Difference between revisions of "1960 AHSME Problems/Problem 34"
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| − | == Problem | + | == Problem== |
Two swimmers, at opposite ends of a <math>90</math>-foot pool, start to swim the length of the pool, | Two swimmers, at opposite ends of a <math>90</math>-foot pool, start to swim the length of the pool, | ||
| Line 15: | Line 15: | ||
<asy> | <asy> | ||
| − | draw((0,0)--(0,90), | + | draw((0,0)--(0,105),EndArrow); |
| − | draw((0,0)--(90,0), | + | draw((0,0)--(105,0),EndArrow); |
| + | for (int i=0; i<6;++i) | ||
| + | { | ||
| + | dot((0,18i)); | ||
| + | } | ||
| + | for (int j=0;j<7;++j) | ||
| + | { | ||
| + | dot((15j,0)); | ||
| + | } | ||
| + | label("0",(0,0),SW); | ||
| + | label("90",(0,90),W); | ||
| + | label("180",(90,0),S); | ||
| + | draw((0,0)--(15,90)--(30,0)--(45,90)--(60,0)--(75,90)--(90,0),red); | ||
| + | draw((0,90)--(22.5,0)--(45,90)--(67.5,0)--(90,90),blue); | ||
</asy> | </asy> | ||
| − | At this point, find the number of meeting points in the first <math>3</math> minutes, then multiply by four to get the answer. | + | At this point, find the number of meeting points in the first <math>3</math> minutes, then multiply by four to get the answer. From the graph (where the x-axis is the time in seconds and the y-axis is distance from one side of the pool), there are five meeting points, so the two swimmers will pass each other <math>20</math> times, which is answer choice <math>\boxed{\textbf{(C)}}</math>. |
==See Also== | ==See Also== | ||
{{AHSME 40p box|year=1960|num-b=33|num-a=35}} | {{AHSME 40p box|year=1960|num-b=33|num-a=35}} | ||
| + | |||
| + | [[Category:Intermediate Algebra Problems]] | ||
Latest revision as of 18:18, 17 May 2018
Problem
Two swimmers, at opposite ends of a 
-foot pool, start to swim the length of the pool,
one at the rate of 
feet per second, the other at 
feet per second.
They swim back and forth for 
minutes. Allowing no loss of times at the turns, find the number of times they pass each other.
Solution
First, note that it will take 
seconds for the first swimmer to reach the other side and 
seconds for the second swimmer to reach the other side. Also, note that after 
seconds (or minutes), both swimmers will complete an even number of laps, essentially returning to their starting point.
![[asy] draw((0,0)--(0,105),EndArrow); draw((0,0)--(105,0),EndArrow); for (int i=0; i<6;++i) { dot((0,18i)); } for (int j=0;j<7;++j) { dot((15j,0)); } label("0",(0,0),SW); label("90",(0,90),W); label("180",(90,0),S); draw((0,0)--(15,90)--(30,0)--(45,90)--(60,0)--(75,90)--(90,0),red); draw((0,90)--(22.5,0)--(45,90)--(67.5,0)--(90,90),blue); [/asy]](http://latex.artofproblemsolving.com/f/8/f/f8f67de47955f00dc268a3d9363c4f4ccc40a57c.png)
At this point, find the number of meeting points in the first minutes, then multiply by four to get the answer. From the graph (where the x-axis is the time in seconds and the y-axis is distance from one side of the pool), there are five meeting points, so the two swimmers will pass each other 
times, which is answer choice 
.
See Also
| 1960 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 33 |
Followed by Problem 35 | |
| 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 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 | ||
| All AHSME Problems and Solutions | ||
