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>\dfrac{10b}{t}</math>, or<math>\boxed{\textbf{(E)}}</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{\textbf{(E)}}</math>.
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==Solution 2==
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We set a proportion by letting the <math>x</math> being the number of feet the truck travels in <math>3</math> minutes.
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<math>\frac{\frac{b}{6}}{t}=\frac{x}{180}</math>
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<math>\frac{b}{6t}=\frac{x}{180}</math>
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 +
<math>\frac{180b}{6t}=x</math>
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 +
<math>x=\frac{30b}{t}</math>
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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.
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<math>\frac{x}{3}=\frac{10b}{t}</math>, which corresponds to <math>\boxed{\text{(E)}}</math>
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==Video Solution==
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https://youtu.be/j5pWLnlQkwE
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~savannahsolver
  
 
==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}}

Latest revision as of 10:25, 16 July 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 $\boxed{\text{(E)}}$

Video Solution

https://youtu.be/j5pWLnlQkwE

~savannahsolver

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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