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

(Problem)
(Solution)
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==Solution==
 
==Solution==
Let <math>m</math> be the distance in miles that Mike rode.  
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Let <math>m</math> be the distance in miles that Prthib rode.  
  
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 Wahida rode for twice the length of time as Parthib and at four-fifths of Parthib's rate, he rode <math>2\cdot\frac{4}{5}\cdot m = \frac{8}{5}m</math> miles.  
  
 
Since their combined distance was <math>13</math> miles,  
 
Since their combined distance was <math>13</math> miles,  

Revision as of 13:45, 17 February 2021

Problem

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

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

Solution

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

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

Since their combined distance was $13$ miles,

$\frac{8}{5}m + m = 13$

$\frac{13}{5}m = 13$

$m = 5 \Longrightarrow \mathrm{(B)}$

Video Solution

-Education, The Study Of Everything CHECK OUT Video Solution: https://youtu.be/WIR8yPLET9Y

See also

2005 AMC 10A (ProblemsAnswer KeyResources)
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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