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

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

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New post 18 Dec 2012, 07:13
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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
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Re: In the expression above, if xn#0, what is the value of S?  [#permalink]

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New post 18 Dec 2012, 07:17
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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?

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

(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.

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

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New post 20 Dec 2013, 09:32
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?

Thanks a lot in advance!
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Re: In the expression above, if xn#0, what is the value of S?  [#permalink]

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New post 21 Dec 2013, 04:39
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?

Thanks a lot in advance!


Your approach is not clear at all. Please show your work. Thank you.
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Re: In the expression above, if xn#0, what is the value of S?  [#permalink]

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New post 21 Dec 2013, 06:54
Hi Bunuel,

sorry for being imprecise and thanks for your quick answer.

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?

Thanks a lot in advance.
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Re: In the expression above, if xn#0, what is the value of S?  [#permalink]

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New post 22 Dec 2013, 05:30
steilbergauf wrote:
Hi Bunuel,

sorry for being imprecise and thanks for your quick answer.

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?

Thanks a lot in advance.


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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PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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

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