View source for 2021 AMC 10A Problems

{{AMC10 Problems|year=2021|ab=A}}



February 4, 2021, is when the AMC 10A starts.

==Problem 1==
What is the value of<cmath>(2^2-2)-(3^2-3)+(4^2-4)?</cmath><math>\textbf{(A)} ~1 \qquad\textbf{(B)} ~2 \qquad\textbf{(C)} ~5 \qquad\textbf{(D)} ~8 \qquad\textbf{(E)} ~12 </math>

[[2021 AMC 10A Problems/Problem 1|Solution]]

==Problem 2==
Portia's high school has <math>3</math> times as many students as Lara's high school. The two high schools have a total of <math>2600</math> students. How many students does Portia's high school have?

<math>\textbf{(A)} ~600 \qquad\textbf{(B)} ~650 \qquad\textbf{(C)} ~1950 \qquad\textbf{(D)} ~2000\qquad\textbf{(E)} ~2050</math>

[[2021 AMC 10A Problems/Problem 2|Solution]]

==Problem 3==
The sum of two natural numbers is <math>17{,}402</math>. One of the two numbers is divisible by <math>10</math>. If the units digit of that number is erased, the other number is obtained. What is the difference of these two numbers?
<math>\textbf{(A)} ~10{,}272\qquad\textbf{(B)} ~11{,}700\qquad\textbf{(C)} ~13{,}362\qquad\textbf{(D)} ~14{,}238\qquad\textbf{(E)} ~15{,}426</math>

[[2021 AMC 10A Problems/Problem 3|Solution]]
==Problem 4==
A cart rolls down a hill, travelling <math>5</math> inches the first second and accelerating so that during each successive <math>1</math>-second time interval, it travels <math>7</math> inches more than during the previous <math>1</math>-second interval. The cart takes <math>30</math> seconds to reach the bottom of the hill. How far, in inches, does it travel?

<math>\textbf{(A)} ~215 \qquad\textbf{(B)} ~360\qquad\textbf{(C)} ~2992\qquad\textbf{(D)} ~3195\qquad\textbf{(E)} ~3242</math>

[[2021 AMC 10A Problems/Problem 4|Solution]]

==Problem 5==
The quiz scores of a class with <math>k > 12</math> students have a mean of <math>8</math>. The mean of a collection of <math>12</math> of these quiz scores is <math>14</math>. What is the mean of the remaining quiz scores of terms of <math>k</math>?

<math>\textbf{(A)} ~\frac{14-8}{k-12} \qquad\textbf{(B)} ~\frac{8k-168}{k-12} \qquad\textbf{(C)} ~\frac{14}{12} - \frac{8}{k} \qquad\textbf{(D)} ~\frac{14(k-12)}{k^2} \qquad\textbf{(E)} ~\frac{14(k-12)}{8k}</math>

[[2021 AMC 10A Problems/Problem 5|Solution]]

==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==