Competition · AMC preparation · step 4 of 4

AMC 8 · 2011 · #4

Grade 6 arithmetic

Archetype Arithmetic Expression Decomposition └ Sub-archetype Weighted Group Aggregation

mean-median-mode-rangefraction-arithmetic identify-subproblems ↑ Prerequisites: fraction-arithmetic

📏 Short solution 💡 2 insights

Problem

Tyler's nine fishing outings yielded the catches 2, 0, 1, 3, 0, 3, 3, 1, 2. Compute the mean, median, and mode of this data set, then pick the inequality that correctly orders all three.

Pick an answer.

(A)

median < mean < mode

(B)

mean < mode < median

(C)

mean < median < mode

(D)

median < mode < mean

(E)

mode < median < mean

AMC 8 2011 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Identify Subproblems

The question asks one thing — an ordering — but answering it cleanly needs three separate calculations. Tool #7 (Identify Subproblems) says: handle mean, median, and mode one at a time, then compare the three numbers at the end. Tool #2 (Re-arrange) does the upfront work that makes the median and mode obvious: sorting the nine values from smallest to largest lets us read the middle value off the list and spot the most frequent value at a glance.

1STEP 1

Sort the nine values

Sort the nine catches from smallest to largest so the median and mode read straight off the list.

{2, 0, 1, 3, 0, 3, 3, 1, 2} ⟶ {0, 0, 1, 1, 2, 2, 3, 3, 3}

2STEP 2

Find the mean

Add the nine values to get 15 and divide by 9, so the mean is $\frac{5}{3}$ ≈ 1.67.

mean = (0+0+1+1+2+2+3+3+3)/9 = 15/9 = 5/3 ≈ 1.67

3STEP 3

Find the median

With nine sorted values the median is the middle, 5th value: median = 2.

Sorted: 0, 0, 1, 1, 2, 2, 3, 3, 3 → median = 2

4STEP 4

Find the mode

Count how often each value appears; 3 shows up three times, more than any other, so mode = 3.

Counts: 0{:} 2, 1{:} 2, 2{:} 2, 3{:} 3 → mode = 3

5STEP 5

Order the three numbers

Line the three up: $\frac{5}{3}$ ≈ 1.67, then 2, then 3, which matches choice (C).

15/9 ＜ 2 ＜ 3 ⟺ mean ＜ median ＜ mode → (C)

Reading 15/9 as 15 ÷ 9 = 1.67 shows it is below 2, so the chain is settled — a Grade 5 "fraction as division" check.

Once each measure of center is worked out, the mean turns out to sit below the median, and the median below the mode.

▸ Why?

The mean equals 15/9, which is below the median value of 2.

▸ Why?

The mean is the whole catch of 15 fish spread equally over the 9 outings, so it is 15 ÷ 9.

▸ Why?

Adding the nine separate catches, with no fish counted twice and none left out, gives the single total of 15 that is being shared.

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?

Spreading that total into 9 equal shares reverses the act of building 9 equal groups, which is exactly dividing by 9.

Inverse operationsSubtraction undoes addition and division undoes multiplication — they reverse each other.

▸ Why?

Nine equal shares of 2 would take 18 fish, but only 15 were caught, so each share must come out under 2.

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

▸ Why?

The median works out to 2 and the mode to 3, and 2 is below 3.

▸ Why?

In the nine sorted values the median is the single center value, the 5th one, which reads as 2.

▸ Why?

Matching each of the four values left of center with one of the four values right of center pairs them off exactly and leaves one lone value in the middle.

One-to-one correspondenceIf each thing here pairs with exactly one thing there, the two groups are the same size.

▸ Why?

The mode is 3 because the value 3 is caught on three outings while each other value appears only twice, so 3 shows up the most.

▸ Why?

Pairing the outings that caught 3 against the outings that caught any other single value always leaves an unmatched 3, so the group of 3s is the larger one.

One-to-one correspondenceIf each thing here pairs with exactly one thing there, the two groups are the same size.

5.NF.B.3Identify Subproblems

Answer

mean ＜ median ＜ mode

Quick sanity check: the data set has a few 3s clustered at the top, which pulls the mode up to 3, while the small values (0s and 1s) pull the mean down. So we expect mode ＞ median ＞ mean, exactly the chain in (C). Also, the mean

\frac{15}{9}

≈ 1.67 sits between the smallest value 0 and the largest value 3, as any mean must.

💡Key takeaway

This AMC 8 problem only needs the Grade 6 ideas of mean, median, and mode — compute each one, then line them up.

Sort the nine values
Find the mean
Find the median
Find the mode
Order the three numbers

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Sort the nine values

Find the mean

Find the median

Find the mode

Order the three numbers

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