AMC 10 · 2005 · #7
Grade 7 rate-ratioJosh 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?
Pick an answer.
AMC 10 2005 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: Two houses are $13$ miles apart. One rider (Josh) leaves toward the other's house; a bit later the other rider (Mike) leaves toward Josh's house. When they meet, Josh has been riding for twice as long as Mike and at four-fifths of Mike's speed. Find how many miles Mike has ridden at the moment they meet.
Givens: The two houses are $13$ miles apart; The riders travel toward each other, so together they cover the whole $13$ miles when they meet; Josh's riding time is twice Mike's riding time; Josh's speed is $\tfrac{4}{5}$ of Mike's speed; Answer choices: (A) $4$, (B) $5$, (C) $6$, (D) $7$, (E) $8$
Unknowns: The number of miles Mike has ridden when the two riders meet
Understand
Restated: Two houses are $13$ miles apart. One rider (Josh) leaves toward the other's house; a bit later the other rider (Mike) leaves toward Josh's house. When they meet, Josh has been riding for twice as long as Mike and at four-fifths of Mike's speed. Find how many miles Mike has ridden at the moment they meet.
Givens: The two houses are $13$ miles apart; The riders travel toward each other, so together they cover the whole $13$ miles when they meet; Josh's riding time is twice Mike's riding time; Josh's speed is $\tfrac{4}{5}$ of Mike's speed; Answer choices: (A) $4$, (B) $5$, (C) $6$, (D) $7$, (E) $8$
Plan
Primary tool: #4 Introduce a Variable
Secondary: #8 Analyze the Units, #13 Convert to Algebra
The problem seems to hide three unknowns — Mike's speed, Mike's time, and the distances — but a single well-chosen variable collapses it. Tool #4 (Introduce a Variable) lets us call Mike's distance $m$; the pair 'twice the time' and 'four-fifths the speed' then pins Josh's distance to $m$ without ever needing the separate speed or time. Tool #8 (Analyze the Units) is what makes that pinning legal: distance $=$ speed $\times$ time, so a time factor of $2$ and a speed factor of $\tfrac{4}{5}$ multiply into a distance factor of $\tfrac{8}{5}$. Tool #13 (Convert to Algebra) then turns 'the two distances add to $13$' into one equation in $m$ that solves in a line.
Execute — Answer: B
6.EE.B.6 Step 1 Name Mike's distance and recall distance = speed × time
- Let $m$ be the number of miles Mike rides.
- For any rider, distance equals speed times time.
- Write Mike's speed as $r$ and Mike's time as $t$, so $m = r\,t$.
- We will not need the actual values of $r$ and $t$; naming them just lets us describe Josh in the same language before everything cancels.
💡 Give the thing you want its own name first, then describe everything else in terms of it.
7.RP.A.2 Step 2 Write Josh's distance as a multiple of Mike's
- Josh rides for twice Mike's time, which is $2t$, and at four-fifths of Mike's speed, which is $\tfrac{4}{5}r$.
- His distance is speed times time: $\left(\tfrac{4}{5}r\right)(2t) = \tfrac{8}{5}\,r\,t$.
- But $r\,t$ is exactly Mike's distance $m$, so Josh rides $\tfrac{8}{5}m$ miles.
- The unknown $r$ and $t$ never appear alone — only their product, which is $m$.
💡 Because distance is speed times time, scaling the time by $2$ and the speed by $\tfrac{4}{5}$ just scales the distance by $2\cdot\tfrac{4}{5}=\tfrac{8}{5}$.
6.EE.B.7 Step 3 Add the two distances to close the 13-mile gap
- The riders head straight at each other, so when they meet their two distances together make up the full $13$ miles between the houses.
- Mike rode $m$ and Josh rode $\tfrac{8}{5}m$, so $m + \tfrac{8}{5}m = 13$.
- Writing $m$ as $\tfrac{5}{5}m$ and adding gives $\tfrac{13}{5}m = 13$.
💡 Meeting head-on means the pieces they each ride must fill the whole distance between them.
6.EE.B.7 Step 4 Solve for Mike's distance
- The equation $\tfrac{13}{5}m = 13$ has the form $p\,m = q$.
- Multiply both sides by $\tfrac{5}{13}$ (or divide by $\tfrac{13}{5}$): $m = 13 \cdot \tfrac{5}{13} = 5$.
- So Mike rode $5$ miles, which is choice (B).
- As a check, Josh then rode $\tfrac{8}{5}\cdot 5 = 8$ miles, and $5 + 8 = 13$.
💡 One clean division undoes the fraction in front of $m$ and reveals the distance.
6.EE.B.6 Let $m$ be the number of miles Mike rides. For any rider, distance equals speed 7.RP.A.2 Josh rides for twice Mike's time, which is $2t$, and at four-fifths of Mike's sp 6.EE.B.7 The riders head straight at each other, so when they meet their two distances to 6.EE.B.7 The equation $\tfrac{13}{5}m = 13$ has the form $p\,m = q$. Multiply both sides Review
Reasonableness: Mike is faster and rides for a shorter time; Josh is slower but rides much longer, so Josh should cover the larger share of the $13$ miles. Our answer gives Mike $5$ miles and Josh $8$ miles, which indeed makes Josh's share bigger and sums to exactly $13$. Mike's $5$ miles is less than half of $13$, matching the fact that his combined 'twice the time at four-fifths the speed' disadvantage means Josh out-distances him. A tempting wrong answer is $8$, which is Josh's distance, not Mike's.
Alternative: Reason with a distance ratio directly. Distance is proportional to speed times time, so the ratio of Josh's distance to Mike's is $\tfrac{4}{5}\times 2 = \tfrac{8}{5}$, i.e. Mike : Josh $= 5 : 8$. These parts sum to $5 + 8 = 13$ parts, and the real total is $13$ miles, so each part is exactly $1$ mile. Mike's $5$ parts are therefore $5$ miles, again choice (B).
CCSS standards used (min grade 7)
6.EE.B.6Use variables to represent numbers and write expressions to solve problems (Naming Mike's distance $m$ and his speed and time $r$, $t$ so that Josh can be described in the same terms.)7.RP.A.2Recognize and represent proportional relationships between quantities (Combining the time factor $2$ and the speed factor $\tfrac{4}{5}$ into a single distance factor $\tfrac{8}{5}$, so Josh's distance is $\tfrac{8}{5}m$.)6.EE.B.7Solve real-world problems by writing and solving equations of the form px = q (Writing $m + \tfrac{8}{5}m = 13$, reducing it to $\tfrac{13}{5}m = 13$, and solving for $m = 5$.)
⭐ Name the distance you want, use distance = speed × time to turn 'twice the time at four-fifths the speed' into one number ($\tfrac{8}{5}$), and the head-on total of $13$ miles solves in a single step.
⭐ Name the distance you want, use distance = speed × time to turn 'twice the time at four-fifths the speed' into one number ($\tfrac{8}{5}$), and the head-on total of $13$ miles solves in a single step.
More like this
Same archetype — closest grade level first.