Difference between revisions of "2014 AMC 10A Problems/Problem 15"
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==Solution== | ==Solution== | ||
| + | Note that he drives at <math>50</math> miles per hour after the first hour and continues doing so until he arrives. | ||
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| + | Let <math>d</math> be the distance still needed to travel after the first <math>1</math> hour. We have that <math>\dfrac{d}{50}+1.5=\dfrac{d}{35}</math>, where the <math>1.5</math> comes from <math>1</math> hour late decreased to <math>0.5</math> hours early. | ||
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| + | Simplifying gives <math>7d+525=10d</math>, or <math>d=175</math>. | ||
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| + | Now, we must add an extra <math>35</math> miles traveled in the first hour, giving a total of <math>\textbf{(C) }210</math> miles. | ||
==See Also== | ==See Also== | ||
Revision as of 19:01, 7 February 2014
Problem
David drives from his home to the airport to catch a flight. He drives 
miles in the first hour, but realizes that he will be 
hour late if he continues at this speed. He increases his speed by 
miles per hour for the rest of the way to the airport and arrives 
minutes early. How many miles is the airport from his home?
Solution
Note that he drives at 
miles per hour after the first hour and continues doing so until he arrives.
Let 
be the distance still needed to travel after the first hour. We have that 
, where the 
comes from hour late decreased to 
hours early.
Simplifying gives 
, or 
.
Now, we must add an extra miles traveled in the first hour, giving a total of 
miles.
See Also
| 2014 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 14 |
Followed by Problem 16 | |
| 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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