# 2005 AMC 12B Problems/Problem 4

The following problem is from both the 2005 AMC 12B #4 and 2005 AMC 10B #6, so both problems redirect to this page.

## Problem

At the beginning of the school year, Lisa's goal was to earn an $A$ on at least $80\%$ of her $50$ quizzes for the year. She earned an $A$ on $22$ of the first $30$ 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 $A$? $\textbf{(A) }\ 1 \qquad \textbf{(B) }\ 2 \qquad \textbf{(C) }\ 3 \qquad \textbf{(D) }\ 4 \qquad \textbf{(E) }\ 5$

## Solution

Lisa's goal was to get an $A$ on $80\% \cdot 50 = 40$ quizzes. She already has $A$'s on $22$ quizzes, so she needs to get $A$'s on $40-22=18$ more. There are $50-30=20$ quizzes left, so she can afford to get less than an $A$ on $20-18=\boxed{\textbf{(B) }2}$ of them.

Here, only the $A$'s matter... No complicated stuff!

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 