2001 AMC 8 Problem 17

Attempt Problem 17 of the 2001 AMC 8 below, then check your answer against the professionally curated solution 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.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

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=1000\text{K}=1000). #123$100200300 \begin{array}{ccccc} \boldsymbol{\#} & 1 & 2 & 3 \\ \boldsymbol{\$} & 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
Answer: B
Concepts:percentage
Small hint:
Most listed increases are doublings
Big hint:
Compare the few increases that are not exactly 100%100\%
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 300−200200=50%\dfrac{300-200}{200}=50\%, 500−300300=6623%\dfrac{500-300}{300}=66\dfrac23\%, and 125000−6400064000\dfrac{125000-64000}{64000}, which is about 95%95\%.
The smallest is from question 22 to question 33.
Thus, B is the correct answer.
LIVE by Po-Shen Loh
Programs

LIVE Classes

Self-Paced Classes

Stars

Scholarship

About Us

Who Is Po-Shen Loh?

Email Contact

Newsletter

Jobs

Become a Star

For a more Thought Full world.

© 2026 Expii, Inc.