In 2012, attendance at an annual sporting event was 5% greater than it was in 2011 and 20% greater than it was in 2010. What was the percent increase in attendance from 2010 to 2011?

Give your answer to the nearest 0.001%.

Go to it! I'm no longer sick, so my answer will definitely appear in the comments sometime during the next 36 hours. As you can tell, this is a "grid-in" problem, i.e., you have to write the correct answer, not select an answer from multiple choices.

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## 4 comments:

14.286?

That's what I got. I did a plug-and-chug:

2010 = x

2011 = 100

2012 = 105

Therefore,

105 = 1.2x

x = 87.5

100/87.5 = 1.142857... (or 8/7)

((8/7) - 1)•100 = approx. 14.286

So: 14.286% increase.

I remember seeing this suggestion in one of MGRE's earlier explanations: it's sometimes easier to solve certain problems with plug-and-chug than through, say, compound ratios or whatever. Use simple, easily manipulable numbers, like 100 (I could've used 10, come to think of it), to figure things out quickly.

Ah, is that what it's called? That's basically what I did, too. Except I used 1,000, because an annual sporting event with only a hundred people attending sounded too sad.

Well, "plug-and-chug" isn't any sort of official terminology; it's just a slangy way of saying that you're using numbers instead of variables. We use the term all the time where I work.

True: 100 is a rather sad attendance figure.

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