1991 AHSME Problems/Problem 11

Revision as of 16:32, 23 February 2018 by Hapaxoromenon (talk | contribs) (Added a solution with explanation)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

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 (ProblemsAnswer KeyResources)
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

Invalid username
Login to AoPS