Difference between revisions of "2025 AMC 8 Problems/Problem 19"
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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>. | ||
~alwaysgonnagiveyouup | ~alwaysgonnagiveyouup | ||
| + | |||
| + | ==Solution 2== | ||
| + | From the answer choices, the cars will meet somewhere along the <math>40</math> mph stretch. Car <math>A</math> travels <math>25</math>mph for <math>5</math> miles, so we can use dimensional analysis to see that it will be <math>\frac{1\ \text{hr}}{25\ \text{mi}}\cdot 5\ \text{mi} = \frac{1}{5}</math> of an hour for this portion. Similarly, car <math>B</math> spends <math>\frac{1}{4}</math> of an hour on the <math>20</math> mph portion. \\ | ||
| + | |||
| + | Suppose that car <math>A</math> travels <math>x</math> miles along the <math>40</math> mph portion-- then car <math>B</math> travels <math>5-x</math> miles along the <math>40</math> mph portion. By identical methods, car <math>A</math> travels for <math>\frac{1}{40}\cdot x = \frac{x}{40}</math> hours, and car <math>B</math> travels for <math>\frac{5-x}{40}</math> hours. \\ | ||
| + | |||
| + | At their meeting point, cars <math>A</math> and <math>B</math> will have traveled for the same amount of time, so we have | ||
| + | \begin{align*} | ||
| + | \frac{1}{5} + \frac{x}{40} &= \frac{1}{4} + \frac{5-x}{40}\\ | ||
| + | 8 + x &= 10 + 5-x, | ||
| + | \end{align*} | ||
| + | so <math>2x = 7</math>, and <math>x = 3.5</math> miles. This means that car <math>A</math> will have traveled <math>5 + 3.5= \boxed{\textbf{(D)\ 8.5}}</math> miles. | ||
| + | |||
| + | -Benedict T (countmath1) | ||
==Vide Solution 1 by SpreadTheMathLove== | ==Vide Solution 1 by SpreadTheMathLove== | ||
https://www.youtube.com/watch?v=jTTcscvcQmI | https://www.youtube.com/watch?v=jTTcscvcQmI | ||
Revision as of 13:22, 30 January 2025
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
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
\begin{align*}
\frac{1}{5} + \frac{x}{40} &= \frac{1}{4} + \frac{5-x}{40}\\
8 + x &= 10 + 5-x,
\end{align*}
so 
, and 
miles. This means that car will have traveled 
miles.
-Benedict T (countmath1)
