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If both x and y are non zero numbers , what is the value of (y/x)

1) x = 6 2) y square = x square

Why isn’t ‘C’ the correct answer? (E is the correct answer here)

If we know that x =6 and y squared is = square root ( 6 squared) = absolute value (6) Then absolute value (6)/6 = 1

If both x and y are nonzero numbers, what is the value of x/y?

Note that \(\sqrt{x^2}=|x|\).

(1) x = 6. Not sufficient.

(2) y^2 = x^2 --> \(|x|=|y|\) --> if \(x\) and \(y\) have the same sign then \(\frac{x}{y}=1\) (for example \(x=y=6\)) but if \(x\) and \(y\) have the opposite signs then \(\frac{x}{y}=-1\) (for example \(x=6\) and \(y=-6\)). Not sufficient.

(1)+(2) We still don't know whether \(x\) and \(y\) have the same sign or not. Not sufficient.

Answer: E.

About \(\sqrt{x^2}=|x|\).

The point here is that as square root function cannot give negative result then \(\sqrt{some \ expression}\geq{0}\).

So \(\sqrt{x^2}\geq{0}\). But what does \(\sqrt{x^2}\) equal to?

Let's consider following examples: If \(x=5\) --> \(\sqrt{x^2}=\sqrt{25}=5=x=positive\); If \(x=-5\) --> \(\sqrt{x^2}=\sqrt{25}=5=-x=positive\).

So we got that: \(\sqrt{x^2}=x\), if \(x\geq{0}\); \(\sqrt{x^2}=-x\), if \(x<0\).

What function does exactly the same thing? The absolute value function: \(|x|=x\), if \(x\geq{0}\) and \(|x|=-x\), if \(x<0\). That is why \(\sqrt{x^2}=|x|\).
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If we know that x =6 and y squared is = square root ( 6 squared) = absolute value (6) Then absolute value (6)/6 = 1

If we know that \(y^2 = 36\), then y can be 6 or -6. On the other hand, if it is given that y = \(\sqrt{36}\), we only consider the positive value i.e. 6. Here y can only be 6.

QR. 30 Number If both x and y are nonzero numbers, what is the value of x/y? (1) x = 6 (2) y^2 = x^2

I did wrong doing following: √y^2 = √x^2 so, y=x and answer c What wrong did i do.

Note that \(\sqrt{x^2}=|x|\).

If both x and y are nonzero numbers, what is the value of x/y?

(1) x = 6. Not sufficient.

(2) y^2 = x^2 --> \(|x|=|y|\) --> if \(x\) and \(y\) have the same sign then \(\frac{x}{y}=1\) (for example \(x=y=6\)) but if \(x\) and \(y\) have the opposite signs then \(\frac{x}{y}=-1\) (for example \(x=6\) and \(y=-6\)). Not sufficient.

(1)+(2) We still don't know whether \(x\) and \(y\) have the same sign or not. Not sufficient.

Answer: E.

About \(\sqrt{x^2}=|x|\).

The point here is that as square root function can not give negative result then \(\sqrt{some \ expression}\geq{0}\).

So \(\sqrt{x^2}\geq{0}\). But what does \(\sqrt{x^2}\) equal to?

Let's consider following examples: If \(x=5\) --> \(\sqrt{x^2}=\sqrt{25}=5=x=positive\); If \(x=-5\) --> \(\sqrt{x^2}=\sqrt{25}=5=-x=positive\).

So we got that: \(\sqrt{x^2}=x\), if \(x\geq{0}\); \(\sqrt{x^2}=-x\), if \(x<0\).

What function does exactly the same thing? The absolute value function: \(|x|=x\), if \(x\geq{0}\) and \(|x|=-x\), if \(x<0\). That is why \(\sqrt{x^2}=|x|\).
_________________

Let's consider following examples: If \(x=5\) --> \(\sqrt{x^2}=\sqrt{25}=5=x=positive\); If \(x=-5\) --> \(\sqrt{x^2}=\sqrt{25}=5=-x=positive\). it is positive or negative?

If \(x=-5=negative\) then \(-x=-negative=positive\).
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Stmt1: x=6. Not sufficient. Stmt2: \(y^2 = x^2\) x can be +6 y can be +6 Value \(y/x = 6/6 =1\) OR x can be -6 y can be +6 Value of \(y/x=-1\) Not sufficient.

Take (1) and (2) together, x=+6. y can be +6 or -6, giving value of y/x as 1 or -1 respectively. Hence not sufficient together.

OA E.
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My dad once said to me: Son, nothing succeeds like success.

Re: If both x and y are nonzero numbers, what is the value of y/x? (1) x [#permalink]

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25 Oct 2014, 08:31

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If both x and y are nonzero numbers, what is the value of [#permalink]

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17 Sep 2016, 05:27

Bunuel -

I'm confused on why B isn't the correct answer. The problem says at the beginning ..."If both x and y are nonzero numbers..."

So when I went to set up (2), it eliminated the negative root possibility on either side of the eq and I arrived at y = x. This makes the main statement = 1

Re: If both x and y are nonzero numbers, what is the value of [#permalink]

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17 Sep 2016, 05:35

lawiniecke wrote:

Bunuel -

I'm confused on why B isn't the correct answer. The problem says at the beginning ..."If both x and y are nonzero numbers..."

So when I went to set up (2), it eliminated the negative root possibility on either side of the eq and I arrived at y = x. This makes the main statement = 1

Did I misread the question?

Thank you in advance for your help!

I think you are missing something here. Note that Non Zero means either positive or negative but not Zero.

So, \(y^2 = x^2\) means y is + or - x. Hence x/y will be either 1 or -1. Hence, B is insufficient.

What do you mean by Negative root possibility. Note that we are not given sqrt(y)=sqrt(x). So, you need to be very careful while solving such type of questions.
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