Problem 17 · 2002 AMC 8
Hard
Algebra & Patterns
work-backward
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?
Show answer
Answer: C — 7.
Show hints
Hint 1 of 2
Skip the algebra: pretend she aced all 10, then ask what flipping one answer to *wrong* costs her.
Still stuck? Show hint 2 →
Hint 2 of 2
Flipping right→wrong is a double hit — you *lose* the 5 you'd have earned *and* drop 2 more, a swing of 5 + 2 = 7 per wrong answer.
Show solution
Approach: start from a perfect score and subtract
- Anchor at a perfect paper: all 10 right scores 5 × 10 = 50.
- Each answer flipped to wrong costs 7 (the 5 forgone *plus* the 2 penalty). She's 50 − 29 = 21 below perfect, and 21 = 3 × 7.
- So 3 were wrong, leaving 7 correct.
- *Why this transfers:* anchoring at an extreme (all right, all wrong, all zero) and counting the cost of each swap turns many "right vs. wrong" point problems into a single division — no equation needed.
Another way — set up an equation:
- Let x = number correct, so 10 − x are wrong: 5x − 2(10 − x) = 29.
- Then 7x − 20 = 29, giving x = 7.
Mark:
· log in to save