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

Mary's top book shelf holds five books with the following widths, in centimeters: $6$ , $\dfrac{1}{2}$ , $1$ , $2.5$ , and $10$ . What is the average 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.

Try it yourself first — the explanation is most useful after you’ve attempted it.

View mode:

Toolkit + CCSS Solution

Understand

Restated: Five books have widths $6$, $\dfrac{1}{2}$, $1$, $2.5$, and $10$ centimeters. Find the average (mean) width.

Givens: The five widths in cm: $6$, $\dfrac{1}{2}$, $1$, $2.5$, $10$; There are $5$ books; Answer choices (in cm): (A) $1$, (B) $2$, (C) $3$, (D) $4$, (E) $5$

Unknowns: The average book width in centimeters

Understand

Restated: Five books have widths $6$, $\dfrac{1}{2}$, $1$, $2.5$, and $10$ centimeters. Find the average (mean) width.

Givens: The five widths in cm: $6$, $\dfrac{1}{2}$, $1$, $2.5$, $10$; There are $5$ books; Answer choices (in cm): (A) $1$, (B) $2$, (C) $3$, (D) $4$, (E) $5$

Plan

Primary tool: #7 Identify Subproblems

Secondary: #15 Organize Information in More Ways, #3 Eliminate Possibilities

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 $\tfrac{1}{2}$ 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.

Execute — Answer: D

#7 Identify Subproblems 6.SP.A.3 Step 1

Recall what "average" means: add up all the widths to get one total, then divide that total by the number of books.
So the whole problem splits into an add step and a divide step.

$$\text{average} = \dfrac{\text{total width}}{\text{number of books}}$$

💡 An average is the one number that would replace every book if they all had the same width.

#15 Organize Information in More Ways 5.NBT.B.7 Step 2

The widths mix a fraction and a decimal.
Rewrite $\dfrac{1}{2}$ as $0.5$ so all five numbers are decimals and line up cleanly for adding.

$$6,\ \tfrac{1}{2} = 0.5,\ 1,\ 2.5,\ 10$$

💡 Putting every value in the same form removes any guesswork when you stack them up.

#7 Identify Subproblems 5.NBT.B.7 Step 3

Add the five widths.
Group the easy ones first: $6 + 10 = 16$, then $0.5 + 2.5 = 3$, then add the leftover $1$.
The total is $16 + 3 + 1 = 20$ centimeters.

$$6 + 0.5 + 1 + 2.5 + 10 = 20$$

💡 Pairing numbers that make round totals keeps the addition quick and error-free.

#7 Identify Subproblems 6.SP.B.5 Step 4

Divide the total by the number of books to get the mean: $20 \div 5 = 4$.
So the average book width is $4$ centimeters, which is choice (D).

$$\dfrac{20}{5} = 4 \;\Rightarrow\; \textbf{(D)}$$

💡 Splitting the total evenly among the 5 books gives each an equal share — that share is the average.

#3 Eliminate Possibilities 3.OA.C.7 Step 5

Check the answer.
Multiply back: $4 \times 5 = 20$, matching the total, so the division was right.
Also $4$ sits between the smallest width $0.5$ and the largest $10$, exactly where an average should land.
This rules out every other choice.

$$4 \times 5 = 20 \checkmark,\quad 0.5 < 4 < 10$$

💡 A correct average must undo the division when you multiply back, and must lie between the extreme values.

[1] #7 6.SP.A.3 Recall what "average" means: add up all the widths to get one total, then divide

[2] #15 5.NBT.B.7 The widths mix a fraction and a decimal. Rewrite $\dfrac{1}{2}$ as $0.5$ so all

[3] #7 5.NBT.B.7 Add the five widths. Group the easy ones first: $6 + 10 = 16$, then $0.5 + 2.5 =

[4] #7 6.SP.B.5 Divide the total by the number of books to get the mean: $20 \div 5 = 4$. So the

[5] #3 3.OA.C.7 Check the answer. Multiply back: $4 \times 5 = 20$, matching the total, so the d

Review

Reasonableness: The five widths add to $20$ cm and $20 \div 5 = 4$, and multiplying back $4 \times 5 = 20$ 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.

Alternative: Instead of adding then dividing, imagine leveling the books. The $10$ is $6$ above a guess of $4$ and the $6$ is $2$ above; together $+8$. The $0.5$, $1$, and $2.5$ sit $3.5 + 3 + 1.5 = 8$ below $4$. The surplus $+8$ and shortfall $-8$ cancel exactly, confirming the balance point — the mean — is $4$.

CCSS standards used (min grade 6)

6.SP.A.3 Recognize that a measure of center summarizes all its values with a single number (Understanding the average as one number (sum divided by count) that stands in for all five widths.)
5.NBT.B.7 Add, subtract, multiply, and divide decimals to hundredths (Rewriting $\tfrac{1}{2}$ as $0.5$ and adding the five decimal widths to get $20$.)
6.SP.B.5 Summarize numerical data sets by reporting number of observations and measures (Forming the mean by dividing the total width $20$ by the $5$ books.)
3.OA.C.7 Fluently multiply and divide within 100 (Computing $20 \div 5 = 4$ and checking it with $4 \times 5 = 20$.)

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

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: D

Review

CCSS standards used (min grade 6)

More like this