At the beginning of the school year, Lisa's goal was to earn an on at least of her quizzes for the year. She earned an on of the first quizzes. If she is to achieve her goal, on at most how many of the remaining quizzes can she earn a grade lower than an ?
- A)
- B)
- C)
- D)
- E)
Answer
B
Key insight
Her goal is 40 A's out of all 50 quizzes. She has 22, so she needs 18 of the last 20 quizzes to be A's. Only 2 can be lower.
Solution
- Turn the goal into a number. of quizzes is . She needs A's for the whole year.
- How many more A's does she need? She already has , so she needs more.
- How many quizzes are left? .
- Room for lower grades. Of those quizzes, must be A's. That leaves quizzes where she can get less than an A.
The answer is .
Why this works
A percent goal is easier to handle once you change it into an actual count (" A's"). Then it is simple bookkeeping: what you need minus what you have, then what is left minus what you still need. Always ask "percent of what?" Here it is of the whole year, not of the remaining quizzes.
The trap
Taking 80% of only the remaining 20 quizzes (16 A's, so 4 misses). The 80% goal is for all 50 quizzes.
Concise solution
Her target is of quizzes, which is A's for the year.
She already has A's, so she still needs more.
There are quizzes left. If of them must be A's, then at most can be below an A.
The answer is .
Review in the AoPS books
- Introduction to Algebra, Chapter 6: Ratios and Percents
- Introduction to Algebra, Chapter 3: One-Variable Linear Equations (§3.3, §3.4)
- Introduction to Algebra, Chapter 7: Proportion (§7.1, §7.2, §7.3)
- Prealgebra, Chapter 13: Data and Statistics (§13.1)
Techniques
Set up the equation/formula and compute; no special trick needed