Difference between revisions of "2008 AMC 8 Problems/Problem 12"
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==Solution== | ==Solution== | ||
| − | Each bounce is 2/3 times the height of the previous bounce | + | Each bounce is <math>2/3</math> times the height of the previous bounce. The first bounce reaches <math>2</math> meters, the second <math>4/3</math>, the third <math>8/9</math>, the fourth <math>16/27</math>, and the fifth <math>32/81</math>. Half of <math>81</math> is <math>40.5</math>, so the ball does not reach the required height on bounce <math>\boxed{\textbf{(C)}\ 5}</math>. |
==See Also== | ==See Also== | ||
{{AMC8 box|year=2008|num-b=11|num-a=13}} | {{AMC8 box|year=2008|num-b=11|num-a=13}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 18:22, 8 August 2021
Problem
A ball is dropped from a height of 
meters. On its first bounce it rises to a height of 
meters. It keeps falling and bouncing to 
of the height it reached in the previous bounce. On which bounce will it not rise to a height of 
meters?
Solution
Each bounce is 
times the height of the previous bounce. The first bounce reaches meters, the second 
, the third 
, the fourth 
, and the fifth 
. Half of 
is 
, so the ball does not reach the required height on bounce 
.
See Also
| 2008 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 11 |
Followed by Problem 13 | |
| 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. 
