Difference between revisions of "1991 AHSME Problems/Problem 11"
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== Problem == | == Problem == | ||
− | Jack and Jill run 10 km. They start at the same point, run 5 km up a hill, and | + | Jack and Jill run 10 km. They start at the same point, run 5 km up a hill, and return to the starting point by the same route. Jack has a 10 minute head start and runs at the rate of 15 km/hr uphill and 20 km/hr downhill. Jill runs 16 km/hr uphill and 22 km/hr downhill. How far from the top of the hill are they when they pass each other going in opposite directions (in km)? |
<math>\text{(A) } \frac{5}{4}\quad | <math>\text{(A) } \frac{5}{4}\quad | ||
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== Solution == | == Solution == | ||
− | <math>\fbox{B}</math> | + | <math>\fbox{B}</math> Consider the distance-time graph and use coordinate geometry, with time on the <math>x</math>-axis and distance on the <math>y</math>-axis. Thus Jack starts at <math>(0,0)</math> and his initial motion, taking time in hours, is <math>y=15x</math>. This ends when <math>y=5</math>, giving the point <math>(\frac{1}{3}, 5)</math>. He now runs in the opposite direction at <math>20</math> km/hr, so the equation is <math>y-5 = -20(x-\frac{1}{3})</math>. Now Jill starts at <math>(\frac{1}{6}, 0)</math> (as Jack has a head start of 10 minutes = <math>\frac{1}{6}</math> hours), so her equation is <math>y=16(x-\frac{1}{6}).</math> Solving simultaneously with Jack's equation gives <math>-20x+\frac{35}{3} = 16x - \frac{8}{3} \implies \frac{43}{3} = 36x \implies x = \frac{43}{108}</math>, so <math>y = 16(\frac{25}{108}) = \frac{400}{108}</math>, and thus the required distance is <math>5 - \frac{400}{108} = \frac{140}{108} = \frac{70}{54} = \frac{35}{27}.</math> |
== See also == | == See also == |
Latest revision as of 17:32, 23 February 2018
Problem
Jack and Jill run 10 km. They start at the same point, run 5 km up a hill, and return to the starting point by the same route. Jack has a 10 minute head start and runs at the rate of 15 km/hr uphill and 20 km/hr downhill. Jill runs 16 km/hr uphill and 22 km/hr downhill. How far from the top of the hill are they when they pass each other going in opposite directions (in km)?
Solution
Consider the distance-time graph and use coordinate geometry, with time on the -axis and distance on the -axis. Thus Jack starts at and his initial motion, taking time in hours, is . This ends when , giving the point . He now runs in the opposite direction at km/hr, so the equation is . Now Jill starts at (as Jack has a head start of 10 minutes = hours), so her equation is Solving simultaneously with Jack's equation gives , so , and thus the required distance is
See also
1991 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
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All AHSME Problems and Solutions |
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