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If both x and y are nonzero numbers, what is the value of y/x? (1) x
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19 Nov 2010, 04:34
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If both x and y are nonzero numbers, what is the value of y/x? (1) x = 6 (2) y^2 = x^2 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 OPEN DISCUSSION OF THIS QUESTION IS HERE: ifbothxandyarenonzeronumberswhatisthevalueofy165767.html
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If both x and y are nonzero numbers, what is the value of y/x? (1) x
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19 Nov 2010, 04:50
shash wrote: 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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Re: If both x and y are nonzero numbers, what is the value of y/x? (1) x
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19 Nov 2010, 06:03
shash wrote: 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. It is all in the notation!
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If both x and y are nonzero numbers, what is the value of
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20 Feb 2011, 10:15



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Re: QR. 30 Number
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20 Feb 2011, 10:21
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
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20 Feb 2011, 10:24
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
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20 Feb 2011, 10:43
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
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Re: QR. 30 Number
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if both x and y are nonzero numbers, what is the value of
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25 Jun 2011, 22:02
if both x and y are nonzero numbers, what is the value of y/x? 1. x=6 2. ysquare = xsquare
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Re: Prep Test DS9
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25 Jun 2011, 22:04
initially i had selected B as the answer... now i realize 1. x=6 this does not give any information about the value of y. hence no 2. ysquare=xsquare ysquare  = 1 xsquare taking squareroots on both sides +or y/x = +or1 hence No Taking together, it does not give better answer either as +or y/6 = +or1 is my understanding correct about the right answer?
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Re: Prep Test DS9
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25 Jun 2011, 23:26
Value of y/x? 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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Re: If both x and y are nonzero numbers, what is the value of
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18 May 2014, 10:00
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
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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
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
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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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If both x and y are nonzero numbers, what is the value of
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17 Sep 2016, 05:43
Hi abhimahna 
I misread the question. I was under the impression we were dealing with nonnegative integers. Problem resolved.
Thank you!



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