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 | ||
+ | ------------------------------------------------------- | ||
+ | Another way to do this (to save on computation) is to compare each of the scores with the average of <math>81</math>: | ||
+ | <math>76</math> --> <math>-5</math> | ||
+ | <math>94</math> --> <math>+13</math> | ||
+ | <math>87</math> --> <math>+6</math> | ||
+ | <math>100</math> --> <math>19</math> | ||
+ | |||
+ | So the last one has to be <math>-33</math> (since all the differences have to sum to <math>0</math>), which corresponds to <math>81-33 = \boxed{48}</math>. | ||
==See Also== | ==See Also== |
Revision as of 22:24, 24 November 2019
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
Another way to do this (to save on computation) is to 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 |
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