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

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<cmath>500-4d_1=100+2d_2=100+4d_1-4d_2</cmath>
 
<cmath>500-4d_1=100+2d_2=100+4d_1-4d_2</cmath>
  
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We have that <math>100+2d_2=100+4d_1-4d_2</math>, so <math>4d_1=6d_2\Rightarrow d_1=\frac{2}{3}d_2</math>. This then shows that <math>500-4d_1=100+\frac{4}{3}d_1\Rightarrow 400=\frac{16}{3}d_1\Rightarrow d_1=75</math>, which in turn gives that <math>d_2=50</math>. Now we only need to solve for <math>T</math>:
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We have that <math>100+2d_2=100+4d_1-4d_2</math>, so <math>4d_1=6d_2\Rightarrow d_1=\frac{3}{2}d_2</math>. This then shows that <math>500-4d_1=100+\frac{4}{3}d_1\Rightarrow 400=\frac{16}{3}d_1\Rightarrow d_1=75</math>, which in turn gives that <math>d_2=50</math>. Now we only need to solve for <math>T</math>:
  
 
<cmath>T=\frac{d_1}{25}+\frac{100-d_1}{5}=\frac{75}{25}+\frac{100-75}{5}=3+5=8</cmath>
 
<cmath>T=\frac{d_1}{25}+\frac{100-d_1}{5}=\frac{75}{25}+\frac{100-75}{5}=3+5=8</cmath>
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[[Category:Intermediate Algebra Problems]]
 
[[Category:Intermediate Algebra Problems]]
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[[Category:Rate Problems]]
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Latest revision as of 14:55, 6 April 2020

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

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

See Also

1951 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 49 		Followed by
Last Question
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All AHSME Problems and Solutions

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