Difference between revisions of "2018 AMC 8 Problems/Problem 13"
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| − | Say | + | Say Laila 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>y</math> has to be greater than 82, because otherwise she would receive the same score on her last test. Additionally, the greatest value for y is 98 (as y=100 would give x as a decimal), so therefore the greatest value <math>x</math> can be is 98. As a result, only <math>4</math> numbers work, <math>86, 90, 94</math> and <math>98</math>. Thus the answer is <math>\boxed{\textbf{(A) }4}</math>. | The value <math>y</math> has to be greater than 82, because otherwise she would receive the same score on her last test. Additionally, the greatest value for y is 98 (as y=100 would give x as a decimal), so therefore the greatest value <math>x</math> can be is 98. As a result, only <math>4</math> numbers work, <math>86, 90, 94</math> and <math>98</math>. Thus the answer is <math>\boxed{\textbf{(A) }4}</math>. | ||
Revision as of 23:02, 8 October 2020
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 Laila gets a value of 
on her first 4 tests, and a value of 
on her last test. Thus, 
The value has to be greater than 82, because otherwise she would receive the same score on her last test. Additionally, the greatest value for y is 98 (as y=100 would give x as a decimal), so therefore the greatest value can be is 98. As a result, only 
numbers work, 
and 
. Thus the answer is 
.
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. 
