Difference between revisions of "2014 AMC 10B Problems/Problem 3"
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==Solution== | ==Solution== | ||
+ | Let the total distance be <math>x</math>. We have <math>\dfrac{x}{3} + 20 + \dfrac{x}{5} = x</math>, or <math>\dfrac{8x}{15} + 20 = x</math>. Subtracting <math>\dfrac{8x}{15}</math> from both sides gives us <math>20 = \dfrac{7x}{15}</math>. Multiplying by <math>\dfrac{15}{7}</math> gives us <math>x = \fbox{E) \dfrac{300}{7}}</math>. | ||
==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 12:16, 20 February 2014
Problem 3
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
Let the total distance be . We have , or . Subtracting from both sides gives us . Multiplying by gives us $x = \fbox{E) \dfrac{300}{7}}$ (Error compiling LaTeX. ! Missing $ inserted.).
See Also
2014 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
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All AMC 10 Problems and Solutions |
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