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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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23 Jan 2017, 00:55
dimitri92 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 & 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? Basically there are two distinct strategies  Number plugging (for example a percentages question in which one set will certainly give the answer) and Number testing (such as this question in which you need to test for various values) I swear by number plugging but number testing is almost never my strategy of choice. I might want to try one set to understand the question/logic better but I will rarely try to deduce my answer from number testing. So when I see x^2 = y^2, I automatically think x = y which implies x = y or x = y or x^2  y^2 = (x + y) * (x  y) which implies x = y or x = y This makes me jump to stmnt 2 immediately since it is clearly insufficient. This brings me to (C) or (E).
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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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27 Jun 2017, 08:49
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. Hello Bunuel, Why did u take only two possibilities? x2=y2x2=y2 > x=yx=y > either y=xy=x or y=−xy=−x we can have x =  y and x = y I'm I missing something? Posted from my mobile device
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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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27 Jun 2017, 09:48
ydmuley 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. Hello Bunuel, Why did u take only two possibilities? x2=y2x2=y2 > x=yx=y > either y=xy=x or y=−xy=−x we can have x =  y and x = y I'm I missing something? Posted from my mobile devicex = y is the same as x = y and x = y is the same as x = y.
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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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27 Jun 2017, 23:55
burnttwinky wrote: If x^2 = y^2, is true that x>0?
(1) x=2y+1
(2) y<= 1 Given : x^2 = y^2 => x=+/y DS : x>0 Option 1 : x = 2y+1 x^2 = [(x1)/2]^2 => 4x^2 = x^2 +1 2x => 3x^2 +2x 1 = 0 => (x+1) (3x1) = 0 => x= 1, 1/3 NOT SUFFICIENT Option 2 : y <= 1 So there can be many values of x . x<=1 or x>=1 NOT SUFFICIENT Combined : x= 1,1/3 so y = 1, 1, 1/3, 1/3 But y <=1 So x = 1 is possible SUFFICIENT Answer 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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01 Jul 2017, 05:03
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. In the following statement, (1)+(2) Since from (2) \(y\leq{1}\) then from (1) \(y=x=1\), so the answer to the question is NO. Sufficient. what if, y=x=1 ? Then x=1>0 and hence (1)+(2) = insufficient
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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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01 Jul 2017, 05:25
rekhabishop 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. In the following statement, (1)+(2) Since from (2) \(y\leq{1}\) then from (1) \(y=x=1\), so the answer to the question is NO. Sufficient. what if, y=x=1 ? Then x=1>0 and hence (1)+(2) = insufficientFrom (1) we have only two possibilities: a. \(y=x=1\) b. \(y=x=\frac{1}{3}\) y = 1 and x = 1 does not satisfy the first statement.
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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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13 Nov 2017, 19:13
Don't we assume x<0 when we infer x=y from x^2=y^2 (and when x>0 x=y)? Isn't this the normal procedure when we have x=something. I got x=y and x=y from the statement, but then got confused by the assumptions made to open the abs value. Thanks.



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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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Re: If x^2 = y^2, is true that x > 0 ? (1) x = 2y+1 (2) y <= 1
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19 Nov 2017, 10:54
Bunuel wrote: alekkx wrote: Don't we assume x<0 when we infer x=y from x^2=y^2 (and when x>0 x=y)? Isn't this the normal procedure when we have x=something. I got x=y and x=y from the statement, but then got confused by the assumptions made to open the abs value. Thanks. You could use number plugging to check whether you are right. x^2 = y^2: The above could be true for example if x = y = 1 or if x = 1 and y = 1. These examples answer no to both of your question. Generally x^2 = y^2 means that x = y, so x and y are at the same distance from 0 on the number line. Thanks. What are the implications of y<x, is it "y<x or y>x" or something different?



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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 Nov 2017, 11:01
alekkx wrote: Bunuel wrote: alekkx wrote: Don't we assume x<0 when we infer x=y from x^2=y^2 (and when x>0 x=y)? Isn't this the normal procedure when we have x=something. I got x=y and x=y from the statement, but then got confused by the assumptions made to open the abs value. Thanks. You could use number plugging to check whether you are right. x^2 = y^2: The above could be true for example if x = y = 1 or if x = 1 and y = 1. These examples answer no to both of your question. Generally x^2 = y^2 means that x = y, so x and y are at the same distance from 0 on the number line. Thanks. What are the implications of y<x, is it "y<x or y>x" or something different? x is the distance from x to 0 on the number line, so y < x means that x if further from 0 than y is. xy0 x0y y0x 0yx
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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 Nov 2017, 11:31
Bunuel wrote: x is the distance from x to 0 on the number line, so y < x means that x if further from 0 than y is.
xy0 x0y y0x 0yx Thanks Bunuel. It looks like I did not capture all cases with my inequalities. For example, when y>0, I have y is always less than x, which is not true based on case 2. When faced with this condition on DS, is the best way to draw out the number line like this? How does this compare to getting a condition that says x5>2 in terms of how you would get the constraints.



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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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Re: If x^2 = y^2, is true that x > 0 ? (1) x = 2y+1 (2) y <= 1
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17 Sep 2018, 14:57
Kris01 wrote: 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! Hi chetan2u, Bunuel, I did understand the solution that you have shared , but what did i do wrong in my approach. I am getting answer E . I understand correct answer is C here is how i tried to solve it So we are given that x^2 =y^2 means we have x=y So we have four cases from it ( understand that it finally transoms to two cases but just listed the cases as i did in my original attempt) \(x\geq{0}\) , y\(\geq{0}\) we have x=y \(x\geq{0}\) , y< 0 we have x=y x<0 , y\(\geq{0}\) we have x=y x<0 , y<0 we have x=y Now statement A says : x= 2y+1 squaring on both sides \(x^2\)= (2y+1)^2 now since we are given that\(x^2\)=\(y^2\)replace \(x^2\) with\(y^2\) we get y=1 or/and y=\(\frac{1}{3}\) now we can have x=y=1 or x=1 and y=1 x=y=\(\frac{1}{3}\) or x=\(\frac{1}{3}\) and y=\(\frac{1}{3}\) Statement 2: \(y\leq{1}\) we will have either x>0 or x<0 because \(x\geq{0}\) , y< 0 or x<0 , y<0 Now combing 1 and 2 we have y=1 so two cases are applicable ( i am referring to 4 cases that show relation between x and y) x<0 , y<0 we have x=y we will get x=y=1 and for case this we get \(x\geq{0}\) , y< 0 we have x=y x=1 and y=1 I did not understand what Kris01 meant to consider true for all statements. I considered both statements and came to conclusion that y=1 Where did i go wrong . Please help me understand my mistake . Probus



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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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17 Sep 2018, 17:47
Probus wrote: Hi chetan2u, Bunuel, I did understand the solution that you have shared , but what did i do wrong in my approach. I am getting answer E . I understand correct answer is C here is how i tried to solve it So we are given that x^2 =y^2 means we have x=y So we have four cases from it ( understand that it finally transoms to two cases but just listed the cases as i did in my original attempt) \(x\geq{0}\) , y\(\geq{0}\) we have x=y \(x\geq{0}\) , y< 0 we have x=y x<0 , y\(\geq{0}\) we have x=y x<0 , y<0 we have x=y Now statement A says : x= 2y+1 squaring on both sides \(x^2\)= (2y+1)^2 now since we are given that\(x^2\)=\(y^2\)replace \(x^2\) with\(y^2\) we get y=1 or/and y=\(\frac{1}{3}\) now we can have x=y=1 or x=1 and y=1 x=y=\(\frac{1}{3}\) or x=\(\frac{1}{3}\) and y=\(\frac{1}{3}\) Statement 2: \(y\leq{1}\) we will have either x>0 or x<0 because \(x\geq{0}\) , y< 0 or x<0 , y<0 Now combing 1 and 2 we have y=1 so two cases are applicable ( i am referring to 4 cases that show relation between x and y) x<0 , y<0 we have x=y we will get x=y=1 and for case this we get \(x\geq{0}\) , y< 0 we have x=y x=1 and y=1 I did not understand what Kris01 meant to consider true for all statements. I considered both statements and came to conclusion that y=1 Where did i go wrong . Please help me understand my mistake . Probus Hi... You have gone wrong in not considering equation 1 while taking value of X.. 1) X=2y+1 You got two values of y as 1/3 and 1 Now substitute y as 1/3 , X=2*(1/3)+1=1/3 has 1, x=2*(1)+1=2+1=1 So only two values are left 1/3 and 1 for X.. x=y may give as 1 as a value but statement I negates it 2) y<=1 Nothing about X Combined it tells you that y=1 and so X=1 Sufficient C
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