Difference between revisions of "2014 AMC 10B Problems/Problem 3"
Royalreter1 (talk | contribs) (→Problem) |
(→Solution) |
||
| Line 6: | Line 6: | ||
==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. Unknown error_msg).
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
