Difference between revisions of "2024 AMC 10A Problems/Problem 12"
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+ | ==Solution 3== | ||
+ | As the scores of each day are dependent on previous days, we get <math>1700 + \dfrac{0\cdot6 + 80\cdot5 + (-90)\cdot4 + (-10)\cdot3 + 60\cdot2 + (-40)\cdot1}{6} = \boxed{\textbf{(E) }1715}</math> | ||
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+ | ~NSAoPS | ||
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== Video Solution by Pi Academy == | == Video Solution by Pi Academy == |
Revision as of 17:16, 9 November 2024
Contents
[hide]Problem
Zelda played the Adventures of Math game on August 1 and scored points. She continued to play daily over the next days. The bar chart below shows the daily change in her score compared to the day before. (For example, Zelda's score on August 2 was points.) What was Zelda's average score in points over the days?
Solution 1
Going through the table, we see her scores over the six days were: , , , , , and .
Taking the average, we get
-i_am_suk_at_math_2
Solution 2
Compared to the first day , her scores change by , , , , and . So, the average is .
-mathfun2012
Solution 3
As the scores of each day are dependent on previous days, we get
~NSAoPS
Video Solution by Pi Academy
https://youtu.be/ABkKz0gS1MU?si=ZQBgDMRaJmMPSSMM
Video Solution 1 by Power Solve
See Also
2024 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.