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If xy > 0, 1/x + 1/y = 5, and 1/xy = 6, then (x+y)/5 = ?

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If xy > 0, 1/x + 1/y = 5, and 1/xy = 6, then (x+y)/5 = ?  [#permalink]

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New post 09 Nov 2015, 23:02
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72% (01:24) correct 28% (02:08) wrong based on 152 sessions

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Re: If xy > 0, 1/x + 1/y = 5, and 1/xy = 6, then (x+y)/5 = ?  [#permalink]

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New post 10 Nov 2015, 08:58
Answer is B.

(1/X+1/Y)=5 canbe solved as {(x+y)/xy}=6. Substituting for 1/xy=6, we get
x+y=5/6

==> (x+y)/5= 5/(6*5)=1/6.
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If xy > 0, 1/x + 1/y = 5, and 1/xy = 6, then (x+y)/5 = ?  [#permalink]

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New post 23 Oct 2017, 10:14
Bunuel wrote:
If xy > 0, 1/x + 1/y = 5, and 1/xy = 6, then (x+y)/5 = ?

A. 1/25
B. 1/6
C. 1/5
D. 5
E. 6

HarrishGowtham wrote:
Answer is B.

(1/X+1/Y)=5 canbe solved as {(x+y)/xy}=6. Substituting for 1/xy=6, we get
x+y=5/6

==> (x+y)/5= 5/(6*5)=1/6.

I am stuck trying to understand this explanation. I got the right answer, but not this way.

\(\frac{1}{x}+ \frac{1}{y} = 5\)

\((\frac{1}{x}+ \frac{1}{y}) = \frac{x + y}{xy}\)

Aren't the two RHS equivalent?

Quote:
(1/X+1/Y)=5 can be solved as {(x+y)/xy}=6.


Can anyone explain the steps from the first part of the above sentence to the last part? If the two RHS above quote are equivalent, where does "= 6" come from?
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Re: If xy > 0, 1/x + 1/y = 5, and 1/xy = 6, then (x+y)/5 = ?  [#permalink]

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New post 02 Apr 2018, 16:10
Bunuel wrote:
If xy > 0, 1/x + 1/y = 5, and 1/xy = 6, then (x+y)/5 = ?

A. 1/25
B. 1/6
C. 1/5
D. 5
E. 6


Bunuel, will you please provide the solution?

I don't understand the solution above either. Thanks in advance.
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Re: If xy > 0, 1/x + 1/y = 5, and 1/xy = 6, then (x+y)/5 = ?  [#permalink]

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New post 02 Apr 2018, 20:52
msurls wrote:
Bunuel wrote:
If xy > 0, 1/x + 1/y = 5, and 1/xy = 6, then (x+y)/5 = ?

A. 1/25
B. 1/6
C. 1/5
D. 5
E. 6


Bunuel, will you please provide the solution?

I don't understand the solution above either. Thanks in advance.


\(\frac{1}{x} + \frac{1}{y} = 5\);

\(\frac{x+y}{xy} = 5\);

Since given that \(\frac{1}{xy}= 6\), then \(6(x+y)=5\).

Divide by 30: \(\frac{x+y}{5}=\frac{1}{6}\).

Answer: B.
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If xy > 0, 1/x + 1/y = 5, and 1/xy = 6, then (x+y)/5 = ?  [#permalink]

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New post 10 Apr 2018, 09:03
Bunuel wrote:
msurls wrote:
Bunuel wrote:
If xy > 0, 1/x + 1/y = 5, and 1/xy = 6, then (x+y)/5 = ?

A. 1/25
B. 1/6
C. 1/5
D. 5
E. 6


Bunuel, will you please provide the solution?

I don't understand the solution above either. Thanks in advance.


\(\frac{1}{x} + \frac{1}{y} = 5\);

\(\frac{x+y}{xy} = 5\);

Since given that \(\frac{1}{xy}= 6\), then \(6(x+y)=5\).

Divide by 30: \(\frac{x+y}{5}=\frac{1}{6}\).

Answer: B.


My approach is pretty similar:

\(\frac{1}{x} + \frac{1}{y} = 5\);

\(\frac{x+y}{xy} = 5\);

\(\frac{x+y}{5} = XY\);

\(\frac{1}{xy} = 6\) --> XY = \(;\frac{1}{6}\);

so

\(\frac{x+y}{5} = XY = 1/6\);
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If xy > 0, 1/x + 1/y = 5, and 1/xy = 6, then (x+y)/5 = ? &nbs [#permalink] 10 Apr 2018, 09:03
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