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> | ||
− | We have that <math>100+2d_2=100+4d_1-4d_2</math>, so <math>4d_1=6d_2\Rightarrow d_1=\frac{ | + | 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]] | ||
+ | [[Category:Rate Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 15:55, 6 April 2020
Problem
Tom, Dick and Harry started out on a -mile journey. Tom and Harry went by automobile at the rate of mph, while Dick walked at the rate of mph. After a certain distance, Harry got off and walked on at 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:
Solution
Let be the distance (in miles) that Harry traveled on car, and let be the distance (in miles) that Tom backtracked to get Dick. Let 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 , so . This then shows that , which in turn gives that . Now we only need to solve for :
The journey took 8 hours, so the correct answer is .
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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