AMC 10 · 2005 · #7

Grade 7 rate-ratio

Archetype Meeting Time with Piecewise Rates └ Sub-archetype Constant-Rate Closing Speed

ratelinear-equations-one-varratio-proportion convert-to-algebradimensional-analysis ↑ Prerequisites: rateratio-proportionfraction-arithmetic

📏 Medium solution 💡 2 insights

Problem

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.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 10 2005 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Introduce a Variable

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 × time, so a time factor of 2 and a speed factor of 4/5 multiply into a distance factor of 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.

1STEP 1

Name Mike's distance and recall distance = speed × time

Let m be Mike's distance in miles. Distance = speed × time, so with Mike's speed r and time t we have m = r t.

m = r t (Mike: speed r, time t)

2STEP 2

Write Josh's distance as a multiple of Mike's

Josh rides 2t at 4/5r, so his distance is 8/5 rt — that is 8/5m, since r and t survive only as their product.

Josh = (4/5r)(2t) = 8/5 r t = 8/5m

3STEP 3

Add the two distances to close the 13-mile gap

Riding head-on, their two distances fill the whole 13 miles: m + 8/5m = 13/5m = 13.

m + 8/5m = 5/5m + 8/5m = 13/5m = 13

4STEP 4

Solve for Mike's distance

Multiply both sides by 5/13 to get m = 5 miles. Check: Josh rode 8 miles, and 5 + 8 = 13, so the answer is (B).

m = 13 · 5/13 = 5 → (B)

Answer

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.

💡Key takeaway

Name the distance you want, use distance = speed × time to turn 'twice the time at four-fifths the speed' into one number (8/5), and the head-on total of 13 miles solves in a single step.

Name Mike's distance and recall distance = speed × time
Write Josh's distance as a multiple of Mike's
Add the two distances to close the 13-mile gap
Solve for Mike's distance

Name Mike's distance and recall distance = speed × time

Write Josh's distance as a multiple of Mike's

Add the two distances to close the 13-mile gap

Solve for Mike's distance

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