AMC 10 · 2005 · #6

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 and two riders set out toward each other. When they meet, one has ridden twice as long as the other and at four-fifths of the other's speed. Find how far the faster rider has ridden.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

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

Naming a speed and a time makes one distance their product.

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

2STEP 2

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

The other rider's distance is a plain multiple of that, so the letters cancel.

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

3STEP 3

Add the two distances to close the 13-mile gap

Riding toward each other, the two distances fill the whole gap.

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

4STEP 4

Solve for Mike's distance

Solving gives 5, and the two distances add back to 13, choice (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

More like this