Competition · AMC preparation · step 4 of 4
AMC 8 · 2013 · #5
Grade 6 arithmeticPick an answer.
AMC 8 2013 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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 3rd 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.
Sort the five weights
Sort the five weights smallest to largest; once ordered, the median is just the value at position 3.
Putting numbers in order is the most basic data-organization move — Grade 6 "summarize numerical data sets" starts here.
6.SP.B.5Make A Systematic ListRead off the median
Subproblem 1: with 5 values the middle is the 3rd sorted entry, 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.
6.SP.A.3Identify SubproblemsAdd the five weights
Subproblem 2: add the five weights to get the total, 130 pounds, then divide by 5.
Adding five whole numbers is a Grade 4 multi-digit addition that lines up by place value.
4.NBT.B.4Identify SubproblemsDivide to get the mean
Divide the total by the 5 children to get the mean, 26 pounds.
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.
4.NBT.B.6Identify SubproblemsSubtract median from mean
Subproblem 3: the mean (26) beats the median (6); subtracting gives a gap of 20 pounds → (E).
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.
For the five children, the average (mean) weight is greater than the median weight, and it is greater by 20 pounds.
▸ Why?
The mean weight is 26 pounds, found by pooling all the weight and sharing it equally among the five children: 130 ÷ 5 = 26.
▸ Why?
The five weights pool to a total of 130 pounds, because the group's whole weight is just each child's weight added together: 5 + 5 + 6 + 8 + 106 = 130.
▸ Why?
Sharing 130 pounds equally among 5 children asks how big each of 5 equal groups is, and since 5 equal groups of 26 rebuild 130, each share must be 26.
▸ Why?
The median weight is 6 pounds, the value sitting in the middle once the five weights are lined up from least to greatest; with an odd count of five, that middle is the single 3rd value, which is 6.
▸ Why?
Since 26 is bigger than 6, the average is the greater of the two summaries, and the amount it leads by is 26 - 6 = 20 pounds.
▸ Why?
The lead is the missing jump from the median up to the mean: 6 plus the lead equals 26, and subtraction recovers that jump as 26 - 6 = 20.
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.
- Sort the five weights
- Read off the median
- Add the five weights
- Divide to get the mean
- Subtract median from mean
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