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

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Re: The operation # is defined for all nonzero x and y by x # y [#permalink]
kartboybo
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 $$2a+1/a+1$$
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Re: The operation # is defined for all nonzero x and y by x # y [#permalink]
Thanks Will definitely check out the other topics
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Re: The operation # is defined for all nonzero x and y by x # y [#permalink]
SaraLotfy
kartboybo
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 $$2a+1/a+1$$?
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Re: The operation # is defined for all nonzero x and y by x # y [#permalink]
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tr1
SaraLotfy
kartboybo
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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Re: The operation # is defined for all nonzero x and y by x # y [#permalink]
Sometimes it's the little things... :D This helped untie the knot in my head, thanks a lot!
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Re: The operation # is defined for all nonzero x and y by x # y [#permalink]
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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

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Re: The operation # is defined for all nonzero x and y by x # y [#permalink]
Bunuel
kartboybo
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!
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Re: The operation # is defined for all nonzero x and y by x # y [#permalink]
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sdlife
Bunuel
kartboybo
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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Re: The operation # is defined for all nonzero x and y by x # y [#permalink]
1
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kartboybo
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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Re: The operation # is defined for all nonzero x and y by x # y [#permalink]
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.
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Re: The operation # is defined for all nonzero x and y by x # y [#permalink]
kartboybo
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
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Re: The operation # is defined for all nonzero x and y by x # y [#permalink]
Bunuel
Bunuel
kartboybo
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)

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

How do we get from the second step where we have $$(1+\frac{1}{a})$$ to the third step where we have $$(1+\frac{a+1}{a})$$ ? Same with the next step....I'm having trouble understanding the sequence of events/operations here. ScottTargetTestPrep can you weigh in by chance?
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Re: The operation # is defined for all nonzero x and y by x # y [#permalink]
HasnainAfxal
The operation $$\triangle$$ is defined for all nonzero $$x$$ and $$y$$ by $$x \triangle y=x+\frac{x}{y}$$. If $$a > 0$$, then $$1\triangle(1 \triangle a) =$$

$$a)$$ $$a$$
$$b)$$ $$a+1$$
$$c)$$ $$\frac{a}{(a+1)}$$
$$d)$$ $$\frac{(a+2)}{(a+1)}$$
$$e)$$ $$\frac{(2a+1)}{(a+1)}$$

Given: $$x \triangle y=x+\frac{x}{y}$$
$$1\triangle(1 \triangle a) =$$?

$$1 \triangle a = 1 + \frac{1}{a} = \frac{a + 1}{a}$$
$$1\triangle(1 \triangle a) =$$ = $$1\triangle(\frac{a + 1}{a}) = = 1 + 1/\frac{a + 1}{a} = 1 + \frac{a}{a + 1} = \frac{a + 1 + a}{a + 1} = \frac{2a + 1}{a + 1}$$

So, correct answer is option E.
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Re: The operation # is defined for all nonzero x and y by x # y [#permalink]
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Top Contributor
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)

#Approach 1

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

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

Option E

#Approach 2: Plugin Values
Dont start with assuming a =1; As you can see when you plugin values in the answer options, Options D and E will give the same answer. So in that case, we dont know which option to eliminate and eventually ends up in assuming another value for a and loose some time. Always keep these checks in mind, when you choose values to plugin.

Assume a=2
1#2= 1+ 1/2 = 3/2
1#(1#2)= 1#3/2 = 1+ 1/(3/2)= 1+ 2/3 = 5/3

Substituting a= 2 in the answer options, Only Option E will give you the answer 5/3.

Option E is the correct answer.

Thanks,
Clifin J Francis,
Gmat SME
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Re: The operation # is defined for all nonzero x and y by x # y [#permalink]
kartboybo
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. $$\frac{2a+1}{a+1}$$

Can someone explain to me this type of questions? I dont really know what "The operation # is defined for all nonzero x and y by x#y..." means and how should i go about solving such questions
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Re: The operation # is defined for all nonzero x and y by x # y [#permalink]
Bunuel
Bunuel
kartboybo
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)

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

can i shorten the fraction (1+a+a)/(1+a) to a/(1+a) ?

Similar questions to practice:
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https://gmatclub.com/forum/if-the-opera ... 44272.html
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Hope it helps.

can i shorten the fraction (1+a+a)/(1+a) to a/(1+a) ?
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Re: The operation # is defined for all nonzero x and y by x # y [#permalink]
No, $$\frac{1+a+a}{1+a}=\frac{1}{1+a}+\frac{a}{1+a}+\frac{a}{1+a}= \frac{1+a}{1+a}+\frac{a}{1+a}= 1+\frac{a}{1+a}$$