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M2625 If x and y are negative numbers [#permalink]
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15 Sep 2013, 19:24
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If x and y are negative numbers, what is the value of √x^2/x −√y∗y? A. 1+y B. 1−y C. −1−y D. y−1 E. x−y I don't understand why do I get a different answer when I use numbers. Bunuel please show the explanation using numbers, not just pure theory. Thanks.
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Re: M2625 If x and y are negative numbers [#permalink]
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16 Sep 2013, 01:36
vaishnogmat wrote: If x and y are negative numbers, what is the value of √x^2/x −√y∗y?
a) 1+y
b) 1−y
c) −1−y
d) y−1
e) x−y
I don't understand why do I get a different answer when I use numbers. Bunuel please show the explanation using numbers, not just pure theory. Thanks. If x and y are negative numbers, what is the value of \(\frac{\sqrt{x^2}}{x}\sqrt{y*y}\)?A. 1+y B. 1y C. 1y D. y1 E. xy Note that \(\sqrt{a^2}=a\). Next, since \(x<0\) and \(y<0\) then \(x=x\) and \(y=y\). So, \(\frac{\sqrt{x^2}}{x}\sqrt{y*y}=\frac{x}{x}\sqrt{(y)*(y)}=\frac{x}{x}\sqrt{y^2}=1y=1+y\) Answer: D.Number plugging:Since given that x and y are negative numbers, say x=1 and y=2. \(\frac{\sqrt{x^2}}{x}\sqrt{y*y}=12=3\). Now, plug x=1 and y=2 into the options to see which one yields 3. Only D fits.
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Re: M2625 If x and y are negative numbers [#permalink]
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03 Jan 2014, 10:53
"Next, since x<0 and y<0 then x=x and y=y."
Hi Bunuel, can you help me with this explanation...I thought absolute values are ALWAYS positive? Using real numbers for x & y, say x = 1 and y = 2, then x = 1 = 1 and y=2 = 2. How could x = x?
Thanks.



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Re: M2625 If x and y are negative numbers [#permalink]
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09 Jan 2014, 06:50
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neehow wrote: "Next, since x<0 and y<0 then x=x and y=y."
Hi Bunuel, can you help me with this explanation...I thought absolute values are ALWAYS positive? Using real numbers for x & y, say x = 1 and y = 2, then x = 1 = 1 and y=2 = 2. How could x = x?
Thanks. Hi Neehow, In the context of the question, we are given that x and y are both negative numbers. The absolute value of a negative number is of course positive. So let's assume x = 2 > x = 2 So what is x? x = 2 Therefore as per first step we know x = x Hope this helps. Kind regards, Spoontrick



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Re: M2625 If x and y are negative numbers [#permalink]
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10 Jan 2014, 13:11
Spoontrick wrote: neehow wrote: "Next, since x<0 and y<0 then x=x and y=y."
Hi Bunuel, can you help me with this explanation...I thought absolute values are ALWAYS positive? Using real numbers for x & y, say x = 1 and y = 2, then x = 1 = 1 and y=2 = 2. How could x = x?
Thanks. Hi Neehow, In the context of the question, we are given that x and y are both negative numbers. The absolute value of a negative number is of course positive. So let's assume x = 2 > x = 2 So what is x? x = 2 Therefore as per first step we know x = x Hope this helps. Kind regards, Spoontrick Got it! Thank you Spoontrick!



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Re: M2625 If x and y are negative numbers [#permalink]
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27 Feb 2014, 08:09
Regarding the question above in Bunnuel, I just don't understand why the y turns positive if the subtract sign is outside the absolute signs? Thanks always for providing the best descriptive answers online.



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Re: M2625 If x and y are negative numbers [#permalink]
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27 Feb 2014, 08:13



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Re: M2625 If x and y are negative numbers [#permalink]
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27 Feb 2014, 08:20
Bunuel wrote: demart78 wrote: Regarding the question above in Bunnuel, I just don't understand why the y turns positive if the subtract sign is outside the absolute signs? Thanks always for providing the best descriptive answers online. We are told that \(y\) is a negative number, thus \(y=y\), so \(1y=1(y)=1+y\). Hope it's clear. I still do not understand. If y is a negative number, once the absolute brackets go up around it, shouldn't it than be converted to a positive number? Or because we are told that it is negative, we automatically assume that y<0 and hit it with the negative sign?



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Re: M2625 If x and y are negative numbers [#permalink]
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27 Feb 2014, 08:30
demart78 wrote: Bunuel wrote: demart78 wrote: Regarding the question above in Bunnuel, I just don't understand why the y turns positive if the subtract sign is outside the absolute signs? Thanks always for providing the best descriptive answers online. We are told that \(y\) is a negative number, thus \(y=y\), so \(1y=1(y)=1+y\). Hope it's clear. I still do not understand. If y is a negative number, once the absolute brackets go up around it, shouldn't it than be converted to a positive number? Or because we are told that it is negative, we automatically assume that y<0 and hit it with the negative sign? It seems that you need to brush up fundamentals on absolute values. When \(x\leq{0}\) then \(x=x\), or more generally when \(some \ expression\leq{0}\) then \(some \ expression={(some \ expression)}\). For example: \(5=5=(5)\); When \(x\geq{0}\) then \(x=x\), or more generally when \(some \ expression\geq{0}\) then \(some \ expression={some \ expression}\). For example: \(5=5\). Hope it helps.
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Re: M2625 If x and y are negative numbers [#permalink]
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05 Mar 2014, 13:18
Bunuel wrote: vaishnogmat wrote: If x and y are negative numbers, what is the value of √x^2/x −√y∗y?
a) 1+y
b) 1−y
c) −1−y
d) y−1
e) x−y
I don't understand why do I get a different answer when I use numbers. Bunuel please show the explanation using numbers, not just pure theory. Thanks. If x and y are negative numbers, what is the value of \(\frac{\sqrt{x^2}}{x}\sqrt{y*y}\)?A. 1+y B. 1y C. 1y D. y1 E. xy Note that \(\sqrt{a^2}=a\). Next, since \(x<0\) and \(y<0\) then \(x=x\) and \(y=y\). So, \(\frac{\sqrt{x^2}}{x}\sqrt{y*y}=\frac{x}{x}\sqrt{(y)*(y)}=\frac{x}{x}\sqrt{y^2}=1y=1+y\) Answer: D.Number plugging:Since given that x and y are negative numbers, say x=1 and y=2. \(\frac{\sqrt{x^2}}{x}\sqrt{y*y}=12=3\). Now, plug x=1 and y=2 into the options to see which one yields 3. Only D fits. Absolute value questions and I don't seem to be getting close friends anytime soon Here's what still doesn't make sense to me: if x=1, then why does \(\frac{\sqrt{x^2}}{x}\) become 1?? i understand the rule that when x<0 then x=x and that the square root of x is equal to x but in this case we already know that x is negative. So when taking the square root out of (1)² then why do we not get x=1 (as we initially said) and hence x/x=(1)/(1)=1? it's not quite intuitive for me to go from \(\frac{\sqrt{(1)^2}}{1}\) to 1. I thought we are taking the square root out of x² in order to determine x, which we already know to be negative, hence it should be 1 in this example. i just don't seem to get it ... or is it just a convention/ rule that when using square roots in value expressions we only consider the absolute values of such expressions and that therefore \(\frac{\sqrt{(1)^2}}{1}\) becomes 1 because we are dividing the absolute value of x=x=1=1 by x=1?



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Re: M2625 If x and y are negative numbers [#permalink]
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06 Mar 2014, 03:19
damamikus wrote: Bunuel wrote: vaishnogmat wrote: If x and y are negative numbers, what is the value of √x^2/x −√y∗y?
a) 1+y
b) 1−y
c) −1−y
d) y−1
e) x−y
I don't understand why do I get a different answer when I use numbers. Bunuel please show the explanation using numbers, not just pure theory. Thanks. If x and y are negative numbers, what is the value of \(\frac{\sqrt{x^2}}{x}\sqrt{y*y}\)?A. 1+y B. 1y C. 1y D. y1 E. xy Note that \(\sqrt{a^2}=a\). Next, since \(x<0\) and \(y<0\) then \(x=x\) and \(y=y\). So, \(\frac{\sqrt{x^2}}{x}\sqrt{y*y}=\frac{x}{x}\sqrt{(y)*(y)}=\frac{x}{x}\sqrt{y^2}=1y=1+y\) Answer: D.Number plugging:Since given that x and y are negative numbers, say x=1 and y=2. \(\frac{\sqrt{x^2}}{x}\sqrt{y*y}=12=3\). Now, plug x=1 and y=2 into the options to see which one yields 3. Only D fits. Absolute value questions and I don't seem to be getting close friends anytime soon Here's what still doesn't make sense to me: if x=1, then why does \(\frac{\sqrt{x^2}}{x}\) become 1?? i understand the rule that when x<0 then x=x and that the square root of x is equal to x but in this case we already know that x is negative. So when taking the square root out of (1)² then why do we not get x=1 (as we initially said) and hence x/x=(1)/(1)=1? it's not quite intuitive for me to go from \(\frac{\sqrt{(1)^2}}{1}\) to 1. I thought we are taking the square root out of x² in order to determine x, which we already know to be negative, hence it should be 1 in this example. i just don't seem to get it ... or is it just a convention/ rule that when using square roots in value expressions we only consider the absolute values of such expressions and that therefore \(\frac{\sqrt{(1)^2}}{1}\) becomes 1 because we are dividing the absolute value of x=x=1=1 by x=1? The point here is that the square root function cannot give negative result: \(\sqrt{some \ expression}\geq{0}\). So, \(\sqrt{(1)^2}=\sqrt{1}=1\). Thus if \(x=1\), then \(\frac{\sqrt{x^2}}{x}=\frac{\sqrt{(1)^2}}{1}=\frac{\sqrt{1}}{1}=\frac{1}{1}=1\). Hope it helps.
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Re: M2625 If x and y are negative numbers [#permalink]
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05 May 2014, 22:11
Bunuel wrote: vaishnogmat wrote: If x and y are negative numbers, what is the value of √x^2/x −√y∗y?
a) 1+y
b) 1−y
c) −1−y
d) y−1
e) x−y
I don't understand why do I get a different answer when I use numbers. Bunuel please show the explanation using numbers, not just pure theory. Thanks. If x and y are negative numbers, what is the value of \(\frac{\sqrt{x^2}}{x}\sqrt{y*y}\)?A. 1+y B. 1y C. 1y D. y1 E. xy Note that \(\sqrt{a^2}=a\). Next, since \(x<0\) and \(y<0\) then \(x=x\) and \(y=y\). So, \(\frac{\sqrt{x^2}}{x}\sqrt{y*y}=\frac{x}{x}\sqrt{(y)*(y)}=\frac{x}{x}\sqrt{y^2}=1y=1+y\) Answer: D.Number plugging:Since given that x and y are negative numbers, say x=1 and y=2. \(\frac{\sqrt{x^2}}{x}\sqrt{y*y}=12=3\). Now, plug x=1 and y=2 into the options to see which one yields 3. Only D fits. Hi Bunnel, I have one doubt on this here x and y is <0 so we are considering x and y as x and y so I just want to know in denominator we have x/x = x/x. why we are not considering x in denominator as x Please clarify Thanks.



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Re: M2625 If x and y are negative numbers [#permalink]
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06 May 2014, 03:13
PathFinder007 wrote: Bunuel wrote: vaishnogmat wrote: If x and y are negative numbers, what is the value of √x^2/x −√y∗y?
a) 1+y
b) 1−y
c) −1−y
d) y−1
e) x−y
I don't understand why do I get a different answer when I use numbers. Bunuel please show the explanation using numbers, not just pure theory. Thanks. If x and y are negative numbers, what is the value of \(\frac{\sqrt{x^2}}{x}\sqrt{y*y}\)?A. 1+y B. 1y C. 1y D. y1 E. xy Note that \(\sqrt{a^2}=a\). Next, since \(x<0\) and \(y<0\) then \(x=x\) and \(y=y\). So, \(\frac{\sqrt{x^2}}{x}\sqrt{y*y}=\frac{x}{x}\sqrt{(y)*(y)}=\frac{x}{x}\sqrt{y^2}=1y=1+y\) Answer: D.Number plugging:Since given that x and y are negative numbers, say x=1 and y=2. \(\frac{\sqrt{x^2}}{x}\sqrt{y*y}=12=3\). Now, plug x=1 and y=2 into the options to see which one yields 3. Only D fits. Hi Bunnel, I have one doubt on this here x and y is <0 so we are considering x and y as x and y so I just want to know in denominator we have x/x = x/x. why we are not considering x in denominator as x Please clarify Thanks. x is a negative number, so x is a positive number. So, knowing that x is negative doesn't mean that you should write x instead of x.
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Re: M2625 If x and y are negative numbers [#permalink]
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05 Aug 2014, 18:48
Bunuel wrote: vaishnogmat wrote: If x and y are negative numbers, what is the value of √x^2/x −√y∗y?
a) 1+y
b) 1−y
c) −1−y
d) y−1
e) x−y
I don't understand why do I get a different answer when I use numbers. Bunuel please show the explanation using numbers, not just pure theory. Thanks. If x and y are negative numbers, what is the value of \(\frac{\sqrt{x^2}}{x}\sqrt{y*y}\)?A. 1+y B. 1y C. 1y D. y1 E. xy Note that \(\sqrt{a^2}=a\). Next, since \(x<0\) and \(y<0\) then \(x=x\) and \(y=y\). So, \(\frac{\sqrt{x^2}}{x}\sqrt{y*y}=\frac{x}{x}\sqrt{(y)*(y)}=\frac{x}{x}\sqrt{y^2}=1y=1+y\) Answer: D.Number plugging:Since given that x and y are negative numbers, say x=1 and y=2. \(\frac{\sqrt{x^2}}{x}\sqrt{y*y}=12=3\). Now, plug x=1 and y=2 into the options to see which one yields 3. Only D fits. Hi Bunuel, Can you please explain one small point in this question. If the expression was √x^2/x −√y∗ y will the answer change? I am only trying to understand whether value will change between sqrt (y^2) and sqrt(y*y), given y is negative.



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M2625 If x and y are negative numbers [#permalink]
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12 Aug 2014, 01:58
hubahuba wrote: Bunuel wrote: vaishnogmat wrote: If x and y are negative numbers, what is the value of √x^2/x −√y∗y?
a) 1+y
b) 1−y
c) −1−y
d) y−1
e) x−y
I don't understand why do I get a different answer when I use numbers. Bunuel please show the explanation using numbers, not just pure theory. Thanks. If x and y are negative numbers, what is the value of \(\frac{\sqrt{x^2}}{x}\sqrt{y*y}\)?A. 1+y B. 1y C. 1y D. y1 E. xy Note that \(\sqrt{a^2}=a\). Next, since \(x<0\) and \(y<0\) then \(x=x\) and \(y=y\). So, \(\frac{\sqrt{x^2}}{x}\sqrt{y*y}=\frac{x}{x}\sqrt{(y)*(y)}=\frac{x}{x}\sqrt{y^2}=1y=1+y\) Answer: D.Number plugging:Since given that x and y are negative numbers, say x=1 and y=2. \(\frac{\sqrt{x^2}}{x}\sqrt{y*y}=12=3\). Now, plug x=1 and y=2 into the options to see which one yields 3. Only D fits. Hi Bunuel, Can you please explain one small point in this question. If the expression was √x^2/x −√y∗ y will the answer change? I am only trying to understand whether value will change between sqrt (y^2) and sqrt(y*y), given y is negative.In this case the question would be flawed: \(\sqrt{(y)*y}=\sqrt{positive*negative}=\sqrt{negative}\) and the square roots from negative numbers are undefined for the GMAT.
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M2625 If x and y are negative numbers
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