AMC 8 · 2013 · #5
Grade 6 arithmeticProblem
Hammie is in 6th grade and weighs 106 pounds. His quadruplet sisters are tiny babies and weigh 5, 5, 6, and 8 pounds. Which is greater, the average (mean) weight of these five children or the median weight, and by how many pounds?
Pick an answer.
AMC 8 2013 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.
Toolkit + CCSS Solution
Understand
Restated: Hammie weighs $106$ pounds and his four quadruplet baby sisters weigh $5$, $5$, $6$, and $8$ pounds. For these five children, decide whether the mean weight or the median weight is larger, and by how many pounds.
Givens: Five weights (pounds): $106, 5, 5, 6, 8$; Median = middle value of a sorted list of $5$ numbers (the $3$rd entry); Mean = (sum of all weights) $\div$ (number of children); Answer choices: (A) median, by $60$; (B) median, by $20$; (C) average, by $5$; (D) average, by $15$; (E) average, by $20$
Unknowns: Which of the two summary numbers (mean or median) is greater, and the size of the gap in pounds
Understand
Restated: Hammie weighs $106$ pounds and his four quadruplet baby sisters weigh $5$, $5$, $6$, and $8$ pounds. For these five children, decide whether the mean weight or the median weight is larger, and by how many pounds.
Givens: Five weights (pounds): $106, 5, 5, 6, 8$; Median = middle value of a sorted list of $5$ numbers (the $3$rd entry); Mean = (sum of all weights) $\div$ (number of children); Answer choices: (A) median, by $60$; (B) median, by $20$; (C) average, by $5$; (D) average, by $15$; (E) average, by $20$
Plan
Primary tool: #2 Make a Systematic List
Secondary: #7 Identify Subproblems
The question bundles two independent summary statistics into one comparison, so Tool #7 (Identify Subproblems) splits it into three clean pieces: find the median, find the mean, then compare. Tool #2 (Make a Systematic List) does the heavy lifting for the median — sort the five weights from smallest to largest and read off the $3$rd entry. The median needs no arithmetic at all once the list is in order, which makes the contrast with the mean (a sum-and-divide computation) very visible: one giant outlier ($106$) doesn't move the middle slot, but it nearly triples the average.
Execute — Answer: E
6.SP.B.5 Step 1 - Sort the five weights from smallest to largest using a single ordering rule (ascending).
- With the data in order, the median is just the value sitting at position $3$.
💡 Putting numbers in order is the most basic data-organization move — Grade 6 "summarize numerical data sets" starts here.
6.SP.A.3 Step 2 - Subproblem 1 — read off the median.
- With $5$ values, the middle is the $3$rd entry in the sorted list.
- That entry is $6$, so the median weight is $6$ pounds.
💡 The Grade 6 definition of median says: order the data, then pick the middle one. No arithmetic required.
4.NBT.B.4 Step 3 - Subproblem 2 — compute the mean.
- Add the five weights to get the total, then divide by $5$.
💡 Adding five whole numbers is a Grade 4 multi-digit addition that lines up by place value.
4.NBT.B.6 Step 4 Divide the sum by the number of children ($5$) to get the mean.
💡 Dividing a three-digit dividend by a one-digit divisor is the Grade 4 whole-number-quotient skill — and $130 / 5 = 26$ comes out clean.
6.SP.A.3 Step 5 - Subproblem 3 — compare the two summaries and report the gap.
- The mean ($26$) is larger than the median ($6$); subtract to get the difference.
💡 When one value is far above the rest, the mean shifts toward it but the median doesn't — Grade 6 "measure of center" intuition about outliers.
6.SP.B.5 Sort the five weights from smallest to largest using a single ordering rule (asc 6.SP.A.3 Subproblem 1 — read off the median. With $5$ values, the middle is the $3$rd ent 4.NBT.B.4 Subproblem 2 — compute the mean. Add the five weights to get the total, then div 4.NBT.B.6 Divide the sum by the number of children ($5$) to get the mean. 6.SP.A.3 Subproblem 3 — compare the two summaries and report the gap. The mean ($26$) is Review
Reasonableness: Four of the five weights are tiny ($5, 5, 6, 8$), all clustered near $6$, so the median should be tiny too — and it is, exactly $6$. The fifth weight ($106$) is huge and pulls the average up; spreading the extra $106 - 6 = 100$ pounds across $5$ children adds $100 / 5 = 20$ pounds to a baseline of $6$, giving a mean of $26$. That's exactly the $20$-pound gap the answer claims, and it lines up with (E).
Alternative: Tool #3 (Eliminate Possibilities) on the choice list: the median is clearly $6$ (a baby weight), so the mean cannot be smaller — that immediately eliminates (A) and (B), which say the median is larger. The total weight $130$ divided by $5$ is $26$, so the difference $26 - 6 = 20$. Only (E) matches, ruling out (C) and (D).
CCSS standards used (min grade 6)
4.NBT.B.4Fluently add and subtract multi-digit whole numbers using the standard algorithm (Computing the total weight $5 + 5 + 6 + 8 + 106 = 130$ pounds.)4.NBT.B.6Find whole-number quotients of whole numbers with up to four-digit dividends and one-digit divisors (Dividing the total by the number of children to get the mean: $130 \div 5 = 26$.)6.SP.A.3Recognize that a measure of center for a numerical data set summarizes all of its values with a single number (Applying the Grade 6 definitions of median (the middle of the sorted list) and mean (sum divided by count), and interpreting why the single large value ($106$) shifts only the mean.)6.SP.B.5Summarize numerical data sets in relation to their context (Ordering the five weights and reporting both summary statistics so the comparison "which is greater, and by how much" is well-defined.)
⭐ This AMC 8 problem only needs Grade 6 mean-vs-median ideas you already know — and one big outlier ($106$ pounds!) shifts the mean way more than the median.
⭐ This AMC 8 problem only needs Grade 6 mean-vs-median ideas you already know — and one big outlier ($106$ pounds!) shifts the mean way more than the median.
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