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Difference between revisions of "1968 AHSME Problems/Problem 25"

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== Solution ==
 
== Solution ==
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<math>\fbox{C}</math>
 
  
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== Solution 2 ==
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Let <math>k</math> denotes the distance Ace needs to run after the <math>y</math> yard. Since the distance they run with same amount of time is proportional to their speed, we have
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<cmath>\frac{1}{x}=\frac{k}{y+k}</cmath>
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<cmath>k=\frac{y}{x-1}</cmath>
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Thus the total distance ran by Flash is
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<cmath>y+k=y+\frac{y}{x-1}=\frac{xy}{x-1}\boxed{C}</cmath>
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~ Nafer
  
 
== See also ==
 
== See also ==
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{{AHSME box|year=1968|num-b=24|num-a=26}}   
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{{AHSME 35p box|year=1968|num-b=24|num-a=26}}   
  
 
[[Category: Intermediate Algebra Problems]]
 
[[Category: Intermediate Algebra Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 20:22, 17 July 2024

Problem

Ace runs with constant speed and Flash runs $x$ times as fast, $x>1$. Flash gives Ace a head start of $y$ yards, and, at a given signal, they start off in the same direction. Then the number of yards Flash must run to catch Ace is:

$\text{(A) } xy\quad \text{(B) } \frac{y}{x+y}\quad \text{(C) } \frac{xy}{x-1}\quad \text{(D) } \frac{x+y}{x+1}\quad \text{(E) } \frac{x+y}{x-1}$

Solution

Let $k$ denotes the distance Ace needs to run after the $y$ yard. Since the distance they run with same amount of time is proportional to their speed, we have \[\frac{1}{x}=\frac{k}{y+k}\] \[k=\frac{y}{x-1}\] Thus the total distance ran by Flash is \[y+k=y+\frac{y}{x-1}=\frac{xy}{x-1}\boxed{C}\]

~ Nafer

See also

1968 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 24 		Followed by
Problem 26
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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