AMC 10 · 2014 · #3

Grade 6 rate-ratio
linear-equations-one-varfraction-arithmetic convert-to-algebra ↑ Prerequisites: fraction-arithmetic
📏 Medium solution 💡 2 insights
Problem
Two stretches are fractions of the whole trip and one is a fixed length. Find the total.

Pick an answer.

(A)
30
(B)
$\frac{400}{11}$
(C)
$\frac{75}{2}$
(D)
40
(E)
$\frac{300}{7}$
How to solve
Strategy Introduce a Variable

The whole trip is the one number everything else is measured against, so Tool #4 (Introduce a Variable) names it x and turns "one-third" and "one-fifth" into x/3 and x/5. Tool #7 (Identify Subproblems) splits the job cleanly: first combine the two fraction parts, then see what fraction the pavement must fill. Tool #13 (Convert to Algebra) writes the sentence "the parts add to the whole" as one equation, so the 20 miles pins down x instead of floating free.

1STEP 1

Name the whole trip

One letter names the whole trip.

gravel = x/3, pavement = 20, dirt = x/5
2STEP 2

Combine the two fraction parts

The two fractions combine to 8/15.

1/3 + 1/5 = 5/15 + 3/15 = 8/15
3STEP 3

Find the pavement's share

What is left is the fixed stretch.

1 - 8/15 = 7/15, so 7/15x = 20
4STEP 4

Solve for the total

Solving gives 300/7, choice (E).

x = 20 · 15/7 = 300/7 → (E)
Answer
300/7
Plug x = 300/7 back in: gravel = 1/3·300/7 = 100/7 and dirt = 1/5·300/7 = 60/7. Their sum is 160/7, and adding the 20 = 140/7 miles of pavement gives 300/7, exactly the whole trip. The value 300/7 ≈ 42.9 miles is a bit larger than 40, which fits: the pavement is only 7/15 (under half) of the trip, so the whole trip must be more than twice 20.
💡Key takeaway

The 20 miles is whatever fraction of the trip the two fractions leave over, so find that leftover fraction first.

  • Name the whole trip
  • Combine the two fraction parts
  • Find the pavement's share
  • Solve for the total