2008 AMC 12B Problems/Problem 20

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 $20$ seconds to go from one pail to the next and then stops for $30$ seconds 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 $\boxed{\textbf{(B)}}$ .

Solution 3

We make a chart by seconds in increments of ten.

$\begin{tabular}{c|c|c}Time (s) &Distance of Michael&Distance of Garbage\\ \hline 0&0&200\\ 10&50&300\\ 20&100&400\\ 30&150&400\\ 40&200&400\\ 50&250&400\\ 60&300&500\\ 70&350&600\\ 80&400&600\\ 90&450&600\\ 100&500&600\\ 110&550&700\\ 120&600&800\\ 130&650&800\\ 140&700&800\\ 150&750&800\\ 160&800&900\\ 170&850&1000\\ 180&900&1000\\ 190&950&1000\\ 200&1000&1000\\ 210&1050&1100\\ 220&1100&1200\\ 230&1150&1200\\ 240&1200&1200\\ 250&1250&1200\\ 260&1300&1300\\ 270&1350&1400\\ 280&1400&1400\\ 290&1450&1400\\ 300&1500&1400\\ 310&1550&1500\\ 320&1600&1600\\ 330&1650&1600\\ 340&1700&1600\\ 350&1750&1600\\ 360&1800&1700\\ 370&1850&1800\\ 380&1900&1800\\ 390&1950&1800\\ 400&2000&1800\\ \end{tabular}$

Notice at 200, 240, 260, 280, and 320 seconds, Michael and the garbage truck meet. It is clear that they met at these times, and will meet no more. Thus the answer is $\boxed{\textbf{(B)}}$ .

~superagh

==Solution 3== (Position Functions and Piecewise) This solution might be time consuming, but it is pretty rigorous. Also, throughout the solution refer to the graph in solution 1 to understand this one more. Lets, first start off by defining the position function for Michael as it is trivial. We let $M(t) = 5t$ , where $t$ is the amount of seconds passed. Now, lets define as the position function of the truck as two function. It is obvious, graphically, that position function of the truck is a piecewise function alternating between linear lines and and constant lines. Lets, focus on the linear pieces of $T$ . Let the $kth$ linear part of the truck's position function be denoted as $L_k(t)$ . Then through algebra, it is found that $L_k(t) = 10t + 500 - 300k$ , ${50k - 50 <= t <= 50k - 30}$ . Now, lets move on to the constant pieces, which is a lot easier in terms of algebra. Let the constant $ith$ constant part of the truck's position function be denoted as $C_i(t)$ . Then, again, through algebra we obtain $C_i(t) = 200 * (i + 1)$ , ${50i - 30 <= t <= 50i}$ . Now, let me stress that $i$ and $k$ are disjoint, which is why I used different variable names. We are interested in where $C_i(t)$ and $L_i(t)$ intersect $M(t)$ ot $5t$ . $L_i(t)$ , then it intersects $M$ at $(60k - 100, 300k - 500)$ . However, only k that actually works is $5 <= k <= 7$ because of the domain restrictions of on $L_i(t)$ . Similarly, we also see that for $C_i(t)$ , the only $i$ that works is $4 <= k <= 7$ . However, some pair $(i, k)$ might yield the same exact intersection point. Checking this through simple algebra we see that $(4,5)$ and $(7,7)$ . Thus, our answer is $(7 - 5 + 1) + (7 - 4 + 1) - 2 = 7 - 2 = 5$ or $\boxed{\textbf{(B)}}$ .

Note: This is probably not the best solution as it seems a lot more tedious, but it still works. I admit this solution isn't elegant at all. ~triggod

2008 AMC 12B (Problems • Answer Key • Resources)
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2008 AMC 12B Problems/Problem 20

Contents

Problem

Solution 1

Solution 2

Solution 3

See also