Difference between revisions of "2021 AMC 10A Problems"
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==See also== | ==See also== | ||
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{{MAA Notice}} | {{MAA Notice}} |
Revision as of 22:11, 29 December 2020
2021 AMC 10A (Answer Key) Printable versions: • AoPS Resources • PDF | ||
Instructions
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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 |
Contents
- 1 Problem 1
- 2 Problem 2
- 3 Problem 3
- 4 Problem 4
- 5 Problem 5
- 6 Problem 6
- 7 Problem 7
- 8 Problem 8
- 9 Problem 9
- 10 Problem 10
- 11 Problem 11
- 12 Problem 12
- 13 Problem 13
- 14 Problem 14
- 15 Problem 15
- 16 Problem 16
- 17 Problem 17
- 18 Problem 18
- 19 Problem 19
- 20 Problem 20
- 21 Problem 21
- 22 Problem 22
- 23 Problem 23
- 24 Problem 24
- 25 Problem 25
- 26 See also
Problem 1
This is a reminder that the 2021 AMC 10A is an unreleased test. The official examination will be on February 4, 2021. For further information, please visit https://www.maa.org/math-competitions/amc-1012.
Another AoPS user wrote: pls do not spam here
Problem 2
The following problem is not an actual AMC 10 Problem but it is good practice. Gordon Ramsey ate 10 donuts on Monday, 5 donuts on Tuesday, 2.5 donuts on Wednesday, and so on. How many donuts did he eat after 100000000 days? Choose the nearest number.
A) 25 B) 20 C) 15 D) 40 E) 41
Secondary Question: Gordon Ramsey has 100 donuts. There are 25 donuts of each type: Rainbow, Fish, Gold, and Brocolli. What is the probability that he will choose at least 1 of each kind if he were to choose 10 donuts?
Fill in the blank: ________
Problem 3
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Problem 4
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Problem 5
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Problem 6
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Problem 7
Problem 8
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Problem 9
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Problem 10
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Problem 11
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Problem 12
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Problem 13
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Problem 14
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Problem 15
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Problem 16
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Problem 17
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Problem 18
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Problem 19
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Problem 20
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Problem 21
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Problem 22
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Problem 23
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Problem 24
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Problem 25
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See also
2021 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by 2020 AMC 10B Problems |
Followed by 2021 AMC 10B Problems | |
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.