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Difference between revisions of "2020 AMC 12A Problems"

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(Just writing the framework)
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These problems will not be available until the 2020 AMC 12A contest is released on Thursday, January 30, 2020.
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{{AMC12 Problems|year=2020|ab=A}}
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==Problem 1==
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Carlos took <math>70\%</math> of a whole pie. Maria took one third of the remainder. What portion of the whole pie was left?
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<math>\textbf{(A)}\ 10\%\qquad\textbf{(B)}\ 15\%\qquad\textbf{(C)}\ 20\%\qquad\textbf{(D)}\ 30\%\qquad\textbf{(E)}\ 35\%</math>
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[[2020 AMC 12A Problems/Problem 1|Solution]]
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==Problem 2==
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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>
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[asy]
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import olympiad;
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unitsize(25);
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for (int i = 0; i < 3; ++i) {
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for (int j = 0; j < 9; ++j) {
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pair A = (j,i);
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}
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}
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for (int i = 0; i < 3; ++i) {
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for (int j = 0; j < 9; ++j) {
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if (j != 8) {
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draw((j,i)--(j+1,i), dashed);
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}
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if (i != 2) {
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draw((j,i)--(j,i+1), dashed);
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}
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}
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}
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draw((0,0)--(2,2),linewidth(2));
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draw((2,0)--(2,2),linewidth(2));
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draw((1,1)--(2,1),linewidth(2));
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draw((3,0)--(3,2),linewidth(2));
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draw((5,0)--(5,2),linewidth(2));
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draw((4,1)--(3,2),linewidth(2));
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draw((4,1)--(5,2),linewidth(2));
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draw((6,0)--(8,0),linewidth(2));
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draw((6,2)--(8,2),linewidth(2));
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draw((6,0)--(6,2),linewidth(2));
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[/asy]
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<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>
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[[2020 AMC 12A Problems/Problem 2|Solution]]
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==Problem 3==
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[[2020 AMC 12A Problems/Problem 3|Solution]]
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==Problem 4==
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[[2020 AMC 12A Problems/Problem 4|Solution]]
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==Problem 5==
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[[2020 AMC 12A Problems/Problem 5|Solution]]
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==Problem 6==
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[[2020 AMC 12A Problems/Problem 6|Solution]]
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==Problem 7==
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[[2020 AMC 12A Problems/Problem 7|Solution]]
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==Problem 8==
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[[2020 AMC 12A Problems/Problem 8|Solution]]
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==Problem 9==
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[[2020 AMC 12A Problems/Problem 9|Solution]]
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==Problem 10==
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[[2020 AMC 12A Problems/Problem 10|Solution]]
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==Problem 11==
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[[2020 AMC 12A Problems/Problem 11|Solution]]
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==Problem 12==
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[[2020 AMC 12A Problems/Problem 12|Solution]]
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==Problem 13==
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[[2020 AMC 12A Problems/Problem 13|Solution]]
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==Problem 14==
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[[2020 AMC 12A Problems/Problem 14|Solution]]
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==Problem 15==
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[[2020 AMC 12A Problems/Problem 15|Solution]]
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==Problem 16==
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[[2020 AMC 12A Problems/Problem 16|Solution]]
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==Problem 17==
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[[2020 AMC 12A Problems/Problem 17|Solution]]
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==Problem 18==
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[[2020 AMC 12A Problems/Problem 18|Solution]]
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==Problem 19==
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[[2020 AMC 12A Problems/Problem 19|Solution]]
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==Problem 20==
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[[2020 AMC 12A Problems/Problem 20|Solution]]
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==Problem 21==
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[[2020 AMC 12A Problems/Problem 21|Solution]]
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==Problem 22==
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[[2020 AMC 12A Problems/Problem 22|Solution]]
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==Problem 23==
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[[2020 AMC 12A Problems/Problem 23|Solution]]
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==Problem 24==
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[[2020 AMC 12A Problems/Problem 24|Solution]]
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==Problem 25==
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[[2020 AMC 12A Problems/Problem 25|Solution]]
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==See also==
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{{AMC12 box|year=2020|ab=A|before=[[2019 AMC 12B Problems]]|after=[[2020 AMC 12B Problems]]}}
 +
{{MAA Notice}}

Revision as of 07:44, 1 February 2020

2020 AMC 12A (Answer Key)
Printable versions: Wiki • AoPS Resources • PDF

Instructions

  1. This is a 25-question, multiple choice test. Each question is followed by answers marked A, B, C, D and E. Only one of these is correct.
  2. You will receive 6 points for each correct answer, 2.5 points for each problem left unanswered if the year is before 2006, 1.5 points for each problem left unanswered if the year is after 2006, and 0 points for each incorrect answer.
  3. No aids are permitted other than scratch paper, graph paper, ruler, compass, protractor and erasers (and calculators that are accepted for use on the test if before 2006. No problems on the test will require the use of a calculator).
  4. Figures are not necessarily drawn to scale.
  5. You will have 75 minutes working time to complete the test.
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

Problem 1

Carlos took $70\%$ of a whole pie. Maria took one third of the remainder. What portion of the whole pie was left?

$\textbf{(A)}\ 10\%\qquad\textbf{(B)}\ 15\%\qquad\textbf{(C)}\ 20\%\qquad\textbf{(D)}\ 30\%\qquad\textbf{(E)}\ 35\%$

Solution

Problem 2

The acronym AMC is shown in the rectangular grid below with grid lines spaced $1$ 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]

$\textbf{(A) } 17 \qquad \textbf{(B) } 15 + 2\sqrt{2} \qquad \textbf{(C) } 13 + 4\sqrt{2} \qquad \textbf{(D) } 11 + 6\sqrt{2} \qquad \textbf{(E) } 21$

Solution


Problem 3

Solution


Problem 4

Solution


Problem 5

Solution


Problem 6

Solution


Problem 7

Solution


Problem 8

Solution


Problem 9

Solution


Problem 10

Solution


Problem 11

Solution


Problem 12

Solution


Problem 13

Solution


Problem 14

Solution


Problem 15

Solution


Problem 16

Solution


Problem 17

Solution


Problem 18

Solution


Problem 19

Solution


Problem 20

Solution


Problem 21

Solution


Problem 22

Solution


Problem 23

Solution


Problem 24

Solution


Problem 25

Solution


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. AMC logo.png

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