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2021 AMC 12A Problems

Revision as of 14:36, 11 February 2021 by Dandeliondreams (talk | contribs) (→‎Problem 5)
2021 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

What is the value of\[2^{1+2+3}-(2^1+2^2+2^3)?\] $\textbf{(A) }0 \qquad \textbf{(B) }50 \qquad \textbf{(C) }52 \qquad \textbf{(D) }54 \qquad \textbf{(E) }57$


Solution

Problem 2

Under what conditions is $\sqrt{a^2+b^2}=a+b$ true, where $a$ and $b$ are real numbers?

$\textbf{(A) }$ It is never true. $\textbf{(B) }$ It is true if and only if $ab=0$. $\textbf{(C) }$ It is true if and only if $a+b\ge 0$. $\textbf{(D) }$ It is true if and only if $ab=0$ and $a+b\ge 0$. $\textbf{(E) }$ It is always true.

Solution

Problem 3

The sum of two natural numbers is $17,402$. One of the two numbers is divisible by $10$. If the units digit of that number is erased, the other number is obtained. What is the difference of these two numbers?

$\textbf{(A) }10,272 \qquad \textbf{(B) }11,700 \qquad \textbf{(C) }13,362 \qquad \textbf{(D) }14,238 \qquad \textbf{(E) }15,462$

Solution

Problem 4

Tom has a collection of $13$ snakes, $4$ of which are purple and $5$ of which are happy. He observes that all of his happy snakes can add, none of his purple snakes can subtract, and all of his snakes that can't subtract also can't add.

Which of these conclusions can be drawn about Tom's snakes?


$\textbf{(A) }$ Purple snakes can add. $\textbf{(B) }$ Purple snakes are happy. $\textbf{(C) }$ Snakes that can add are purple. $\textbf{(D) }$ Happy snakes are not purple. $\textbf{(E) }$ Happy snakes can't subtract.

Solution

Problem 5

When a student multiplied the number $66$ by the repeating decimal\[\underline{1}.\underline{a}\underline{b}\underline{a}\underline{b}...=\underline{1}.\overline{\underline{a}\underline{b}},\]where $a$ and $b$ are digits, he did not notice the notation and just multiplied $66$ times $\underline{1}.\underline{a}\underline{b}$. Later he found that his answer is $0.5$ less than the correct answer. What is the $2$-digit number $\underline{a}\underline{b}?$

$\textbf{(A) }15 \qquad \textbf{(B) }30 \qquad \textbf{(C) }45 \qquad \textbf{(D) }60 \qquad \textbf{(E) }75$


Solution

Problem 6

These problems will not be posted until the 2021 AMC12A is released on Thursday, February 4, 2021.

Solution

Problem 7

These problems will not be posted until the 2021 AMC12A is released on Thursday, February 4, 2021.

Solution

Problem 8

These problems will not be posted until the 2021 AMC12A is released on Thursday, February 4, 2021.

Solution

Problem 9

These problems will not be posted until the 2021 AMC12A is released on Thursday, February 4, 2021.

Solution

Problem 10

These problems will not be posted until the 2021 AMC12A is released on Thursday, February 4, 2021.

Solution

Problem 11

These problems will not be posted until the 2021 AMC12A is released on Thursday, February 4, 2021.

Solution

Problem 12

These problems will not be posted until the 2021 AMC12A is released on Thursday, February 4, 2021.

Solution

Problem 13

These problems will not be posted until the 2021 AMC12A is released on Thursday, February 4, 2021.

Solution

Problem 14

These problems will not be posted until the 2021 AMC12A is released on Thursday, February 4, 2021.

Solution

Problem 15

A choir director must select a group of singers from among his $6$ tenors and $8$ basses. The only requirements are that the difference between the numbers of tenors and basses must be a multiple of $4$, and the group must have at least one singer. Let $N$ be the number of groups that could be selected. What is the remainder when $N$ is divided by $100$?

$\textbf{(A) } 47\qquad\textbf{(B) } 48\qquad\textbf{(C) } 83\qquad\textbf{(D) } 95\qquad\textbf{(E) } 96\qquad$

Solution

Problem 16

These problems will not be posted until the 2021 AMC12A is released on Thursday, February 4, 2021.

Solution

Problem 17

These problems will not be posted until the 2021 AMC12A is released on Thursday, February 4, 2021.

Solution

Problem 18

These problems will not be posted until the 2021 AMC12A is released on Thursday, February 4, 2021.

Solution

Problem 19

These problems will not be posted until the 2021 AMC12A is released on Thursday, February 4, 2021.

Solution

Problem 20

These problems will not be posted until the 2021 AMC12A is released on Thursday, February 4, 2021.

Solution

Problem 21

These problems will not be posted until the 2021 AMC12A is released on Thursday, February 4, 2021.

Solution

Problem 22

These problems will not be posted until the 2021 AMC12A is released on Thursday, February 4, 2021.

Solution

Problem 23

These problems will not be posted until the 2021 AMC12A is released on Thursday, February 4, 2021.

Solution

Problem 24

These problems will not be posted until the 2021 AMC12A is released on Thursday, February 4, 2021.

Solution

Problem 25

These problems will not be posted until the 2021 AMC12A is released on Thursday, February 4, 2021.

Solution

See also

2021 AMC 12A (Problems • Answer Key • Resources)
Preceded by
2020 AMC 12B Problems 		Followed by
2021 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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