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Difference between revisions of "2008 AMC 12B Problems/Problem 20"
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==Solution 2==
 
==Solution 2==
  
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The truck takes 20 sec to go from one pail to the next and then stops for 30 sec at the new pail. Thus it sets off from a pail every 50 sec. Let <math>t</math> denote the time elapsed and write <math>t=50k + \Delta</math>, where <math>\Delta \in [0,50)</math>. In  this time Michael has traveled <math>5t = 250k+5\Delta</math> feet. What about the truck? In the first <math>50k</math> seconds the truck covers <math>k</math> pails, i.e. <math>200k</math> feet so it moves <math>200+200k</math> feet from Michael's starting point. Then we have two cases:
  
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Number the trash pails <math>0, 1, 2, 3, \ldots</math>, with 0 being the pail where the truck is at the beginning. For <math>t \in (50k, 50k+20)</math> the truck is traveling between the <math>k</math>th and <math>(k+1)</math>th trash pails at 10 ft/sec. Let <math>\Delta = t-50k \in (0,20)</math>. The truck's position at time <math>t</math> (relative to Michael's starting point) is given by <math>d = 200 + 200k + 10\Delta</math>. Michael's position at this time is of course <math>5t=5\Delta + 250k</math>. If Michael passes the truck at this time then we must have <math>200 + 200k + 10\Delta = 5\Delta + 250k</math>. Solving for <math>\Delta</math>, we get <math>\Delta = 10k - 40</math>. Since <math>\Delta</math> must lie in the interval <math>(0,20)</math> we get <math>k =5</math>.  
+
'''(a)''' if <math>\Delta < 20</math>, then the truck travels an additional <math>10\Delta</math> feet. For them to intersect we must have <math>200+200k + 10\Delta = 250k + 5\Delta</math>. Solving, we get <math>\Delta = 10k - 40</math>. Since <math>\Delta</math> must lie in the interval <math>[0,20)</math> we get <math>k \in \{4, 5\}</math>.
  
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For <math>t \in (50k+20, 50(k+1))</math> the truck stops at the <math>(k+1)</math>th trash pail and its position is thus given by <math>200+200(k+1)</math>. Let <math>\Delta = t-50k-20 \in [0,30]</math>. If Michael passes the truck at this time then we must have <math>400 + 200k = 5t = 5\Delta + 250k + 100</math>. Solving for <math>\Delta</math>, we get <math>\Delta = 10(6-k)</math>. Since <math>\Delta</math> must lie in the interval <math>[0,30]</math> we get <math>k \in \{3, 4, 5, 6\}</math>.  
+
'''(b)''' if <math>\Delta \in [20, 50)</math>, then the truck travels an additional <math>200</math> feet. For them to intersect we must have <math>200+200k + 200 = 250k + 5\Delta</math>. Solving, we get <math>\Delta = 10(8-k)</math>. Since <math>\Delta</math> must lie in the interval <math>[20,50)</math> we get <math>k \in \{4, 5, 6\}</math>.  
  
 
Thus Michael intersects with the truck 5 times, which is option <math>\boxed{\textbf{(B)}}</math>.
 
Thus Michael intersects with the truck 5 times, which is option <math>\boxed{\textbf{(B)}}</math>.
 +
  
 
==See also==
 
==See also==

Revision as of 16:43, 17 December 2019

The following problem is from both the 2008 AMC 12B #20 and 2008 AMC 10B #25, so both problems redirect to this page.

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 traveling at $10$ feet per second in the same direction as Michael stops for $30$ seconds at each pail. As Michael passes a pail, he notices the truck ahead of him just leaving the next pail. How many times will Michael and the truck intersect?

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

Solution 1

Pick a coordinate system where Michael's starting pail is $0$ and the one where the truck starts is $200$. Let $M(t)$ and $T(t)$ be the coordinates of Michael and the truck after $t$ seconds. Let $D(t)=T(t)-M(t)$ be their (signed) distance after $t$ seconds. Meetings occur whenever $D(t)=0$. We have $D(0)=200$.

The truck always moves for $20$ seconds, then stands still for $30$. During the first $20$ seconds of the cycle the truck moves by $200$ feet and Michael by $100$, hence during the first $20$ seconds of the cycle $D(t)$ increases by $100$. During the remaining $30$ seconds $D(t)$ decreases by $150$.

From this observation it is obvious that after four full cycles, i.e. at $t=200$, we will have $D(t)=0$ for the first time.

During the fifth cycle, $D(t)$ will first grow from $0$ to $100$, then fall from $100$ to $-50$. Hence Michael overtakes the truck while it is standing at the pail.

During the sixth cycle, $D(t)$ will first grow from $-50$ to $50$, then fall from $50$ to $-100$. Hence the truck starts moving, overtakes Michael on their way to the next pail, and then Michael overtakes the truck while it is standing at the pail.

During the seventh cycle, $D(t)$ will first grow from $-100$ to $0$, then fall from $0$ to $-150$. Hence the truck meets Michael at the moment when it arrives to the next pail.

Obviously, from this point on $D(t)$ will always be negative, meaning that Michael is already too far ahead. Hence we found all $\boxed{5 \Longrightarrow B}$ meetings.

The movement of Michael and the truck is plotted below: Michael in blue, the truck in red.

[asy] import graph; size(400,300,IgnoreAspect); real[] xt = new real[21]; real[] yt = new real[21]; for (int i=0; i<11; ++i) xt[2*i]=50*i; for (int i=0; i<10; ++i) xt[2*i+1]=50*i+20; for (int i=0; i<11; ++i) yt[2*i]=200*(i+1); for (int i=0; i<10; ++i) yt[2*i+1]=200*(i+2); real[] xm={0,500}; real[] ym={0,2500}; draw(graph(xt,yt),red); draw(graph(xm,ym),blue); xaxis("$\mathrm{time}$",Bottom,LeftTicks); yaxis("$\mathrm{position}$",Left,LeftTicks); [/asy]


Solution 2

The truck takes 20 sec to go from one pail to the next and then stops for 30 sec at the new pail. Thus it sets off from a pail every 50 sec. Let $t$ denote the time elapsed and write $t=50k + \Delta$, where $\Delta \in [0,50)$. In this time Michael has traveled $5t = 250k+5\Delta$ feet. What about the truck? In the first $50k$ seconds the truck covers $k$ pails, i.e. $200k$ feet so it moves $200+200k$ feet from Michael's starting point. Then we have two cases:

(a) if $\Delta < 20$, then the truck travels an additional $10\Delta$ feet. For them to intersect we must have $200+200k + 10\Delta = 250k + 5\Delta$. Solving, we get $\Delta = 10k - 40$. Since $\Delta$ must lie in the interval $[0,20)$ we get $k \in \{4, 5\}$.

(b) if $\Delta \in [20, 50)$, then the truck travels an additional $200$ feet. For them to intersect we must have $200+200k + 200 = 250k + 5\Delta$. Solving, we get $\Delta = 10(8-k)$. Since $\Delta$ must lie in the interval $[20,50)$ we get $k \in \{4, 5, 6\}$.

Thus Michael intersects with the truck 5 times, which is option $\boxed{\textbf{(B)}}$.


See also

2008 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 19 		Followed by
Problem 21
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 12 Problems and Solutions
2008 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 24 		Followed by
Last Problem
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

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