Difference between revisions of "2025 AMC 8 Problems/Problem 19"
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==Solution 1== | ==Solution 1== | ||
| − | The first car, moving from town <math>A</math> at <math>25</math> miles per hour, takes <math>\frac{5}{25} = \frac{1}{5} \text{hours} = 12</math> minutes. The second car, traveling another <math>5</math> miles from town <math>B</math>, takes <math>\frac{5}{20} = \frac{1}{4} \text{hours} = 15</math> minutes. The first car has traveled for 3 minutes or <math>\frac{1}{20}</math>th of an hour at <math>40</math> miles per hour when the second car has traveled 5 miles. The first car has traveled <math>40 \cdot \frac{1}{20} = 2</math> miles from the previous <math>5</math> miles it traveled at <math>25</math> miles per hour. They have <math>3</math> miles left, and they travel at the same speed, so they meet <math>1.5</math> miles through, so they are <math>5 + 2 + 1.5 = \boxed{\textbf{(D)}8.5}</math> miles from town <math>A</math>. | + | The first car, moving from town <math>A</math> at <math>25</math> miles per hour, takes <math>\frac{5}{25} = \frac{1}{5} \text{hours} = 12</math> minutes. The second car, traveling another <math>5</math> miles from town <math>B</math>, takes <math>\frac{5}{20} = \frac{1}{4} \text{hours} = 15</math> minutes. The first car has traveled for 3 minutes or <math>\frac{1}{20}</math>th of an hour at <math>40</math> miles per hour when the second car has traveled 5 miles. The first car has traveled <math>40 \cdot \frac{1}{20} = 2</math> miles from the previous <math>5</math> miles it traveled at <math>25</math> miles per hour. They have <math>3</math> miles left, and they travel at the same speed, so they meet <math>1.5</math> miles through, so they are <math>5 + 2 + 1.5 = \boxed{\textbf{(D) }8.5}</math> miles from town <math>A</math>. |
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~alwaysgonnagiveyouup | ~alwaysgonnagiveyouup | ||
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| + | ~ Edited by [[User:Aoum|Aoum]] | ||
==Solution 2== | ==Solution 2== | ||
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-Benedict T (countmath1) | -Benedict T (countmath1) | ||
| − | = Video Solution by Pi Academy = | + | ==Video Solution by Pi Academy== |
https://youtu.be/Iv_a3Rz725w?si=E0SI_h1XT8msWgkK | https://youtu.be/Iv_a3Rz725w?si=E0SI_h1XT8msWgkK | ||
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==Video Solution 1 by SpreadTheMathLove== | ==Video Solution 1 by SpreadTheMathLove== | ||
Latest revision as of 20:03, 21 February 2025
Contents
Problem
Two towns, 
and 
, are connected by a straight road, 
miles long. Traveling from town to town , the speed limit changes every 
miles: from 
to 
to 
miles per hour (mph). Two cars, one at town and one at town , start moving toward each other at the same time. They drive at exactly the speed limit in each portion of the road. How far from town , in miles, will the two cars meet?
Solution 1
The first car, moving from town at miles per hour, takes 
minutes. The second car, traveling another miles from town , takes 
minutes. The first car has traveled for 3 minutes or 
th of an hour at miles per hour when the second car has traveled 5 miles. The first car has traveled 
miles from the previous miles it traveled at miles per hour. They have 
miles left, and they travel at the same speed, so they meet 
miles through, so they are 
miles from town .
~alwaysgonnagiveyouup
~ Edited by Aoum
Solution 2
From the answer choices, the cars will meet somewhere along the mph stretch. Car travels mph for miles, so we can use dimensional analysis to see that it will be 
of an hour for this portion. Similarly, car spends 
of an hour on the mph portion.
Suppose that car travels 
miles along the mph portion-- then car travels 
miles along the mph portion. By identical methods, car travels for 
hours, and car travels for 
hours.
At their meeting point, cars and will have traveled for the same amount of time, so we have
so 
, and 
miles. This means that car will have traveled 
miles.
-Benedict T (countmath1)
Video Solution by Pi Academy
https://youtu.be/Iv_a3Rz725w?si=E0SI_h1XT8msWgkK
Video Solution 1 by SpreadTheMathLove
https://www.youtube.com/watch?v=jTTcscvcQmI
Video Solution (A Clever Explanation You’ll Get Instantly)
https://youtu.be/VP7g-s8akMY?si=Y7swThPvf2WCCGxM&t=2394 ~hsnacademy
Video Solution by Thinking Feet
See Also
| 2025 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 18 |
Followed by Problem 20 | |
| 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 | ||
| All AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
