Difference between revisions of "2019 AMC 8 Problems/Problem 7"
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| − | Right now, she scored <math>76, 94,</math> and <math>87</math> points, with a total of <math>257</math> points. She wants her average to be <math>81</math> for her <math>5</math> tests so she needs to score <math>405</math> points in total. She needs to score a total of <math>(405-257) 148</math> points in her <math>2</math> tests. So the minimum score she can get is when one of her <math>2</math> scores is <math>100</math>. So the least possible score she can get is <math>\boxed{\textbf{(A)}\ 48}</math>. | + | Right now, she scored <math>76, 94,</math> and <math>87</math> points, with a total of <math>257</math> points. She wants her average to be <math>81</math> for her <math>5</math> tests so she needs to score <math>405</math> points in total. She needs to score a total of <math>(405-257) |
| + | 148</math> points in her <math>2</math> tests. So the minimum score she can get is when one of her <math>2</math> scores is <math>100</math>. So the least possible score she can get is <math>\boxed{\textbf{(A)}\ 48}</math>. | ||
~heeeeeeeheeeeee | ~heeeeeeeheeeeee | ||
Revision as of 10:06, 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
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. 
