Competition · AMC preparation · step 4 of 4

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

Turn the graph into a table

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

Count all the students

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

Total the exercise days

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 to get the mean

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 = 109/25 = (109 × 4)/(25 × 4) = 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.

The class's mean number of exercise days is 109 ÷ 25, which equals 4.36.

▸ Why?

The mean is the class's total exercise days shared out equally among all the students, so it is the 109 total days divided into 25 equal parts.

▸ Why?

The class logged 109 exercise days in all: each bar contributes (its day count) × (how many students reported it), and the seven results join together.

▸ Why?

Multiplying a day count by its number of students is correct because each of those students exercised that same number of days, so it is that many equal groups counted together.

Multiplication as equal groupsa times b means a equal groups of b — it is just repeated adding of the same amount.

▸ Why?

Adding the seven products gives the class total because the seven student groups do not overlap and leave no one out, so their days rebuild the whole.

Whole is the sum of its partsBreak a thing into pieces with no gaps or overlaps and the pieces add back to the whole.

▸ Why?

There are 25 students because the seven bar heights count separate groups that together make up the entire class, so summing them counts everyone once.

Whole is the sum of its partsBreak a thing into pieces with no gaps or overlaps and the pieces add back to the whole.

▸ Why?

Splitting the 109 total days equally among the 25 students is exactly what the mean is: the average of a data set is defined as the sum of its values divided by how many values there are.

The mean is total over countThe arithmetic mean is the total shared equally among the values: mean = sum ÷ count (so sum = mean × count).

▸ Why?

That quotient 109 ÷ 25 equals 4.36 because rewriting it with a denominator of 100 gives 436/100, which is 436 hundredths.

▸ Why?

Multiplying the top and bottom both by 4 does not change the value because it is the same as multiplying the fraction by 4/4, and 4/4 is one.

Multiplicative identity (one)One group of a number is just that number — multiplying by one changes nothing.

▸ Why?

436/100 is written 4.36 because every place holds ten of the next, so a denominator of 100 puts the digits two places past the decimal point.

Place-value groupingTen ones make one ten and ten tens make one hundred — each place bundles ten of the last.

6.SP.B.5Identify Subproblems

5STEP 5

Match against the choices

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!

Turn the graph into a table
Count all the students
Total the exercise days
Divide to get the mean
Match against the choices

A parent dashboard for the family lives at sensimlab.com.

Turn the graph into a table

Count all the students

Total the exercise days

Divide to get the mean

Match against the choices

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