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Difference between revisions of "1968 AHSME Problems/Problem 32"

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== Solution ==
 
== Solution ==
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<math>\fbox{E}</math>
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Let the speed of <math>A</math> be <math>a</math> and the speed of <math>B</math> be <math>b</math>. The first time that <math>A</math> and <math>B</math> will be equidistant from <math>O</math>, <math>B</math> will have not yet reached <math>O</math>. Thus, after two minutes, <math>B</math>'s distance from <math>O</math> will be <math>500-2b</math>, and <math>A</math>'s distance from <math>O</math> will be <math>2a</math>. Setting these expressions equal to each other and dividing by 2, we see that <math>a=250-b</math>.
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After another eight minutes (or after a total of ten minutes since <math>A</math> was at <math>O</math>), <math>A</math> and <math>B</math> will again be equidistant from <math>O</math>, but this time <math>B</math> will have passed <math>O</math>. The distance <math>A</math> will be from <math>O</math> is <math>10a</math>, and the distance <math>B</math> will be from <math>O</math> is <math>10b-500</math>. Setting these expressions equal to each other and dividing by 10, we see that <math>a=b-50</math>.
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Adding the two equations that we have obtained above, we see that <math>2a=250-b+b-50</math>, and so <math>a=100</math>. Substituting this value of <math>a</math> into the second equation, we see that <math>100=b-50</math>, or <math>b=150</math>. Then, <math>\frac{a}{b}=\frac{100}{150}=\frac{2}{3}</math>, so the ratio of <math>A</math>'s speed to that of <math>B</math> is <math>\fbox{(C) 2:3}</math>.
  
 
== See also ==
 
== See also ==
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{{AHSME box|year=1968|num-b=31|num-a=33}}   
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{{AHSME 35p box|year=1968|num-b=31|num-a=33}}   
  
 
[[Category: Intermediate Geometry Problems]]
 
[[Category: Intermediate Geometry Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 16:01, 17 July 2024

Problem

$A$ and $B$ move uniformly along two straight paths intersecting at right angles in point $O$. When $A$ is at $O$, $B$ is $500$ yards short of $O$. In two minutes they are equidistant from $O$, and in $8$ minutes more they are again equidistant from $O$. Then the ratio of $A$'s speed to $B$'s speed is:

$\text{(A) } 4:5\quad \text{(B) } 5:6\quad \text{(C) } 2:3\quad \text{(D) } 5:8\quad \text{(E) } 1:2$


Solution

Let the speed of $A$ be $a$ and the speed of $B$ be $b$. The first time that $A$ and $B$ will be equidistant from $O$, $B$ will have not yet reached $O$. Thus, after two minutes, $B$'s distance from $O$ will be $500-2b$, and $A$'s distance from $O$ will be $2a$. Setting these expressions equal to each other and dividing by 2, we see that $a=250-b$.

After another eight minutes (or after a total of ten minutes since $A$ was at $O$), $A$ and $B$ will again be equidistant from $O$, but this time $B$ will have passed $O$. The distance $A$ will be from $O$ is $10a$, and the distance $B$ will be from $O$ is $10b-500$. Setting these expressions equal to each other and dividing by 10, we see that $a=b-50$.

Adding the two equations that we have obtained above, we see that $2a=250-b+b-50$, and so $a=100$. Substituting this value of $a$ into the second equation, we see that $100=b-50$, or $b=150$. Then, $\frac{a}{b}=\frac{100}{150}=\frac{2}{3}$, so the ratio of $A$'s speed to that of $B$ is $\fbox{(C) 2:3}$.

See also

1968 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 31 		Followed by
Problem 33
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
All AHSME Problems and Solutions

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