Difference between revisions of "2008 AMC 12B Problems/Problem 20"
m (→See also: the amc redirect box at the bottom wasn't included) |
|||
Line 6: | Line 6: | ||
<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 50: | Line 50: | ||
yaxis("$\mathrm{position}$",Left,LeftTicks); | yaxis("$\mathrm{position}$",Left,LeftTicks); | ||
</asy> | </asy> | ||
+ | |||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | |||
+ | 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>. | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | Thus Michael intersects with the truck 5 times, which is option <math>\boxed{\textbf{(B)}}</math>. | ||
==See also== | ==See also== |
Revision as of 10:43, 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.
Contents
[hide]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 intersect?
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
Number the trash pails , with 0 being the pail where the truck is at the beginning. For the truck is traveling between the th and th trash pails at 10 ft/sec. Let . The truck's position at time (relative to Michael's starting point) is given by . Michael's position at this time is of course . If Michael passes the truck at this time then we must have . Solving for , we get . Since must lie in the interval we get .
For the truck stops at the th trash pail and its position is thus given by . Let . If Michael passes the truck at this time then we must have . Solving for , we get . Since must lie in the interval we get .
Thus Michael intersects with the truck 5 times, which is option .
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.