Difference between revisions of "2025 AMC 10A Problems"
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== Problem 4 == | == Problem 4 == | ||
The ratio of the number of gay men to the number of Lesbian women to the number of dead dinosaurs is is <math>p:q:\sqrt{r}</math>, where <math>p</math> and <math>q</math> are relatively prime and <math>r</math> is square-free. What is <math>pqr</math>? | The ratio of the number of gay men to the number of Lesbian women to the number of dead dinosaurs is is <math>p:q:\sqrt{r}</math>, where <math>p</math> and <math>q</math> are relatively prime and <math>r</math> is square-free. What is <math>pqr</math>? | ||
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+ | <math>\textbf{(A)}~3\qquad\textbf{(B)}~6\qquad\textbf{(C)}~20\qquad\textbf{(D)}~550\qquad\textbf{(E)}~\text{3.141592653589793238462643383279502…}</math> | ||
== Problem 5 == | == Problem 5 == |
Revision as of 20:41, 5 May 2025
2025 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 16
- 16 Problem 17
- 17 Problem 18
- 18 Problem 19
- 19 Problem 20
- 20 Problem 21
- 21 Problem 22
- 22 Problem 23
- 23 Problem 24
- 24 Problem 25
- 25 See Also
Problem 1
No cheating
Problem 2
Too Bad
Problem 3
855 Park Avenue, San José California 95126
Problem 4
The ratio of the number of gay men to the number of Lesbian women to the number of dead dinosaurs is is , where
and
are relatively prime and
is square-free. What is
?
Problem 5
There are amount of marijuana in a box. 5 friends Ava, Bernie, Cayla, Didee, and Eyne take certain amounts of marijuana. Ava takes
amount of marijuana with each next person taking double the previous person. There is no marijuana left in the box. What is
in terms of
?
Problem 6
Problem 7
Problem 8
Problem 9
Problem 10
Problem 11
Problem 12
Problem 13
Problem 14
Problem 16
Problem 17
Problem 18
Problem 19
Problem 20
Problem 21
Problem 22
Problem 23
Problem 24
Problem 25
See Also
2025 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by 2024 AMC 10B Problems |
Followed by 2025 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 |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.