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If x/3 - x/6 + x/9 - x/12 = 1 - 1/2

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If x/3 - x/6 + x/9 - x/12 = 1 - 1/2 [#permalink] New post 08 Jan 2019, 10:29
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If \(\frac{x}{3} - \frac{x}{6} + \frac{x}{9} - \frac{x}{12} = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4}\) , then \(x =\)

A. \(3\)

B. \(1\)

C. \(\frac{1}{3}\)

D. \(-\frac{1}{3}\)

E. \(-3\)
[Reveal] Spoiler: OA

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Re: If x/3 - x/6 + x/9 - x/12 = 1 - 1/2 [#permalink] New post 09 Jan 2019, 04:35
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Carcass wrote:
If \(\frac{x}{3} - \frac{x}{6} + \frac{x}{9} - \frac{x}{12} = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4}\) , then \(x =\)

A. \(3\)

B. \(1\)

C. \(\frac{1}{3}\)

D. \(-\frac{1}{3}\)

E. \(-3\)


Explanation::

\(\frac{x}{3} - \frac{x}{6} + \frac{x}{9} - \frac{x}{12} = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4}\)

\(\frac{7x}{36} = \frac{7}{12}\)

or x = 3
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Re: If x/3 - x/6 + x/9 - x/12 = 1 - 1/2 [#permalink] New post 09 Jan 2019, 06:29
Expert's post
Carcass wrote:
If \(\frac{x}{3} - \frac{x}{6} + \frac{x}{9} - \frac{x}{12} = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4}\) , then \(x =\)

A. \(3\)

B. \(1\)

C. \(\frac{1}{3}\)

D. \(-\frac{1}{3}\)

E. \(-3\)


\(\frac{x}{3} - \frac{x}{6} + \frac{x}{9} - \frac{x}{12} = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4}\) .....
We can see the formation of the terms is similar on both sides, so let us get the left hand side in terms similar to RHS, by taking out \(\frac{x}{3}\)..
\(\frac{x}{3} - \frac{x}{6} + \frac{x}{9} - \frac{x}{12} =\frac{x}{3}*1 - \frac{x}{3}*\frac{1}{2} + \frac{x}{3}*\frac{1}{3} - \frac{x}{3}*\frac{1}{4} = \frac{x}{3}(1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4})\)
This should be equal to RHS..
\(\frac{x}{3}(1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4}) = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4}........\frac{x}{3}=1....x=3\) .....

A
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Re: If x/3 - x/6 + x/9 - x/12 = 1 - 1/2   [#permalink] 09 Jan 2019, 06:29
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