AMC 10 Step by Step
Log inSign up

2005 AMC 10B Problem 6

P1-10~2 minPrint
Ratios, Percents & AveragesLinear Equations & Word ProblemsSet up the equation/formula and compute; no special trick needed

At the beginning of the school year, Lisa's goal was to earn an AA on at least 80%80\% of her 5050 quizzes for the year. She earned an AA on 2222 of the first 3030 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 AA ?

Log in to save your progress.
Show solution without answering

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

  1. Turn the goal into a number. 80%80\% of 5050 quizzes is 0.8⋅50=400.8 \cdot 50 = 40. She needs 4040 A's for the whole year.
  2. How many more A's does she need? She already has 2222, so she needs 40−22=1840 - 22 = 18 more.
  3. How many quizzes are left? 50−30=2050 - 30 = 20.
  4. Room for lower grades. Of those 2020 quizzes, 1818 must be A's. That leaves 20−18=220 - 18 = 2 quizzes where she can get less than an A.

The answer is (B) 2\boxed{\textbf{(B)}\ 2}.

Why this works

A percent goal is easier to handle once you change it into an actual count ("4040 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 80%80\% of the whole year, not of the remaining quizzes.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

Her target is 80%80\% of 5050 quizzes, which is 0.8⋅50=400.8 \cdot 50 = 40 A's for the year.

She already has 2222 A's, so she still needs 40−22=1840 - 22 = 18 more.

There are 50−30=2050 - 30 = 20 quizzes left. If 1818 of them must be A's, then at most 20−18=220 - 18 = 2 can be below an A.

The answer is (B) 2\boxed{\textbf{(B)}\ 2}.

Review in the AoPS books

Similar problems

Problem © Mathematical Association of America (MAA), American Mathematics Competitions. Reproduced for non-commercial educational use. Solution and commentary are original to this site. Also appeared as 2005 12B #4.