AMC 10 · 2005 · #6
Grade 6 arithmeticPick an answer.
AMC 10 2005 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The question asks for below-A quizzes, but the goal is stated in A's, so I flip focus with Tool #16: first pin down how many A's she must still earn, then the below-A quizzes are simply whatever is left over among the remaining quizzes. To get that A-count I break the problem into small pieces with Tool #7 (Identify Subproblems): total A's needed, A's already banked, A's still owed, and finally the leftover. Each piece is one short arithmetic step, so no algebra is required.
Turn 80% into a count
80% of the 50 quizzes is 0.80 × 50 = 40, so she needs A's on at least 40 quizzes for the year.
A percent of a total is just a fraction of it, so 80% of 50 is a concrete count of quizzes.
Eighty percent of a whole number of quizzes is a concrete count, not a rate.
▸ Why?
A percent of a total is that many hundredths of it, which here comes out a whole number.
▸ Why?
The quizzes already taken and those still to come make up the whole term, so the goal splits between them.
Subtract the A's already earned
She already has A's on 22 quizzes, so she still owes 40 - 22 = 18 more.
The A's she already banked shrink the goal by exactly that many.
4.OA.A.3Identify SubproblemsCount the quizzes still left
She has taken 30 of the 50 quizzes, so 50 - 30 = 20 quizzes remain to hold those 18 A's.
Only the quizzes she has not taken yet can still change her total.
4.OA.A.3Identify SubproblemsThe leftover is the below-A cap
Of the 20 remaining she must ace 18, so the spares are 20 - 18 = 2, which is choice (B).
Every remaining quiz is either a needed A or a spare, and the spares are what may drop below an A.
4.OA.A.3Change Focus Count The ComplementWhen a goal is set in one kind of thing, count that thing first — the answer you want is often just whatever is left over.
- Turn 80% into a count
- Subtract the A's already earned
- Count the quizzes still left
- The leftover is the below-A cap