Difference between revisions of "1955 AHSME Problems/Problem 10"

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==Solution==
 
==Solution==
 
The train will take <math>\frac{a}{40}</math> hours to travel <math>a</math> miles, and it takes <math>\frac{nm}{60}</math>. The LCM of <math>40</math> and <math>60</math> is <math>120</math>, which allows for the addition of the fractions <math>\frac{3a}{120}</math> and <math>\frac{2mn}{120}</math>. The end result is <math>\boxed{\textbf{(A)}\frac{3a+2mn}{120}}</math>.
 
The train will take <math>\frac{a}{40}</math> hours to travel <math>a</math> miles, and it takes <math>\frac{nm}{60}</math>. The LCM of <math>40</math> and <math>60</math> is <math>120</math>, which allows for the addition of the fractions <math>\frac{3a}{120}</math> and <math>\frac{2mn}{120}</math>. The end result is <math>\boxed{\textbf{(A)}\frac{3a+2mn}{120}}</math>.
==See Also==
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== See Also ==
 
== See Also ==
 
{{AHSME 50p box|year=1955|num-b=9|num-a=1}}
 
{{AHSME 50p box|year=1955|num-b=9|num-a=1}}
  
 
{{MAA Notice}
 
{{MAA Notice}

Revision as of 10:41, 15 February 2021

How many hours does it take a train traveling at an average rate of 40 mph between stops to travel a miles it makes n stops of m minutes each?

$\textbf{(A)}\ \frac{3a+2mn}{120}\qquad\textbf{(B)}\ 3a+2mn\qquad\textbf{(C)}\ \frac{3a+2mn}{12}\qquad\textbf{(D)}\ \frac{a+mn}{40}\qquad\textbf{(E)}\ \frac{a+40mn}{40}$

Solution

The train will take $\frac{a}{40}$ hours to travel $a$ miles, and it takes $\frac{nm}{60}$. The LCM of $40$ and $60$ is $120$, which allows for the addition of the fractions $\frac{3a}{120}$ and $\frac{2mn}{120}$. The end result is $\boxed{\textbf{(A)}\frac{3a+2mn}{120}}$.

See Also

1955 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Problem 1
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All AHSME Problems and Solutions


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