2012 AMC 8 Problems/Problem 7

Problem

Isabella must take four 100-point tests in her math class. Her goal is to achieve an average grade of 95 on the tests. Her first two test scores were 97 and 91. After seeing her score on the third test, she realized she can still reach her goal. What is the lowest possible score she could have made on the third test?

$\textbf{(A)}\hspace{.05in}90\qquad\textbf{(B)}\hspace{.05in}92\qquad\textbf{(C)}\hspace{.05in}95\qquad\textbf{(D)}\hspace{.05in}96\qquad\textbf{(E)}\hspace{.05in}97$

Solution

Since Isabella wants an average grade of $95$ on her tests, we can say she wants the sum of here test scores to be $95 \times 4 = 380$ . We already have two test scores which sum to $188$ , which means she needs $192$ more points to achieve her desired average. In order to minimize the score on the third test, we assume that Isabella received all $100$ points on the fourth test. Therefore, the lowest Isabella could have scored on the third test would be $192-100 = \boxed{\textbf{(B)}\ 92}$ .

2012 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AJHSME/AMC 8 Problems and Solutions

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

Page

Toolbox

Search

2012 AMC 8 Problems/Problem 7

Problem

Solution

See Also