AMC 10 · 2010 · #7

Grade 7 rate-ratio
ratelinear-equations-one-varunit-conversion convert-to-algebra ↑ Prerequisites: rate
📏 Medium solution 💡 2 insights
Problem
A trip runs at one speed then a slower one, with the distance and total time known. Find the slow portion's time.

Pick an answer.

(A)
18
(B)
21
(C)
24
(D)
27
(E)
30
How to solve
Strategy Introduce a Variable

Two unknown time chunks share one total time and one total distance, so Tool #4 (Introduce a Variable) is the natural fit: name the rain time, write the sun time as "what's left," and turn the distance fact into one equation. Tool #8 (Analyze the Units) keeps the arithmetic honest — speeds are per hour but the answer is wanted in minutes, so units must be converted at the start and again at the end.

1STEP 1

Match the units

The total time becomes a fraction of an hour.

40 min = 40/60 hr = 2/3 hr
2STEP 2

Name the unknown and set up

One unknown names both stretches.

30(2/3 - r) + 20r = 16
3STEP 3

Solve for the rain time

The distances add to a single equation.

20 - 30r + 20r = 16 → 20 - 10r = 16 → 10r = 4 → r = 2/5 hr
4STEP 4

Convert back to minutes

Converting back gives 24 minutes, choice (C).

2/5 hr × 60 = 24 min → (C)
Answer
24
Check both totals. Rain time = 2/5 hr, so sun time = 2/3 - 2/5 = 10/15 - 6/15 = 4/15 hr. Distances: sun = 30 × 4/15 = 8 miles, rain = 20 × 2/5 = 8 miles, total 8 + 8 = 16 miles. Times in minutes: sun = 4/15 × 60 = 16 min, rain = 24 min, total 40 min. Both the distance and the time come out exactly right, and 24 minutes is choice (C).
💡Key takeaway

Name the rain time as one variable, make the sun time the leftover, and add up the distances — the whole race problem shrinks to a single Grade 7 equation.

  • Match the units
  • Name the unknown and set up
  • Solve for the rain time
  • Convert back to minutes