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

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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
What wrong did i do.
[Reveal] Spoiler: OA

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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
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.

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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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
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) (y-x) (y+x) = 0
y-x=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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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
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) (y-x) (y+x) = 0
y-x=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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Collections of work/rate problems with solutions http://gmatclub.com/forum/collections-of-work-rate-problem-with-solutions-118919.html
Mixture problems in a file with best solutions: http://gmatclub.com/forum/mixture-problems-with-best-and-easy-solutions-all-together-124644.html

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20 Feb 2011, 11:49
Baten80 wrote:
But as x= 6 (positive) so why i will consider √y^2 i.e., √36=-6 other than 6. [for ans c.] please help.

Read my post above: $$\sqrt{x^2}=|x|$$.

$$x^2=y^2$$ means $$|x|=|y|$$ as from (1) $$x=6$$ then $$|y|=6$$ --> $$y=6$$ or $$y=-6$$.
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20 Feb 2011, 11:53
Baten80 wrote:
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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Re: If both x and y are nonzero numbers, what is the value of [#permalink]

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18 May 2014, 11:00
1
KUDOS
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

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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

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 non-negative integers. Problem resolved.

Thank you!
If both x and y are nonzero numbers, what is the value of   [#permalink] 17 Sep 2016, 06:43
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