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# In the expression above, if xn#0, what is the value of S?

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Joined: 02 Dec 2012
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In the expression above, if xn#0, what is the value of S? [#permalink]

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18 Dec 2012, 08:13
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$$S=\frac{\frac{2}{n}}{\frac{1}{x}+\frac{2}{3x}}$$

In the expression above, if xn#0, what is the value of S?

(1) x = 2n
(2) n = 1/2
[Reveal] Spoiler: OA

Kudos [?]: 3462 [2], given: 0

Math Expert
Joined: 02 Sep 2009
Posts: 41893

Kudos [?]: 128826 [1], given: 12183

Re: In the expression above, if xn#0, what is the value of S? [#permalink]

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18 Dec 2012, 08:17
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Expert's post
$$S=\frac{\frac{2}{n}}{\frac{1}{x}+\frac{2}{3x}}$$

In the expression above, if xn#0, what is the value of S?

$$S=\frac{\frac{2}{n}}{\frac{1}{x}+\frac{2}{3x}}=\frac{\frac{2}{n}}{\frac{5}{3x}}=\frac{6x}{5n}$$

(1) x = 2n --> $$S=\frac{6x}{5n}=\frac{6*2n}{5n}=\frac{12}{5}$$. Sufficient.

(2) n = 1/2. We need the value of x to answer the question. Not sufficient.

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Kudos [?]: 128826 [1], given: 12183

Intern
Joined: 28 Jun 2011
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Kudos [?]: 10 [0], given: 10

Re: In the expression above, if xn#0, what is the value of S? [#permalink]

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20 Dec 2013, 10:32
Hi guys,

quick question:

I started to simplify the equation by multiplying the denominator such as 2/n (x+3x/2) and did not get the same solution. What is wrong with my approach?

Kudos [?]: 10 [0], given: 10

Math Expert
Joined: 02 Sep 2009
Posts: 41893

Kudos [?]: 128826 [0], given: 12183

Re: In the expression above, if xn#0, what is the value of S? [#permalink]

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21 Dec 2013, 05:39
steilbergauf wrote:
Hi guys,

quick question:

I started to simplify the equation by multiplying the denominator such as 2/n (x+3x/2) and did not get the same solution. What is wrong with my approach?

_________________

Kudos [?]: 128826 [0], given: 12183

Intern
Joined: 28 Jun 2011
Posts: 13

Kudos [?]: 10 [0], given: 10

Re: In the expression above, if xn#0, what is the value of S? [#permalink]

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21 Dec 2013, 07:54
Hi Bunuel,

I used the following approach:

$$S=\frac{\frac{2}{n}}{\frac{1}{x}+\frac{2}{3x}}$$

$$S= \frac{2}{n}(x+\frac{3x}{2})$$

$$S= \frac{2x}{n}+\frac{6x}{2n}$$

$$S= \frac{4x}{2n}+\frac{6x}{2n}=\frac{5x}{n}$$

Can you please tell me what is wrong with my approach?

Kudos [?]: 10 [0], given: 10

Math Expert
Joined: 02 Sep 2009
Posts: 41893

Kudos [?]: 128826 [0], given: 12183

Re: In the expression above, if xn#0, what is the value of S? [#permalink]

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22 Dec 2013, 06:30
steilbergauf wrote:
Hi Bunuel,

I used the following approach:

$$S=\frac{\frac{2}{n}}{\frac{1}{x}+\frac{2}{3x}}$$

$$S= \frac{2}{n}(x+\frac{3x}{2})$$

$$S= \frac{2x}{n}+\frac{6x}{2n}$$

$$S= \frac{4x}{2n}+\frac{6x}{2n}=\frac{5x}{n}$$

Can you please tell me what is wrong with my approach?

The point is that reciprocal of the denominator is not $$x+\frac{3x}{2}$$.

The denominator is $$\frac{1}{x}+\frac{2}{3x}=\frac{3+2}{3x}=\frac{5}{3x}$$, thus its reciprocal is $$\frac{3x}{5}$$.
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Kudos [?]: 128826 [0], given: 12183

Re: In the expression above, if xn#0, what is the value of S?   [#permalink] 22 Dec 2013, 06:30
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