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

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

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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? [#permalink]

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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: Is x > 0? [#permalink]

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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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New post 29 Aug 2012, 01:48
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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

am i wrong somewhere...?


-1/3 is greater than -1, so y cannot be -1/3.

Hope it helps.
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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.

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

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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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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New post 21 May 2013, 00:56
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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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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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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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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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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 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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New post 26 May 2012, 05:09
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.

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

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

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

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New post 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= -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..........
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Re: If x^2 = y^2, is true that x>0?   [#permalink] 31 Aug 2012, 08:41

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