Difference between revisions of "2016 AMC 8 Problems/Problem 4"
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<math>\textbf{(A) }6\qquad\textbf{(B) }10\qquad\textbf{(C) }15\qquad\textbf{(D) }18\qquad \textbf{(E) }30</math> | <math>\textbf{(A) }6\qquad\textbf{(B) }10\qquad\textbf{(C) }15\qquad\textbf{(D) }18\qquad \textbf{(E) }30</math> | ||
| − | ==Solution== | + | == Solution 1 == |
When Cheenu was a boy, he could run <math>15</math> miles in <math>3</math> hours and <math>30</math> minutes <math>= 3\times60 + 30</math> minutes <math>= 210</math> minutes, thus running <math>\frac{210}{15} = 14</math> minutes per mile. Now that he is an old man, he can walk <math>10</math> miles in <math>4</math> hours <math>= 4 \times 60</math> minutes <math>= 240</math> minutes, thus walking <math>\frac{240}{10} = 24</math> minutes per mile. Therefore, it takes him <math>\boxed{\textbf{(B)}\ 10}</math> minutes longer to walk a mile now compared to when he was a boy. | When Cheenu was a boy, he could run <math>15</math> miles in <math>3</math> hours and <math>30</math> minutes <math>= 3\times60 + 30</math> minutes <math>= 210</math> minutes, thus running <math>\frac{210}{15} = 14</math> minutes per mile. Now that he is an old man, he can walk <math>10</math> miles in <math>4</math> hours <math>= 4 \times 60</math> minutes <math>= 240</math> minutes, thus walking <math>\frac{240}{10} = 24</math> minutes per mile. Therefore, it takes him <math>\boxed{\textbf{(B)}\ 10}</math> minutes longer to walk a mile now compared to when he was a boy. | ||
~CHECKMATE2021 | ~CHECKMATE2021 | ||
| − | ==Video Solution== | + | == Solution 2 == |
| + | From the question, the old man can travel five miles in two hours, so we can set both speeds to <math>15</math> miles. We can see that Cheenu as an old man takes <math>2</math> hours and <math>30</math> minutes for him to travel <math>15</math> miles, which is also <math>150</math> minutes. We can then divide this by fifteen, which gives us <math>10</math>, thus the answer is <math>\boxed{\textbf{B) 10}}</math> | ||
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| + | == Video Solution == | ||
https://youtu.be/I9neY-xoG90?si=tDSd8my8W8Mb7Lqp | https://youtu.be/I9neY-xoG90?si=tDSd8my8W8Mb7Lqp | ||
| − | + | ~Elijahman~ | |
| − | + | == Video Solution (THINKING CREATIVELY!!!) == | |
| − | ==Video Solution (THINKING CREATIVELY!!!)== | ||
https://youtu.be/r92zdVMTamI | https://youtu.be/r92zdVMTamI | ||
~Education, the Study of Everything | ~Education, the Study of Everything | ||
| − | ==Video Solution== | + | == Video Solution == |
https://youtu.be/gKjWxvyNSjk | https://youtu.be/gKjWxvyNSjk | ||
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~savannahsolver | ~savannahsolver | ||
| − | ==See Also== | + | == See Also == |
{{AMC8 box|year=2016|num-b=3|num-a=5}} | {{AMC8 box|year=2016|num-b=3|num-a=5}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
| − | + | [[Category:Introductory Algebra Problems]] | |
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Latest revision as of 18:00, 25 June 2025
Contents
Problem
When Cheenu was a boy, he could run 
miles in 
hours and 
minutes. As an old man, he can now walk 
miles in 
hours. How many minutes longer does it take for him to walk a mile now compared to when he was a boy?
Solution 1
When Cheenu was a boy, he could run miles in hours and minutes 
minutes 
minutes, thus running 
minutes per mile. Now that he is an old man, he can walk miles in hours 
minutes 
minutes, thus walking 
minutes per mile. Therefore, it takes him 
minutes longer to walk a mile now compared to when he was a boy.
~CHECKMATE2021
Solution 2
From the question, the old man can travel five miles in two hours, so we can set both speeds to miles. We can see that Cheenu as an old man takes 
hours and minutes for him to travel miles, which is also 
minutes. We can then divide this by fifteen, which gives us , thus the answer is 
Video Solution
https://youtu.be/I9neY-xoG90?si=tDSd8my8W8Mb7Lqp
~Elijahman~
Video Solution (THINKING CREATIVELY!!!)
~Education, the Study of Everything
Video Solution
~savannahsolver
See Also
| 2016 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 3 |
Followed by Problem 5 | |
| 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 AJHSME/AMC 8 Problems and Solutions | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
