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If x^2 = y^2, is true that x>0? [#permalink]
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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: Is x > 0? [#permalink]
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18 Jun 2012, 17:16
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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? [#permalink]
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29 Aug 2012, 01:48



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Re: If x^2 = y^2, is true that x>0? [#permalink]
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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 it true that x > 0? [#permalink]
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19 May 2013, 11:08
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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 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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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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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? [#permalink]
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23 May 2012, 12:02
picked C. if x^2 = y^2 means that x = y with statement 1 considering when y is negative only if y =1 mod x and mod y will be equal (really both will be 1). (if y is positive x will not be equal to y) statement B says y less than or equal to 1 i.e. x =1 hence we have a unique ans.



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Re: If x^2 = y^2, is true that x>0? [#permalink]
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26 May 2012, 05:09
for those of you who hate absolute value: Given x^2=y^2 Is x>0? 1. y=(x1)/2, x^=(x1)^2/4. Solving for x [1/3, 1]. Not sufficient 2. y<=1 Nothing about x. Not Sufficient. Together: (x1)/2 <= 1, x<= 1 Answer is NO. C.



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If x^2 = y^2, is it true that x > 0? [#permalink]
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18 Jun 2012, 16:44
If x^2 = y^2, is it true that x > 0? (1) x = 2y + 1 (2) y<= 1 I got the correct answer, but I got stuck. Below is my solution. When \(x ^ 2 = Y ^2\) then \(x = y\) and \(x = y\) Considering statement 1 \(x = 2y + 1\)when x = y then y = 1 and x = 1 i.e negative. When y =1 then x is positive. Therefore, insufficient. Considering statement 2 if x = y then x < = 1 i.e. negative if x = y then x <=1 or x >=1 i.e. positive . Therefore, insufficient. Now I am strugling to combine the two and get the answer. can someone please help?
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Re: Is x > 0? [#permalink]
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19 Jun 2012, 01:10
Here's my take on the problem Statement (1) is clearly insufficient. Matter of fact, it tells us that x is odd. LOL wtf? Statement (2) is clearly insufficient since there's no mention of x. Now let's see what we have: X = 2y + 1 and y <  1 I then make y <  1 into y = Lessthan(1) < I am just translating it. Don't panic. Now: plug in y into x = 2y + 1 we have x = 2(lessthan 1) + 1 x = (less than 2) + 1 < by simple arithmetic x = less than 1 which means that x < 1 Now we can conclude that x is indeed less than 0 Sufficient. *by the way, vandygrad11's way is also superb. need to familiarize ourselves in both ways. P.S. haven't replied to vandygrad11's reply yet in re: Big4 reputation. will do so when I get home. Cheers!
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Re: If x^2 = y^2, is true that x>0? [#permalink]
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28 Aug 2012, 10:13
Sorry to open an old thread... if we solve like this...
X^2 =Y^2
from 1st x=2y+1 if we plug in the value in the above equation for x we get (2y+1)^2=y^2 4y^2 +4y+1 =y^2 3y^2 +4y +1=0 3y^2 +3y+y+1=0 3y(y+1)+(y+1)=0 (3y+1)(y+1)=0 we get 2 values of y for which x=4y+1 and x^2=y^2 i.e y=1/3 and 1 and x = 1/3 and 1 respectively
the second condition tells us that y<= 1 hence y can be both 1/3 and 1 and corresponding x values would be 1 and 1/3 a positive and negative Hence, answer should be E
am i wrong somewhere...?



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Re: If x^2 = y^2, is true that x>0? [#permalink]
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30 Aug 2012, 20:17
i am struggling with dsinequalities i have gone through the gmat club quant all the concepts were very clear in that except inequalities i didnot understand the concept n that book can anyone suggest me some other good resource
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Re: If x^2 = y^2, is true that x>0? [#permalink]
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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? [#permalink]
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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? [#permalink]
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31 Aug 2012, 08:41
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. Hi Bunnel I made a blunder by not considering 1/3, anyway thanks for the quick response..........
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Re: If x^2 = y^2, is true that x>0? [#permalink]
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20 Jan 2013, 21:14
great solution, thanks bunuel.
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Re: If x^2 = y^2, is true that x>0? [#permalink]
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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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