GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 20 Oct 2019, 08:05 GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  M26-25

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:

Hide Tags

Math Expert V
Joined: 02 Sep 2009
Posts: 58434

Show Tags

2
24 00:00

Difficulty:   65% (hard)

Question Stats: 45% (01:13) correct 55% (01:18) wrong based on 214 sessions

HideShow timer Statistics

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. $$x-y$$

_________________
Math Expert V
Joined: 02 Sep 2009
Posts: 58434

Show Tags

3
5
Official Solution:

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 $$x \lt 0$$ and $$y \lt 0$$ then $$|x|=-x$$ and $$|y|=-y$$.

So, $$\frac{\sqrt{x^2}}{x}-\sqrt{-y*|y|}=\frac{|x|}{x}-\sqrt{(-y)*(-y)}=\frac{-x}{x}-\sqrt{y^2}=-1-|y|=-1+y$$

_________________
Intern  Joined: 02 Aug 2014
Posts: 1
GPA: 3.79
WE: Military Officer (Military & Defense)

Show Tags

1
How does abs(x) = -x, and abs(y) = -y when x<0 and y<0? I am do not understand this concept
Math Expert V
Joined: 02 Sep 2009
Posts: 58434

Show Tags

2
4
delm2010 wrote:
How does abs(x) = -x, and abs(y) = -y when x<0 and y<0? I am do not understand this concept

_________________
Intern  B
Joined: 11 Jul 2012
Posts: 47
GMAT 1: 650 Q49 V29 Show Tags

Hi Bunuel,
I have a query.... wouldn't \sqrt{x^2}/ x would be equal to +1?

I say so since it's already given in the question that x is a -ve number, so by \sqrt{x^2} we know that we need to consider only the -ve Root.

But in the denominator the variable x would be a -ve number. Hence after the resolution of \sqrt{x^2} one part of the question would be |x|/x, where |x| would give us -x(since the -ve root needs to be considered) and the denominator in isolation is again a -ve number... so the fraction becomes \frac{-x}{-x} = 1.

Kindly clarify
Math Expert V
Joined: 02 Sep 2009
Posts: 58434

Show Tags

avaneeshvyas wrote:
Hi Bunuel,
I have a query.... wouldn't \sqrt{x^2}/ x would be equal to +1?

I say so since it's already given in the question that x is a -ve number, so by \sqrt{x^2} we know that we need to consider only the -ve Root.

But in the denominator the variable x would be a -ve number. Hence after the resolution of \sqrt{x^2} one part of the question would be |x|/x, where |x| would give us -x(since the -ve root needs to be considered) and the denominator in isolation is again a -ve number... so the fraction becomes \frac{-x}{-x} = 1.

Kindly clarify

First of all, read this: rules-for-posting-please-read-this-before-posting-133935.html#p1096628 (Writing Mathematical Formulas on the Forum)

Next, the point is that the square root function cannot give negative result ($$\sqrt{some \ expression}\geq{0}$$) and $$\sqrt{x^2}=|x|$$, which means that $$\sqrt{x^2}/ x=\frac{|x|}{x}=\frac{-x}{x}=-1$$.

Please follow the links in my previous post for more on absolute values.
_________________
Manager  S
Joined: 11 Sep 2013
Posts: 132
Concentration: Finance, Finance

Show Tags

[quote="Bunuel"]Official Solution:

-1-|y|=-1+y

I have not understood the above portion. Could you please tell me how to get this?

I have done it by taking numbers.

Let x=-1 and y=-1 Here i have taken integer value thinking that I will get an answer which can be in option also.

Now putting the value of x and y I am getting the value of question is -2.

Only option D gives me the value. Is my approach correct?
Math Expert V
Joined: 02 Sep 2009
Posts: 58434

Show Tags

2
2
Raihanuddin wrote:
Bunuel wrote:
Official Solution:

-1-|y|=-1+y

I have not understood the above portion. Could you please tell me how to get this?

I have done it by taking numbers.

Let x=-1 and y=-1 Here i have taken integer value thinking that I will get an answer which can be in option also.

Now putting the value of x and y I am getting the value of question is -2.

Only option D gives me the value. Is my approach correct?

-1 - |y| = -1 + y because we are given that y is negative. When y is negative |y| = -y. Hence -1 - |y| = -1 - (-y) = -1 + y.
_________________
Intern  Joined: 14 Oct 2015
Posts: 29
GMAT 1: 640 Q45 V33 Show Tags

There is a step that is off in this question. If you substitute negative y from the get go -(-y) = +y * (-y) leaves you with a negative inside the square root and this cannot happen. Should be written as square root of y*lyl
Math Expert V
Joined: 02 Sep 2009
Posts: 58434

Show Tags

danjbon wrote:
There is a step that is off in this question. If you substitute negative y from the get go -(-y) = +y * (-y) leaves you with a negative inside the square root and this cannot happen. Should be written as square root of y*lyl

That's wrong. $$\sqrt{-y*|y|}=\sqrt{-(negative)*positive}=\sqrt{positive*positive}$$
_________________
Intern  Joined: 13 Jun 2016
Posts: 18

Show Tags

having a hard time understanding this basic rule: $$\sqrt{x^2}=|x|$$

Assuming it's an integer, shouldn't this be the positive integer of x? why |x|
Math Expert V
Joined: 02 Sep 2009
Posts: 58434

Show Tags

2
1
xnthic wrote:
having a hard time understanding this basic rule: $$\sqrt{x^2}=|x|$$

Assuming it's an integer, shouldn't this be the positive integer of x? why |x|

MUST KNOW: $$\sqrt{x^2}=|x|$$:

The point here is that since 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 $$\sqrt{x^2}=|x|$$.

For example if x = -5, then $$\sqrt{x^2}=\sqrt{25}=5=|-5|=|x|$$

Theory on Abolute Values: math-absolute-value-modulus-86462.html
Absolute value tips: absolute-value-tips-and-hints-175002.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

_________________
SVP  V
Joined: 26 Mar 2013
Posts: 2345

Show Tags

Bunuel wrote:
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. $$x-y$$

Alternate approach is to testing some easy values:

As per question stem, x & y are negative number..

Let x= -1 & y=-1

Substituting the in equation

$$\frac{\sqrt{1}}{-1}-\sqrt{-(-1)*|-1|}$$ = -2

y-1=-1-1=-2 only answer that matches what we get above

Intern  Joined: 11 Aug 2016
Posts: 9
Location: United States (PA)
Concentration: Finance, General Management
GMAT 1: 710 Q48 V40 GPA: 2.42
WE: Other (Mutual Funds and Brokerage)

Show Tags

1
Hi,

Having trouble wrapping my head around why $$\sqrt{x^2}$$ is equal to |x|

I looked at the post on this page, but did not see any explanation.

For example, how is the equation $$\sqrt{4}$$ = {-2, 2} any different from $$\sqrt{2^2}$$ = ? Is it because the solution of $$\sqrt{x^2}$$ must be equal to the solution of $$(\sqrt{x})^2$$? if so, why?

Thanks in advance for any help!

(with regard to bunnel's explanation to xnthic, throughout the explanation bunnel writes "$$\sqrt{25}$$ = 5" as though this is given, but my understanding is that $$\sqrt{25}$$ could be either positive or negative 5, as either positive or negative 5 squared result in 25. If I am incorrect, then that is most likely the root of my confusion)
Math Expert V
Joined: 02 Sep 2009
Posts: 58434

Show Tags

blitheclyde wrote:
Hi,

Having trouble wrapping my head around why $$\sqrt{x^2}$$ is equal to |x|

I looked at the post on this page, but did not see any explanation.

For example, how is the equation $$\sqrt{4}$$ = {-2, 2} any different from $$\sqrt{2^2}$$ = ? Is it because the solution of $$\sqrt{x^2}$$ must be equal to the solution of $$(\sqrt{x})^2$$? if so, why?

Thanks in advance for any help!

(with regard to bunnel's explanation to xnthic, throughout the explanation bunnel writes "$$\sqrt{25}$$ = 5" as though this is given, but my understanding is that $$\sqrt{25}$$ could be either positive or negative 5, as either positive or negative 5 squared result in 25. If I am incorrect, then that is most likely the root of my confusion)

$$\sqrt{4}=2$$ ONLY, not 2 or -2. The square root function cannot give negative result. Check here: m26-184464.html#p1729559
_________________
Intern  B
Joined: 09 Jun 2016
Posts: 24

Show Tags

Bunuel wrote:
blitheclyde wrote:
Hi,

Having trouble wrapping my head around why $$\sqrt{x^2}$$ is equal to |x|

I looked at the post on this page, but did not see any explanation.

For example, how is the equation $$\sqrt{4}$$ = {-2, 2} any different from $$\sqrt{2^2}$$ = ? Is it because the solution of $$\sqrt{x^2}$$ must be equal to the solution of $$(\sqrt{x})^2$$? if so, why?

Thanks in advance for any help!

(with regard to bunnel's explanation to xnthic, throughout the explanation bunnel writes "$$\sqrt{25}$$ = 5" as though this is given, but my understanding is that $$\sqrt{25}$$ could be either positive or negative 5, as either positive or negative 5 squared result in 25. If I am incorrect, then that is most likely the root of my confusion)

$$\sqrt{4}=2$$ ONLY, not 2 or -2. The square root function cannot give negative result. Check here: http://gmatclub.com/forum/m26-184464.html#p1729559

Hi Bunuel, certain application of the sqrt function seem to be contradictory. If sqrt function can't give negative numbers, why do we substitute both -ve and +ve values of $$\sqrt{expression}$$ while solving algebraic equations.
_________________
..................................................................................................................
micro-level speed, macro-level patience | KUDOS for the post ,shall be appreciated the most
Math Expert V
Joined: 02 Sep 2009
Posts: 58434

Show Tags

abhishek911 wrote:
Bunuel wrote:
blitheclyde wrote:
Hi,

Having trouble wrapping my head around why $$\sqrt{x^2}$$ is equal to |x|

I looked at the post on this page, but did not see any explanation.

For example, how is the equation $$\sqrt{4}$$ = {-2, 2} any different from $$\sqrt{2^2}$$ = ? Is it because the solution of $$\sqrt{x^2}$$ must be equal to the solution of $$(\sqrt{x})^2$$? if so, why?

Thanks in advance for any help!

(with regard to bunnel's explanation to xnthic, throughout the explanation bunnel writes "$$\sqrt{25}$$ = 5" as though this is given, but my understanding is that $$\sqrt{25}$$ could be either positive or negative 5, as either positive or negative 5 squared result in 25. If I am incorrect, then that is most likely the root of my confusion)

$$\sqrt{4}=2$$ ONLY, not 2 or -2. The square root function cannot give negative result. Check here: http://gmatclub.com/forum/m26-184464.html#p1729559

Hi Bunuel, certain application of the sqrt function seem to be contradictory. If sqrt function can't give negative numbers, why do we substitute both -ve and +ve values of $$\sqrt{expression}$$ while solving algebraic equations.

When the GMAT provides the square root sign for an even root, such as a square root, fourth root, etc. then the only accepted answer is the positive root. That is:

$$\sqrt{9} = 3$$, NOT +3 or -3;
$$\sqrt{16} = 2$$, NOT +2 or -2;

Notice that in contrast, the equation $$x^2 = 9$$ has TWO solutions, +3 and -3. Because $$x^2 = 9$$ means that $$x =-\sqrt{9}=-3$$ or $$x=\sqrt{9}=3$$.
_________________
Intern  B
Joined: 09 Sep 2015
Posts: 17

Show Tags

Hi,

Since x is negative, \sqrt{x^2}/-x - \sqrt{-y*-y}
= $$-x/-x$$ - \sqrt{y^2}
= 1-(-y)
= 1+y

Can someone please explain what wrong am I doing here?
Math Expert V
Joined: 02 Sep 2009
Posts: 58434

Show Tags

Hi,

Since x is negative, \sqrt{x^2}/-x - \sqrt{-y*-y}
= $$-x/-x$$ - \sqrt{y^2}
= 1-(-y)
= 1+y

Can someone please explain what wrong am I doing here?

Why are you changing x to -x there? The stem just says that x is negative, so x stands for some negative number, changing x to -x does not make sense.
_________________
Intern  B
Joined: 09 Apr 2018
Posts: 35
Location: India
GMAT 1: 690 Q47 V38 GMAT 2: 710 Q49 V36 GPA: 3.7

Show Tags

Hi,

In one of the other questions, the explanation is provided as follows

If p=−1
then (|p|!)^p=(|−1|!)^−1=1^−1=1

Here |-1| = 1

But in this problem |y| = -y since y is negative

Can you please explain the discrepancy or the misunderstanding in my thought Re: M26-25   [#permalink] 11 Jul 2018, 08:26

Go to page    1   2    Next  [ 26 posts ]

Display posts from previous: Sort by

M26-25

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics

Moderators: chetan2u, Bunuel Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne  