Difference between revisions of "2018 AMC 8 Problems/Problem 13"
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==Solution 1== | ==Solution 1== | ||
| − | 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=82 \cdot | + | 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=82 \cdot 5=410</math>. |
Because <math>x</math> and <math>y</math> are different, <math>x</math> must be less than <math>82</math> and <math>y</math> must be greater than <math>82</math>. When <math>x</math> decreases by <math>1</math>, <math>y</math> must increase by <math>4</math> to keep the total constant. | Because <math>x</math> and <math>y</math> are different, <math>x</math> must be less than <math>82</math> and <math>y</math> must be greater than <math>82</math>. When <math>x</math> decreases by <math>1</math>, <math>y</math> must increase by <math>4</math> to keep the total constant. | ||
Revision as of 20:02, 7 September 2024
Contents
Problem
Lailai took five math tests, each worth a maximum of 100 points. Lailai'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 Lailai's score on the last test?
Solution 1
Say Laila gets a value of 
on her first 4 tests, and a value of 
on her last test. Thus, 
.
Because and are different, must be less than 
and must be greater than . When decreases by 
, must increase by 
to keep the total constant.
The greatest value for is 
(as 
would make non-integer). In the range 
, only values for result in integer values for : 86, 90, 94 and 98. Thus, the answer is 
.
Solution 2
The average score is which leads us to suppose that Laila got all points for the tests. We know that Laila got the same points in the first four tests and they are all lower than the last test. Let the first four tests be 
points, then the last test must be 
points to keep the average fixed. Continue to decrement the first four tests to identify other possible combinations. The possible points for the fifth test are , 
, 
, . The answer is .
Video Solution (CREATIVE ANALYSIS!!!)
~Education, the Study of Everything
Video Solution by OmegaLearn
https://youtu.be/HISL2-N5NVg?t=3251
~ pi_is_3.14
Video Solutions
~savannahsolver
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. 
