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

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Bunuel

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22 Oct 2013, 09:07

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)

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

Answer: E.

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SaraLotfy

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22 Oct 2013, 09:13

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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kartboybo

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22 Oct 2013, 22:43

Thanks

Will definitely check out the other topics

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tr1

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23 Feb 2014, 02:25

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)

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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Bunuel

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23 Feb 2014, 02:29

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)

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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tr1

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23 Feb 2014, 02:52

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

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Lars1988

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19 Jan 2015, 13:31

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.

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01 May 2018, 14:44

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)

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

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!

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Bunuel

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01 May 2018, 20:47

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)

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

Answer: E.

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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03 May 2018, 09:51

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)

Answer: E

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mborski

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13 Dec 2019, 17:20

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.

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..and stopped.JPG [ 13.61 KiB | Viewed 33176 times ]

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14 Dec 2019, 21:34

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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18 May 2020, 18:19

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)

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

Answer: E.

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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05 May 2021, 22:11

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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