AMC 8 · 2002 · #17
Grade 4 arithmeticalgebraPick an answer.
AMC 8 2002 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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.
Start from the all-correct baseline: answering all 10 questions correctly would score 50.
Picking the easiest extreme first gives a number to compare 29 against.
3.OA.A.3Guess And CheckThe target is 29, so from the baseline 50 the score must drop by 21.
The gap between the baseline guess and the actual score is what each swap has to close.
4.OA.A.3Look For A PatternEach right-to-wrong swap loses the earned +5 and adds the -2 penalty, dropping the score by 7.
A clean rate of -7 per swap reduces the rest of the problem to one division.
4.OA.A.3Look For A PatternDivide the gap: 21 ÷ 7 = 3 wrong, so 10 - 3 = 7 answers were correct — choice (C).
Closing a 21-point gap at 7 points per swap takes exactly 3 swaps.
3.OA.A.3Guess And CheckStart 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.