Competition · AMC preparation · step 4 of 4
AMC 8 · 2022 · #16
Grade 4 arithmeticalgebraPick an answer.
AMC 8 2022 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Trying to solve for a, b, c, d individually is hopeless — three equations, four unknowns. But we do not need the individual values; we only need a+d. Tool #16 (Change Focus / Complement) is the key insight: inside the whole sum a+b+c+d, the pair a+d is exactly the complement of the middle pair b+c. If we know the whole and we know the middle, the outside is forced. Tool #7 (Identify Subproblems) breaks the work into three clean subproblems: (1) turn each given average into a pair sum by multiplying by 2, (2) build the whole-sum from the first and last pair sums, (3) subtract the middle pair sum, then halve to get the desired average. No algebra is needed — only the definition of average and four-operation arithmetic.
Turn averages into sums
Each given average is a pair's sum halved, so doubling it recovers that pair's sum.
Reading "the average of two numbers is 21" as "the two numbers add to 42" is a Grade 3 multiplication word-problem move.
3.OA.A.3Identify SubproblemsFind the total of all four
Adding the outer pair sums a+b and c+d uses each number once, giving the whole a+b+c+d = 102.
Combining pair sums whose terms don't overlap is the Tool #7 "add the subproblem answers" move — pure multi-step arithmetic.
4.OA.A.3Identify SubproblemsSubtract the middle pair
The outer pair a+d is the complement of the middle: subtract b+c = 52 from the whole to get a+d = 50.
When you know the whole and one part, the other part is just whole - part — that's exactly the Tool #16 complement trick, dressed as Grade 4 subtraction.
The first and last numbers together must sum to 50, because inside the total of all four numbers the outer pair is exactly what is left once the middle pair (sum 52) is set aside.
▸ Why?
The total of all four numbers is 102: adding the first pair's sum 42 to the last pair's sum 60 counts each of the four numbers once, with none left out and none counted twice.
▸ Why?
That same four-number total also breaks up the other way — into the middle pair and the outer pair — so the outer pair is the whole with the middle pair taken away.
▸ Why?
The middle pair b+c and the outer pair a+d cover all four numbers with no gaps and no overlaps, so together they equal the whole 102.
▸ Why?
When two parts make a whole, removing one part from the whole leaves the other part — subtraction reverses the adding that built the whole.
Halve to get the average
Halving the pair sum a+d turns it back into the requested average of the first and last numbers.
Same definition as Step 1, in reverse: "sum is 50, so the average of the two numbers is 25" — Grade 3 division.
3.OA.A.3Identify SubproblemsThis AMC 8 problem only needs Grade 4 multi-step arithmetic — add the outer pair sums, subtract the middle pair sum — that you already know!
- Turn averages into sums
- Find the total of all four
- Subtract the middle pair
- Halve to get the average
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