AMC 8 · 2018 · #8

Grade 6 arithmetic

Archetype Arithmetic Expression Decomposition └ Sub-archetype Fraction / Expression Sum Evaluation

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

📏 Medium solution 💡 2 insights 📊 Diagram

Problem

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.

Pick an answer.

(A)

3.50

(B)

3.57

(C)

4.36

(D)

4.50

(E)

5.00

AMC 8 2018 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Organize Information in More Ways

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 × 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).

1STEP 1

Reorganize the bar graph into a frequency table — each bar's height is the number of students who reported that many days.

Days & Students ; 1 & 1 ; 2 & 3 ; 3 & 2 ; 4 & 6 ; 5 & 8 ; 6 & 3 ; 7 & 2

2STEP 2

Subproblem 1 — add up the frequency column to get 25 students in the class.

1 + 3 + 2 + 6 + 8 + 3 + 2 = 25 students

3STEP 3

Subproblem 2 — weight each day-count by its frequency and sum the products: the class logs 109 exercise days total.

1 · 1 + 2 · 3 + 3 · 2 + 4 · 6 + 5 · 8 + 6 · 3 + 7 · 2 = 1 + 6 + 6 + 24 + 40 + 18 + 14 = 109 days

4STEP 4

Divide days by students: scaling $\frac{109}{25}$ to a denominator of 100 gives $\frac{436}{100}$ = 4.36, read straight off as hundredths.

mean =

\frac{109}{25}

\frac{109 × 4}{25 × 4}

\frac{436}{100}

= 4.36

5STEP 5

Only choice (C) equals 4.36; since the data clusters at 4–5 days the mean must fall between 4 and 5, killing the rest.

4.36 → (C)

Answer

4.36

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.

💡Key takeaway

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

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