Last visit was: 14 Jul 2024, 06:41 It is currently 14 Jul 2024, 06:41
Close
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
Your Progress

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.
Close
Request Expert Reply
Confirm Cancel
SORT BY:
Date
Math Expert
Joined: 02 Sep 2009
Posts: 94342
Own Kudos [?]: 640692 [40]
Given Kudos: 85011
Send PM
Most Helpful Reply
RC & DI Moderator
Joined: 02 Aug 2009
Status:Math and DI Expert
Posts: 11470
Own Kudos [?]: 34305 [6]
Given Kudos: 322
Send PM
General Discussion
Manager
Manager
Joined: 23 Jul 2015
Posts: 66
Own Kudos [?]: 29 [0]
Given Kudos: 297
Send PM
Intern
Intern
Joined: 08 Oct 2019
Posts: 1
Own Kudos [?]: 5 [0]
Given Kudos: 36
Send PM
Re: The operator @ is defined by the following expression: a@b = |(a + 1)| [#permalink]
So \(\frac{1}{2}*(|\frac{x+1}{x}|)-|\frac{2+1}{2}|=0\)......
\(|\frac{x+1}{x}|=3\)....

Can anyone explain how \(|\frac{x+1}{x}|=3\)?
Manager
Manager
Joined: 10 Mar 2018
Posts: 60
Own Kudos [?]: 86 [1]
Given Kudos: 44
Location: India
Concentration: Entrepreneurship, Marketing
GMAT 1: 680 Q44 V38
WE:Design (Retail)
Send PM
Re: The operator @ is defined by the following expression: a@b = |(a + 1)| [#permalink]
1
Kudos
harshit121 wrote:
So \(\frac{1}{2}*(|\frac{x+1}{x}|)-|\frac{2+1}{2}|=0\)......
\(|\frac{x+1}{x}|=3\)....

Can anyone explain how \(|\frac{x+1}{x}|=3\)?



\(\frac{1}{2}*(|\frac{x+1}{x}|)-|\frac{2+1}{2}|=0\)......

=> \(\frac{1}{2}*(|\frac{x+1}{x}|) = \frac{3}{2}\)

Multiply L.H.S. and R.H.S. by 2

\(|\frac{x+1}{x}|=3\)
Manager
Manager
Joined: 21 Aug 2019
Posts: 101
Own Kudos [?]: 42 [1]
Given Kudos: 353
Send PM
Re: The operator @ is defined by the following expression: a@b = |(a + 1)| [#permalink]
1
Kudos
I didnt understand this question
Re: The operator @ is defined by the following expression: a@b = |(a + 1)| [#permalink]
|(X+1)/X|=3

If (X+1)/X < 0, then X<-1
On that basis I started solving
(X+1)/X = -3
X + 1 = -3X
X = -1/4 or -0.25

this solution did not match the criteria as -0.25 is not less than -1
So I rejected it. Where am I going wrong? Bunuel
RC & DI Moderator
Joined: 02 Aug 2009
Status:Math and DI Expert
Posts: 11470
Own Kudos [?]: 34305 [0]
Given Kudos: 322
Send PM
Re: The operator @ is defined by the following expression: a@b = |(a + 1)| [#permalink]
Expert Reply
smw wrote:
|(X+1)/X|=3

If (X+1)/X < 0, then X<-1
On that basis I started solving
(X+1)/X = -3
X + 1 = -3X
X = -1/4 or -0.25

this solution did not match the criteria as -0.25 is not less than -1
So I rejected it. Where am I going wrong? Bunuel



If \(|\frac{x+1}{x}|=3\)

\(|\frac{x+1}{x}|<0\)

So, both x+1 and x are of opposite sign.

Two cases
1) x>0, then x+1<0 or x<-1…….NO
2) x<0, then x+1>0 or x>-1…..YES
Tutor
Joined: 05 Apr 2011
Status:Tutor - BrushMyQuant
Posts: 1797
Own Kudos [?]: 2140 [2]
Given Kudos: 100
Location: India
Concentration: Finance, Marketing
Schools: XLRI (A)
GMAT 1: 700 Q51 V31
GPA: 3
WE:Information Technology (Computer Software)
Send PM
Re: The operator @ is defined by the following expression: a@b = |(a + 1)| [#permalink]
2
Kudos
Expert Reply
Top Contributor
Given that \(a@b = |\frac{a+1}{a}| - \frac{b+1}{b}\) where \(ab\neq{0}\) and we need to find the sum of all solutions of the equation \(x@2 = \frac{x@(-1)}{2}\)

Let's start by finding the value of x@2 first
To find the value of x@2, compare x@2 with a@b
=> a=x and b=2
So, to find x@2 substitute a=x and b=2 in \(a@b = |\frac{a+1}{a}| - \frac{b+1}{b}\)
=> \(x@2 = |\frac{x+1}{x}| - \frac{2+1}{2}\) =\( |\frac{x+1}{x}| - \frac{3}{2}\)

Similarly, \(x@(-1) = |\frac{x+1}{x}| - \frac{-1+1}{-1}\) =\(| \frac{x+1}{x}| - 0\) = \(|\frac{x+1}{x}|\)

Given that \(x@2 = \frac{x@(-1)}{2}\)
=> \( |\frac{x+1}{x}| - \frac{3}{2}\) = \(|\frac{x+1}{x}|\) / 2
=> \(| \frac{x+1}{x}| - |\frac{x+1}{x}|\) / 2 =\( \frac{3}{2}\)
Multiply both the sides by 2 we get
\( 2*|\frac{x+1}{x}| - |\frac{x+1}{x}|\) = 3
=> \(|\frac{x+1}{x}|\) = 3

We will get two cases (Watch this video to know about the basic of Absolute Value)
Case 1: \(\frac{x+1}{x }\)>= 0 (0r x+1 > 0 => x > -1 and x≠0
=> \(|\frac{x+1}{x}|\) = \(\frac{x+1}{x }\)
=> \(\frac{x+1}{x }\) = 3
=> x + 1 = 3x
=> 2x = 1 => x = \(\frac{1}{2}\) which is > -1 so this is one solution

Case 2: \(\frac{x+1}{x }\)< 0
=> \(|\frac{x+1}{x}|\) = - \(\frac{x+1}{x }\)
=> -\(\frac{x+1}{x }\) = 3
=> -x - 1 = 3x => 4x = -1
or x = \(\frac{-1}{4}\)
Let's check if this is a right answer by substituting the value in \(\frac{x+1}{x }\)< 0
((-1/4) + 1) / (-1/4) = (3/4) / (-1/4) < 0 so this is also a solution

=> Sum of values of the solution = \(\frac{1}{2}\) + \(\frac{-1}{4}\) = \(\frac{1}{4}\) = 0.25

So, Answer will be D
Hope it helps!

Watch the following video to learn the Basics of Functions and Custom Characters

Intern
Intern
Joined: 13 Dec 2021
Posts: 17
Own Kudos [?]: 4 [0]
Given Kudos: 118
Send PM
Re: The operator @ is defined by the following expression: a@b = |(a + 1)| [#permalink]
Hi BrushMyQuant,

Thank you for the solution! Can you kindly elaborate more on the steps before Case 1 and 2? Why was this necessary? \(|\frac{x+1}{x}|\) = 3

Many thanks,
Em
Tutor
Joined: 26 Jun 2014
Status:Mentor & Coach | GMAT Q51 | CAT 99.98
Posts: 450
Own Kudos [?]: 809 [1]
Given Kudos: 8
Send PM
Re: The operator @ is defined by the following expression: a@b = |(a + 1)| [#permalink]
1
Kudos
Expert Reply
Bunuel wrote:
The operator \(@\) is defined by the following expression: \(a@b = |\frac{a+1}{a}| - \frac{b+1}{b}\) where \(ab\neq{0}\). What is the sum of the solutions to the equation \(x@2 = \frac{x@(-1)}{2}\) ?


A. -1

B. -0.75

C. -0.25

D. 0.25

E. 0.75



a@b = |(a+1)/a| - (b+1)/b
We need to solve: x@2 = 1/2 * [x@(-1)]
i.e. 2 * x@2 = x@(-1)

We have: 2 * x@2 = 2 * [|(x+1)/x| - (2+1)/2] = 2 * |(x+1)/x| - 3
Also: x@(-1) = |(x+1)/x| - (-1+1)/2 = |(x+1)/x| - 0

Thus, we have: 2 * |(x+1)/x| - 3 = |(x+1)/x|
=> |(x+1)/x| = 3
=> (x+1)/x = 3 or -3
=> x+1 = 3x or -3x
=> x = 1/2 or -1/4

Sum of the solutions = 1/2 + (-1/4) = 1/4

Answer D
GMAT Club Bot
Re: The operator @ is defined by the following expression: a@b = |(a + 1)| [#permalink]
Moderator:
Math Expert
94342 posts