2001 AMC 8 Problem 17

Below is the professionally curated solution for Problem 17 of the 2001 AMC 8, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2001 AMC 8 solutions, or check the answer key.

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Concepts:percentage

Difficulty rating: 1380

17.

For the game show Who Wants To Be a Millionaire?, the dollar values of each question are shown in the following table (where K\text{K} = 1000). Question123Value100200300 \begin{array}{ccccc} \text{Question} & 1 & 2 & 3 \\ \text{Value} & 100 & 200 & 300 \end{array} 456785001K2K4K8K \begin{array}{cccccc} 4 & 5 & 6 & 7 & 8 \\ 500 & 1\text{K} & 2\text{K} & 4\text{K} & 8\text{K} \end{array} 910111216K32K64K125K \begin{array}{ccccc} 9 & 10 & 11 & 12 \\ 16\text{K} & 32\text{K} & 64\text{K} & 125\text{K} \end{array} 131415250K500K1000K \begin{array}{ccc} 13 & 14 & 15 \\ 250\text{K} & 500\text{K} & 1000\text{K} \end{array} Between which two questions is the percent increase of the value the smallest?

From 11 to 22

From 22 to 33

From 33 to 44

From 1111 to 1212

From 1414 to 1515

Solution:

Most of the listed increases are doublings, which are 100%100\% increases.

The exceptions among the answer choices are from 22 to 33, from 33 to 44, and from 1111 to 1212.

These percent increases are 300200200=50%\dfrac{300-200}{200}=50\%, 500300300=6623%\dfrac{500-300}{300}=66\dfrac23\%, and 1250006400064000\dfrac{125000-64000}{64000}, which is about 95%95\%.

The smallest is from question 22 to question 33.

Thus, B is the correct answer.

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