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# 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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22 Oct 2013, 09:58
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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
Joined: 02 Sep 2009
Posts: 60694
Re: The operation # is defined for all nonzero x and y by x # y  [#permalink]

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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)

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

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Hope it helps.
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Posts: 60694
Re: The operation # is defined for all nonzero x and y by x # y  [#permalink]

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22 Oct 2013, 10:07
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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Re: The operation # is defined for all nonzero x and y by x # y  [#permalink]

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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
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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22 Oct 2013, 23:43
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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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
Joined: 02 Sep 2009
Posts: 60694
Re: The operation # is defined for all nonzero x and y by x # y  [#permalink]

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

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23 Feb 2014, 03:52
Sometimes it's the little things... :D This helped untie the knot in my head, thanks a lot!
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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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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

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Re: The operation # is defined for all nonzero x and y by x # y  [#permalink]

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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)

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

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
Joined: 02 Sep 2009
Posts: 60694
Re: The operation # is defined for all nonzero x and y by x # y  [#permalink]

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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)

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

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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Posts: 2806
Re: The operation # is defined for all nonzero x and y by x # y  [#permalink]

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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)

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Re: The operation # is defined for all nonzero x and y by x # y  [#permalink]

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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

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Math Expert
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Posts: 8328
Re: The operation # is defined for all nonzero x and y by x # y  [#permalink]

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