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If x^2 = y^2, is true that x > 0 ? (1) x = 2y+1 (2) y <= 1
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20 May 2012, 19:02
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If x^2 = y^2, is true that x > 0 ? (1) x = 2y+1 (2) y <= 1
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Re: If x^2 = y^2, is true that x > 0 ? (1) x = 2y+1 (2) y <= 1
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21 May 2012, 00:38
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.
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Re: If x^2 = y^2, is true that x > 0 ? (1) x = 2y+1 (2) y <= 1
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18 Jun 2012, 17:16
The way I usually approach these problems is with plugging in examples, like 1, 0, and 1 to see when the equations hold true.
1) x = 2y +1
Here, we actually don't need to do anything. Obviously we can't know if x>0 if we don't know the value of y.
2) y<=1
Alone, this obviously tells us nothing about x! Insufficient. Answer is either C or E.
1+2)
Let's plug a few values of y that are <= 1 into the equation from situation #1 to see how they affect x:
x=2y+1 x=2(1)+1 = 1 x=2(2)+1 = 3 x=2(3)+1 = 5 Obviously, this will continue as a series.
Therefore, we clearly know that x will never be >0 and therefore, C is the answer.




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Re: If x^2 = y^2, is true that x > 0 ? (1) x = 2y+1 (2) y <= 1
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31 Aug 2012, 07:48
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 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?
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Re: If x^2 = y^2, is true that x > 0 ? (1) x = 2y+1 (2) y <= 1
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31 Aug 2012, 08:17
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 ? (1) x = 2y+1 (2) y <= 1
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02 Mar 2013, 10:38
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! I did a bit of a unnecessarily long way but got to the wrong answer and was hoping someone could tell me what my logical error is. i squared both sides of statement 1 and got to \(x^2 = 4y^2 + 4y +1\) then I replaced \(x^2\) with \(y^2\) and got the same as everyone else that y = 1 or y = (1/3) NOT sufficient Using statement B i eliminated y=(1/3) but where I got it wrong is that I thought that since y = 1 then x could be equal to +/ 1 so I chose E Am i missing something? Thank you in advance or any responses!



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Re: If x^2 = y^2, is true that x > 0 ? (1) x = 2y+1 (2) y <= 1
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02 Mar 2013, 14:10
Hello Alexpavlos, You are right in assuming that since y=1 and x^2=y^2, x=+/1. However, remember that equation in statement 1 also gives you a relationship between statement x and y. The correct answer should satisfy all the data. Now the question you should ask yourself is whether it is possible to use the information in statement 2 and 1 together to get a single value for x. Statement 1 mentions that x=2y+1 Substituting y=1 in the this equation, we get x=1 and hence, x<0. Hence, together the two statements suffice. AnswerC Hope this helps! Let me know if you need any further clarification. alexpavlos 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! I did a bit of a unnecessarily long way but got to the wrong answer and was hoping someone could tell me what my logical error is. i squared both sides of statement 1 and got to \(x^2 = 4y^2 + 4y +1\) then I replaced \(x^2\) with \(y^2\) and got the same as everyone else that y = 1 or y = (1/3) NOT sufficient Using statement B i eliminated y=(1/3) but where I got it wrong is that I thought that since y = 1 then x could be equal to +/ 1 so I chose E Am i missing something? Thank you in advance or any responses!



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Re: If x^2 = y^2, is true that x > 0 ? (1) x = 2y+1 (2) y <= 1
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10 May 2013, 22:50
burnttwinky wrote: If x^2 = y^2, is true that x>0?
(1) x=2y+1
(2) y<= 1 Nothing new to add. Maybe another approach: From F.S 1, we know that x = 2y+1. Also \(x^2 = y^2\) Upon adding,\(x^2+x = (y+1)^2\).Thus, x(x+1) > 0> x >0 OR x <1.Insufficient. From F.S 2, we know that y <1. Clearly Insufficient. On combining both, we know that y = (x1)/2 > (x1)/2 <1>x <1. Sufficient. C.
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Re: If x^2 = y^2, is true that x > 0 ? (1) x = 2y+1 (2) y <= 1
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19 May 2013, 11:08
If x^2 = y^2, is it true that x > 0?
1. x = 2y + 1 2. y <= 1
from 1
x^2 = 4y^2 + 4y+1 so now 4y^2 + 4y+1 = y^2 therefore 3y^2 + 4y+1 = 0 ..... (3y + 1) (y+1) = 0 and thus y is either 1/3 or y = 1
subst in 1
x = 2/3 + 1 > 0 or x = 2+1<0 ......insuff
from 2 alone obviously not suff
both ........ y = 1 ........x = 1 too ....suff.......C



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Re: If x^2 = y^2, is true that x > 0 ? (1) x = 2y+1 (2) y <= 1
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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 ? (1) x = 2y+1 (2) y <= 1
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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 ? (1) x = 2y+1 (2) y <= 1
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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 ? (1) x = 2y+1 (2) y <= 1
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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 ? (1) x = 2y+1 (2) y <= 1
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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 ? (1) x = 2y+1 (2) y <= 1
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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 ? (1) x = 2y+1 (2) y <= 1
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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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Re: If x^2 = y^2, is true that x > 0 ? (1) x = 2y+1 (2) y <= 1
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12 Nov 2014, 10:48
sinhap07 wrote: Bunuel wrote: sinhap07 wrote: 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. That's not correct. For (1) it's possible that x = 1 and y = 1 or x = 1/3 and y = 1/3. The correct answer is C, not B or A.
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Re: If x^2 = y^2, is true that x > 0 ? (1) x = 2y+1 (2) y <= 1
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29 Nov 2014, 06:57
VeritasPrepKarishma wrote: 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 Hi Karishma I did get C as the answer applying the same method that you have indicated here. But if we solve through using variables, are we not getting x in the positive range? Applying the scenario where x=y or y=x



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Re: If x^2 = y^2, is true that x > 0 ? (1) x = 2y+1 (2) y <= 1
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30 Nov 2014, 22:24
sinhap07 wrote: Hi Karishma
I did get C as the answer applying the same method that you have indicated here. But if we solve through using variables, are we not getting x in the positive range? Applying the scenario where x=y or y=x
x = y implies either x = y or x = y. Which is the actual case, we do not know. Stmnt 2 gives us y is negative. But do we know whether x = y or x = y? No. If x = y, x is negative. If x = y, x is positive. So stmnt 2 alone is not sufficient to say whether x is positive or negative. It could be either.
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Re: If x^2 = y^2, is true that x > 0 ? (1) x = 2y+1 (2) y <= 1
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20 Jan 2017, 03:42
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 & Karishma, I will appreciate if you can comment: 1 How can we see this question and decide whether we need to pick numbers or not? 2 In other words, combining the two statements here is fairly easy, but how do we get it to a 5050 between C and E in 5 to 10 seconds? 3  In other words, how do we ensure that we don't waste time plugging in numbers in statement 1?




Re: If x^2 = y^2, is true that x > 0 ? (1) x = 2y+1 (2) y <= 1
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