Difference between revisions of "2008 AMC 12B Problems/Problem 20"
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==Problem== | ==Problem== | ||
− | Michael walks at the rate of <math>5</math> feet per second on a long straight path. Trash pails are located every <math>200</math> feet along the path. A garbage truck | + | Michael walks at the rate of <math>5</math> feet per second on a long straight path. Trash pails are located every <math>200</math> feet along the path. A garbage truck traveling at <math>10</math> feet per second in the same direction as Michael stops for <math>30</math> 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 meet? |
<math>\mathrm{(A)}\ 4\qquad\mathrm{(B)}\ 5\qquad\mathrm{(C)}\ 6\qquad\mathrm{(D)}\ 7\qquad\mathrm{(E)}\ 8</math> | <math>\mathrm{(A)}\ 4\qquad\mathrm{(B)}\ 5\qquad\mathrm{(C)}\ 6\qquad\mathrm{(D)}\ 7\qquad\mathrm{(E)}\ 8</math> | ||
− | ==Solution== | + | ==Solution 1== |
Pick a coordinate system where Michael's starting pail is <math>0</math> and the one where the truck starts is <math>200</math>. | Pick a coordinate system where Michael's starting pail is <math>0</math> and the one where the truck starts is <math>200</math>. | ||
Line 14: | Line 14: | ||
We have <math>D(0)=200</math>. | We have <math>D(0)=200</math>. | ||
− | The truck always moves for <math>20</math> seconds, then stands still for <math>30</math>. During the first <math>20</math> seconds of the cycle the truck moves by <math>200</math> | + | The truck always moves for <math>20</math> seconds, then stands still for <math>30</math>. During the first <math>20</math> seconds of the cycle the truck moves by <math>200</math> feet and Michael by <math>100</math>, hence during the first <math>20</math> seconds of the cycle <math>D(t)</math> increases by <math>100</math>. |
During the remaining <math>30</math> seconds <math>D(t)</math> decreases by <math>150</math>. | During the remaining <math>30</math> seconds <math>D(t)</math> decreases by <math>150</math>. | ||
Line 25: | Line 25: | ||
During the seventh cycle, <math>D(t)</math> will first grow from <math>-100</math> to <math>0</math>, then fall from <math>0</math> to <math>-150</math>. Hence the truck meets Michael at the moment when it arrives to the next pail. | During the seventh cycle, <math>D(t)</math> will first grow from <math>-100</math> to <math>0</math>, then fall from <math>0</math> to <math>-150</math>. Hence the truck meets Michael at the moment when it arrives to the next pail. | ||
− | Obviously, from this point on <math>D(t)</math> will always be negative, meaning that Michael is already too far ahead. Hence we found all <math>\boxed{5}</math> meetings. | + | Obviously, from this point on <math>D(t)</math> will always be negative, meaning that Michael is already too far ahead. Hence we found all <math>\boxed{5 \Longrightarrow B}</math> meetings. |
The movement of Michael and the truck is plotted below: Michael in blue, the truck in red. | The movement of Michael and the truck is plotted below: Michael in blue, the truck in red. | ||
Line 47: | Line 47: | ||
draw(graph(xm,ym),blue); | draw(graph(xm,ym),blue); | ||
− | xaxis("$time$",Bottom,LeftTicks); | + | xaxis("$\mathrm{time}$",Bottom,LeftTicks); |
− | yaxis("$position$",Left,LeftTicks); | + | yaxis("$\mathrm{position}$",Left,LeftTicks); |
</asy> | </asy> | ||
+ | ==Solution 2== | ||
+ | |||
+ | The truck takes <math>20</math> seconds to go from one pail to the next and then stops for <math>30</math> seconds 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: | ||
+ | |||
+ | '''(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>. | ||
+ | |||
+ | '''(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 <math>5</math> times, which is option <math>\boxed{\textbf{(B)}}</math>. | ||
+ | |||
+ | ==Solution 3== | ||
+ | |||
+ | We make a chart by seconds in increments of ten. | ||
+ | |||
+ | <math>\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}</math> | ||
+ | |||
+ | |||
+ | 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 <math>\boxed{\textbf{(B)}}</math>. | ||
+ | |||
+ | |||
+ | |||
+ | ~superagh | ||
+ | |||
+ | |||
+ | ==Solution 4 (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. We let <math>M(t) = 5t</math>, where <math>t</math> is the amount of seconds passed. Now, lets define the position function of the truck as two independent functions. It is obvious, graphically, that the position function of the truck is a piecewise function alternating between linear lines and constant lines. Lets focus on the linear pieces of the truck's position function. Let the <math>kth</math> linear part of the truck's position function be denoted as <math>L_k(t)</math>. Then through algebra, it is found that <math>L_k(t) = 10t + 500 - 300k</math>, <math>{50k - 50 <= t <= 50k - 30}</math>. Now, lets move on to the constant pieces, which is a lot easier in terms of algebra. Let the <math>ith</math> constant part of the truck's position function be denoted as <math>C_i(t)</math>. Then, again, through algebra we obtain <math>C_i(t) = 200 * (i + 1)</math>, <math>{50i - 30 <= t <= 50i}</math>. Now, let me stress that <math>i</math> and <math>k</math> are disjoint, which is why I used different variable names. We are interested in where <math>C_i(t)</math> and <math>L_k(t)</math> intersect <math>M(t)</math> or <math>5t</math>. <math>L_k(t)</math> intersects <math>M</math> at <math>(60k - 100, 300k - 500)</math>. However, the only <math>k</math> values that actually work are <math>5 <= k <= 7</math> because of the domain restrictions on <math>L_k(t)</math>. Similarly, we also see that for <math>C_i(t)</math> it intersects at <math>(40i + 40, 200i + 200)</math>. The only <math>i</math> values that work is <math>4 <= i <= 7</math>. However, some pair <math>(i, k)</math> might yield the same exact intersection point. Checking this through simple algebra we see that <math>(4,5)</math> and <math>(7,7)</math> do indeed yield the same intersection point. Thus, our answer is <math>(7 - 5 + 1) + (7 - 4 + 1) - 2 = 7 - 2 = 5</math> or <math>\boxed{\textbf{(B)}}</math>. | ||
+ | |||
+ | 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 | ||
==See also== | ==See also== | ||
{{AMC12 box|year=2008|ab=B|num-b=19|num-a=21}} | {{AMC12 box|year=2008|ab=B|num-b=19|num-a=21}} | ||
− | {{AMC10 box|year=2008|ab=B|num-b=24|after=Last | + | {{AMC10 box|year=2008|ab=B|num-b=24|after=Last Problem}} |
+ | {{MAA Notice}} |
Latest revision as of 17:55, 11 August 2023
- The following problem is from both the 2008 AMC 12B #20 and 2008 AMC 10B #25, so both problems redirect to this page.
Contents
Problem
Michael walks at the rate of feet per second on a long straight path. Trash pails are located every
feet along the path. A garbage truck traveling at
feet per second in the same direction as Michael stops for
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 meet?
Solution 1
Pick a coordinate system where Michael's starting pail is and the one where the truck starts is
.
Let
and
be the coordinates of Michael and the truck after
seconds.
Let
be their (signed) distance after
seconds.
Meetings occur whenever
.
We have
.
The truck always moves for seconds, then stands still for
. During the first
seconds of the cycle the truck moves by
feet and Michael by
, hence during the first
seconds of the cycle
increases by
.
During the remaining
seconds
decreases by
.
From this observation it is obvious that after four full cycles, i.e. at , we will have
for the first time.
During the fifth cycle, will first grow from
to
, then fall from
to
. Hence Michael overtakes the truck while it is standing at the pail.
During the sixth cycle, will first grow from
to
, then fall from
to
. 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, will first grow from
to
, then fall from
to
. Hence the truck meets Michael at the moment when it arrives to the next pail.
Obviously, from this point on will always be negative, meaning that Michael is already too far ahead. Hence we found all
meetings.
The movement of Michael and the truck is plotted below: Michael in blue, the truck in red.
Solution 2
The truck takes seconds to go from one pail to the next and then stops for
seconds at the new pail. Thus it sets off from a pail every 50 sec. Let
denote the time elapsed and write
, where
. In this time Michael has traveled
feet. What about the truck? In the first
seconds the truck covers
pails, i.e.
feet so it moves
feet from Michael's starting point. Then we have two cases:
(a) if , then the truck travels an additional
feet. For them to intersect we must have
. Solving, we get
. Since
must lie in the interval
we get
.
(b) if , then the truck travels an additional
feet. For them to intersect we must have
. Solving, we get
. Since
must lie in the interval
we get
.
Thus Michael intersects with the truck times, which is option
.
Solution 3
We make a chart by seconds in increments of ten.
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 .
~superagh
Solution 4 (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. We let , where
is the amount of seconds passed. Now, lets define the position function of the truck as two independent functions. It is obvious, graphically, that the position function of the truck is a piecewise function alternating between linear lines and constant lines. Lets focus on the linear pieces of the truck's position function. Let the
linear part of the truck's position function be denoted as
. Then through algebra, it is found that
,
. Now, lets move on to the constant pieces, which is a lot easier in terms of algebra. Let the
constant part of the truck's position function be denoted as
. Then, again, through algebra we obtain
,
. Now, let me stress that
and
are disjoint, which is why I used different variable names. We are interested in where
and
intersect
or
.
intersects
at
. However, the only
values that actually work are
because of the domain restrictions on
. Similarly, we also see that for
it intersects at
. The only
values that work is
. However, some pair
might yield the same exact intersection point. Checking this through simple algebra we see that
and
do indeed yield the same intersection point. Thus, our answer is
or
.
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
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.