Difference between revisions of "2014 AMC 10B Problems/Problem 8"

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==Solution==
 
==Solution==
  
Converting feet to yards and minutes to second, we see that the truck travels <math>\dfrac{b}{18}</math> yards every <math>t</math> seconds for <math>180</math> seconds. We see that he does <math>\dfrac{180}{t}</math> cycles of <math>\dfrac{b}{18}</math> yards. Multiplying, we get <math>\dfrac{180b}{18t}</math>, or <math>(E)\dfrac{10b}{t}</math>
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Converting feet to yards and minutes to second, we see that the truck travels <math>\dfrac{b}{18}</math> yards every <math>t</math> seconds for <math>180</math> seconds. We see that he does <math>\dfrac{180}{t}</math> cycles of <math>\dfrac{b}{18}</math> yards. Multiplying, we get <math>\dfrac{180b}{18t}</math>, or <math>\dfrac{10b}{t}</math>, or <math>\boxed{(E)}</math>
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2014|ab=B|num-b=7|num-a=9}}
 
{{AMC10 box|year=2014|ab=B|num-b=7|num-a=9}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 15:16, 20 February 2014

Problem

A truck travels $\dfrac{b}{6}$ feet every $t$ seconds. There are $3$ feet in a yard. How many yards does the truck travel in $3$ minutes?

$\textbf {(A) } \frac{b}{1080t} \qquad \textbf {(B) } \frac{30t}{b} \qquad \textbf {(C) } \frac{30b}{t}\qquad \textbf {(D) } \frac{10t}{b} \qquad \textbf {(E) } \frac{10b}{t}$

Solution

Converting feet to yards and minutes to second, we see that the truck travels $\dfrac{b}{18}$ yards every $t$ seconds for $180$ seconds. We see that he does $\dfrac{180}{t}$ cycles of $\dfrac{b}{18}$ yards. Multiplying, we get $\dfrac{180b}{18t}$, or $\dfrac{10b}{t}$, or $\boxed{(E)}$

See Also

2014 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 7
Followed by
Problem 9
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
All AMC 10 Problems and Solutions

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