AMC 10 · 2016 · #5
Grade 6 arithmeticPick an answer.
AMC 10 2016 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
We are never asked for the four ages one by one, only for the two ends added together. Line the ages up youngest to oldest: the total splits into the two ends plus the two in the middle. Tool #16 (Change Focus / Count the Complement) says the two ends are just the total minus the middle pair, so we never need the individual ages. Tool #7 (Identify Subproblems) breaks the work into two easy pieces — get the total from the mean, get the middle pair from the median — and Tool #4 (Introduce a Variable) lets us treat each of those pairs as one quantity.
Total age from the mean
Total = mean × count, so the four ages add up to 8 × 4 = 32.
If the average is 8, you can rebuild the whole pile by giving every one of the four cousins 8 and adding them up.
6.SP.B.5Identify SubproblemsMiddle pair from the median
The median of four ages is the average of the two middle ones, so the middle pair sums to 2 × 5 = 10.
An average of 5 for two numbers is the same as those two numbers sharing a total of 10.
6.SP.A.3Use Matrix LogicEnds are the total minus the middle
The two ends are the total minus the middle pair: 32 - 10 = 22, which is (D).
If four numbers add to 32 and the inner two add to 10, whatever is left over must belong to the outer two.
2.NBT.B.5Count The ComplementMean times the count gives the total; the rest you don't need to find one by one — just subtract the part you already know.
- Total age from the mean
- Middle pair from the median
- Ends are the total minus the middle