1995 AJHSME Problems/Problem 11
Problem
Jane can walk any distance in half the time it takes Hector to walk the same distance. They set off in opposite directions around the outside of the 18-block area as shown. When they meet for the first time, they will be closest to
Solution
Counting around, when Jane walks steps, she will be at . When Hector walks steps, he will also be at . Since Jane has walked twice as many steps as Hector, they will reach this spot at the same time. Thus, the answer is .
See Also
1995 AJHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
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All AJHSME/AMC 8 Problems and Solutions |
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