1991 AHSME Problems/Problem 11: Difference between revisions
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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 16:32, 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 | |
| 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 • 26 • 27 • 28 • 29 • 30 | ||
| All AHSME Problems and Solutions | ||
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