AMC 10 · 2014 · #11

Grade 7 rate-ratio
ratelinear-equations-one-var convert-to-algebradimensional-analysis ↑ Prerequisites: rate
📏 Medium solution 💡 2 insights
Problem
Keeping the first speed would arrive late, and speeding up arrives early. Find the whole distance.

Pick an answer.

(A)
140
(B)
175
(C)
210
(D)
245
(E)
280
How to solve
Strategy Introduce a Variable

The whole problem turns on the distance still left after the first hour, so Tool #4 (Introduce a Variable) names that leftover distance d and lets me write its travel time at each speed. Tool #8 (Analyze the Units) keeps the rates honest: miles divided by miles-per-hour gives hours, so d/35 and d/50 are times I can compare. Tool #13 (Convert to Algebra) turns the phrase "from 1 hour late to 30 minutes early" into a single equation about saved time.

1STEP 1

Read off the two speeds

The second speed is the first plus 15.

35+15=50 mph
2STEP 2

Name the leftover distance

One letter names the distance still to go.

t₃₅=d/35, t₅₀=d/50
3STEP 3

Turn the timing into an equation

The two timings differ by 1.5 hours.

d/35-d/50=1.5
4STEP 4

Solve for the leftover distance

Solving gives the leftover distance 175.

3d/350=3/2 → d=175
5STEP 5

Add back the first hour

Adding the first hour gives 210, choice (C).

35+175=210 → (C)
Answer
210
Check 210 directly. After the first 35 miles, 175 miles remain. At 50 mph that takes 175/50=3.5 hours, so the whole trip is 1+3.5=4.5 hours. At the old 35 mph the leftover would take 175/35=5 hours, a total of 6 hours; being 1 hour late means the flight time is 5 hours after leaving. Arriving in 4.5 hours is exactly 0.5 hour before 5 — that is 30 minutes early, just as stated. Everything lines up, so (C) is correct.
💡Key takeaway

Going from late to early just means saved time, so compare how long the leftover road takes at each speed, solve for that leftover distance, then add back the first-hour miles.

  • Read off the two speeds
  • Name the leftover distance
  • Turn the timing into an equation
  • Solve for the leftover distance
  • Add back the first hour