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Difference between revisions of "2025 AMC 8 Problems/Problem 19"

(→‎Solution 3 Recorrected)
(→‎Solution 3 Recorrected)
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== Solution 3 ==
 
== Solution 3 ==
−
Instead of using algebra, we can visualize how far each car has traveled every hour. Let us divide each portion of the road into distances traveled each hour: For the <math>25</math> mph portion, we divide it up into 5 sections, because <math>\frac{25\ \text{mph}}{5\ \text{mi}} = 1/5\ \text{hr}</math>. Similarly, we divide the <math>40</math> mph portion into 8 sections (<math>\frac{40\ \text{mph}}{5\ \text{mi}} = 1/8\ \text{hr}</math>) and the <math>20</math> mph portion into 4 sections (<math>\frac{20\ \text{mph}}{5\ \text{mi}} = 1/4\ \text{hr}</math>). Thus, we have the following diagram.
+
Instead of using algebra, we can visualize how far each car has traveled every hour. Let us divide each portion of the road into distances traveled each hour: For the <math>25</math> mph portion, we divide it up into 5 sections, because <math>\frac{25\ \text{mph}}{5\ \text{mi}} = 5\ \text{hr}</math>. Similarly, we divide the <math>40</math> mph portion into 8 sections (<math>\frac{40\ \text{mph}}{5\ \text{mi}} = 8\ \text{hr}</math>) and the <math>20</math> mph portion into 4 sections (<math>\frac{20\ \text{mph}}{5\ \text{mi}} = 4\ \text{hr}</math>). Thus, we have the following diagram.
  
 
<asy>
 
<asy>
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</asy>
 
</asy>
  
−
After four hours, '''car A''' has reached the end of the <math>25</math> mph portion, while '''car B''' has traveled <math>\frac{4}{5}</math> of the <math>20</math> mph portion. We can plot the amount of distance traveled, with the '''red dot''' representing car A and the '''blue dot''' representing car B.
+
After four hours, '''car A''' has reached the end of the <math>20</math> mph portion, while '''car B''' has traveled <math>\frac{4}{5}</math> of the <math>25</math> mph portion. We can plot the amount of distance traveled, with the '''red dot''' representing car A and the '''blue dot''' representing car B.
  
 
<asy>
 
<asy>
Line 134: Line 134:
  
 
// Red dot in the middle of the 4th line in the 25 mph portion
 
// Red dot in the middle of the 4th line in the 25 mph portion
−
dot((1, h/2), red);
+
dot((4/5.0, h/2), red);
  
 
// Blue dot in the middle of the line between the 40 mph and 20 mph portions
 
// Blue dot in the middle of the line between the 40 mph and 20 mph portions
−
dot((2.2, h/2), blue);
+
dot((2, h/2), blue);
 
</asy>
 
</asy>
  
−
If we keep moving the dots until Car B reaches the boundary between 20 and 40mph, we observe Car A has already moved 2 miles in the 40mph sector.
+
If we keep moving the dots, they will eventually meet at '''segment 3.5''' of the <math>40</math> mph portion of the road (note that each segment represents 1 hour of time traveled):
  
 
<asy>
 
<asy>
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// Red dot slightly to the left of the 3.5th segment in the 40 mph portion
 
// Red dot slightly to the left of the 3.5th segment in the 40 mph portion
−
dot((1.4 - 0.02, h/2), red);
+
dot((1 + 3.5/8.0 - 0.02, h/2), red);
  
 
// Blue dot slightly to the right of the 3.5th segment in the 40 mph portion
 
// Blue dot slightly to the right of the 3.5th segment in the 40 mph portion
−
dot((2 + 0.02, h/2), blue);
+
dot((1 + 3.5/8.0 + 0.02, h/2), blue);
  
 
// Dashed line at segment 3.5 in the 40 mph section
 
// Dashed line at segment 3.5 in the 40 mph section
−
draw((2 + 3.5/8.0, 0)--(1 + 3.5/8.0, h), dashed);
+
draw((1 + 3.5/8.0, 0)--(1 + 3.5/8.0, h), dashed);
 
</asy>
 
</asy>
  
−
This we see the cars have now both 40mph speed, 3 miles to meeting, thus they meet 1.5 Miles leftwards from the start of the 20mph sector, which is 10 mi away from A.
+
Now, we must account for the <math>5</math> miles in the <math>25</math> mph portion. Since the two cars meet at segment 3.5 of the <math>40</math> mph portion, we add the <math>5</math> miles traveled in the <math>25</math> mph section:
  
−
So, the distance from A is 10 - 1.5 = 8.5mi
+
<cmath> 5\ \text{mi} + 3.5\ \text{mi} = \boxed{\textbf{(D) 8.5}} </cmath> miles. <math>\square</math>
  
 
~ [[User:Aoum|aoum]]
 
~ [[User:Aoum|aoum]]
−
~ Fundamental Error corrected by MUIR (MathUnderInverseRatios)
 
  
 
==Video Solution by Pi Academy==
 
==Video Solution by Pi Academy==

Revision as of 08:18, 12 October 2025

Problem

Two towns, $A$ and $B$, are connected by a straight road, $15$ miles long. Traveling from town $A$ to town $B$, the speed limit changes every $5$ miles: from $25$ to $40$ to $20$ miles per hour (mph). Two cars, one at town $A$ and one at town $B$, 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 $A$, in miles, will the two cars meet?

[asy] // Asymptote code by aoum size(10cm); real h = 0.1; real s = 0.07; path b = brace((1,0),(0,0),amplitude=s); filldraw((0,0)--(3,0)--(3,h)--(0,h)--cycle,lightgray,black+1bp); draw((1,0)--(1,h),dashed); draw((2,0)--(2,h),dashed); label("$A$",(0,h/2),W); label("$B$",(3,h/2),E); draw(scale(0.7)*"$25\,\textrm{mph}$",(1,h+s)--(0,h+s),Bars); draw(scale(0.7)*"$40\,\textrm{mph}$",(2,h+s)--(1,h+s),Bars); draw(scale(0.7)*"$20\,\textrm{mph}$",(3,h+s)--(2,h+s),Bars); draw(b); draw(shift(1,0)*b); draw(shift(2,0)*b); label(scale(0.7)*"$5\,\textrm{mi}$",(0.5,-s),S); label(scale(0.7)*"$5\,\textrm{mi}$",(1.5,-s),S); label(scale(0.7)*"$5\,\textrm{mi}$",(2.5,-s),S); [/asy]

$\textbf{(A)}\ 7.75\qquad \textbf{(B)}\ 8\qquad \textbf{(C)}\ 8.25\qquad \textbf{(D)}\ 8.5\qquad \textbf{(E)}\ 8.75$

Solution 1

The first car, moving from town $A$ at $25$ miles per hour, takes $\frac{5}{25} = \frac{1}{5} \text{hours} = 12$ minutes. The second car, traveling another $5$ miles from town $B$, takes $\frac{5}{20} = \frac{1}{4} \text{hours} = 15$ minutes. The first car has traveled for 3 minutes or $\frac{1}{20}$th of an hour at $40$ miles per hour when the second car has traveled 5 miles. The first car has traveled $40 \cdot \frac{1}{20} = 2$ miles from the previous $5$ miles it traveled at $25$ miles per hour. They have $3$ miles left, and they travel at the same speed, so they meet $1.5$ miles through, so they are $5 + 2 + 1.5 = \boxed{\textbf{(D) }8.5}$ miles from town $A$.

~alwaysgonnagiveyouup

Solution 2

From the answer choices, the cars will meet somewhere along the $40$ mph stretch. Car $A$ travels $25$mph for $5$ miles, so we can use dimensional analysis to see that it will be $\frac{1\ \text{hr}}{25\ \text{mi}}\cdot 5\ \text{mi} = \frac{1}{5}$ of an hour for this portion. Similarly, car $B$ spends $\frac{1}{4}$ of an hour on the $20$ mph portion.

Suppose that car $A$ travels $x$ miles along the $40$ mph portion-- then car $B$ travels $5-x$ miles along the $40$ mph portion. By identical methods, car $A$ travels for $\frac{1}{40}\cdot x = \frac{x}{40}$ hours, and car $B$ travels for $\frac{5-x}{40}$ hours.

At their meeting point, cars $A$ and $B$ 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 $2x = 7$, and $x = 3.5$ miles. This means that car $A$ will have traveled $5 + 3.5= \boxed{\textbf{(D)\ 8.5}}$ miles.

-Benedict T (countmath1)

Solution 3

Instead of using algebra, we can visualize how far each car has traveled every hour. Let us divide each portion of the road into distances traveled each hour: For the $25$ mph portion, we divide it up into 5 sections, because $\frac{25\ \text{mph}}{5\ \text{mi}} = 5\ \text{hr}$. Similarly, we divide the $40$ mph portion into 8 sections ($\frac{40\ \text{mph}}{5\ \text{mi}} = 8\ \text{hr}$) and the $20$ mph portion into 4 sections ($\frac{20\ \text{mph}}{5\ \text{mi}} = 4\ \text{hr}$). Thus, we have the following diagram.

[asy] // Asymptote code by aoum size(10cm); real h = 0.1; real s = 0.07; path b = brace((1,0),(0,0),amplitude=s); // Draw the background and labeled bars filldraw((0,0)--(3,0)--(3,h)--(0,h)--cycle,lightgray,black+1bp); draw((1,0)--(1,h),dashed); draw((2,0)--(2,h),dashed); label("$A$",(0,h/2),W); label("$B$",(3,h/2),E); draw(scale(0.7)*"$25\,\textrm{mph}$",(1,h+s)--(0,h+s),Bars); draw(scale(0.7)*"$40\,\textrm{mph}$",(2,h+s)--(1,h+s),Bars); draw(scale(0.7)*"$20\,\textrm{mph}$",(3,h+s)--(2,h+s),Bars); draw(b); draw(shift(1,0)*b); draw(shift(2,0)*b); // Labels for the 5-mile segments label(scale(0.7)*"$5\,\textrm{mi}$",(0.5,-s),S); label(scale(0.7)*"$5\,\textrm{mi}$",(1.5,-s),S); label(scale(0.7)*"$5\,\textrm{mi}$",(2.5,-s),S); // Divisions for the 25 mph section (5 parts of 1 mile each) for (int i = 1; i < 5; ++i) { draw((i/5.0,0)--(i/5.0,h), dashed); } // Divisions for the 40 mph section (8 parts of 0.625 miles each) for (int i = 1; i < 8; ++i) { draw((i/8.0 + 1,0)--(i/8.0 + 1,h), dashed); } // Divisions for the 20 mph section (4 parts of 1.25 miles each) for (int i = 1; i < 4; ++i) { draw((i/4.0 + 2,0)--(i/4.0 + 2,h), dashed); } [/asy]

After four hours, car A has reached the end of the $20$ mph portion, while car B has traveled $\frac{4}{5}$ of the $25$ mph portion. We can plot the amount of distance traveled, with the red dot representing car A and the blue dot representing car B.

[asy] // Asymptote code by aoum size(10cm); real h = 0.1; real s = 0.07; path b = brace((1,0),(0,0),amplitude=s); // Draw the background and labeled bars filldraw((0,0)--(3,0)--(3,h)--(0,h)--cycle,lightgray,black+1bp); draw((1,0)--(1,h),dashed); draw((2,0)--(2,h),dashed); label("$A$",(0,h/2),W); label("$B$",(3,h/2),E); draw(scale(0.7)*"$25\,\textrm{mph}$",(1,h+s)--(0,h+s),Bars); draw(scale(0.7)*"$40\,\textrm{mph}$",(2,h+s)--(1,h+s),Bars); draw(scale(0.7)*"$20\,\textrm{mph}$",(3,h+s)--(2,h+s),Bars); draw(b); draw(shift(1,0)*b); draw(shift(2,0)*b); // Labels for the 5-mile segments label(scale(0.7)*"$5\,\textrm{mi}$",(0.5,-s),S); label(scale(0.7)*"$5\,\textrm{mi}$",(1.5,-s),S); label(scale(0.7)*"$5\,\textrm{mi}$",(2.5,-s),S); // Divisions for the 25 mph section (5 parts of 1 mile each) for (int i = 1; i < 5; ++i) { draw((i/5.0,0)--(i/5.0,h), dashed); } // Divisions for the 40 mph section (8 parts of 0.625 miles each) for (int i = 1; i < 8; ++i) { draw((i/8.0 + 1,0)--(i/8.0 + 1,h), dashed); } // Divisions for the 20 mph section (4 parts of 1.25 miles each) for (int i = 1; i < 4; ++i) { draw((i/4.0 + 2,0)--(i/4.0 + 2,h), dashed); } // Red dot in the middle of the 4th line in the 25 mph portion dot((4/5.0, h/2), red); // Blue dot in the middle of the line between the 40 mph and 20 mph portions dot((2, h/2), blue); [/asy]

If we keep moving the dots, they will eventually meet at segment 3.5 of the $40$ mph portion of the road (note that each segment represents 1 hour of time traveled):

[asy] // Asymptote code by aoum size(10cm); real h = 0.1; real s = 0.07; path b = brace((1,0),(0,0),amplitude=s); // Draw the background and labeled bars filldraw((0,0)--(3,0)--(3,h)--(0,h)--cycle,lightgray,black+1bp); draw((1,0)--(1,h),dashed); draw((2,0)--(2,h),dashed); label("$A$",(0,h/2),W); label("$B$",(3,h/2),E); draw(scale(0.7)*"$25\,\textrm{mph}$",(1,h+s)--(0,h+s),Bars); draw(scale(0.7)*"$40\,\textrm{mph}$",(2,h+s)--(1,h+s),Bars); draw(scale(0.7)*"$20\,\textrm{mph}$",(3,h+s)--(2,h+s),Bars); draw(b); draw(shift(1,0)*b); draw(shift(2,0)*b); // Labels for the 5-mile segments label(scale(0.7)*"$5\,\textrm{mi}$",(0.5,-s),S); label(scale(0.7)*"$5\,\textrm{mi}$",(1.5,-s),S); label(scale(0.7)*"$5\,\textrm{mi}$",(2.5,-s),S); // Divisions for the 25 mph section (5 parts of 1 mile each) for (int i = 1; i < 5; ++i) { draw((i/5.0,0)--(i/5.0,h), dashed); } // Divisions for the 40 mph section (8 parts of 0.625 miles each) for (int i = 1; i < 8; ++i) { draw((i/8.0 + 1,0)--(i/8.0 + 1,h), dashed); } // Divisions for the 20 mph section (4 parts of 1.25 miles each) for (int i = 1; i < 4; ++i) { draw((i/4.0 + 2,0)--(i/4.0 + 2,h), dashed); } // Red dot slightly to the left of the 3.5th segment in the 40 mph portion dot((1 + 3.5/8.0 - 0.02, h/2), red); // Blue dot slightly to the right of the 3.5th segment in the 40 mph portion dot((1 + 3.5/8.0 + 0.02, h/2), blue); // Dashed line at segment 3.5 in the 40 mph section draw((1 + 3.5/8.0, 0)--(1 + 3.5/8.0, h), dashed); [/asy]

Now, we must account for the $5$ miles in the $25$ mph portion. Since the two cars meet at segment 3.5 of the $40$ mph portion, we add the $5$ miles traveled in the $25$ mph section:

\[5\ \text{mi} + 3.5\ \text{mi} = \boxed{\textbf{(D) 8.5}}\] miles. $\square$

~ aoum

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

https://youtu.be/PKMpTS6b988

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

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