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Re: If x^2 = y^2, is true that x>0? [#permalink]
20 May 2012, 23:38
9
This post received KUDOS
Expert's post
8
This post was BOOKMARKED
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.
Re: If x^2 = y^2, is true that x>0? [#permalink]
23 May 2012, 11:02
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.
Re: If x^2 = y^2, is true that x>0? [#permalink]
26 May 2012, 04:09
for those of you who hate absolute value: Given x^2=y^2 Is x>0? 1. y=(x-1)/2, x^=(x-1)^2/4. Solving for x [1/3, -1]. Not sufficient 2. y<=-1 Nothing about x. Not Sufficient. Together: (x-1)/2 <= -1, x<= -1 Answer is NO. C.
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! _________________
Far better is it to dare mighty things, to win glorious triumphs, even though checkered by failure... than to rank with those poor spirits who neither enjoy nor suffer much, because they live in a gray twilight that knows not victory nor defeat. - T. Roosevelt
Re: If x^2 = y^2, is true that x>0? [#permalink]
28 Aug 2012, 09:13
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
Re: If x^2 = y^2, is true that x>0? [#permalink]
29 Aug 2012, 00:48
1
This post received KUDOS
Expert's post
venom2330 wrote:
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
Re: If x^2 = y^2, is true that x>0? [#permalink]
31 Aug 2012, 06:48
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= -4Y-1,
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? _________________
Re: If x^2 = y^2, is true that x>0? [#permalink]
31 Aug 2012, 07:17
Expert's post
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= -4Y-1,
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. _________________
Re: If x^2 = y^2, is true that x>0? [#permalink]
31 Aug 2012, 07:41
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= -4Y-1,
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.......... _________________
Re: If x^2 = y^2, is true that x>0? [#permalink]
02 Mar 2013, 09:38
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
Re: If x^2 = y^2, is true that x>0? [#permalink]
02 Mar 2013, 13:10
1
This post received KUDOS
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.
Answer-C
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
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