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If both x and y are nonzero numbers, what is the value of [#permalink]
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20 Feb 2011, 11:15
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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.
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Re: QR. 30 Number [#permalink]
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20 Feb 2011, 11:21
Baten80 wrote: 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 \(\sqrt{x^2}=x\).
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Re: QR. 30 Number [#permalink]
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20 Feb 2011, 11:24
Baten80 wrote: 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. \(sqrt(y^2)=\sqrt(x^2)\) may not always mean  y=x Say x=6; y=6; x^2=36, y=36; but x and y are different. x=6; y=6; x^2=36, y=36; and here x and y are same. (1) x=6; Nothing about y. Not sufficient. (2) (yx) (y+x) = 0 yx=0 or y+x=0 If former; y=x Latter; y=x Not sufficient. Combining both; x=6 y can be 6 or 6. Ans: "E"
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Re: QR. 30 Number [#permalink]
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20 Feb 2011, 11:43
fluke wrote: Baten80 wrote: 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. \(sqrt(y^2)=\sqrt(x^2)\) may not always mean  y=x Say x=6; y=6; x^2=36, y=36; but x and y are different. x=6; y=6; x^2=36, y=36; and here x and y are same. (1) x=6; Nothing about y. Not sufficient. (2) (yx) (y+x) = 0 yx=0 or y+x=0 If former; y=x Latter; y=x Not sufficient. Combining both; x=6 y can be 6 or 6. Ans: "E" But as x= 6 (positive) so why i will consider √y^2 i.e., √36=6 other than 6. [for ans c.] please help.
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Re: QR. 30 Number [#permalink]
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20 Feb 2011, 11:49



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Re: QR. 30 Number [#permalink]
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20 Feb 2011, 11:53



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Re: If both x and y are nonzero numbers, what is the value of [#permalink]
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18 May 2014, 11:00
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st(1) no info about y Insuff st(2) x=y Insuff combined st(1&2): 6=6 Insuff Hence E



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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, 06:27
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
Did I misread the question?
Thank you in advance for your help!



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Re: If both x and y are nonzero numbers, what is the value of [#permalink]
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17 Sep 2016, 06:35
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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If both x and y are nonzero numbers, what is the value of [#permalink]
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17 Sep 2016, 06:43
Hi abhimahna 
I misread the question. I was under the impression we were dealing with nonnegative integers. Problem resolved.
Thank you!




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