Difference between revisions of "1951 AHSME Problems/Problem 50"

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[[Category:Intermediate Algebra Problems]]
[[Category:Intermediate Algebra Problems]]
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Revision as of 12:27, 5 July 2013


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:

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


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




We combine these three equations:


After multiplying everything by 25 and simplifying, we get


We have that $100+2d_2=100+4d_1-4d_2$, so $4d_1=6d_2\Rightarrow d_1=\frac{2}{3}d_2$. 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 $T$:


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

See Also

1951 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 49
Followed by
Last Question
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