Difference between revisions of "2015 AMC 10A Problems/Problem 5"
Pi is 3.14 (talk | contribs) (→Video Solution) |
(→Video Solution) |
||
| (3 intermediate revisions by 2 users not shown) | |||
| Line 8: | Line 8: | ||
==Solution== | ==Solution== | ||
| − | If the average of the first <math>14</math> peoples' scores was <math>80</math>, then the sum of all of their tests is <math>14 | + | If the average of the first <math>14</math> peoples' scores was <math>80</math>, then the sum of all of their tests is <math>14 \cdot 80 = 1120</math>. When Payton's score was added, the sum of all of the scores became <math>15 \cdot 81 = 1215</math>. So, Payton's score must be <math>1215-1120 = \boxed{\textbf{(E) }95}</math> |
| − | |||
==Alternate Solution== | ==Alternate Solution== | ||
The average of a set of numbers is the value we get if we evenly distribute the total across all entries. So assume that the first <math>14</math> students each scored <math>80</math>. If Payton also scored an <math>80</math>, the average would still be <math>80</math>. In order to increase the overall average to <math>81</math>, we need to add one more point to all of the scores, including Payton's. This means we need to add a total of <math>15</math> more points, so Payton needs <math>80+15 = \boxed{\textbf{(E) }95}</math> | The average of a set of numbers is the value we get if we evenly distribute the total across all entries. So assume that the first <math>14</math> students each scored <math>80</math>. If Payton also scored an <math>80</math>, the average would still be <math>80</math>. In order to increase the overall average to <math>81</math>, we need to add one more point to all of the scores, including Payton's. This means we need to add a total of <math>15</math> more points, so Payton needs <math>80+15 = \boxed{\textbf{(E) }95}</math> | ||
| + | |||
| + | ==Video Solution (CREATIVE THINKING)== | ||
| + | https://youtu.be/TZtbmvFyGic | ||
| + | |||
| + | ~Education, the Study of Everything | ||
| + | |||
| + | |||
| + | |||
| + | |||
==Video Solution== | ==Video Solution== | ||
| Line 20: | Line 28: | ||
~savannahsolver | ~savannahsolver | ||
| − | == Video Solution == | + | == Video Solution by OmegaLearn == |
https://youtu.be/HISL2-N5NVg?t=3425 | https://youtu.be/HISL2-N5NVg?t=3425 | ||
Latest revision as of 22:06, 26 June 2023
- The following problem is from both the 2015 AMC 12A #3 and 2015 AMC 10A #5, so both problems redirect to this page.
Contents
Problem
Mr. Patrick teaches math to 
students. He was grading tests and found that when he graded everyone's test except Payton's, the average grade for the class was 
. After he graded Payton's test, the test average became 
. What was Payton's score on the test?
Solution
If the average of the first 
peoples' scores was , then the sum of all of their tests is 
. When Payton's score was added, the sum of all of the scores became 
. So, Payton's score must be 
Alternate Solution
The average of a set of numbers is the value we get if we evenly distribute the total across all entries. So assume that the first students each scored . If Payton also scored an , the average would still be . In order to increase the overall average to , we need to add one more point to all of the scores, including Payton's. This means we need to add a total of more points, so Payton needs 
Video Solution (CREATIVE THINKING)
https://youtu.be/TZtbmvFyGic
~Education, the Study of Everything
Video Solution
~savannahsolver
Video Solution by OmegaLearn
https://youtu.be/HISL2-N5NVg?t=3425
~ pi_is_3.14
See also
| 2015 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 4 |
Followed by Problem 6 | |
| 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 AMC 10 Problems and Solutions | ||
| 2015 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 2 |
Followed by Problem 4 |
| 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 AMC 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
