AMC 8 · 2002 · #17
Grade 4 arithmeticalgebraProblem
In a mathematics contest with ten problems, a student gains 5 points for a correct answer and loses 2 points for an incorrect answer. If Olivia answered every problem and her score was 29, how many correct answers did she have?
Pick an answer.
AMC 8 2002 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: On a $10$-question contest, each correct answer earns $+5$ points and each incorrect answer costs $-2$ points. Olivia answered every question and scored $29$ total. How many of her answers were correct?
Givens: There are $10$ questions in all; Olivia answered every question, so correct $+$ incorrect $= 10$; Each correct answer is worth $+5$ points; each incorrect is worth $-2$ points; Olivia's final score is exactly $29$; Answer choices: (A) $5$, (B) $6$, (C) $7$, (D) $8$, (E) $9$
Unknowns: The number of questions Olivia answered correctly
Understand
Restated: On a $10$-question contest, each correct answer earns $+5$ points and each incorrect answer costs $-2$ points. Olivia answered every question and scored $29$ total. How many of her answers were correct?
Givens: There are $10$ questions in all; Olivia answered every question, so correct $+$ incorrect $= 10$; Each correct answer is worth $+5$ points; each incorrect is worth $-2$ points; Olivia's final score is exactly $29$; Answer choices: (A) $5$, (B) $6$, (C) $7$, (D) $8$, (E) $9$
Plan
Primary tool: #6 Guess and Check
Secondary: #5 Look for a Pattern
Only $11$ possible correct counts ($0$ through $10$) exist, and the five answer choices narrow that further, so testing values beats algebra. Tool #6 (Guess and Check) makes that direct: pick a correct count, pair it with the matching incorrect count, and compute the score. Tool #5 (Look for a Pattern) shortens the search — swapping one correct for one incorrect changes the score by $-5 - 2 = -7$ each time. That "$-7$ per swap" rule turns the search into one division: divide the score gap by $7$ to find how many swaps separate the all-correct baseline from Olivia's score.
Execute — Answer: C
3.OA.A.3 Step 1 - Start from the all-correct baseline.
- If Olivia had answered every one of the $10$ questions correctly, her score would be $10 \times 5$.
💡 Picking the easiest extreme first gives a number to compare $29$ against.
4.OA.A.3 Step 2 - Measure the score gap.
- The target is $29$ and the baseline is $50$, so the score has to come down by $50 - 29$.
💡 The gap between the baseline guess and the actual score is what each swap has to close.
4.OA.A.3 Step 3 - Find the per-swap pattern.
- Turning one correct answer into an incorrect one removes $5$ points (no longer earned) and adds $-2$ points (the penalty), so the score drops by $5 + 2 = 7$ per swap.
💡 A clean rate of $-7$ per swap reduces the rest of the problem to one division.
3.OA.A.3 Step 4 - Divide the gap by the per-swap rate to count the swaps.
- The score must drop by $21$ at $7$ per swap, so make $21 \div 7 = 3$ swaps.
- That turns $3$ of the $10$ correct answers into incorrect ones, leaving $10 - 3 = 7$ correct.
💡 Closing a $21$-point gap at $7$ points per swap takes exactly $3$ swaps.
3.OA.A.3 Start from the all-correct baseline. If Olivia had answered every one of the $10 4.OA.A.3 Measure the score gap. The target is $29$ and the baseline is $50$, so the score 4.OA.A.3 Find the per-swap pattern. Turning one correct answer into an incorrect one remo 3.OA.A.3 Divide the gap by the per-swap rate to count the swaps. The score must drop by $ Review
Reasonableness: Check the totals directly. With $7$ correct and $3$ incorrect: answers $= 7 + 3 = 10$ (matches), and score $= 7 \times 5 - 3 \times 2 = 35 - 6 = 29$ (matches). The score must fall between the all-incorrect total of $-20$ and the all-correct total of $50$, and $29$ sits inside that range — so a valid mix exists. The trap choices fail the score check: $5$ correct gives $25 - 10 = 15$, $6$ correct gives $30 - 8 = 22$, $8$ correct gives $40 - 4 = 36$, and $9$ correct gives $45 - 2 = 43$. Only $7$ hits $29$.
Alternative: Tool #2 (Make a Systematic List): tabulate the score for every value of $c$ from $5$ to $9$. With $c$ correct and $10 - c$ incorrect, the score is $5c - 2(10 - c) = 7c - 20$. That gives $15, 22, 29, 36, 43$ for $c = 5, 6, 7, 8, 9$. The hit at $29$ is $c = 7$, same answer (C). The formula $7c - 20$ also encodes the "$+7$ per correct" pattern from the main solution.
CCSS standards used (min grade 4)
3.OA.A.3Use multiplication and division within 100 to solve word problems involving equal groups, arrays, and measurement quantities (Computing the all-correct baseline $10 \times 5 = 50$, the swap count $21 \div 7 = 3$, and the final check $7 \times 5 - 3 \times 2 = 29$ as equal-group multiplications and divisions.)4.OA.A.3Solve multistep word problems with whole numbers using the four operations, including problems with remainders (Finding the score gap $50 - 29 = 21$ and dividing by the per-swap drop of $7$ points to count the incorrect answers.)
⭐ Start with the simplest guess (all correct), then notice that each right-to-wrong swap drops the score by exactly $7$ points — the score gap divided by $7$ gives the number of wrong answers directly. This AMC 8 problem becomes a Grade 4 multistep word problem, no algebra needed.
⭐ Start with the simplest guess (all correct), then notice that each right-to-wrong swap drops the score by exactly $7$ points — the score gap divided by $7$ gives the number of wrong answers directly. This AMC 8 problem becomes a Grade 4 multistep word problem, no algebra needed.
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