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M26-25 If x and y are negative numbers

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M26-25 If x and y are negative numbers [#permalink] New post 15 Sep 2013, 18:24
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: M26-25 If x and y are negative numbers [#permalink] New post 16 Sep 2013, 00:36
Expert's post
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. 1-y
C. -1-y
D. y-1
E. x-y

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}=-1-|y|=-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|}=-1-2=-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: M26-25 If x and y are negative numbers [#permalink] New post 03 Jan 2014, 09:53
"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: M26-25 If x and y are negative numbers [#permalink] New post 09 Jan 2014, 05:50
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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,
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Re: M26-25 If x and y are negative numbers [#permalink] New post 10 Jan 2014, 12:11
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: M26-25 If x and y are negative numbers [#permalink] New post 27 Feb 2014, 07:09
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: M26-25 If x and y are negative numbers [#permalink] New post 27 Feb 2014, 07:13
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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 -1-y=-1-(-y)=-1+y.

Hope it's clear.
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Re: M26-25 If x and y are negative numbers [#permalink] New post 27 Feb 2014, 07:20
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 -1-y=-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: M26-25 If x and y are negative numbers [#permalink] New post 27 Feb 2014, 07:30
Expert's post
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 -1-y=-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.

Theory on Abolute Values: math-absolute-value-modulus-86462.html

DS Abolute Values Questions to practice: search.php?search_id=tag&tag_id=37
PS Abolute Values Questions to practice: search.php?search_id=tag&tag_id=58

Hard set on Abolute Values: inequality-and-absolute-value-questions-from-my-collection-86939.html


Hope it helps.
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Re: M26-25 If x and y are negative numbers [#permalink] New post 05 Mar 2014, 12:18
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. 1-y
C. -1-y
D. y-1
E. x-y

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}=-1-|y|=-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|}=-1-2=-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 :lol:

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: M26-25 If x and y are negative numbers [#permalink] New post 06 Mar 2014, 02:19
Expert's post
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. 1-y
C. -1-y
D. y-1
E. x-y

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}=-1-|y|=-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|}=-1-2=-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 :lol:

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.

Theory on Abolute Values: math-absolute-value-modulus-86462.html

DS Abolute Values Questions to practice: search.php?search_id=tag&tag_id=37
PS Abolute Values Questions to practice: search.php?search_id=tag&tag_id=58

Hard set on Abolute Values: inequality-and-absolute-value-questions-from-my-collection-86939.html


Hope it helps.
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PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: M26-25 If x and y are negative numbers [#permalink] New post 05 May 2014, 21:11
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. 1-y
C. -1-y
D. y-1
E. x-y

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}=-1-|y|=-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|}=-1-2=-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: M26-25 If x and y are negative numbers [#permalink] New post 06 May 2014, 02:13
Expert's post
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. 1-y
C. -1-y
D. y-1
E. x-y

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}=-1-|y|=-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|}=-1-2=-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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COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: M26-25 If x and y are negative numbers [#permalink] New post 05 Aug 2014, 17:48
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. 1-y
C. -1-y
D. y-1
E. x-y

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}=-1-|y|=-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|}=-1-2=-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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M26-25 If x and y are negative numbers [#permalink] New post 12 Aug 2014, 00:58
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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. 1-y
C. -1-y
D. y-1
E. x-y

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}=-1-|y|=-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|}=-1-2=-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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M26-25 If x and y are negative numbers   [#permalink] 12 Aug 2014, 00:58
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