Difference between revisions of "2012 AMC 8 Problems/Problem 7"
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<math> \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 </math> | <math> \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 </math> | ||
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| + | ==Solution== | ||
| + | Since Isabella wants an average grade of <math> 95 </math> on her tests, we can say she wants the sum of here test scores to be <math>95 \times 4 = 380</math>. We already have two test scores which sum to <math>188</math>, which means she needs <math>192</math> more points to achieve her desired average. In order to minimize the score on the third test, we assume that Isabella received all <math>100</math> points on the fourth test. Therefore, the lowest Isabella could have scored on the third test would be <math> 192-100 = \boxed{\textbf{(B)}\ 92} </math>. | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2012|num-b=6|num-a=8}} | {{AMC8 box|year=2012|num-b=6|num-a=8}} | ||
Revision as of 10:20, 24 November 2012
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?
Solution
Since Isabella wants an average grade of 
on her tests, we can say she wants the sum of here test scores to be 
. We already have two test scores which sum to 
, which means she needs 
more points to achieve her desired average. In order to minimize the score on the third test, we assume that Isabella received all 
points on the fourth test. Therefore, the lowest Isabella could have scored on the third test would be 
.
See Also
| 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 | ||
