Difference between revisions of "2019 AMC 8 Problems/Problem 7"
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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>. | 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>. | ||
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+ | == Video Solution == | ||
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+ | Solution detailing how to solve the problem: https://www.youtube.com/watch?v=mwHrUESo2_A&list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&index=8 | ||
==See also== | ==See also== |
Latest revision as of 13:48, 23 April 2021
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
We should notice that we can turn the information we are given into a linear equation and just solve for our set variables. I'll use the variables and for the scores on the last two tests. We can now cross multiply to get rid of the denominator. Now that we have this equation, we will assign as the lowest score of the two other tests, and so: Now we know that the lowest score on the two other tests is .
~ aopsav
Solution 2
Right now, she scored and points, for a total of points. She wants her average to be for her tests, so she needs to score points in total. This means she needs to score a total of points in her next tests. Since the maximum score she can get on one of her tests is , the least possible score she can get is .
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.
Solution 3
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 .
Video Solution
Solution detailing how to solve the problem: https://www.youtube.com/watch?v=mwHrUESo2_A&list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&index=8
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.