Difference between revisions of "2008 AMC 10B Problems/Problem 25"

Line 8: Line 8:
 
The truck always moves for <math>20</math> seconds, then stands still for <math>30</math>. In these <math>50</math> seconds, the truck will drive <math>200+0=200</math> meters. In those <math>50</math> seconds Michael will walk <math>250</math> meters. So ultimately Michael will be way too far ahead of the truck for any more meetings to happen.
 
The truck always moves for <math>20</math> seconds, then stands still for <math>30</math>. In these <math>50</math> seconds, the truck will drive <math>200+0=200</math> meters. In those <math>50</math> seconds Michael will walk <math>250</math> meters. So ultimately Michael will be way too far ahead of the truck for any more meetings to happen.
  
The movement of Michael and the truck is plotted below: Michael in blue, the truck in red. We can easily verify that indeed there will be <math>\boxed{4}</math> more meetings:  
+
The movement of Michael and the truck is plotted below: Michael in blue, the truck in red. We can easily verify that indeed there will be <math>\boxed{5}</math> more meetings: (the graph of the truck should start 200 units higher)
 
* Michael will catch and overtake the truck while it is standing at the first pail.
 
* Michael will catch and overtake the truck while it is standing at the first pail.
 
* The truck will start moving again and on its way to the second pail it will overtake Michael.
 
* The truck will start moving again and on its way to the second pail it will overtake Michael.
Line 31: Line 31:
 
yaxis("$position$",Left,LeftTicks);
 
yaxis("$position$",Left,LeftTicks);
 
</asy>
 
</asy>
 +
  
 
==See also==
 
==See also==
 
{{AMC10 box|year=2008|ab=B|num-b=24|after=Last Question}}
 
{{AMC10 box|year=2008|ab=B|num-b=24|after=Last Question}}

Revision as of 19:22, 31 January 2009

Problem

Michael walks at the rate of $5$ feet per second on a long straight path. Trash pails are located every $200$ feet along the path. A garbage truck travels at $10$ feet per second in the same direction as Michael and stops for $30$ seconds at each pail. As Michael passes a pail, he notices the truck ahead of him just leaving for the next pail. How many times will Michael and the truck meet?

$\mathrm{(A)}\ 4\qquad\mathrm{(B)}\ 5\qquad\mathrm{(C)}\ 6\qquad\mathrm{(D)}\ 7\qquad\mathrm{(E)}\ 8$

Solution

The truck always moves for $20$ seconds, then stands still for $30$. In these $50$ seconds, the truck will drive $200+0=200$ meters. In those $50$ seconds Michael will walk $250$ meters. So ultimately Michael will be way too far ahead of the truck for any more meetings to happen.

The movement of Michael and the truck is plotted below: Michael in blue, the truck in red. We can easily verify that indeed there will be $\boxed{5}$ more meetings: (the graph of the truck should start 200 units higher)

  • Michael will catch and overtake the truck while it is standing at the first pail.
  • The truck will start moving again and on its way to the second pail it will overtake Michael.
  • While the truck is standing at the second pail, Michael will walk past it.
  • The last meeting will occur exactly when both Michael and the truck arrive at the same time to the third pail.

[asy] import graph;  size(400,300,IgnoreAspect);  real[] xt={0,20,50,70,100,120,150,170,200}; real[] yt={0,200,200,400,400,600,600,800,800};  real[] xm={0,200}; real[] ym={0,1000};  draw(graph(xt,yt),red); draw(graph(xm,ym),blue);  xaxis("$time$",Bottom,LeftTicks); yaxis("$position$",Left,LeftTicks); [/asy]


See also

2008 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 24
Followed by
Last Question
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 AMC 10 Problems and Solutions
Invalid username
Login to AoPS