Tori's mathematics test had 75 problems: 10 arithmetic, 30 algebra, and 35 geometry problems. Although she answered 70% of the arithmetic, 40% of the algebra, and 60% of the geometry problems correctly, she did not pass the test because she got less than 60% of the problems right. How many more problems would she have needed to answer correctly to earn a 60% passing grade?
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Toolkit + CCSS Solution
Understand
Restated: Tori's test had $75$ problems: $10$ arithmetic, $30$ algebra, $35$ geometry. She got $70\%$ of the arithmetic, $40\%$ of the algebra, and $60\%$ of the geometry right. A passing grade is $60\%$ of the whole test. How many more correct answers would she have needed to pass?
Givens: Total problems: $75$ (split as $10 + 30 + 35$); Arithmetic correct rate: $70\%$ of $10$; Algebra correct rate: $40\%$ of $30$; Geometry correct rate: $60\%$ of $35$; Passing threshold: $60\%$ of $75$; Answer choices: (A) $1$, (B) $5$, (C) $7$, (D) $9$, (E) $11$
Unknowns: The number of additional correct answers needed to reach the passing total
Understand
Restated: Tori's test had $75$ problems: $10$ arithmetic, $30$ algebra, $35$ geometry. She got $70\%$ of the arithmetic, $40\%$ of the algebra, and $60\%$ of the geometry right. A passing grade is $60\%$ of the whole test. How many more correct answers would she have needed to pass?
Givens: Total problems: $75$ (split as $10 + 30 + 35$); Arithmetic correct rate: $70\%$ of $10$; Algebra correct rate: $40\%$ of $30$; Geometry correct rate: $60\%$ of $35$; Passing threshold: $60\%$ of $75$; Answer choices: (A) $1$, (B) $5$, (C) $7$, (D) $9$, (E) $11$
Plan
Primary tool: #9 Solve an Easier Problem
Secondary: #2 Make an Organized List
One big question ("how many more?") splits cleanly into smaller, easier questions — tool #9. The three categories are independent: each contributes its own count of correct answers. We solve three Grade 6 "percent of a number" sub-problems, organize the results in a short list (tool #2), add them up, then compare with the passing total of $60\%$ of $75$. The final answer is one subtraction.
Execute — Answer: B
#9 Solve an Easier Problem 6.RP.A.3Step 1
Compute correct answers in each category.
Each is a Grade 6 "percent of a quantity" calculation: convert the percent to a decimal, then multiply.
💡 Solving the test "piece by piece" is much easier than handling all $75$ at once. Each piece is just one decimal multiplication.
#2 Make an Organized List 4.NBT.B.4Step 2
List the three sub-counts together and add them to get the total she actually answered correctly.
$$7 + 12 + 21 = 40 \text{ correct}$$
💡 Three numbers, one sum — keeping them in a short list makes sure no category is forgotten.
#9 Solve an Easier Problem 6.RP.A.3Step 3
Compute the passing total: $60\%$ of the full $75$ problems.
$$0.60 \times 75 = 45 \text{ needed to pass}$$
💡 Same "percent of a quantity" idea as the category sub-problems, just applied to the full test.
#9 Solve an Easier Problem 4.NBT.B.4Step 4
Subtract her actual correct count from the passing total.
That difference is how many more she would have needed.
$$45 - 40 = 5 \;\Rightarrow\; \textbf{(B)}$$
💡 "How many more" is a comparison subtraction: target minus actual.
[1]
#9 6.RP.A.3Compute correct answers in each category. Each is a Grade 6 "percent of a quanti
[2]
#2 4.NBT.B.4List the three sub-counts together and add them to get the total she actually an
[3]
#9 6.RP.A.3Compute the passing total: $60\%$ of the full $75$ problems.
[4]
#9 4.NBT.B.4Subtract her actual correct count from the passing total. That difference is how
Review
Reasonableness: Quick percent sanity check: $40$ correct out of $75$ is about $40/75 \approx 53\%$, which is below $60\%$ — consistent with the problem saying she failed. The gap between $53\%$ and $60\%$ is roughly $7\%$ of $75$, or about $5$ problems. That matches answer (B). Also, of the other choices, (A) $1$ would mean she was almost passing (a gap of about $1.3\%$, too small), and (D) $9$ or (E) $11$ would put her near $48\%$ correct, well below what $40$ out of $75$ actually gives. Only (B) $5$ fits.
Alternative: Tool #1 (Draw a Diagram): build a $3$-row table with columns "category, total, percent right, correct." Fill in the products $7, 12, 21$ on the right, sum the column to get $40$. Beside it, write "passing $= 60\%$ of $75 = 45$." The visual gap of $5$ between the two column totals is the answer (B). The table also makes it obvious which category cost her the most: algebra ($40\%$ of $30 = 12$ instead of $18$) — a $6$-problem shortfall that more than accounts for the failing margin.
CCSS standards used (min grade 6)
6.RP.A.3 Use ratio and rate reasoning to solve real-world and mathematical problems, including finding a percent of a quantity as a rate per 100 (Computing each category's correct count ($0.70 \times 10$, $0.40 \times 30$, $0.60 \times 35$) and the passing total ($0.60 \times 75 = 45$).)
4.NBT.B.4 Fluently add and subtract multi-digit whole numbers using the standard algorithm (Summing the three category counts to $40$ and subtracting $45 - 40 = 5$ to find how many more answers were needed.)
⭐ Break the test into its three categories, compute each "percent of a quantity" separately, add, and compare with the passing total — a Grade 6 percent problem dressed up as an AMC 8 word problem.
⭐ Break the test into its three categories, compute each "percent of a quantity" separately, add, and compare with the passing total — a Grade 6 percent problem dressed up as an AMC 8 word problem.