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# M18-01

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Math Expert
Joined: 02 Sep 2009
Posts: 44421

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16 Sep 2014, 01:02
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5% (low)

Question Stats:

97% (00:21) correct 3% (00:20) wrong based on 121 sessions

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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. $$2 \frac{1}{2}$$
[Reveal] Spoiler: OA

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Math Expert
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16 Sep 2014, 01:02
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 $$x = -1$$.

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GMAT 1: 640 Q48 V29

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01 Jun 2017, 05:29
Solving we get x/(2x+1) = 1 ; i.e., x=-1. The answer is option A
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Joined: 05 Sep 2016
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01 Jun 2017, 06:29
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.

Posted from my mobile device
Math Expert
Joined: 02 Sep 2009
Posts: 44421

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01 Jun 2017, 06:35
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.

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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Joined: 16 Jan 2013
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GMAT 1: 490 Q41 V18
GMAT 2: 610 Q45 V28
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29 Nov 2017, 08:34
I think this is a high-quality question and I agree with explanation.
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17 Dec 2017, 05:48
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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17 Dec 2017, 05:56
1
KUDOS
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?

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.
Math Expert
Joined: 02 Sep 2009
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17 Dec 2017, 05:58
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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17 Dec 2017, 06:24
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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Math Expert
Joined: 02 Sep 2009
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17 Dec 2017, 06:33
1
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Expert's post
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.

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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17 Dec 2017, 06:40
Bunuel

I understood my mistake. I over complicated simple proportion by taking reciprocals
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Re: M18-01   [#permalink] 17 Dec 2017, 06:40
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# M18-01

Moderators: chetan2u, Bunuel

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