Difference between revisions of "1998 AHSME Problems/Problem 21"

Latest revision as of 13:29, 5 July 2013

Problem

In an $h$ -meter race, Sunny is exactly $d$ meters ahead of Windy when Sunny finishes the race. The next time they race, Sunny sportingly starts $d$ meters behind Windy, who is at the starting line. Both runners run at the same constant speed as they did in the first race. How many meters ahead is Sunny when Sunny finishes the second race?

$\mathrm{(A) \ } \frac dh \qquad \mathrm{(B) \ } 0 \qquad \mathrm{(C) \ } \frac {d^2}h \qquad \mathrm{(D) \ } \frac {h^2}d \qquad \mathrm{(E) \ } \frac{d^2}{h-d}$

Solution

Let $s$ and $w$ be the speeds of Sunny and Windy. From the first race we know that $\frac sw = \frac h{h-d}$ . In the second race, Sunny's track length is $h+d$ . She will finish this track in $\frac{h+d}s$ . In this time, Windy will run the distance $w\cdot \frac{h+d}s = \frac{(h+d)(h-d)}h$ . This is less than $h$ , therefore Sunny is ahead. The exact distance between Windy and the finish is $h - \frac{(h+d)(h-d)}h = \frac{ h^2 - (h^2-d^2) }h = \boxed{\frac{d^2}h}$ .

1998 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
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All AHSME Problems and Solutions

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

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 == See also ==
 {{AHSME box|year=1998|num-b=20|num-a=22}}
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Difference between revisions of "1998 AHSME Problems/Problem 21"

Latest revision as of 13:29, 5 July 2013

Problem

Solution

See also