 It is currently 21 Mar 2019, 05:55 ### 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. # Q02-38 Question # 08 Section # 09  Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS: Intern Joined: 04 Jun 2016
Posts: 2
Followers: 0

Kudos [?]: 2  , given: 0

Q02-38 Question # 08 Section # 09 [#permalink]
1
KUDOS 00:00

Question Stats: 37% (01:59) correct 62% (02:59) wrong based on 8 sessions
Given the question 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
[Reveal] Spoiler: OA

Last edited by Carcass on 25 Jun 2016, 00:50, edited 1 time in total.
Editing the post and adding the tags
Moderator  Joined: 18 Apr 2015
Posts: 5845
Followers: 94

Kudos [?]: 1144 , given: 5448

Re: Q02-38 Question # 08 Section # 09 [#permalink]
Expert's post
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.
_________________
Manager Joined: 27 Sep 2017
Posts: 112
Followers: 1

Kudos [?]: 30 , given: 4

Re: Q02-38 Question # 08 Section # 09 [#permalink]
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.

I still don't know why $$\sqrt{\frac{-y}{|y|}}$$ =-y NOT -1? Intern Joined: 05 Jan 2018
Posts: 32
Followers: 0

Kudos [?]: 16  , given: 8

Re: Q02-38 Question # 08 Section # 09 [#permalink]
1
KUDOS
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.

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] 24 May 2018, 11:07
Display posts from previous: Sort by

# Q02-38 Question # 08 Section # 09  Question banks Downloads My Bookmarks Reviews Important topics Powered by phpBB © phpBB Group Kindly note that the GRE® test is a registered trademark of the Educational Testing Service®, and this site has neither been reviewed nor endorsed by ETS®.