AMC 8 · 1999 · #16

Grade 6 rate-ratio
percentagemulti-digit-arithmeticfraction-arithmetic identify-subproblems ↑ Prerequisites: percentagemulti-digit-arithmetic
📏 Medium solution 💡 3 insights
Problem
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?

Pick an answer.

(A)
1
(B)
5
(C)
7
(D)
9
(E)
11

AMC 8 1999 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Solve an Easier Problem

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.

1STEP 1

Find each category's correct count — take that percent of that category's total.

Arith: 0.70 × 10 = 7; Alg: 0.40 × 30 = 12; Geom: 0.60 × 35 = 21
2STEP 2

Add the three category counts to get the total she answered correctly: 40.

7 + 12 + 21 = 40 correct
3STEP 3

The passing mark is 60% of all 75 problems: 45.

0.60 × 75 = 45 needed to pass
4STEP 4

Subtract actual from passing: the number of extra correct answers she needed is 5 → (B).

45 - 40 = 5 → (B)
Answer
5
Quick percent sanity check: 40 correct out of 75 is about 4075\frac{40}{75} ≈ 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.
💡Key takeaway

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.