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# M17-35

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Math Expert
Joined: 02 Sep 2009
Posts: 58396

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16 Sep 2014, 01:02
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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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Math Expert
Joined: 02 Sep 2009
Posts: 58396

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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 = $$\frac{1 + \frac{x}{100}}{2(1 - \frac{x}{200})} = \frac{1 + \frac{x}{100}}{2 - \frac{x}{100}} = \frac{100 + x}{200 - x}$$.

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Joined: 19 Dec 2015
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Location: United States
GMAT 1: 720 Q50 V38
GPA: 3.8
WE: Information Technology (Computer Software)

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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).

Manager
Joined: 26 Feb 2018
Posts: 51
Location: India
GMAT 1: 640 Q45 V34
GPA: 3.9
WE: Web Development (Computer Software)

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19 Jun 2018, 13:17
I think this is a high-quality question and I agree with explanation.
Director
Joined: 12 Feb 2015
Posts: 917

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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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# M17-35

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