Difference between revisions of "1960 AHSME Problems/Problem 34"

(Created page with "== Problem 34== Two swimmers, at opposite ends of a <math>90</math>-foot pool, start to swim the length of the pool, one at the rate of <math>3</math> feet per second, the o...")
 
(Solution to Problem 34)
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<asy>
 
<asy>
draw((0,0)--(0,90),EndArrow);
+
draw((0,0)--(0,105),EndArrow);
draw((0,0)--(90,0),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>
 
</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}}

Revision as of 11:57, 17 May 2018

Problem 34

Two swimmers, at opposite ends of a $90$-foot pool, start to swim the length of the pool, one at the rate of $3$ feet per second, the other at $2$ feet per second. They swim back and forth for $12$ minutes. Allowing no loss of times at the turns, find the number of times they pass each other.

$\textbf{(A)}\ 24\qquad \textbf{(B)}\ 21\qquad \textbf{(C)}\ 20\qquad \textbf{(D)}\ 19\qquad \textbf{(E)}\ 18$

Solution

First, note that it will take $30$ seconds for the first swimmer to reach the other side and $45$ seconds for the second swimmer to reach the other side. Also, note that after $180$ seconds (or $3$ 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]

At this point, find the number of meeting points in the first $3$ 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 $20$ times, which is answer choice $\boxed{\textbf{(C)}}$.

See Also

1960 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 33
Followed by
Problem 35
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