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If x^2 = y^2, is true that x>0?

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Re: If x^2 = y^2, is true that x>0? [#permalink]

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New post 28 Nov 2014, 13:04
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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.


thank you for this.
Strangely enough, GMAT Prep states x=2/3 (when y=(-1/3)!

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Re: If x^2 = y^2, is true that x>0? [#permalink]

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New post 29 Nov 2014, 05:57
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? [#permalink]

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New post 30 Nov 2014, 21:24
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? [#permalink]

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New post 03 Jan 2015, 03:54
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.



would I be making a mistake if I had plugged in values.
The first one statement was insufficient because we had two values of X based on whether Y was positive or Negative.

The second statement was insufficient because it had no information about x.

Combining both, we can easily observe for all negative values of Y, X is also Negative.

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If x^2 = y^2, is true that x>0? [#permalink]

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New post 20 Jan 2017, 02:42
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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 50-50 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?
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Re: If x^2 = y^2, is true that x>0? [#permalink]

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New post 22 Jan 2017, 23:55
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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 50-50 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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New post 27 Jun 2017, 07:49
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|=|y||x|=|y| --> either y=xy=x or y=−xy=−x

we can have -x = - y and -x = y

I'm I missing something?

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Re: If x^2 = y^2, is true that x>0? [#permalink]

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New post 27 Jun 2017, 08:48
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|=|y||x|=|y| --> either y=xy=x or y=−xy=−x

we can have -x = - y and -x = y

I'm I missing something?

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-x = -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? [#permalink]

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New post 27 Jun 2017, 09:51
Bunuel wrote:
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|=|y||x|=|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


-x = -y is the same as x = y and -x = y is the same as x = -y.


My bad, so dumb of me not to think that, looks like my brain was freezed after seeing this question. Thank you Bunuel.
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Re: If x^2 = y^2, is true that x>0? [#permalink]

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New post 27 Jun 2017, 22:55
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 = [(x-1)/2]^2
=> 4x^2 = x^2 +1 -2x
=> 3x^2 +2x -1 = 0
=> (x+1) (3x-1) = 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? [#permalink]

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New post 01 Jul 2017, 04:03
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? [#permalink]

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New post 01 Jul 2017, 04:25
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) = insufficient


From (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? [#permalink]

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New post 01 Jul 2017, 05:04
Bunuel wrote:
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) = insufficient


From (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.


Oh, my bad! Thanks a ton, Bunuel :)
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If x^2 = y^2, is true that x>0? [#permalink]

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New post 13 Nov 2017, 18:13
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? [#permalink]

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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.
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Re: If x^2 = y^2, is true that x>0? [#permalink]

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New post 13 Nov 2017, 22:49
Test for -1, 0, 1 condition

x = 2y +1

Here, we actually don't need to do anything. we can't know if x>0 if we don't know the value of y.

2) y<=-1

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:
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Re: If x^2 = y^2, is true that x>0? [#permalink]

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New post 19 Nov 2017, 09:54
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? [#permalink]

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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.

----x----y----0--------------
----x---------0----y---------
---------y----0---------x----
--------------0----y----x----
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Re: If x^2 = y^2, is true that x>0? [#permalink]

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New post 19 Nov 2017, 10:31
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.

----x----y----0--------------
----x---------0----y---------
---------y----0---------x----
--------------0----y----x----


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 |x-5|>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? [#permalink]

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New post 19 Nov 2017, 10:35
alekkx wrote:
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.

----x----y----0--------------
----x---------0----y---------
---------y----0---------x----
--------------0----y----x----


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 |x-5|>2 in terms of how you would get the constraints.


10. Absolute Value



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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: If x^2 = y^2, is true that x>0?   [#permalink] 19 Nov 2017, 10:35

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