However, I believe that 1/3 is not a possible value for y because the question states x^2 = y^2. Explanation: If y = 1/3 then x = 2/3, which contradicts the question. Thus 1/3 is rejected, leaving only y = 1. This would make Statement 1 sufficient. Official Answer: C My Answer: A Please help me to understand why is my position not correct.



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Re: If x^2 = y^2, is it true that x > 0? [#permalink]
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20 May 2013, 10:50
stormbind wrote: Bunuel wrote: Please refer to the solutions provided. Explanation: If y = 1/3 then x = 2/3, which contradicts the question. Thus 1/3 is rejected, leaving only y = 1. This would make Statement 1 sufficient. This part is not correct. x = 2y+1. If y = 1/3, x is not equal to 2/3.
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Re: If x^2 = y^2, is true that x>0? [#permalink]
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20 May 2013, 13:20
Thanks vinaymimani!
Not seeing that was driving me round the bend.



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Re: If x^2 = y^2, is true that x>0? [#permalink]
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20 May 2013, 15:43
(1) X = 2y + 1 \(x^2 = 4y^2 + 1 + 4y\) \(y^2 = 4y^2 + 1 + 4y\) \(3y^2 + 4y + 1 = 0\) \(3y^2 + 3y +y +1 = 0\) \((3y+1)(y+1) = 0\) y = 1/3 or 1 x = 1/3 (Yes) or 1 (No) – Insufficient (2) Alone is insuff – Combining – we know Y = 1 Therefore x = 1 and thus NO it is not greater than 0 IMO (C)
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Re: If x^2 = y^2, is it true that x > 0? [#permalink]
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21 May 2013, 00:56
stormbind wrote: Bunuel wrote: Please refer to the solutions provided. That is what I am doing, and the solutions provided sometimes appear to me to be incorrect. This is one of those times. Here is my working: stormbind wrote: If x^2 = y^2, is it true that x > 0?
1. x = 2y + 1 2. y <= 1
I answered "Statement 1 alone is sufficient"; Can you please show why my working is not correct?
x^2 = y^2 x^2 = (2y + 1)(2y + 1)
If x=1 or x=1 then x^2 = 1.
1 = (2y + 1)(2y + 1)
Using number substitution, only 1 is substitutable for y without breaking the x^2 = y^2 statement in the question:
i.e. (2*1 + 1)(2*1 + 1) = (4  2 2 + 1) = 1 = x^2.
Therefore y = 1
x = 2y + 1; so x = 1 also.
Statement 1 is sufficient. Where did I go wrong?
Thanks.
Official Solution states Statement 1 is insufficient because it renders y = 1/3 or y = 1. However, I believe that 1/3 is not a possible value for y because the question states x^2 = y^2. Explanation: If y = 1/3 then x = 2/3, which contradicts the question. Thus 1/3 is rejected, leaving only y = 1. This would make Statement 1 sufficient. Official Answer: C My Answer: A Please help me to understand why is my position not correct. Consider y=x=1/3 > x^2=y^2 and x=2y+1. Hope it helps.
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Re: If x^2 = y^2, is true that x>0? [#permalink]
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11 Jun 2013, 16:57
Why do we have to solve for 'y' when we are looking for the value of x? I know how to get x=1 and x=1/3 but I'm still not sure why the values of y are relevant. Bunuel wrote: kotela wrote: Bunuel wrote: If x^2 = y^2, is true that x>0?
\(x^2 = y^2\) > \(x=y\) > either \(y=x\) or \(y=x\).
(1) x=2y+1 > if \(y=x\) then we would have: \(x=2x+1\) > \(x=1<0\) (notice that in this case \(y=x=1\)) but if \(y=x\) then we would have: \(x=2x+1\) > \(x=\frac{1}{3}>0\) (notice that in this case \(y=x=\frac{1}{3}\)). Not sufficient.
(2) y<= 1. Clearly insufficient.
(1)+(2) Since from (2) \(y\leq{1}\) then from (1) \(y=x=1\), so the answer to the question is NO. Sufficient.
Answer: C.
Hope it's clear. Hi Bunnel, I need a clarification here, when i put statement 1 in X^2 = Y^2 i get the below 3Y^2= 4Y1, and since LHS is positive so the RHS needs to be positive so in order to RHS to be positive Y should be "ve" and when Y is negative, Now considering statement 1 we can get to know that X is negative..... So my point is again statement 2 needed? Please clarify me if i am wrong? How did you get that if y is negative x must be positive? For (1) we have: \(x^2 = y^2\) and \(x=2y+1\). Solving gives: \(x=1\) and \(y=1\) OR \(x=\frac{1}{3}\) and \(y=\frac{1}{3}\), just substitute these values to check that they satisfy both equations.



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Re: If x^2 = y^2, is true that x>0? [#permalink]
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11 Jun 2013, 18:03
Can someone tell me if my reasoning is sound?
x^2=y^2 and thus, x=y or x=y
1.)
I.) x=2y+1
y=2y+1 y=1 y=1
y=2(1)+1 x=1
II.) x=2y+1
y=2y+1 3y=1 y=1/3
x=2(1/3)+1 x=1/3
Insufficient because we have two values for x. (however, using the same reasoning to arrive at c. cant we say that both answers we arrive at show that x ≠ 0 and thus, a is sufficient?)
2. y≤1 Not sufficent
1+2 2. says that y≤1. In #1, the only case where y≤1 is y=2y+1
y=1 y=1
y=2(1)+1 x=1
Here we get an answer of x=1 which is obviously ≠ to 0.



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Re: If x^2 = y^2, is true that x>0? [#permalink]
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11 Jun 2013, 22:33
WholeLottaLove wrote: Can someone tell me if my reasoning is sound?
x^2=y^2 and thus, x=y or x=y
1.)
I.) x=2y+1
y=2y+1 y=1 y=1
y=2(1)+1 x=1
II.) x=2y+1
y=2y+1 3y=1 y=1/3
x=2(1/3)+1 x=1/3
Insufficient because we have two values for x. (however, using the same reasoning to arrive at c. cant we say that both answers we arrive at show that x ≠ 0 and thus, a is sufficient?)
2. y≤1 Not sufficent
1+2 2. says that y≤1. In #1, the only case where y≤1 is y=2y+1
y=1 y=1
y=2(1)+1 x=1
Here we get an answer of x=1 which is obviously ≠ to 0. Everything you did is correct except that you misunderstood the question. The question is: Is x positive? Is x > 0? It does not ask you whether x is equal to 0. Statement I tells you that x could be positive or negative. So not sufficient. Both statements together tell you that x is negative. Hence it is not positive. It answers the question with 'No'. Sufficient.
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Re: If x^2 = y^2, is true that x>0? [#permalink]
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18 Jun 2013, 10:33
burnttwinky wrote: If x^2 = y^2, is true that x>0?
(1) x=2y+1
(2) y<= 1 S1: put x= y once and y once so we get 1/2 and 1 two values hence NS now s2: NS s1 and s2 : sufficient and answer is C Questions above 700 are either direct math or pure logic



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Re: If x^2 = y^2, is true that x>0? [#permalink]
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20 Jun 2013, 16:45
Question time:
for #2.) we are given y<=1. This is not sufficient because of the following:
x = y
x=y OR x=y
x=y OR x=(y) x=y
Correct?
Another thing that has bothered me is this. If x=y, and for example, y=5, then would x=(5)?
As always, thanks for the help!



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Re: If x^2 = y^2, is true that x>0? [#permalink]
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20 Jun 2013, 20:55
WholeLottaLove wrote: Question time:
for #2.) we are given y<=1. This is not sufficient because of the following:
x = y
x=y OR x=y
x=y OR x=(y) x=y
Correct?
I am not really sure what you have done here. The 4 cases will be x = y x = y x = y x = y which are equivalent to just two cases: x = y or x = y. Statement 2 is not sufficient because all we know now is that y is negative. IF x = y, x is negative. If x = y, x is positive. So we still don't know whether x is positive or not. WholeLottaLove wrote: Another thing that has bothered me is this. If x=y, and for example, y=5, then would x=(5)?
As always, thanks for the help! Yes. If x = y and y = 5, then x = 5
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Re: If x^2 = y^2, is true that x>0? [#permalink]
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21 Jun 2013, 20:49
Correct answer is E not C
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Re: If x^2 = y^2, is true that x>0? [#permalink]
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22 Jun 2013, 03:40



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Re: If x^2 = y^2, is true that x>0? [#permalink]
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27 Jun 2013, 13:05
If x^2 = y^2, is true that x>0?
If x^2 = y^2, then we can get the square root of both sides: √x^2 = √y^2 x = y x=y OR x=y
1) x=2y+1 (2) y<= 1
1) x=2y+1 So, we are looking for the value of x. x=y or x=y
I.) x=2y+1 x=2x+1 x=1 x=1
II.) x=2y+1 x=2x+1 3x=1 x=1/3
x could be greater than zero or less than zero. NOT SUFFICIENT
(2) y<= 1
x could = y or y. For example. x = y 2 = 2 OR 2 = 2
NOT SUFFICIENT
1 + 2) x=2y+1 and y<= 1 In 1) we have two cases, where x is negative and where x is positive. However, when x is negative y is positive and when x is positive y is negative. 2) tells us that y is negative meaning x must be positive or greater than zero. SUFFICIENT
(c)



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Re: If x^2 = y^2, is true that x>0? [#permalink]
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06 Sep 2014, 13:04
If x^2=y^2, is it true that X>0? x=y > x=y or x=y
1. x=2y+1, x is also equal to y. > 2y+1=y or y=1. If y=1, x=2(1)+1=2+1=1 x=2y+1, x is also equal to y. > 2y+1=y or y=(1/3). If y=(1/3), x=(2/3)+1=(2+3)/3=1/3 we see that x is positive in second case and not in the first. NOT SUFFICIENT
2. y<=1. This one is easy to notice that x can be anything as long as its magnitude is same as y's.
1+2: y can only be 1 from combining these two statement. This is also our case 1 from statement 1. Hence, x=1<0. SUfficient



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Re: If x^2 = y^2, is true that x>0? [#permalink]
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12 Nov 2014, 05:48
Bunuel wrote: If x^2 = y^2, is true that x>0?
\(x^2 = y^2\) > \(x=y\) > either \(y=x\) or \(y=x\).
(1) x=2y+1 > if \(y=x\) then we would have: \(x=2x+1\) > \(x=1<0\) (notice that in this case \(y=x=1\)) but if \(y=x\) then we would have: \(x=2x+1\) > \(x=\frac{1}{3}>0\) (notice that in this case \(y=x=\frac{1}{3}\)). Not sufficient.
(2) y<= 1. Clearly insufficient.
(1)+(2) Since from (2) \(y\leq{1}\) then from (1) \(y=x=1\), so the answer to the question is NO. Sufficient.
Answer: C.
Hope it's clear. Hi Bunuel Why cant answer be B From x^2=y^2 x and y can take the following values: 1 and 1 or 1 and 1 as x and y different so not considering same integers like both positive or both negative from 1st statement none of the values satisfy the given equation from 2nd statement, y is negative which makes x as positive. Where am I wrong?



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Re: If x^2 = y^2, is true that x>0? [#permalink]
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12 Nov 2014, 06:29
sinhap07 wrote: Bunuel wrote: If x^2 = y^2, is true that x>0?
\(x^2 = y^2\) > \(x=y\) > either \(y=x\) or \(y=x\).
(1) x=2y+1 > if \(y=x\) then we would have: \(x=2x+1\) > \(x=1<0\) (notice that in this case \(y=x=1\)) but if \(y=x\) then we would have: \(x=2x+1\) > \(x=\frac{1}{3}>0\) (notice that in this case \(y=x=\frac{1}{3}\)). Not sufficient.
(2) y<= 1. Clearly insufficient.
(1)+(2) Since from (2) \(y\leq{1}\) then from (1) \(y=x=1\), so the answer to the question is NO. Sufficient.
Answer: C.
Hope it's clear. Hi Bunuel Why cant answer be B From x^2=y^2 x and y can take the following values: 1 and 1 or 1 and 1 as x and y different so not considering same integers like both positive or both negative from 1st statement none of the values satisfy the given equation from 2nd statement, y is negative which makes x as positive. Where am I wrong? x and y can be the same number. Generally, unless it is explicitly stated otherwise, different variables CAN represent the same number. So, for (2) it's possible that x = y = 1, or x= 1 and y = 1, or x = y = 2, or x = y = 1.5, or x = 100 and y = 100...
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Re: If x^2 = y^2, is true that x>0? [#permalink]
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12 Nov 2014, 10:39
Bunuel wrote: sinhap07 wrote: Bunuel wrote: If x^2 = y^2, is true that x>0?
\(x^2 = y^2\) > \(x=y\) > either \(y=x\) or \(y=x\).
(1) x=2y+1 > if \(y=x\) then we would have: \(x=2x+1\) > \(x=1<0\) (notice that in this case \(y=x=1\)) but if \(y=x\) then we would have: \(x=2x+1\) > \(x=\frac{1}{3}>0\) (notice that in this case \(y=x=\frac{1}{3}\)). Not sufficient.
(2) y<= 1. Clearly insufficient.
(1)+(2) Since from (2) \(y\leq{1}\) then from (1) \(y=x=1\), so the answer to the question is NO. Sufficient.
Answer: C.
Hope it's clear. Hi Bunuel Why cant answer be B From x^2=y^2 x and y can take the following values: 1 and 1 or 1 and 1 as x and y different so not considering same integers like both positive or both negative from 1st statement none of the values satisfy the given equation from 2nd statement, y is negative which makes x as positive. Where am I wrong? x and y can be the same number. Generally, unless it is explicitly stated otherwise, different variables CAN represent the same number. So, for (2) it's possible that x = y = 1, or x= 1 and y = 1, or x = y = 2, or x = y = 1.5, or x = 100 and y = 100... Agreed Bunuel. Tried that too. But by that logic, Statement 1 holds true only when both x and y are negative and hence we could get A as the answer as we would know that y is negative along with x.




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