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

1951 AHSME Problems/Problem 50

Revision as of 19:57, 29 September 2025 by Juniorcole (talk | contribs) (simplify)

Problem

Tom, Dick and Harry started out on a $100$-mile journey. Tom and Harry went by automobile at the rate of $25$ mph, while Dick walked at the rate of $5$ mph. After a certain distance, Harry got off and walked on at $5$ mph, while Tom went back for Dick and got him to the destination at the same time that Harry arrived. The number of hours required for the trip was:

$\textbf{(A)}\ 5\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 7\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ \text{none of these answers}$

Solution 1

Let $d_1$ be the distance (in miles) that Harry traveled on car, and let $d_2$ be the distance (in miles) that Tom backtracked to get Dick. Let $T$ be the time (in hours) that it took the three to complete the journey. We now examine Harry's journey, Tom's journey, and Dick's journey. These yield, respectively, the equations

\[\frac{d_1}{25}+\frac{100-d_1}{5}=T,\]

\[\frac{d_1}{25}+\frac{d_2}{25}+\frac{100-(d_1-d_2)}{25}=T,\]

\[\frac{d_1-d_2}{5}+\frac{100-(d_1-d_2)}{25}=T.\]

We combine these three equations:

\[\frac{d_1}{25}+\frac{100-d_1}{5}=\frac{d_1}{25}+\frac{d_2}{25}+\frac{100-(d_1-d_2)}{25}=\frac{d_1-d_2}{5}+\frac{100-(d_1-d_2)}{25}\]

After multiplying everything by 25 and simplifying, we get

\[500-4d_1=100+2d_2=100+4d_1-4d_2\]

We have that $100+2d_2=100+4d_1-4d_2$, so $4d_1=6d_2\Rightarrow d_1=\frac{3}{2}d_2$. This then shows that $500-4d_1=100+\frac{4}{3}d_1\Rightarrow 400=\frac{16}{3}d_1\Rightarrow d_1=75$, which in turn gives that $d_2=50$. Now we only need to solve for $T$:

\[T=\frac{d_1}{25}+\frac{100-d_1}{5}=\frac{75}{25}+\frac{100-75}{5}=3+5=8\]

The journey took 8 hours, so the correct answer is $\boxed{\textbf{(D)}}$.

Solution 2

As before let $d_1$ be the distance that Harry traveled on car, and let $d_2$ be the distance in miles that Tom backtracked to get Dick and let $T$ be the time (in hours) to complete the journey.

For Harry we have \[T=\frac{d_1}{25} + \frac{100-d_1}{5}\] \[T=\frac{500-4d_1}{25}\]

for Tom we have \[T=\frac{d_1}{25}+\frac{d_2}{25}+ \frac{100-d_1+d_2}{25}\] \[T=\frac{100+2d_2}{25}\] for Dick we have \[T=\frac{d_1-d_2}{5}+\frac{100-(d_1-d_2)}{25}\] \[T=\frac{100+4(d_2-d_1)}{25}\]

our unknowns then are $T,d_1,d_2$ We first get rid of $d_1$ by adding the equations for Harry and Dick and divide by 2 to get \[T=\frac{300-2d_2}{25}\] now we can add this to Tom's time and divide by 2 to get rid of $d_2$ and simultaneously solve for $T$

\[T=\frac{200}{25}=8\]

which is $\boxed{\textbf{(D)}}$.

See Also

1951 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 49 		Followed by
Last Question
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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50
All AHSME Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1951_AHSME_Problems/Problem_50&oldid=261831"