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

(Solution)
(9 intermediate revisions by 8 users not shown)
Line 1: Line 1:
4. When Cheenu was a boy he could run <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?
+
== Problem ==
 +
 
 +
When Cheenu was a boy he could run <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>
  
 
==Solution==
 
==Solution==
When Cheenu was a boy, he could run <math>15</math> miles in <math>210</math> minutes, thus running <math>14</math> minutes per mile. When he is an old man, he can walk <math>10</math> miles in <math>240</math> minutes, thus walking <math>24</math> minutes per mile. Therefore it takes him <math>\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.
 +
 
 +
==Video Solution==
 +
 
 +
https://www.youtube.com/watch?v=LqnQQcUVJmA (has questions 1-5)
  
  
 
{{AMC8 box|year=2016|num-b=3|num-a=5}}
 
{{AMC8 box|year=2016|num-b=3|num-a=5}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 03:33, 16 January 2021

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://www.youtube.com/watch?v=LqnQQcUVJmA (has questions 1-5)


2016 AMC 8 (ProblemsAnswer KeyResources)
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. AMC logo.png