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Difference between revisions of "2005 AMC 10A Problems/Problem 7"

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Josh and Mike live <math>13</math> miles apart. Yesterday Josh started to ride his bicycle toward Mike's house. A little later Mike started to ride his bicycle toward Josh's house. When they met, Josh had ridden for twice the length of time as Mike and at four-fifths of Mike's rate. How many miles had Mike ridden when they met?  
 
Josh and Mike live <math>13</math> miles apart. Yesterday Josh started to ride his bicycle toward Mike's house. A little later Mike started to ride his bicycle toward Josh's house. When they met, Josh had ridden for twice the length of time as Mike and at four-fifths of Mike's rate. How many miles had Mike ridden when they met?  
  
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<math> \mathrm{(A) \ } 4\qquad \mathrm{(B) \ } 5\qquad \mathrm{(C) \ } 6\qquad \mathrm{(D) \ } 7\qquad \mathrm{(E) \ } 8 </math>
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<math> \textbf{(A) } 4\qquad \textbf{(B) } 5\qquad \textbf{(C) } 6\qquad \textbf{(D) } 7\qquad \textbf{(E) } 8 </math>
  
 
==Solution==
 
==Solution==
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Since Josh rode for twice the length of time as Mike and at four-fifths of Mike's rate, he rode <math>2\cdot\frac{4}{5}\cdot m = \frac{8}{5}m</math> miles.  
 
Since Josh rode for twice the length of time as Mike and at four-fifths of Mike's rate, he rode <math>2\cdot\frac{4}{5}\cdot m = \frac{8}{5}m</math> miles.  
  
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Since their combined distance was <math>13</math> miles,  
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Now, as their combined distance was <math>13</math> miles, we deduce that <math>\frac{8}{5}m + m = 13 \iff \frac{13}{5}m = 13 \iff m = \boxed{\textbf{(B) }5}</math>.
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<math> \frac{8}{5}m + m = 13 </math>
 
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<math> \frac{13}{5}m = 13 </math>
 
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<math> m = 5 \Longrightarrow \mathrm{(B)} </math>
 
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CHECK OUT Video Solution: https://youtu.be/WIR8yPLET9Y
 
  
 
==See also==
 
==See also==

Latest revision as of 17:05, 1 July 2025

Problem

Josh and Mike live $13$ miles apart. Yesterday Josh started to ride his bicycle toward Mike's house. A little later Mike started to ride his bicycle toward Josh's house. When they met, Josh had ridden for twice the length of time as Mike and at four-fifths of Mike's rate. How many miles had Mike ridden when they met?

$\textbf{(A) } 4\qquad \textbf{(B) } 5\qquad \textbf{(C) } 6\qquad \textbf{(D) } 7\qquad \textbf{(E) } 8$

Solution

Let $m$ be the distance in miles that Mike rode.

Since Josh rode for twice the length of time as Mike and at four-fifths of Mike's rate, he rode $2\cdot\frac{4}{5}\cdot m = \frac{8}{5}m$ miles.

Now, as their combined distance was $13$ miles, we deduce that $\frac{8}{5}m + m = 13 \iff \frac{13}{5}m = 13 \iff m = \boxed{\textbf{(B) }5}$.

See also

2005 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
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 AMC 10 Problems and Solutions

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