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

Revision as of 10:57, 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 $\boxed{\textbf{(C)}}$ .

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

Revision as of 10:57, 17 May 2018

Problem 34

Solution

See Also

@@ Line 15: / Line 15: @@
 <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>
-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==
 {{AHSME 40p box|year=1960|num-b=33|num-a=35}}