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M18-01

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Joined: 02 Sep 2009

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M18-01 [#permalink]

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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. $2 \frac{1}{2}$

Spoiler: OA

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Re M18-01 [#permalink]

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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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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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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.

Posted from my mobile device

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Re: M18-01 [#permalink]

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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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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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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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17 Dec 2017, 04:56

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 $x = -1$.

Hi adkikani

Your approach is perfectly correct.

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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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 $x = -1$.

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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M18-01 [#permalink]

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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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17 Dec 2017, 05:33

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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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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