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

m (simplify)
(further explanation)
 
Line 40: Line 40:
 
for Dick we have
 
for Dick we have
 
<cmath>T=\frac{d_1-d_2}{5}+\frac{100-(d_1-d_2)}{25}</cmath>
 
<cmath>T=\frac{d_1-d_2}{5}+\frac{100-(d_1-d_2)}{25}</cmath>
−
<cmath>T=\frac{100+4(d_2-d_1)}{25}</cmath>
+
<cmath>T=\frac{100+4(d_1-d_2)}{25}</cmath>
  
 
our unknowns then are <math>T,d_1,d_2</math>
 
our unknowns then are <math>T,d_1,d_2</math>
 
We first get rid of <math>d_1</math> by adding the equations for Harry and Dick and divide by 2 to get  
 
We first get rid of <math>d_1</math> by adding the equations for Harry and Dick and divide by 2 to get  
 +
<cmath>2T=\frac{500-4d_1+100+4d_1-4d_2}{25}</cmath>
 
<cmath>T=\frac{300-2d_2}{25}</cmath>
 
<cmath>T=\frac{300-2d_2}{25}</cmath>
−
now we can add this to Tom's time and divide by 2 to get rid of <math>d_2</math> and simultaneously solve for <math>T</math>
+
now we can add above to Tom and divide by 2 to get rid of <math>d_2</math> and simultaneously solving for <math>T</math>
−
 
+
<cmath>2T=\frac{300-2d_2+100+2d_2}{25}</cmath>
 
<cmath>T=\frac{200}{25}=8</cmath>
 
<cmath>T=\frac{200}{25}=8</cmath>
−
 
 
which is <math>\boxed{\textbf{(D)}}</math>.
 
which is <math>\boxed{\textbf{(D)}}</math>.
  

Latest revision as of 20:11, 29 September 2025

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 \[T=\frac{d_1-d_2}{5}+\frac{100-(d_1-d_2)}{25}\] \[T=\frac{100+4(d_1-d_2)}{25}\]

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 \[2T=\frac{500-4d_1+100+4d_1-4d_2}{25}\] \[T=\frac{300-2d_2}{25}\] now we can add above to Tom and divide by 2 to get rid of $d_2$ and simultaneously solving for $T$ \[2T=\frac{300-2d_2+100+2d_2}{25}\] \[T=\frac{200}{25}=8\] which is $\boxed{\textbf{(D)}}$.

See Also

1951 AHSC (Problems • Answer Key • Resources)
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