It is currently 05 Jun 2023, 09:07 |

Customized

for You

Track

Your Progress

Practice

Pays

- Jun
**05**### Free GRE Practice Test and Personal Score Assessment

03:00 PM PDT

-04:00 PM PDT

Magoosh is excited to offer you a free GRE practice test (online) with video answers and explanations. The assessment will take a bit over an hour and you’ll also get access to: - Jun
**13**### Join the MBA Spotlight

06:00 AM PDT

-04:00 PM PDT

Join GMAT Club Live in a Q&A with Current Students from the Top 50 MBA and MiM programs, and more. Register by May 15th to Receive Free Access to GMAT Club Tests for 2 Weeks.

Q02-38 Question # 08 Section # 09
[#permalink]
Updated on: 04 Jul 2019, 23:42

1

1

4

Bookmarks

Question Stats:

Given \(x\), \(y<0\), what is the value of \(\frac{\sqrt{x^2}}{x} - \sqrt{\frac{-y}{|y|}}\) ?

A. \(1+y\)

B. \(1-y\)

C. \(-1-y\)

D. \(y-1\)

E. \(x-y\)

Hello,

I was looking at the explanation of this question (ID Q02-38, Question2 section 9) and it quite does not make any sense in the very first step (or maybe I am overlooking something). Kindly check (or possibly explain it if possible/and-is-correct).

Thanks

Arsh

A. \(1+y\)

B. \(1-y\)

C. \(-1-y\)

D. \(y-1\)

E. \(x-y\)

Hello,

I was looking at the explanation of this question (ID Q02-38, Question2 section 9) and it quite does not make any sense in the very first step (or maybe I am overlooking something). Kindly check (or possibly explain it if possible/and-is-correct).

Thanks

Arsh

Re: Q02-38 Question # 08 Section # 09
[#permalink]
25 Jun 2016, 01:16

1

Expert Reply

Hi,

well I have to admit that the explanation is not the top-notch but it is not wrong.

Starting from what we do know: X could be positive or negative: we do not know at the moment AND y is negative.

Now look at the first part of the equation \(\sqrt{x^2}\) is equal to |x|. From this we also know that |x| = -x. Which means that on the left hand side X is positive, always. On the right hand side for X to be always positive X must be negative. This are math rules that is better for you to know as cold during the exam.

As such, actually we do have that X/X is = 1 and is negative. So -1

Going to the second square root |y| = -y so -y * - y = y^2 that under square root becomes |y|.

At this point we have -1 - |y| AND we already know that |y| = -y .

-1 - (-y) = -1 +y

Hope is clear this. Waiting, though, math expert for further clarification.

_________________

well I have to admit that the explanation is not the top-notch but it is not wrong.

Starting from what we do know: X could be positive or negative: we do not know at the moment AND y is negative.

Now look at the first part of the equation \(\sqrt{x^2}\) is equal to |x|. From this we also know that |x| = -x. Which means that on the left hand side X is positive, always. On the right hand side for X to be always positive X must be negative. This are math rules that is better for you to know as cold during the exam.

As such, actually we do have that X/X is = 1 and is negative. So -1

Going to the second square root |y| = -y so -y * - y = y^2 that under square root becomes |y|.

At this point we have -1 - |y| AND we already know that |y| = -y .

-1 - (-y) = -1 +y

Hope is clear this. Waiting, though, math expert for further clarification.

_________________

Re: Q02-38 Question # 08 Section # 09
[#permalink]
28 Feb 2018, 21:33

Carcass wrote:

Hi,

well I have to admit that the explanation is not the top-notch but it is not wrong.

Starting from what we do know: X could be positive or negative: we do not know at the moment AND y is negative.

Now look at the first part of the equation \(\sqrt{x^2}\) is equal to |x|. From this we also know that |x| = -x. Which means that on the left hand side X is positive, always. On the right hand side for X to be always positive X must be negative. This are math rules that is better for you to know as cold during the exam.

As such, actually we do have that X/X is = 1 and is negative. So -1

Going to the second square root |y| = -y so -y * - y = y^2 that under square root becomes |y|.

At this point we have -1 - |y| AND we already know that |y| = -y .

-1 - (-y) = -1 +y

Hope is clear this. Waiting, though, math expert for further clarification.

well I have to admit that the explanation is not the top-notch but it is not wrong.

Starting from what we do know: X could be positive or negative: we do not know at the moment AND y is negative.

Now look at the first part of the equation \(\sqrt{x^2}\) is equal to |x|. From this we also know that |x| = -x. Which means that on the left hand side X is positive, always. On the right hand side for X to be always positive X must be negative. This are math rules that is better for you to know as cold during the exam.

As such, actually we do have that X/X is = 1 and is negative. So -1

Going to the second square root |y| = -y so -y * - y = y^2 that under square root becomes |y|.

At this point we have -1 - |y| AND we already know that |y| = -y .

-1 - (-y) = -1 +y

Hope is clear this. Waiting, though, math expert for further clarification.

I still don't know why \(\sqrt{\frac{-y}{|y|}}\) =-y NOT -1?

Re: Q02-38 Question # 08 Section # 09
[#permalink]
24 May 2018, 11:07

1

Peter wrote:

Carcass wrote:

Hi,

well I have to admit that the explanation is not the top-notch but it is not wrong.

Starting from what we do know: X could be positive or negative: we do not know at the moment AND y is negative.

Now look at the first part of the equation \(\sqrt{x^2}\) is equal to |x|. From this we also know that |x| = -x. Which means that on the left hand side X is positive, always. On the right hand side for X to be always positive X must be negative. This are math rules that is better for you to know as cold during the exam.

As such, actually we do have that X/X is = 1 and is negative. So -1

Going to the second square root |y| = -y so -y * - y = y^2 that under square root becomes |y|.

At this point we have -1 - |y| AND we already know that |y| = -y .

-1 - (-y) = -1 +y

Hope is clear this. Waiting, though, math expert for further clarification.

well I have to admit that the explanation is not the top-notch but it is not wrong.

Starting from what we do know: X could be positive or negative: we do not know at the moment AND y is negative.

Now look at the first part of the equation \(\sqrt{x^2}\) is equal to |x|. From this we also know that |x| = -x. Which means that on the left hand side X is positive, always. On the right hand side for X to be always positive X must be negative. This are math rules that is better for you to know as cold during the exam.

As such, actually we do have that X/X is = 1 and is negative. So -1

Going to the second square root |y| = -y so -y * - y = y^2 that under square root becomes |y|.

At this point we have -1 - |y| AND we already know that |y| = -y .

-1 - (-y) = -1 +y

Hope is clear this. Waiting, though, math expert for further clarification.

I still don't know why \(\sqrt{\frac{-y}{|y|}}\) =-y NOT -1?

Consider y = (-2), as y<0. hence -y = -(-2) = 2.

|y| = |-2| = 2.

Re: Q02-38 Question # 08 Section # 09
[#permalink]
04 Jul 2019, 15:54

1

I was thinking another way. Let me correct if I am horribly wrong! Let x and y both equal -1. from the first, sqrt of (x square)/x we get -1 and also form second part we get -1. So adding, -2. Consider all options!. Option D fits!

Re: Q02-38 Question # 08 Section # 09
[#permalink]
05 Jul 2019, 04:49

@AlaminMolla is correct. For ALL negative values of x and y, the expression will ALWAYS evaluate to be -2

So, the question is flawed. At the very least, the question should read "Which of the following COULD BE the value of ...", in which case A, D and E COULD equal -2

@arsh are you sure you transcribed the question correctly?

Cheers,

Brent

_________________

So, the question is flawed. At the very least, the question should read "Which of the following COULD BE the value of ...", in which case A, D and E COULD equal -2

@arsh are you sure you transcribed the question correctly?

Cheers,

Brent

_________________

Re: Q02-38 Question # 08 Section # 09
[#permalink]
28 Jul 2021, 14:25

Hello from the GRE Prep Club BumpBot!

Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.

Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.

gmatclubot

Moderators:

Multiple-choice Questions — Select One or More Answer Choices |
||

## Hi Guest,Here are updates for you:## ANNOUNCEMENTS✅ Diana Economy, former Ross AdCom, provides strategic advice on how to make the best use of MBA spotlight fair ✅ Not Registered for MBA Spotlight yet? Spot are Limited. Reserve your spot now ✅ Subscribe to us on YouTube AND Get FREE Access to Premium GMAT Question Bank for 7 Days ## Prep Club for GRE REWARDS |