Difference between revisions of "2018 AMC 8 Problems/Problem 13"
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Say Lalia gets a value of <math>x</math> on her first 4 tests, and a value of <math>y</math> on her last test. Thus, <math>4x+y=410.</math> | Say Lalia gets a value of <math>x</math> on her first 4 tests, and a value of <math>y</math> on her last test. Thus, <math>4x+y=410.</math> | ||
| + | The value <math>x</math> has to be less than 82, because then she would receive a lower score on her last test. Additionaly, the greatest value for y is 100, so therefore the smallest value <math>x</math> can be is 78. As a result, only <math>4</math> numbers work, <math>78, 79,80</math> and <math>81</math>. Thus the answer is <math>\boxed{\textbf{(A) }4}</math>. - song2sons | ||
| − | + | ==See Also== | |
| + | {{AMC8 box|year=2018|num-b=12|num-a=14}} | ||
| − | + | {{MAA Notice}} | |
| − | {{ | ||
Revision as of 18:50, 21 November 2018
Problem 13
Laila took five math tests, each worth a maximum of 100 points. Laila's score on each test was an integer between 0 and 100, inclusive. Laila received the same score on the first four tests, and she received a higher score on the last test. Her average score on the five tests was 82. How many values are possible for Laila's score on the last test?
Solution
Say Lalia gets a value of 
on her first 4 tests, and a value of 
on her last test. Thus, 
The value has to be less than 82, because then she would receive a lower score on her last test. Additionaly, the greatest value for y is 100, so therefore the smallest value can be is 78. As a result, only 
numbers work, 
and 
. Thus the answer is 
. - song2sons
See Also
| 2018 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 12 |
Followed by Problem 14 | |
| 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. 
