Difference between revisions of "2001 AMC 8 Problems/Problem 17"
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==Solution== | ==Solution== | ||
− | Notice that in | + | Notice that in two of the increases, the dollar amount doubles. The increases in which this is not true is <math> 2 </math> to <math> 3 </math>, <math> 3 </math> to <math> 4 </math>, and <math> 11 </math> to <math> 12 </math>. We can disregard <math> 11 </math> to <math> 12 </math> since that increase is almost <math> 2 </math> times. The increase from <math> 2 </math> to <math> 3 </math> is <math> \frac{300-200}{200}=\frac{1}{2} </math>, so it's only multiplied by a factor of <math> 1.5 </math>. The increase from <math> 3 </math> to <math> 4 </math> is <math> \frac{500-300}{300}=\frac{2}{3} </math>, so it's only multiplied by a factor of <math> 1.\bar{6} </math>. Therefore, the smallest percent increase is <math> \text{From 2 to 3}, \boxed{\text{B}} </math> |
==See Also== | ==See Also== | ||
{{AMC8 box|year=2001|num-b=16|num-a=18}} | {{AMC8 box|year=2001|num-b=16|num-a=18}} |
Revision as of 14:57, 23 December 2012
Problem
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).
Between which two questions is the percent increase of the value the smallest?
Solution
Notice that in two of the increases, the dollar amount doubles. The increases in which this is not true is to , to , and to . We can disregard to since that increase is almost times. The increase from to is , so it's only multiplied by a factor of . The increase from to is , so it's only multiplied by a factor of . Therefore, the smallest percent increase is
See Also
2001 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 16 |
Followed by Problem 18 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AJHSME/AMC 8 Problems and Solutions |