AMC 10 · 2007 · #2

Grade 6 rate-ratio
rateunit-conversiondimensional-analysis dimensional-analysisidentify-subproblems ↑ Prerequisites: rate
📏 Medium solution 💡 2 insights
Problem
Equal distances are driven out and back at two different fuel efficiencies. Find the average efficiency for the whole round trip.

Pick an answer.

(A)
22
(B)
24
(C)
25
(D)
26
(E)
28
How to solve
Strategy Analyze the Units

Miles per gallon is a rate — miles divided by gallons — so Tool #8 (Analyze the Units) tells you that overall mileage must be total miles over total gallons, not the average of 30 and 20. Tool #7 (Identify Subproblems) splits the work into finding the gas each leg burns before combining. Tool #3 (Eliminate Possibilities) flags the trap answer: naively averaging the two mileages gives (30+20)/2=25, which is choice (C) and is wrong.

1STEP 1

Gas used driving home

The outbound leg burns 4.

(120 mi)/(30 mi/gal) = 4 gal
2STEP 2

Gas used driving back

The return leg burns 6, more than the first.

(120 mi)/(20 mi/gal) = 6 gal
3STEP 3

Total miles and total gallons

Distances and fuel both add directly.

240 mi total, 4+6=10 gal total
4STEP 4

Average mileage for the round trip

Dividing gives 24, below the naive midpoint, choice (B).

(240 mi)/(10 gal) = 24 mi/gal → (B)
Answer
24
The answer 24 lands between the two mileages 20 and 30, which it must, and it sits below the plain midpoint 25. That is exactly right: the low-mileage return leg uses more gallons (6 versus 4), so it carries more weight and drags the average toward 20. A quick check confirms the totals: 4 gal+6 gal=10 gal carries the car 240 miles, and 240/10=24.
💡Key takeaway

Average mileage is all your miles divided by all your gallons, not the average of the two speeds — so the gas-guzzling leg pulls it down.

  • Gas used driving home
  • Gas used driving back
  • Total miles and total gallons
  • Average mileage for the round trip