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

(added problem and solution)
 
m (Video Solution)
 
(18 intermediate revisions by 9 users not shown)
Line 2: Line 2:
 
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?  
  
<math> \mathrm{(A) \ } 4\qquad \mathrm{(B) \ } 5\qquad \mathrm{(C) \ } 6\qquad \mathrm{(D) \ } 7\qquad \mathrm{(E) \ } 8 </math>
+
<math> \mathrm{(A) \ } 2\qquad \mathrm{(B) \ } 5\qquad \mathrm{(C) \ } 3\qquad \mathrm{(D) \ } 6\qquad \mathrm{(E) \ } 4 </math>
  
 
==Solution==
 
==Solution==
Line 15: Line 15:
 
<math> \frac{13}{5}m = 13 </math>  
 
<math> \frac{13}{5}m = 13 </math>  
  
<math> m = 5 \Rightarrow B </math>
+
<math> m = 5 \Longrightarrow \mathrm{(B)} </math>
 
==See Also==
 
*[[2005 AMC 10A Problems]]
 
  
*[[2005 AMC 10A Problems/Problem 6|Previous Problem]]
+
==Video Solution==
 +
- Wrong link was added here
  
*[[2005 AMC 10A Problems/Problem 8|Next Problem]]
+
- No video link as of now
 +
 
 +
==See also==
 +
{{AMC10 box|year=2005|ab=A|num-b=6|num-a=8}}
 +
 
 +
{{MAA Notice}}

Latest revision as of 11:46, 22 July 2021

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?

$\mathrm{(A) \ } 2\qquad \mathrm{(B) \ } 5\qquad \mathrm{(C) \ } 3\qquad \mathrm{(D) \ } 6\qquad \mathrm{(E) \ } 4$

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.

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

- Wrong link was added here

- No video link as of now

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

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

Invalid username
Login to AoPS