Difference between revisions of "2016 AMC 8 Problems/Problem 4"
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== Problem == | == Problem == | ||
− | When Cheenu was a boy, he could | + | When Cheenu was a boy, he could walk <math>15</math> miles in <math>3</math> hours and <math>30</math> minutes. As an old man, he can now walk <math>10</math> miles in <math>4</math> hours. How many minutes longer does it take for him to walk a mile now compared to when he was a boy? |
<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> |
Revision as of 19:59, 8 January 2024
Problem
When Cheenu was a boy, he could walk 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
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.
Video Solution (THINKING CREATIVELY!!!)
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Video Solution
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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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
Another Solution From the question, the old man can travel five miles in two hours, so we can set both speeds to 15 miles. We can see that Cheenu as an old man takes 2hrs 30mins for him to travel 15 miles, which is also 150 minutes. We can then divide this by fifteen, which gives us 10, so the answwer is B) 10