AMC 10 · 2010 · #1

Grade 6 arithmetic
Archetype Arithmetic Expression Decomposition └ Sub-archetype Fraction / Expression Sum Evaluation
mean-median-mode-rangefraction-arithmeticdecimal-arithmetic identify-subproblems ↑ Prerequisites: mean-median-mode-range
📏 Medium solution 💡 1 insight
Problem
A shelf holds five books whose widths are 6, 1/2, 1, 2.5, and 10 centimeters. Find the average (mean) book width, in centimeters.

Pick an answer.

(A)
$\ 1$
(B)
$\ 2$
(C)
$\ 3$
(D)
$\ 4$
(E)
$\ 5$

AMC 10 2010 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Identify Subproblems

Tool #7 (Subproblems): an average is really two smaller jobs — first add all the widths into one total, then divide that total by how many books there are. Tool #15 (Organize in More Ways): the widths are a mix of a fraction and a decimal, so rewrite 1/2 as 0.5 to put everything in decimals and add without confusion. Tool #3 (Eliminate): a quick check that the total divides evenly and that the mean sits between the smallest and largest widths pins down the single correct choice.

1STEP 1

Recall what average means

Average is one job split in two: add all five widths into a total, then divide by the number of books.

average = (total width)/(number of books)
2STEP 2

Turn the fraction into a decimal

The widths mix a fraction and a decimal, so rewrite 1/2 as 0.5 — now all five line up as decimals.

6, 1/2 = 0.5, 1, 2.5, 10
3STEP 3

Add the five widths

Group the friendly pairs: 6 + 10 = 16 and 0.5 + 2.5 = 3, plus the leftover 1 — the total is 20 centimeters.

6 + 0.5 + 1 + 2.5 + 10 = 20
4STEP 4

Divide by the book count

Divide the total by the number of books: 20 ÷ 5 = 4 centimeters, the average width — choice (D).

20/5 = 4 → (D)
5STEP 5

Check the mean

Check it: 4 × 5 = 20 matches the total, and 4 sits between the smallest width 0.5 and the largest 10.

4 × 5 = 20 ✓, 0.5 ＜ 4 ＜ 10
Answer
4
The five widths add to 20 cm and 20 ÷ 5 = 4, and multiplying back 4 × 5 = 20 confirms it. The mean 4 lies between the smallest width 0.5 and the largest 10, as any average must. Choices (A) 1, (B) 2, (C) 3 are too small to balance the large 10, and (E) 5 would need a total of 25, which is too high — only (D) 4 fits.
💡Key takeaway

To average, add everything up then split it evenly by how many you have: 20 cm shared among 5 books is 4 cm each.

  • Recall what average means
  • Turn the fraction into a decimal
  • Add the five widths
  • Divide by the book count
  • Check the mean

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