GMAT Question of the Day: Daily via email | Daily via Instagram New to GMAT Club? Watch this Video

It is currently 29 Jan 2020, 07:09

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

The operation # is defined for all nonzero x and y by x # y

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

Hide Tags

Find Similar Topics 
Intern
Intern
avatar
Joined: 16 Oct 2013
Posts: 6
The operation # is defined for all nonzero x and y by x # y  [#permalink]

Show Tags

New post 22 Oct 2013, 09:58
2
10
00:00
A
B
C
D
E

Difficulty:

  15% (low)

Question Stats:

81% (01:31) correct 19% (01:52) wrong based on 386 sessions

HideShow timer Statistics

The operation # is defined for all nonzero x and y by \(x#y = x + \frac{x}{y}\). If \(a>0\), then \(1#(1#a) =\)


A. \(a\)

B. \(a+1\)

C. \(\frac{a}{a+1}\)

D. \(\frac{a+2}{a+1}\)

E. \(\frac{2a+1}{a+1}\)
Most Helpful Expert Reply
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 60778
Re: The operation # is defined for all nonzero x and y by x # y  [#permalink]

Show Tags

New post 22 Oct 2013, 10:14
3
5
Bunuel wrote:
kartboybo wrote:
The operation # is defined for all nonzero x and y by x#y = x + x/y. If a>0, then 1#(1#a) =

A. a
B. a+1
C. a/(a+1)
D. (a+2)/(a+1)
E. (2a+1)/(a+1)



\(1#(1#a) = 1#(1+\frac{1}{a}) = 1#(\frac{a+1}{a})= 1+ \frac{1}{(\frac{a+1}{a})}=1+\frac{a}{a+1}=\frac{a+1+a}{a+1}=\frac{2a+1}{a+1}\).

Answer: E.


Similar questions to practice:
the-operation-is-defined-by-x-y-1-x-1-y-for-all-67650.html
gmat-prep-question-101282.html
if-the-operation-is-defined-by-x-y-xy-1-2-for-all-144272.html
an-operation-is-defined-by-the-equation-a-b-a-b-a-144074.html
let-denote-a-mathematical-operation-is-it-true-that-x-y-131347.html
for-any-operation-that-acts-on-two-numbers-x-and-y-the-127024.html
the-operation-is-defined-for-all-nonzero-numbers-a-and-b-106495.html
the-operation-x-n-for-all-positive-integers-greater-than-99064.html
for-all-integers-x-and-y-the-operation-is-defined-by-x-y-86208.html

Hope it helps.
_________________
General Discussion
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 60778
Re: The operation # is defined for all nonzero x and y by x # y  [#permalink]

Show Tags

New post 22 Oct 2013, 10:07
1
4
Manager
Manager
avatar
Joined: 24 Apr 2013
Posts: 50
Location: United States
Re: The operation # is defined for all nonzero x and y by x # y  [#permalink]

Show Tags

New post 22 Oct 2013, 10:13
kartboybo wrote:
The operation # is defined for all nonzero x and y by x#y = x + x/y. If a>0, then 1#(1#a) =

A. a
B. a+1
C. a/(a+1)
D. (a+2)/(a+1)
E. (2a+1)/(a+1)


You should first start with the parenthesis and mimic exactly the operation given

(1#a) = 1+\(1/a\)
and re-apply the equation to get 1#(1#a)
1#(1#a)= 1 + \(1/1+1/a\)
unify the denominator and multiply with the numerator, you will finally get 1 + \(a/a+1\)
simplify again to get \(2a+1/a+1\)
Intern
Intern
avatar
Joined: 16 Oct 2013
Posts: 6
Re: The operation # is defined for all nonzero x and y by x # y  [#permalink]

Show Tags

New post 22 Oct 2013, 23:43
Thanks :) Will definitely check out the other topics
Intern
Intern
avatar
Joined: 22 Feb 2014
Posts: 5
Re: The operation # is defined for all nonzero x and y by x # y  [#permalink]

Show Tags

New post 23 Feb 2014, 03:25
SaraLotfy wrote:
kartboybo wrote:
The operation # is defined for all nonzero x and y by x#y = x + x/y. If a>0, then 1#(1#a) =

A. a
B. a+1
C. a/(a+1)
D. (a+2)/(a+1)
E. (2a+1)/(a+1)


You should first start with the parenthesis and mimic exactly the operation given

(1#a) = 1+\(1/a\)
and re-apply the equation to get 1#(1#a)
1#(1#a)= 1 + \(1/1+1/a\)
unify the denominator and multiply with the numerator, you will finally get 1 + \(a/a+1\)
simplify again to get \(2a+1/a+1\)


Thank you for this explanation! I'm struggling with understanding your last step. Could you please explain to me how you got from 1 + \(a/a+1\) to \(2a+1/a+1\)?
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 60778
Re: The operation # is defined for all nonzero x and y by x # y  [#permalink]

Show Tags

New post 23 Feb 2014, 03:29
1
tr1 wrote:
SaraLotfy wrote:
kartboybo wrote:
The operation # is defined for all nonzero x and y by x#y = x + x/y. If a>0, then 1#(1#a) =

A. a
B. a+1
C. a/(a+1)
D. (a+2)/(a+1)
E. (2a+1)/(a+1)


You should first start with the parenthesis and mimic exactly the operation given

(1#a) = 1+\(1/a\)
and re-apply the equation to get 1#(1#a)
1#(1#a)= 1 + \(1/1+1/a\)
unify the denominator and multiply with the numerator, you will finally get 1 + \(a/a+1\)
simplify again to get \(2a+1/a+1\)


Thank you for this explanation! I'm struggling with understanding your last step. Could you please explain to me how you got from 1 + \(a/a+1\) to \(2a+1/a+1\)?


\(1+\frac{a}{a+1}=\frac{a+1}{a+1}+\frac{a}{a+1}=\frac{(a+1)+a}{a+1}=\frac{2a+1}{a+1}\).

Check properly formatted solution here: the-operation-is-defined-for-all-nonzero-x-and-y-by-x-y-162019.html#p1281893

Hope it helps.
_________________
Intern
Intern
avatar
Joined: 22 Feb 2014
Posts: 5
Re: The operation # is defined for all nonzero x and y by x # y  [#permalink]

Show Tags

New post 23 Feb 2014, 03:52
Sometimes it's the little things... :D This helped untie the knot in my head, thanks a lot!
Intern
Intern
avatar
Joined: 17 Dec 2014
Posts: 11
Re: The operation # is defined for all nonzero x and y by x # y  [#permalink]

Show Tags

New post 19 Jan 2015, 14:31
3
Is my calculation correct?

x#y = x + x/y. If a>0, then 1#(1#a) =

You can see that a stands for y and 1 stands for x. I like to work with numbers so I did x=1 a=2 and a=y so y=2
1#a= 1+(1/2)=1.5 so now a is 1.5 --> 1+(1/1.5)=1 2/3 = 5/3

So now you can check by filling in a, which is 2:
A. a --> 2
B. a+1 --> 2+1=3
C. a/(a+1)-->2/(2+1)=2/3
D. (a+2)/(a+1)-->(2+2)/(2+1)=4/3
E. (2a+1)/(a+1)-->(2*2+1)/(2+1)=5/3

So E is the answer.
Manager
Manager
avatar
G
Joined: 21 Jul 2015
Posts: 185
GMAT 1: 710 Q49 V38
GMAT 2: 730 Q50 V38
Reviews Badge
Re: The operation # is defined for all nonzero x and y by x # y  [#permalink]

Show Tags

New post 01 May 2018, 15:44
Bunuel wrote:
kartboybo wrote:
The operation # is defined for all nonzero x and y by x#y = x + x/y. If a>0, then 1#(1#a) =

A. a
B. a+1
C. a/(a+1)
D. (a+2)/(a+1)
E. (2a+1)/(a+1)



\(1#(1#a) = 1#(1+\frac{1}{a}) = 1#(\frac{a+1}{a})= 1+ \frac{1}{(\frac{a+1}{a})}=1+\frac{a}{a+1}=\frac{a+1+a}{a+1}=\frac{2a+1}{a+1}\).

Answer: E.


Hi Bunuel / Experts,

I solved this by taking a=1, but ended up with a wrong answer.

1#(1#1) = 1#(1+1/1) = 1#2 = 1+1/2 = 1+.5 = 1.5

if you put a=1 in choice D, you get 1.5, while E also gives you the same 1.5.

Anyway, when I got to D and got 1.5, I marked it and moved on. Can you tell me what to keep in mind while solving these type of questions through plugin numbers? Thanks so much!
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 60778
Re: The operation # is defined for all nonzero x and y by x # y  [#permalink]

Show Tags

New post 01 May 2018, 21:47
1
sdlife wrote:
Bunuel wrote:
kartboybo wrote:
The operation # is defined for all nonzero x and y by x#y = x + x/y. If a>0, then 1#(1#a) =

A. a
B. a+1
C. a/(a+1)
D. (a+2)/(a+1)
E. (2a+1)/(a+1)



\(1#(1#a) = 1#(1+\frac{1}{a}) = 1#(\frac{a+1}{a})= 1+ \frac{1}{(\frac{a+1}{a})}=1+\frac{a}{a+1}=\frac{a+1+a}{a+1}=\frac{2a+1}{a+1}\).

Answer: E.


Hi Bunuel / Experts,

I solved this by taking a=1, but ended up with a wrong answer.

1#(1#1) = 1#(1+1/1) = 1#2 = 1+1/2 = 1+.5 = 1.5

if you put a=1 in choice D, you get 1.5, while E also gives you the same 1.5.

Anyway, when I got to D and got 1.5, I marked it and moved on. Can you tell me what to keep in mind while solving these type of questions through plugin numbers? Thanks so much!


For plug-in method it might happen that for some particular number(s) more than one option may give "correct" answer. In this case just pick some other numbers and check again these "correct" options only.
_________________
Target Test Prep Representative
User avatar
G
Status: Head GMAT Instructor
Affiliations: Target Test Prep
Joined: 04 Mar 2011
Posts: 2804
Re: The operation # is defined for all nonzero x and y by x # y  [#permalink]

Show Tags

New post 03 May 2018, 10:51
1
kartboybo wrote:
The operation # is defined for all nonzero x and y by x#y = x + x/y. If a>0, then 1#(1#a) =

A. a
B. a+1
C. a/(a+1)
D. (a+2)/(a+1)
E. (2a+1)/(a+1)


As with any composite function, we work from the inside out. So we first must calculate 1 # a.

1 # a is:

1 + 1/a

a/a + 1/a = (a + 1)/a

Now we see that 1#(1#a) = 1#[(a + 1)/a]. We have:


1 # (a + 1)/a is:

1 + 1/[(a + 1)/a]

1 + a/(a + 1)

(a + 1)/(a + 1) + a/(a + 1)

(2a + 1)/(a + 1)

Answer: E
_________________

Jeffrey Miller

Head of GMAT Instruction

Jeff@TargetTestPrep.com
TTP - Target Test Prep Logo
181 Reviews

5-star rated online GMAT quant
self study course

See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews

If you find one of my posts helpful, please take a moment to click on the "Kudos" button.

Intern
Intern
avatar
B
Joined: 26 Oct 2019
Posts: 11
Location: United States (TX)
GPA: 3.93
WE: Sales (Manufacturing)
Re: The operation # is defined for all nonzero x and y by x # y  [#permalink]

Show Tags

New post 13 Dec 2019, 18:20
How do you know to convert 1+(1/a)?

If you keep going and don't convert, is there a way to get to the answer?

Thanks.
Attachments

How do you know..JPG
How do you know..JPG [ 15.86 KiB | Viewed 936 times ]

..and stopped.JPG
..and stopped.JPG [ 13.61 KiB | Viewed 936 times ]

Math Expert
avatar
V
Joined: 02 Aug 2009
Posts: 8322
Re: The operation # is defined for all nonzero x and y by x # y  [#permalink]

Show Tags

New post 14 Dec 2019, 22:34
kartboybo wrote:
The operation # is defined for all nonzero x and y by \(x#y = x + \frac{x}{y}\). If \(a>0\), then \(1#(1#a) =\)


A. \(a\)

B. \(a+1\)

C. \(\frac{a}{a+1}\)

D. \(\frac{a+2}{a+1}\)

E. \(\frac{2a+1}{a+1}\)



x#y=\(x+\frac{x}{y}\), and we are looking for 1#(1#a)
Now 1#a=\(1+\frac{1}{a}=\frac{1+a}{a}\)
Now 1#(1#a) becomes 1#\(\frac{1+a}{a}\)
Also 1#\(\frac{1+a}{a}\)=\(1+\frac{a}{1+a}=\frac{1+2a}{1+a}\)

E
_________________
GMAT Club Bot
Re: The operation # is defined for all nonzero x and y by x # y   [#permalink] 14 Dec 2019, 22:34
Display posts from previous: Sort by

The operation # is defined for all nonzero x and y by x # y

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





Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne