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 1

Isabella wants an average grade of $95$ on her 4 tests; this also means that she wants the sum of her test scores to be at least $95 \times 4 = 380$ (if she goes over this number, she'll be over her goal!). She's already taken two tests, which sum to $97+91 = 188$ , 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 will receive all $100$ points on the fourth test. Therefore, the lowest Isabella could have scored on the third test would be $192-100 = \boxed{\textbf{(B)}\ 92}$ .

Solution 2

Isabella needs to lose a max of $20$ . When she got $97$ and $91$ , the max goes down to $20 - 3 - 9 = 8$ . If we get $100$ on her last test, then she will get $\boxed{\textbf{(B)}\ 92}$ on the third test.

Video Solution by OmegaLearn

https://youtu.be/HISL2-N5NVg?t=3126

~ pi_is_3.14

Video Solution

https://youtu.be/W_BFw23Ca8c ~savannahsolver

2012 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
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All AJHSME/AMC 8 Problems and Solutions

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

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