Difference between revisions of "2020 AMC 12A Problems"
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− | + | {{AMC12 Problems|year=2020|ab=A}} | |
+ | |||
+ | ==Problem 1== | ||
+ | |||
+ | Carlos took <math>70\%</math> of a whole pie. Maria took one third of the remainder. What portion of the whole pie was left? | ||
+ | |||
+ | <math>\textbf{(A)}\ 10\%\qquad\textbf{(B)}\ 15\%\qquad\textbf{(C)}\ 20\%\qquad\textbf{(D)}\ 30\%\qquad\textbf{(E)}\ 35\%</math> | ||
+ | |||
+ | [[2020 AMC 12A Problems/Problem 1|Solution]] | ||
+ | |||
+ | ==Problem 2== | ||
+ | |||
+ | The acronym AMC is shown in the rectangular grid below with grid lines spaced <math>1</math> unit apart. In units, what is the sum of the lengths of the line segments that form the acronym AMC<math>?</math> | ||
+ | |||
+ | [asy] | ||
+ | import olympiad; | ||
+ | unitsize(25); | ||
+ | for (int i = 0; i < 3; ++i) { | ||
+ | for (int j = 0; j < 9; ++j) { | ||
+ | pair A = (j,i); | ||
+ | |||
+ | } | ||
+ | } | ||
+ | for (int i = 0; i < 3; ++i) { | ||
+ | for (int j = 0; j < 9; ++j) { | ||
+ | if (j != 8) { | ||
+ | draw((j,i)--(j+1,i), dashed); | ||
+ | } | ||
+ | if (i != 2) { | ||
+ | draw((j,i)--(j,i+1), dashed); | ||
+ | } | ||
+ | } | ||
+ | } | ||
+ | draw((0,0)--(2,2),linewidth(2)); | ||
+ | draw((2,0)--(2,2),linewidth(2)); | ||
+ | draw((1,1)--(2,1),linewidth(2)); | ||
+ | draw((3,0)--(3,2),linewidth(2)); | ||
+ | draw((5,0)--(5,2),linewidth(2)); | ||
+ | draw((4,1)--(3,2),linewidth(2)); | ||
+ | draw((4,1)--(5,2),linewidth(2)); | ||
+ | draw((6,0)--(8,0),linewidth(2)); | ||
+ | draw((6,2)--(8,2),linewidth(2)); | ||
+ | draw((6,0)--(6,2),linewidth(2)); | ||
+ | [/asy] | ||
+ | |||
+ | <math>\textbf{(A) } 17 \qquad \textbf{(B) } 15 + 2\sqrt{2} \qquad \textbf{(C) } 13 + 4\sqrt{2} \qquad \textbf{(D) } 11 + 6\sqrt{2} \qquad \textbf{(E) } 21</math> | ||
+ | |||
+ | [[2020 AMC 12A Problems/Problem 2|Solution]] | ||
+ | |||
+ | |||
+ | ==Problem 3== | ||
+ | |||
+ | [[2020 AMC 12A Problems/Problem 3|Solution]] | ||
+ | |||
+ | |||
+ | ==Problem 4== | ||
+ | |||
+ | [[2020 AMC 12A Problems/Problem 4|Solution]] | ||
+ | |||
+ | |||
+ | ==Problem 5== | ||
+ | |||
+ | [[2020 AMC 12A Problems/Problem 5|Solution]] | ||
+ | |||
+ | |||
+ | ==Problem 6== | ||
+ | |||
+ | [[2020 AMC 12A Problems/Problem 6|Solution]] | ||
+ | |||
+ | |||
+ | ==Problem 7== | ||
+ | |||
+ | [[2020 AMC 12A Problems/Problem 7|Solution]] | ||
+ | |||
+ | |||
+ | ==Problem 8== | ||
+ | |||
+ | [[2020 AMC 12A Problems/Problem 8|Solution]] | ||
+ | |||
+ | |||
+ | ==Problem 9== | ||
+ | |||
+ | [[2020 AMC 12A Problems/Problem 9|Solution]] | ||
+ | |||
+ | |||
+ | ==Problem 10== | ||
+ | |||
+ | [[2020 AMC 12A Problems/Problem 10|Solution]] | ||
+ | |||
+ | |||
+ | ==Problem 11== | ||
+ | |||
+ | [[2020 AMC 12A Problems/Problem 11|Solution]] | ||
+ | |||
+ | |||
+ | ==Problem 12== | ||
+ | |||
+ | [[2020 AMC 12A Problems/Problem 12|Solution]] | ||
+ | |||
+ | |||
+ | ==Problem 13== | ||
+ | |||
+ | [[2020 AMC 12A Problems/Problem 13|Solution]] | ||
+ | |||
+ | |||
+ | ==Problem 14== | ||
+ | |||
+ | [[2020 AMC 12A Problems/Problem 14|Solution]] | ||
+ | |||
+ | |||
+ | ==Problem 15== | ||
+ | |||
+ | [[2020 AMC 12A Problems/Problem 15|Solution]] | ||
+ | |||
+ | |||
+ | ==Problem 16== | ||
+ | |||
+ | [[2020 AMC 12A Problems/Problem 16|Solution]] | ||
+ | |||
+ | |||
+ | ==Problem 17== | ||
+ | |||
+ | [[2020 AMC 12A Problems/Problem 17|Solution]] | ||
+ | |||
+ | |||
+ | ==Problem 18== | ||
+ | |||
+ | [[2020 AMC 12A Problems/Problem 18|Solution]] | ||
+ | |||
+ | |||
+ | ==Problem 19== | ||
+ | |||
+ | [[2020 AMC 12A Problems/Problem 19|Solution]] | ||
+ | |||
+ | |||
+ | ==Problem 20== | ||
+ | |||
+ | [[2020 AMC 12A Problems/Problem 20|Solution]] | ||
+ | |||
+ | |||
+ | ==Problem 21== | ||
+ | |||
+ | [[2020 AMC 12A Problems/Problem 21|Solution]] | ||
+ | |||
+ | |||
+ | ==Problem 22== | ||
+ | |||
+ | [[2020 AMC 12A Problems/Problem 22|Solution]] | ||
+ | |||
+ | |||
+ | ==Problem 23== | ||
+ | |||
+ | [[2020 AMC 12A Problems/Problem 23|Solution]] | ||
+ | |||
+ | |||
+ | ==Problem 24== | ||
+ | |||
+ | [[2020 AMC 12A Problems/Problem 24|Solution]] | ||
+ | |||
+ | |||
+ | ==Problem 25== | ||
+ | |||
+ | [[2020 AMC 12A Problems/Problem 25|Solution]] | ||
+ | |||
+ | |||
+ | ==See also== | ||
+ | {{AMC12 box|year=2020|ab=A|before=[[2019 AMC 12B Problems]]|after=[[2020 AMC 12B Problems]]}} | ||
+ | {{MAA Notice}} |
Revision as of 06:44, 1 February 2020
2020 AMC 12A (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
Carlos took of a whole pie. Maria took one third of the remainder. What portion of the whole pie was left?
Problem 2
The acronym AMC is shown in the rectangular grid below with grid lines spaced unit apart. In units, what is the sum of the lengths of the line segments that form the acronym AMC
[asy] import olympiad; unitsize(25); for (int i = 0; i < 3; ++i) { for (int j = 0; j < 9; ++j) { pair A = (j,i);
} } for (int i = 0; i < 3; ++i) { for (int j = 0; j < 9; ++j) { if (j != 8) { draw((j,i)--(j+1,i), dashed); } if (i != 2) { draw((j,i)--(j,i+1), dashed); } } } draw((0,0)--(2,2),linewidth(2)); draw((2,0)--(2,2),linewidth(2)); draw((1,1)--(2,1),linewidth(2)); draw((3,0)--(3,2),linewidth(2)); draw((5,0)--(5,2),linewidth(2)); draw((4,1)--(3,2),linewidth(2)); draw((4,1)--(5,2),linewidth(2)); draw((6,0)--(8,0),linewidth(2)); draw((6,2)--(8,2),linewidth(2)); draw((6,0)--(6,2),linewidth(2)); [/asy]
Problem 3
Problem 4
Problem 5
Problem 6
Problem 7
Problem 8
Problem 9
Problem 10
Problem 11
Problem 12
Problem 13
Problem 14
Problem 15
Problem 16
Problem 17
Problem 18
Problem 19
Problem 20
Problem 21
Problem 22
Problem 23
Problem 24
Problem 25
See also
2020 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by 2019 AMC 12B Problems |
Followed by 2020 AMC 12B 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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.