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M18-01

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M18-01  [#permalink]

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New post 16 Sep 2014, 00:02
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If \(\frac{1}{2 + \frac{1}{x}} = 1\), what is the value of \(x\)?

A. \(-1\)
B. \(-\frac{1}{2}\)
C. \(\frac{1}{2}\)
D. \(1 \frac{1}{2}\)
E. \(2 \frac{1}{2}\)

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Re M18-01  [#permalink]

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New post 16 Sep 2014, 00:02
Official Solution:

If \(\frac{1}{2 + \frac{1}{x}} = 1\), what is the value of \(x\)?

A. \(-1\)
B. \(-\frac{1}{2}\)
C. \(\frac{1}{2}\)
D. \(1 \frac{1}{2}\)
E. \(2 \frac{1}{2}\)

\(\frac{1}{2 + \frac{1}{x}} = 1\) simplifies to \(2 + \frac{1}{x} = 1\) or \(\frac{1}{x} = -1\) from where \(x = -1\).

Answer: A
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Re: M18-01  [#permalink]

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New post 01 Jun 2017, 04:29
Solving we get x/(2x+1) = 1 ; i.e., x=-1. The answer is option A
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  [#permalink]

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New post 01 Jun 2017, 05:29
What am I doing wrong?

If 1/(2+1x)=1
1/2 + 1/(1/x) = 1
1/2 + x/1 = 1
Then x/1 must equal 1/2 in order to maintain the equality.
Thx in advance.

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Re: M18-01  [#permalink]

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New post 01 Jun 2017, 05:35
cocojiz wrote:
What am I doing wrong?

If 1/(2+1x)=1
1/2 + 1/(1/x) = 1
1/2 + x/1 = 1
Then x/1 must equal 1/2 in order to maintain the equality.
Thx in advance.

Posted from my mobile device


\(\frac{1}{a+b}\neq \frac{1}{a} + \frac{1}{b}\).

Does \(\frac{1}{1+1}=\frac{1}{2}\) equal to \(\frac{1}{1}+\frac{1}{1}=2\)?
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Re M18-01  [#permalink]

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New post 29 Nov 2017, 07:34
I think this is a high-quality question and I agree with explanation.
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M18-01  [#permalink]

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New post 17 Dec 2017, 04:48
Bunuel

Quote:
\(\frac{1}{2 + \frac{1}{x}} = 1\) simplifies to \(2 + \frac{1}{x} = 1\) or \(\frac{1}{x} = -1\) from where \(x = -1\).


I agree with OE but ended up simplifying \(\frac{1}{2 + \frac{1}{x}} = 1\) to \(\frac{1}{ \frac{2x+1}{x}} = 1\) finally ending up with same answer in
form of : 2x + 1 = x
or
x=-1.

Is this correct too?

Let me know if you assumed x to be positive integer in your approach?
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Re: M18-01  [#permalink]

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New post 17 Dec 2017, 04:56
1
adkikani wrote:
Bunuel

Quote:
\(\frac{1}{2 + \frac{1}{x}} = 1\) simplifies to \(2 + \frac{1}{x} = 1\) or \(\frac{1}{x} = -1\) from where \(x = -1\).


I agree with OE but ended up simplifying \(\frac{1}{2 + \frac{1}{x}} = 1\) to \(\frac{1}{ \frac{2x+1}{x}} = 1\) finally ending up with same answer in
form of : 2x + 1 = x
or
x=-1.

Is this correct too?

Let me know if you assumed x to be positive integer in your approach?


Hi adkikani

Your approach is perfectly correct. :thumbup:

as this is an "Equality" and not an "Inequality" so you need not worry about of x being positive or negative while cross-multiplying variables. In case of inequality sign changes when you cross-multiply with negative number.
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Re: M18-01  [#permalink]

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New post 17 Dec 2017, 04:58
adkikani wrote:
Bunuel

Quote:
\(\frac{1}{2 + \frac{1}{x}} = 1\) simplifies to \(2 + \frac{1}{x} = 1\) or \(\frac{1}{x} = -1\) from where \(x = -1\).


I agree with OE but ended up simplifying \(\frac{1}{2 + \frac{1}{x}} = 1\) to \(\frac{1}{ \frac{2x+1}{x}} = 1\) finally ending up with same answer in
form of : 2x + 1 = x
or
x=-1.

Is this correct too?

Let me know if you assumed x to be positive integer in your approach?


Yes, that's correct. Why do you think that I assumed that x is a positive integer? Why should I've assumed that? How could I assumed that if I got x = -1?
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New post 17 Dec 2017, 05:24
Bunuel

Quote:
Yes, that's correct.Why do you think that I assumed that x is a positive integer? Why should I've assumed that? How could I assumed that if I got x = -1?


Let me take numbers show two different results for LHS of \(\frac{1}{2 + \frac{1}{x}} = 1\)

x = 1 the fraction becomes 1/3 (overall fraction is now positive)

x = 0 (fraction is not a positive integer)

What I inferred from your OE is that you reversed the fraction but I also need to know the sign of denominator.

I think this Q can be solved using simple common sense w/o touching a pen and paper:

1. I need the ratio to be positive.
2. Denominator has to be positive since 1 is positive
3. To satisfy LHS = RHS I need x = -1

I think my original query was based more on ratio than equating LHS and RHS.
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Re: M18-01  [#permalink]

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New post 17 Dec 2017, 05:33
1
adkikani wrote:
Bunuel

Quote:
Yes, that's correct.Why do you think that I assumed that x is a positive integer? Why should I've assumed that? How could I assumed that if I got x = -1?


Let me take numbers show two different results for LHS of \(\frac{1}{2 + \frac{1}{x}} = 1\)

x = 1 the fraction becomes 1/3 (overall fraction is now positive)

x = -3 fraction becomes -1 (overall fraction is now negative)

What I inferred from your OE is that you reversed the fraction but I also need to know the sign of denominator.

I think this Q can be solved using simple common sense w/o touching a pen and paper:

1. I need the ratio to be positive.
2. Denominator has to be positive since 1 is positive
3. To satisfy LHS = RHS I need x = -1

I think my original query was based more on ratio than equating LHS and RHS.


I do not follow you...

1. What does x = 1 and x = -3 has to do here? Neither of them satisfy the equation.

2. You did not understand the solution. \(\frac{1}{2 + \frac{1}{x}} = 1\), from here we cross multiply to get \(1=2 + \frac{1}{x}\).

3. We have an EQUATION not INEQUALITY. The signs of denominator/numerator are absolutely irrelevant when we manipulate the equations.
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Re: M18-01  [#permalink]

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New post 17 Dec 2017, 05:40
Bunuel

I understood my mistake. I over complicated simple proportion by taking reciprocals
instead of cross-multiplication. :-)
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Re: M18-01   [#permalink] 17 Dec 2017, 05:40
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