Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1991 AHSME Problems/Problem 11"

m (→‎Problem)
(Added a solution with explanation)
 
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
  
−
Jack and Jill run 10 km. They start at the same point, run 5 km up a hill, and reurn 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)?
+
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
Line 10: Line 10:
  
 
== 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)?

$\text{(A) } \frac{5}{4}\quad \text{(B) } \frac{35}{27}\quad \text{(C) } \frac{27}{20}\quad \text{(D) } \frac{7}{3}\quad \text{(E) } \frac{28}{49}$

Solution

$\fbox{B}$ Consider the distance-time graph and use coordinate geometry, with time on the $x$-axis and distance on the $y$-axis. Thus Jack starts at $(0,0)$ and his initial motion, taking time in hours, is $y=15x$. This ends when $y=5$, giving the point $(\frac{1}{3}, 5)$. He now runs in the opposite direction at $20$ km/hr, so the equation is $y-5 = -20(x-\frac{1}{3})$. Now Jill starts at $(\frac{1}{6}, 0)$ (as Jack has a head start of 10 minutes = $\frac{1}{6}$ hours), so her equation is $y=16(x-\frac{1}{6}).$ Solving simultaneously with Jack's equation gives $-20x+\frac{35}{3} = 16x - \frac{8}{3} \implies \frac{43}{3} = 36x \implies x = \frac{43}{108}$, so $y = 16(\frac{25}{108}) = \frac{400}{108}$, and thus the required distance is $5 - \frac{400}{108} = \frac{140}{108} = \frac{70}{54} = \frac{35}{27}.$

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1991_AHSME_Problems/Problem_11&oldid=92278"