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

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==Solution==
 
==Solution==
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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.
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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.
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~CHECKMATE2021
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==Video Solution==
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https://youtu.be/I9neY-xoG90?si=tDSd8my8W8Mb7Lqp
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 +
A solution so simple a 12-year-old made it!
 +
 
 +
~Elijahman~
 +
==Video Solution (THINKING CREATIVELY!!!)==
 +
https://youtu.be/r92zdVMTamI
 +
 
 +
~Education, the Study of Everything
  
 
==Video Solution==
 
==Video Solution==
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{{AMC8 box|year=2016|num-b=3|num-a=5}}
 
{{AMC8 box|year=2016|num-b=3|num-a=5}}
 
{{MAA Notice}}
 
{{MAA Notice}}
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Another Solution
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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

Latest revision as of 10:42, 20 June 2024

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.

~CHECKMATE2021

Video Solution

https://youtu.be/I9neY-xoG90?si=tDSd8my8W8Mb7Lqp

A solution so simple a 12-year-old made it!

~Elijahman~

Video Solution (THINKING CREATIVELY!!!)

https://youtu.be/r92zdVMTamI

~Education, the Study of Everything

Video Solution

https://youtu.be/gKjWxvyNSjk

~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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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

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