Difference between revisions of "2019 AMC 8 Problems/Problem 7"
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Note: You can verify that <math>\boxed{48}</math> is the right answer because it is the lowest answer out of the 5. Since it is possible to get 48, we are guaranteed that that is the right answer. | Note: You can verify that <math>\boxed{48}</math> is the right answer because it is the lowest answer out of the 5. Since it is possible to get 48, we are guaranteed that that is the right answer. | ||
~~ gorefeebuddie | ~~ gorefeebuddie | ||
| − | + | ==Solution 2== | |
| − | + | We can compare each of the scores with the average of <math>81</math>: | |
<math>76</math> --> <math>-5</math> | <math>76</math> --> <math>-5</math> | ||
<math>94</math> --> <math>+13</math> | <math>94</math> --> <math>+13</math> | ||
Revision as of 11:05, 30 November 2019
Contents
Problem 7
Shauna takes five tests, each worth a maximum of 
points. Her scores on the first three tests are 
, 
, and 
. In order to average 
for all five tests, what is the lowest score she could earn on one of the other two tests?
Solution 1
Right now, she scored 
and points, with a total of 
points. She wants her average to be for her 
tests so she needs to score 
points in total. She needs to score a total of 
points in her 
tests. So the minimum score she can get is when one of her scores is . So the least possible score she can get is 
.
~heeeeeeeheeeeee
Note: You can verify that 
is the right answer because it is the lowest answer out of the 5. Since it is possible to get 48, we are guaranteed that that is the right answer.
~~ gorefeebuddie
Solution 2
We can compare each of the scores with the average of :
--> 
--> 
--> 
--> 
So the last one has to be 
(since all the differences have to sum to 
), which corresponds to 
.
See Also
| 2019 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. 
