Difference between revisions of "2014 AMC 10B Problems/Problem 3"
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==Problem== | ==Problem== | ||
| − | Randy drove the first third of his trip on a gravel road, the next <math>20</math> miles on pavement, and the remaining one-fifth on a dirt road. In miles how long was Randy's trip? | + | Randy drove the first third of his trip on a gravel road, the next <math>20</math> miles on pavement, and the remaining one-fifth on a dirt road. In miles, how long was Randy's trip? |
<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 1== | ==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 = \ | + | 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 = \boxed{{\textbf{(E) }\dfrac{300}{7}}}</math> |
==Solution 2== | ==Solution 2== | ||
The first third of his distance added to the last one-fifth of his distance equals <math>\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</math>. | The first third of his distance added to the last one-fifth of his distance equals <math>\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</math>. | ||
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| + | ==Video Solution (CREATIVE THINKING)== | ||
| + | https://youtu.be/Aaa8O3ko6pQ | ||
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| + | ~Education, the Study of Everything | ||
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| + | ==Video Solution== | ||
| + | https://youtu.be/C4zRNsTVMtY | ||
| + | |||
| + | ~savannahsolver | ||
==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}} | ||
Latest revision as of 14:09, 20 March 2024
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 
. Therefore, 
of his distance is . Let be his total distance, and solve for . Therefore, is equal to 
, or 
.
Video Solution (CREATIVE THINKING)
~Education, the Study of Everything
Video Solution
~savannahsolver
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. 
