Competition · AMC preparation · step 4 of 4
AMC 8 · 2014 · #1
Grade 7 arithmeticPick an answer.
AMC 8 2014 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The question packages three small calculations into one: compute H, compute T, then subtract. Tool #7 (Identify Subproblems) splits these so the parenthesis rule and the left-to-right rule do not get tangled. Tool #5 (Look for a Pattern) names the trap behind the problem — dropping parentheses around (2+5) flips the sign of the 5 inside, so T overshoots H by exactly 2 × 5 = 10. Spotting this pattern is a fast sanity check on the final H-T.
Compute Harry's value
Subproblem 1 — parentheses first: 2+5=7, then 8-7, so Harry's H = 1.
Parentheses are a Grade 5 grouping symbol — do what is inside first, then continue.
Harry's value is 8-(2+5)=8-7=1, because the parentheses bind 2 and 5 into a single amount that must be formed before it is taken away from 8.
▸ Why?
Inside the parentheses, 2 and 5 merge into the single amount 7, since 7 is the whole you get by combining the parts 2 and 5.
▸ Why?
Taking that amount from 8 leaves 8-7=1, because subtraction reverses addition and 1 is exactly the number that adds back with 7 to make 8.
Compute Terry's value
Subproblem 2 — no parentheses, so go left to right: 8-2=6, then +5, giving Terry's T = 11.
Same Grade 5 standard, opposite trap: with no parentheses the +5 stays as addition, not subtraction.
5.OA.A.1Identify SubproblemsSubtract the two values
Subproblem 3 — combine: T is larger, so H-T = 1-11 = -10, i.e. choice (A).
Subtracting a larger positive from a smaller one lands on the negative side of zero — Grade 7 integer subtraction.
7.NS.A.1Identify SubproblemsThis AMC 8 problem only needs the Grade 5 parenthesis rule plus Grade 7 integer subtraction — no algebra at all.
- Compute Harry's value
- Compute Terry's value
- Subtract the two values
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