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

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Remember <math>x</math> is the number of feet the truck travels, so we divide by <math>3</math> to convert to yards.  
 
Remember <math>x</math> is the number of feet the truck travels, so we divide by <math>3</math> to convert to yards.  
  
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<math>\frac{x}{3}=\frac{10b}{6}</math>, which corresponds to <math>\boxed{\text{(E)}}</math>
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<math>\frac{x}{3}=\frac{10b}{t}</math>, which corresponds to <math>\boxed{\text{(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 17:07, 10 January 2020

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 $\boxed{\textbf{(E)}}$.

Solution 2

We set a proportion by letting the $x$ being the number of feet the truck travels in $3$ minutes.

$\frac{\frac{b}{6}}{t}=\frac{x}{180}$

$\frac{b}{6t}=\frac{x}{180}$

$\frac{180b}{6t}=x$

$x=\frac{30b}{t}$

Remember $x$ is the number of feet the truck travels, so we divide by $3$ to convert to yards.

$\frac{x}{3}=\frac{10b}{t}$, which corresponds to $\boxed{\text{(E)}}$

See Also

2014 AMC 10B (Problems • Answer Key • Resources)
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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