AMC 8 · 2018 · #8

Grade 6 arithmetic

Archetype Arithmetic Expression Decomposition

mean-median-mode-rangegraph-readingfraction-arithmetic identify-subproblems ↑ Prerequisites: fraction-arithmeticmulti-digit-arithmetic

📏 Medium solution 💡 2 insights 📊 Diagram

Problem

Mr. Garcia asked the members of his health class how many days last week they exercised for at least 30 minutes. The results are summarized in the following bar graph, where the heights of the bars represent the number of students.

What was the mean number of days of exercise last week, rounded to the nearest hundredth, reported by the students in Mr. Garcia's class?

$\textbf{(A) } 3.50 \qquad \textbf{(B) } 3.57 \qquad \textbf{(C) } 4.36 \qquad \textbf{(D) } 4.50 \qquad \textbf{(E) } 5.00$

Pick an answer.

(A)

3.50

(B)

3.57

(C)

4.36

(D)

4.50

(E)

5.00

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Toolkit + CCSS Solution

Understand

Restated: Mr. Garcia's health class reported how many days each student exercised at least $30$ minutes last week. A bar graph shows, for each day-count $1$ through $7$, how many students gave that answer. Compute the mean (average) number of days per student, rounded to the nearest hundredth, and match to one of the five choices.

Givens: Frequency data from the bar graph: $1$ day $\to 1$ student, $2$ days $\to 3$, $3$ days $\to 2$, $4$ days $\to 6$, $5$ days $\to 8$, $6$ days $\to 3$, $7$ days $\to 2$; The mean is total exercise days divided by total students; Answer choices: (A) $3.50$, (B) $3.57$, (C) $4.36$, (D) $4.50$, (E) $5.00$

Unknowns: The mean number of exercise days per student, rounded to the nearest hundredth

Understand

Plan

Primary tool: #15 Organize Information in More Ways

Secondary: #7 Identify Subproblems, #3 Eliminate Possibilities

The data lives in a picture (bar graph), which is hard to compute with directly. Tool #15 (Reorganize) converts the graph into a frequency table — the same information in a layout where arithmetic is easy. Tool #7 (Subproblems) then splits the mean calculation into two clean pieces: (a) total number of students = sum of frequencies, (b) total exercise days = weighted sum (days $\times$ students). Dividing the two gives the mean. Tool #3 (Eliminate) is a fast sanity check on the answer choices: the mean of values from $1$ to $7$ weighted toward $4$ and $5$ must land between $4$ and $5$, instantly killing (A), (B), (D), (E) and leaving only (C).

Execute — Answer: C

#15 Organize Information in More Ways 3.MD.B.3 Step 1

Reorganize the bar graph into a frequency table.
Read each bar's height as the number of students who reported that many days.
This makes the next two arithmetic steps mechanical.

$$\begin{array}{c|c} \text{Days} & \text{Students} \\ \hline 1 & 1 \\ 2 & 3 \\ 3 & 2 \\ 4 & 6 \\ 5 & 8 \\ 6 & 3 \\ 7 & 2 \end{array}$$

💡 Reading a scaled bar graph and recording each category's count is exactly the Grade 3 scaled-bar-graph standard.

#7 Identify Subproblems 3.OA.D.8 Step 2

Subproblem 1 — total number of students.
Add up the frequency column.

$$1 + 3 + 2 + 6 + 8 + 3 + 2 = 25 \text{ students}$$

💡 Adding a short list of small whole numbers in one go is a Grade 3 multi-step word-problem skill.

#7 Identify Subproblems 4.OA.A.3 Step 3

Subproblem 2 — total exercise days for the whole class.
For each row, multiply (days) $\times$ (students) to get how many exercise days that row contributes, then add all seven products.
This is a weighted sum: a student who exercised $5$ days contributes $5$ to the total, and there are $8$ such students.

$$1 \cdot 1 + 2 \cdot 3 + 3 \cdot 2 + 4 \cdot 6 + 5 \cdot 8 + 6 \cdot 3 + 7 \cdot 2 = 1 + 6 + 6 + 24 + 40 + 18 + 14 = 109 \text{ days}$$

💡 A multi-step word problem that mixes multiplication and addition with whole numbers is the Grade 4 four-operation standard.

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

Divide total days by total students to get the mean.
To turn $\tfrac{109}{25}$ into a decimal without long division, multiply numerator and denominator by $4$ so the denominator becomes $100$ — then the digits read straight off as hundredths.

$$\text{mean} = \dfrac{109}{25} = \dfrac{109 \times 4}{25 \times 4} = \dfrac{436}{100} = 4.36$$

💡 Computing the mean of a numerical data set as (sum of values) / (number of values) is the Grade 6 summarize-data standard; expressing the result as a decimal to hundredths leans on Grade 5 decimal arithmetic too.

#3 Eliminate Possibilities 6.SP.A.3 Step 5

Match $4.36$ to the answer choices.
Only choice (C) equals $4.36$.
Confirmation by Tool #3: the data is skewed toward $4$ and $5$ days, so the mean must lie strictly between $4$ and $5$, ruling out (A), (B), (D), (E) by inspection.

$$4.36 \Rightarrow \textbf{(C)}$$

💡 Recognizing that the mean is a single number that summarizes the whole data set is the Grade 6 measure-of-center idea — and it sits where we expect it to.

[1] #15 3.MD.B.3 Reorganize the bar graph into a frequency table. Read each bar's height as the n

[2] #7 3.OA.D.8 Subproblem 1 — total number of students. Add up the frequency column.

[3] #7 4.OA.A.3 Subproblem 2 — total exercise days for the whole class. For each row, multiply (

[4] #7 6.SP.B.5 Divide total days by total students to get the mean. To turn $\tfrac{109}{25}$ i

[5] #3 6.SP.A.3 Match $4.36$ to the answer choices. Only choice (C) equals $4.36$. Confirmation

Review

Reasonableness: The data has $14$ of $25$ students reporting either $4$ or $5$ days, so the mean has to be very close to the midpoint $4.5$. A small left-tail (a few students at $1$, $2$, $3$ days) pulls the average slightly below $4.5$, landing at $4.36$ — exactly what we computed. The mean is also between the smallest value ($1$) and the largest ($7$), as any mean must be.

Alternative: Tool #3 (Eliminate Possibilities) alone almost solves it: choices (A) $3.50$ and (B) $3.57$ are too low because more than half the students reported $4$ or more days, while (D) $4.50$ and (E) $5.00$ are too high because three students reported only $1$ or $2$ days. Only (C) $4.36$ fits the rough position of the bulge in the bar graph, so a student short on time could pick (C) without ever computing $109 / 25$.

CCSS standards used (min grade 6)

3.MD.B.3 Draw and interpret scaled picture graphs and bar graphs (Reading each bar's height in the graph to recover the frequency table of (days, number of students).)
3.OA.D.8 Solve two-step word problems using four operations within 100 (Adding the seven frequencies $1+3+2+6+8+3+2 = 25$ to get the total number of students.)
4.OA.A.3 Solve multi-step word problems using four operations with whole numbers (Computing the weighted sum $1\cdot 1 + 2\cdot 3 + \dots + 7\cdot 2 = 109$ for the total exercise days.)
6.SP.B.5 Summarize numerical data sets by reporting number of observations and measures (Calculating the mean as (sum of values) / (number of values) $= 109 / 25 = 4.36$.)
6.SP.A.3 Recognize that a measure of center summarizes all its values with a single number (Interpreting $4.36$ as the single number that summarizes the whole class's exercise-day distribution and matching it to choice (C).)

⭐ This AMC 8 problem only needs Grade 6 'mean = total divided by how many' that you already know!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Review

CCSS standards used (min grade 6)

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