AMC 10 · 2010 · #15
Grade 7 number-theoryOn a 50-question multiple choice math contest, students receive 4 points for a correct answer, 0 points for an answer left blank, and −1 point for an incorrect answer. Jesse’s total score on the contest was 99. What is the maximum number of questions that Jesse could have answered correctly?
Pick an answer.
AMC 10 2010 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: A math contest has $50$ multiple-choice questions. Each correct answer is worth $+4$ points, each blank is worth $0$ points, and each wrong answer is worth $-1$ point. Jesse's total score was $99$. We want the largest possible number of questions Jesse answered correctly.
Givens: Total questions: $50$; Scoring: correct $= +4$, blank $= 0$, wrong $= -1$; Jesse's total score: $99$; Five answer choices: (A) 25, (B) 27, (C) 29, (D) 31, (E) 33
Unknowns: The maximum number of correct answers consistent with a score of $99$ on $50$ questions
Understand
Restated: A math contest has $50$ multiple-choice questions. Each correct answer is worth $+4$ points, each blank is worth $0$ points, and each wrong answer is worth $-1$ point. Jesse's total score was $99$. We want the largest possible number of questions Jesse answered correctly.
Givens: Total questions: $50$; Scoring: correct $= +4$, blank $= 0$, wrong $= -1$; Jesse's total score: $99$; Five answer choices: (A) 25, (B) 27, (C) 29, (D) 31, (E) 33
Plan
Primary tool: #4 Introduce a Variable
Secondary: #14 Extreme Principle, #13 Convert to Algebra, #3 Eliminate Possibilities
The three kinds of answers — correct, wrong, and blank — are unknown counts tied together by two facts (they add to $50$, and they produce a score of $99$). Naming them with letters (Tool #4, Introduce a Variable) turns the word problem into two clean equations. The question asks for a maximum, which is the signal for Tool #14 (Extreme Principle): push the number of correct answers as high as possible while the leftover counts stay non-negative. The key lever is that blanks cannot be negative, which caps how many questions correct and wrong can use up. We finish by checking the winning value against the answer choices (Tool #3, Eliminate Possibilities). No advanced machinery is needed — just setting up and solving one linear inequality.
Execute — Answer: C
6.EE.B.6 Step 1 Name the three counts
- Let $c$ be the number of correct answers, $w$ the number of wrong answers, and $b$ the number of blanks.
- Two facts pin them down.
- First, every one of the $50$ questions is correct, wrong, or blank, so the three counts add to $50$.
- Second, the score comes from $+4$ per correct and $-1$ per wrong (blanks add nothing), and that score is $99$.
💡 Giving each unknown count its own letter lets the two sentences in the problem become two equations you can actually work with.
6.EE.B.8 Step 2 Blanks can't be negative
- We want $c$ as large as possible.
- The counts $c$, $w$, $b$ are all $\geq 0$.
- In particular the number of blanks cannot be negative, so $b \geq 0$.
- Rearranging the first equation gives $b = 50 - c - w$, and $b \geq 0$ means the correct and wrong answers together cannot use up more than all $50$ questions.
💡 You can't leave a negative number of questions blank, so 'correct plus wrong' is boxed in at $50$ — that ceiling is what limits $c$.
7.EE.B.4 Step 3 Replace wrong with correct
- From the score equation, solve for $w$: $w = 4c - 99$.
- Substitute this into the ceiling $c + w \leq 50$ so the inequality is written in terms of $c$ alone.
💡 Trading $w$ for its formula in $c$ collapses two unknowns into one, so the bound is now purely about how big $c$ can be.
7.EE.B.4 Step 4 Take the biggest whole number
- Divide by $5$: $c \leq 149 / 5 = 29.8$.
- Since $c$ must be a whole number of questions, the largest value it can take is $29$.
💡 A count of questions has to be a whole number, so the fractional cap $29.8$ rounds down to $29$.
6.EE.B.5 Step 5 Check that 29 really works
- Test $c = 29$.
- Then $w = 4(29) - 99 = 116 - 99 = 17$, and $b = 50 - 29 - 17 = 4$.
- Every count is a non-negative whole number, and the score is $4(29) - 17 = 116 - 17 = 99$, exactly as required.
- So $29$ correct answers is achievable, and by the inequality it is the most possible.
- The answer is $29$, choice $\textbf{(C)}$.
- Choices (D) $31$ and (E) $33$ fail because they would force $c + w$ above $50$; (A) $25$ and (B) $27$ are valid but not the maximum.
💡 Finding a real breakdown of $29$ correct, $17$ wrong, $4$ blank that hits $99$ proves the ceiling is actually reachable, not just a limit on paper.
6.EE.B.6 Let $c$ be the number of correct answers, $w$ the number of wrong answers, and $ 6.EE.B.8 We want $c$ as large as possible. The counts $c$, $w$, $b$ are all $\geq 0$. In 7.EE.B.4 From the score equation, solve for $w$: $w = 4c - 99$. Substitute this into the 7.EE.B.4 Divide by $5$: $c \leq 149 / 5 = 29.8$. Since $c$ must be a whole number of ques 6.EE.B.5 Test $c = 29$. Then $w = 4(29) - 99 = 116 - 99 = 17$, and $b = 50 - 29 - 17 = 4$ Review
Reasonableness: The bound $5c \leq 149$ gives $c \leq 29.8$, and $29$ is the largest whole number under that cap — and it is realized exactly by $29$ correct, $17$ wrong, $4$ blank, scoring $4(29) - 17 = 99$. Pushing to $c = 30$ would need $w = 4(30) - 99 = 21$, so $c + w = 51 > 50$, which is impossible on a $50$-question test. That confirms $29$ is both attainable and maximal, matching choice (C).
Alternative: Tool #3 (Eliminate Possibilities): test the choices from largest down. For (E) $33$ correct: $w = 4(33) - 99 = 33$, so $c + w = 66 > 50$ — impossible. For (D) $31$: $w = 4(31) - 99 = 25$, so $c + w = 56 > 50$ — impossible. For (C) $29$: $w = 17$, $c + w = 46 \leq 50$, leaving $4$ blanks — possible. Since (C) is the first choice that fits, it is the maximum, agreeing with the algebra.
CCSS standards used (min grade 7)
6.EE.B.6Use variables to represent numbers and write expressions to solve problems (Naming the correct, wrong, and blank counts as $c$, $w$, $b$ and writing the two governing equations $c + w + b = 50$ and $4c - w = 99$.)6.EE.B.8Write an inequality of the form x > c or x < c and graph on a number line (Turning the fact that blanks cannot be negative ($b \geq 0$) into the inequality $c + w \leq 50$ that caps the correct answers.)7.EE.B.4Use variables to represent quantities and construct simple equations and inequalities (Substituting $w = 4c - 99$ to get $5c \leq 149$ and solving that inequality to bound $c$ at $29.8$.)6.EE.B.5Understand solving an equation or inequality as a process of finding values (Checking that $c = 29$ produces valid non-negative counts ($w = 17$, $b = 4$) and the exact score $99$, confirming the bound is reachable.)
⭐ Name the unknown counts, use 'blanks can't be negative' to build one inequality, and the biggest whole number that fits is your answer.
⭐ Name the unknown counts, use 'blanks can't be negative' to build one inequality, and the biggest whole number that fits is your answer.
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