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M17-35

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M17-35  [#permalink]

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New post 16 Sep 2014, 01:02
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A
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Difficulty:

  45% (medium)

Question Stats:

68% (01:33) correct 32% (02:01) wrong based on 247 sessions

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The ratio of \(A\) to \(B\) is \(\frac{1}{2}\). If \(A\) is increased by \(x\%\) and \(B\) is decreased by \(\frac{x}{2}\%\), which of the following represents the new value of the ratio of \(A\) to \(B\)?

A. \(\frac{100 + x}{200 - x}\)
B. \(\frac{1 + \frac{x}{100}}{1 - \frac{x}{200}}\)
C. \(\frac{x + 1}{2x + 2}\)
D. \(\frac{2x + 1}{200 - x}\)
E. \(\frac{200 + x}{1 - x}\)

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Re M17-35  [#permalink]

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New post 16 Sep 2014, 01:02
Official Solution:

The ratio of \(A\) to \(B\) is \(\frac{1}{2}\). If \(A\) is increased by \(x\%\) and \(B\) is decreased by \(\frac{x}{2}\%\), which of the following represents the new value of the ratio of \(A\) to \(B\)?

A. \(\frac{100 + x}{200 - x}\)
B. \(\frac{1 + \frac{x}{100}}{1 - \frac{x}{200}}\)
C. \(\frac{x + 1}{2x + 2}\)
D. \(\frac{2x + 1}{200 - x}\)
E. \(\frac{200 + x}{1 - x}\)

The new ratio = \(\frac{1 + \frac{x}{100}}{2(1 - \frac{x}{200})} = \frac{1 + \frac{x}{100}}{2 - \frac{x}{100}} = \frac{100 + x}{200 - x}\).

Answer: A
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New post 09 Feb 2016, 18:32
Alternate approach :

Let A and B be 100 and 200. Ratio 1:2. Now think, x% of 100 is actually x/2% of 200. 10% of 100 ( which is 10 in this case) is 5% of 200. So, basically we are adding and subtracting the same amount from numerator and denominator. Option A does exactly that (100+x)/(200-x).

Thus, answer A.
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Re M17-35  [#permalink]

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New post 19 Jun 2018, 13:17
I think this is a high-quality question and I agree with explanation.
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New post 23 Oct 2018, 00:10
Let A be Z and B be 2Z. Since Z is appearing in numerator and denominator after increasing the numberator by x% and decreasing by \(\frac{x}{2}\)% therefore z will cancel out.

The remaining workings remain same as aforesaid!
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M17-35   [#permalink] 23 Oct 2018, 00:10
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