Difference between revisions of "2014 AMC 10B Problems/Problem 8"
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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>. | 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== | ||
| + | |||
| + | We set a proportion by letting the <math>x</math> being the number of feet the truck travels in <math>3</math> minutes. | ||
| + | |||
| + | <math>\frac{\frac{b}{6}}{t}=\frac{x}{180}</math> | ||
| + | |||
| + | <math>\frac{b}{6t}=\frac{x}{180}</math> | ||
| + | |||
| + | <math>\frac{180b}{6t}=x</math> | ||
| + | |||
| + | <math>x=\frac{30b}{t}</math> | ||
| + | |||
| + | Remember <math>x</math> is the number of feet the truck travels, so we divide by <math>3</math> to convert to yards. | ||
| + | |||
| + | <math>\frac{x}{3}=\frac{10b}{6}</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 17:06, 10 January 2020
Contents
Problem
A truck travels 
feet every 
seconds. There are 
feet in a yard. How many yards does the truck travel in minutes?
Solution
Converting feet to yards and minutes to second, we see that the truck travels 
yards every seconds for 
seconds. We see that he does 
cycles of yards. Multiplying, we get 
, or 
, or 
.
Solution 2
We set a proportion by letting the 
being the number of feet the truck travels in minutes.
Remember is the number of feet the truck travels, so we divide by to convert to yards.
, which corresponds to 
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
