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 is 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> |
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+ | ==Video Solution== | ||
+ | https://www.youtube.com/watch?v=S8hyVeObqUs Soo, DRMS, NM | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2001|num-b=16|num-a=18}} | {{AMC8 box|year=2001|num-b=16|num-a=18}} | ||
+ | {{MAA Notice}} |
Latest revision as of 18:16, 10 January 2024
Contents
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 is multiplied by a factor of
. Therefore, the smallest percent increase is
Video Solution
https://www.youtube.com/watch?v=S8hyVeObqUs Soo, DRMS, NM
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.