AMC 10 · 2005 · #6
Grade 6 arithmeticAt the beginning of the school year, Lisa's goal was to earn an A on at least 80% of her 50 quizzes for the year. She earned an A on 22 of the first 30 quizzes. If she is to achieve her goal, on at most how many of the remaining quizzes can she earn a grade lower than an A?
Pick an answer.
AMC 10 2005 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: Lisa wants an A on at least 80% of her 50 quizzes for the year. She has A's on 22 of the first 30 quizzes. Find the largest number of the remaining quizzes on which she can score below an A and still reach her goal.
Givens: There are 50 quizzes in all; Her goal is an A on at least 80% of them; She earned an A on 22 of the first 30 quizzes; Answer choices: (A) 1, (B) 2, (C) 3, (D) 4, (E) 5
Unknowns: At most how many of the remaining quizzes can score below an A
Understand
Restated: Lisa wants an A on at least 80% of her 50 quizzes for the year. She has A's on 22 of the first 30 quizzes. Find the largest number of the remaining quizzes on which she can score below an A and still reach her goal.
Givens: There are 50 quizzes in all; Her goal is an A on at least 80% of them; She earned an A on 22 of the first 30 quizzes; Answer choices: (A) 1, (B) 2, (C) 3, (D) 4, (E) 5
Plan
Primary tool: #16 Change Focus / Count the Complement
Secondary: #7 Identify Subproblems
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.
Execute — Answer: B
6.RP.A.3 Step 1 Turn 80% into a count
- Her goal is an A on at least 80% of the 50 quizzes.
- To turn that percent into an actual number of quizzes, take 80% of 50: $0.80 \times 50 = 40$.
- So she needs A's on at least 40 quizzes for the whole year.
💡 A percent of a total is just a fraction of it, so 80% of 50 is a concrete count of quizzes.
4.OA.A.3 Step 2 Subtract the A's already earned
- She already has A's on 22 quizzes.
- Since she needs 40 A's in total, the number she still has to earn is the difference: $40 - 22 = 18$.
- She must get A's on 18 more quizzes.
💡 The A's she already banked shrink the goal by exactly that many.
4.OA.A.3 Step 3 Count the quizzes still left
- She has taken 30 of the 50 quizzes, so the number of quizzes remaining is $50 - 30 = 20$.
- These 20 quizzes are all she has left to reach 18 more A's.
💡 Only the quizzes she has not taken yet can still change her total.
4.OA.A.3 Step 4 The leftover is the below-A cap
- Of the 20 remaining quizzes she must turn 18 into A's.
- Whatever is left over may be below an A.
- That leftover is $20 - 18 = 2$.
- So she can afford a grade lower than an A on at most 2 of the remaining quizzes, 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.
6.RP.A.3 Her goal is an A on at least 80% of the 50 quizzes. To turn that percent into an 4.OA.A.3 She already has A's on 22 quizzes. Since she needs 40 A's in total, the number s 4.OA.A.3 She has taken 30 of the 50 quizzes, so the number of quizzes remaining is $50 - 4.OA.A.3 Of the 20 remaining quizzes she must turn 18 into A's. Whatever is left over may Review
Reasonableness: So far she has A's on 22 of 30, about 73%, which is under her 80% target — so she must do noticeably better on the rest, and needing 18 A's out of 20 (90%) fits that. That leaves almost no slack, so a small answer like 2 is sensible; a 4 or 5 would be too generous. The tempting wrong answers come from stopping early: 8 is how many below-A quizzes she already had in the first 30, and 18 is the A's still owed — neither is what the question asks. The leftover 2 is choice (B).
Alternative: Count the below-A quizzes for the whole year instead. To hit 40 A's out of 50, she may score below an A on at most $50 - 40 = 10$ quizzes all year. She has already used $30 - 22 = 8$ of those on the first 30. So the below-A quizzes still available among the rest are $10 - 8 = 2$ — the same (B).
CCSS standards used (min grade 6)
6.RP.A.3Use ratio and rate reasoning to solve real-world problems, including finding a percent of a quantity (Converting the '80% of 50 quizzes' goal into the concrete count of 40 required A's.)4.OA.A.3Solve multistep word problems posed with whole numbers using the four operations (Subtracting to find the A's still owed (40 - 22), the quizzes remaining (50 - 30), and the leftover below-A cap (20 - 18).)
⭐ When a goal is set in one kind of thing, count that thing first — the answer you want is often just whatever is left over.
⭐ When a goal is set in one kind of thing, count that thing first — the answer you want is often just whatever is left over.
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