Difference between revisions of "2014 AMC 10B Problems/Problem 3"
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<math> \textbf {(A) } 30 \qquad \textbf {(B) } \frac{400}{11} \qquad \textbf {(C) } \frac{75}{2} \qquad \textbf {(D) } 40 \qquad \textbf {(E) } \frac{300}{7}</math> | <math> \textbf {(A) } 30 \qquad \textbf {(B) } \frac{400}{11} \qquad \textbf {(C) } \frac{75}{2} \qquad \textbf {(D) } 40 \qquad \textbf {(E) } \frac{300}{7}</math> | ||
− | ==Solution== | + | ==Solution 1== |
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>. | 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>. | ||
Revision as of 20:35, 11 March 2014
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 8/15. Therefore, 7/15 of his distance is 20. Let x be his total distance, and solve for x. Therefore, x is equal to 300/7, or 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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