Difference between revisions of "2014 AMC 10B Problems/Problem 3"
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==Solution 2== | ==Solution 2== | ||
| − | The first third of his distance added to the last one-fifth of his distance equals 8 | + | The first third of his distance added to the last one-fifth of his distance equals \frac{8}{15}<math>. Therefore, </math>\frac{7}{15}<math> of his distance is </math>20<math>. Let </math>x<math> be his total distance, and solve for </math>x<math>. Therefore, </math>x<math> is equal to </math>\frac{300}{7}<math>, or </math>E$. |
==See Also== | ==See Also== | ||
{{AMC10 box|year=2014|ab=B|num-b=2|num-a=4}} | {{AMC10 box|year=2014|ab=B|num-b=2|num-a=4}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 14:05, 28 December 2015
Contents
Problem
Randy drove the first third of his trip on a gravel road, the next 
miles on pavement, and the remaining one-fifth on a dirt road. In miles how long was Randy's trip?
Solution 1
Let the total distance be 
. We have 
, or 
. Subtracting 
from both sides gives us 
. Multiplying by 
gives us 
.
Solution 2
The first third of his distance added to the last one-fifth of his distance equals \frac{8}{15}
\frac{7}{15}
20
x
xx
\frac{300}{7}
E$.
See Also
| 2014 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 2 |
Followed by Problem 4 | |
| 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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