Difference between revisions of "2011 AMC 12A Problems/Problem 12"
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size(8cm,8cm); | size(8cm,8cm); | ||
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− | A=(-3, | + | A=(-3,4); |
− | B=(3, | + | B=(3,4); |
draw(A--B); | draw(A--B); | ||
label("$A$",A,S); | label("$A$",A,S); | ||
label("$B$",B,S); | label("$B$",B,S); | ||
pair A, B; | pair A, B; | ||
− | arrow((-2.5, | + | arrow((-2.5,6),dir(180),blue); |
− | arrow((-2.5, | + | arrow((-2.5,6.2),dir(180),red); |
A=(-3,5); | A=(-3,5); | ||
B=(3,5); | B=(3,5); |
Revision as of 20:30, 14 February 2016
Problem
A power boat and a raft both left dock on a river and headed downstream. The raft drifted at the speed of the river current. The power boat maintained a constant speed with respect to the river. The power boat reached dock downriver, then immediately turned and traveled back upriver. It eventually met the raft on the river 9 hours after leaving dock How many hours did it take the power boat to go from to ?
Solution
Solution 1
Since the speed of the river is not specified, the outcome of the problem must be independent of this speed. We may thus trivially assume that the river has a speed of . In this case, when the powerboat travels from to , the raft remains at . Thus the trip from to takes the same time as the trip from to the raft. Since these times are equal and sum to hours, the trip from to must take half this time, or hours. The answer is thus .
Solution 2
What's important in this problem is to consider everything in terms of the power boat and the raft, since that is how the problem is given to us. Let us denote the speed of the boat as and the speed of the raft at .
See also
2011 AMC 12A (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 AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.