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#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # The operation # is defined for all nonzero x and y by x # y

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

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10 00:00

Difficulty:   15% (low)

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

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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. $$\frac{2a+1}{a+1}$$
Math Expert V
Joined: 02 Sep 2009
Posts: 60778
Re: The operation # is defined for all nonzero x and y by x # y  [#permalink]

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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)

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

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

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1
4
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)

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

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Manager  Joined: 24 Apr 2013
Posts: 50
Location: United States
Re: The operation # is defined for all nonzero x and y by x # y  [#permalink]

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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  Joined: 16 Oct 2013
Posts: 6
Re: The operation # is defined for all nonzero x and y by x # y  [#permalink]

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Thanks Will definitely check out the other topics
Intern  Joined: 22 Feb 2014
Posts: 5
Re: The operation # is defined for all nonzero x and y by x # y  [#permalink]

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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 V
Joined: 02 Sep 2009
Posts: 60778
Re: The operation # is defined for all nonzero x and y by x # y  [#permalink]

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

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

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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

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

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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)

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

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 V
Joined: 02 Sep 2009
Posts: 60778
Re: The operation # is defined for all nonzero x and y by x # y  [#permalink]

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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)

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

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.
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Target Test Prep Representative G
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]

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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)

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

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

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

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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
_________________ Re: The operation # is defined for all nonzero x and y by x # y   [#permalink] 14 Dec 2019, 22:34
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