Difference between revisions of "2012 AMC 8 Problems/Problem 7"
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==Solution 2== | ==Solution 2== | ||
| − | Isabella needs to lose a max of <math>20</math>. When she got <math>97</math> and <math>91</math>, the max goes down to <math>20 - 3 - 9 = 8</math>. If we get 100 on her last test, then she will get <math>\boxed{\textbf{(B)}\ 92}</math> on the | + | Isabella needs to lose a max of <math>20</math>. When she got <math>97</math> and <math>91</math>, the max goes down to <math>20 - 3 - 9 = 8</math>. If we get <math>100</math> on her last test, then she will get <math>\boxed{\textbf{(B)}\ 92}</math> on the third test. |
| + | |||
| + | == Video Solution by OmegaLearn== | ||
| + | https://youtu.be/HISL2-N5NVg?t=3126 | ||
| + | |||
| + | ~ pi_is_3.14 | ||
| + | |||
| + | == Video Solution by WhyMath== | ||
| + | https://youtu.be/W_BFw23Ca8c ~ savannahsolver | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2012|num-b=6|num-a=8}} | {{AMC8 box|year=2012|num-b=6|num-a=8}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 17:52, 10 October 2023
Contents
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?
Solution 1
Isabella wants an average grade of 
on her 4 tests; this also means that she wants the sum of her test scores to be at least 
(if she goes over this number, she'll be over her goal!). She's already taken two tests, 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 will receive all 
points on the fourth test. Therefore, the lowest Isabella could have scored on the third test would be 
.
Solution 2
Isabella needs to lose a max of 
. When she got 
and 
, the max goes down to 
. If we get on her last test, then she will get 
on the third test.
Video Solution by OmegaLearn
https://youtu.be/HISL2-N5NVg?t=3126
~ pi_is_3.14
Video Solution by WhyMath
https://youtu.be/W_BFw23Ca8c ~ savannahsolver
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
