Difference between revisions of "2016 AMC 8 Problems/Problem 4"

Revision as of 05:22, 17 January 2023

Problem

When Cheenu was a boy, he could run $15$ miles in $3$ hours and $30$ minutes. As an old man, he can now walk $10$ miles in $4$ hours. How many minutes longer does it take for him to walk a mile now compared to when he was a boy?

$\textbf{(A) }6\qquad\textbf{(B) }10\qquad\textbf{(C) }15\qquad\textbf{(D) }18\qquad \textbf{(E) }30$

Solution

When Cheenu was a boy, he could run $15$ miles in $3$ hours and $30$ minutes $= 3\times60 + 30$ minutes $= 210$ minutes, thus running $\frac{210}{15} = 14$ minutes per mile. Now that he is an old man, he can walk $10$ miles in $4$ hours $= 4 \times 60$ minutes $= 240$ minutes, thus walking $\frac{240}{10} = 24$ minutes per mile. Therefore it takes him $\boxed{\textbf{(B)}\ 10}$ minutes longer to walk a mile now compared to when he was a boy.

Video Solution

https://youtu.be/gKjWxvyNSjk

~savannahsolver

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.

@@ Line 9: / Line 9: @@
 ==Video Solution==
-https://www.youtube.com/watch?v=LqnQQcUVJmA (has questions 1-5)
 https://youtu.be/gKjWxvyNSjk

Page

Toolbox

Search

Difference between revisions of "2016 AMC 8 Problems/Problem 4"

Revision as of 05:22, 17 January 2023

Contents

Problem

Solution

Video Solution

See Also