2020 AMC 10A Real Questions and Analysis
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April 20, 2020·Pengi AI Team

2020 AMC 10A Real Questions and Analysis

This article presents four representative problems from the 2020 AMC 10A, one from each major topic area — Algebra, Number Theory, Geometry, and Combinatorics — with full solutions and common mistake analysis. The complete answer key and a module breakdown table are included for structured review. Students preparing for AMC 10 will find this a focused resource for understanding the 2020 exam's structure and difficulty.

AMC 10AMC 10Amath competition2020problem solving

Pengi Editor's Note

This article presents four worked problems from the 2020 AMC 10A alongside the complete answer key and topic distribution breakdown. The Pengi editorial team selected this resource for students who want hands-on practice with real 2020 AMC 10A problems and a clear understanding of which topics to prioritize in preparation.

Source: Think Academy Blog


2020 AMC 10A Real Questions and Analysis

In this article, you'll find:

  • Representative real questions from each module with detailed solutions.
  • The complete 2020 AMC 10A Answer Key.
  • The best resources to prepare effectively for the AMC 10.
  • A concise topic distribution chart showing which areas appeared most in the 2020 AMC 10A.
  • A module-to-question mapping table highlighting the core concepts tested in each module for the 2020 AMC 10A.

Real Question and Solutions Explained

Algebra Example – Problem 3

Question:

Assuming (a\neq3), (b\neq4), and (c\neq5), what is the value in simplest form of
[ \frac{a-3}{5-c} \cdot \frac{b-4}{3-a} \cdot \frac{c-5}{4-b} ]

(A) (-1) (B) (1) (C) (\frac{abc}{60}) (D) (\frac{1}{abc}-\frac{1}{60}) (E) (\frac{1}{60}-\frac{1}{abc})

Solution:

Pair terms to simplify sign changes:
[ \frac{a-3}{3-a}=-1,\quad \frac{b-4}{4-b}=-1,\quad \frac{c-5}{5-c}=-1 ]
Thus,
[ (-1)\times(-1)\times(-1)=-1 ]

Answer (A)

Common Mistakes:

  • Canceling mismatched terms like (a-3) with (5-c).
  • Missing the negative sign when flipping (x-y) and (y-x).
  • Assuming remaining factors of (a,b,c) do not cancel.

Number Theory Example – Problem 6

Question:

How many 4-digit positive integers (from 1000 to 9999) having only even digits are divisible by 5?

(A) 80 (B) 100 (C) 123 (D) 200 (E) 500

Solution:

Divisibility by 5 with even digits means the last digit must be 0.

  • Thousands digit: ({2,4,6,8}) → 4 choices
  • Hundreds and tens digits: each ({0,2,4,6,8}) → (5\times5) choices
  • Total numbers: (4\times5\times5=100)

Answer (B)

Common Mistakes:

  • Using 5 choices for the first digit (cannot be 0).
  • Allowing the last digit 5 (not even).
  • Forgetting each middle digit is independent.

Geometry Example – Problem 10

Question:

Seven cubes, whose volumes are (1,8,27,64,125,216,\text{ and }343) cubic units, are stacked vertically with volumes decreasing from bottom to top. Except for the bottom cube, the bottom face of each cube lies completely on top of the cube below it. What is the total surface area of the tower (including the bottom) in square units?

(A) 644 (B) 658 (C) 664 (D) 720 (E) 749

Solution:

Side lengths: (1,2,3,4,5,6,7).
Total surface area without overlaps:
[ 6\sum_{n=1}^{7}n^2=6(140)=840 ]
Subtract overlapped faces between stacked cubes (each hidden pair removes (2n^2) for (n=1,2,\dots,6)):
[ 2(1^2+2^2+3^2+4^2+5^2+6^2)=2(91)=182 ]
Net exposed area: (840-182=658).

Answer (B)

Common Mistakes:

  • Subtracting the entire top face of the larger cube instead of only the covered portion.
  • Forgetting to include the bottom face of the lowest cube.
  • Using cube volumes instead of side areas.

Combinatorics Example – Problem 9

Question:

A single bench section at a school event can hold either 7 adults or 11 children. When (N) bench sections are connected end to end, an equal number of adults and children together will occupy all the bench space. What is the least possible positive integer value of (N)?

(A) 9 (B) 18 (C) 27 (D) 36 (E) 77

Solution:

Let (k) be the number of adults (and also (k) children).
Each adult occupies (\frac{1}{7}) of a bench, each child (\frac{1}{11}).
Total benches used:
[ \frac{k}{7}+\frac{k}{11}=N=\frac{18k}{77} ]
For (N) to be an integer, (77\mid18k). Since (\gcd(18,77)=1), the smallest (k=77), giving
[ N=\frac{18\times77}{77}=18 ]

Answer (B)

Common Mistakes:

  • Setting (7k=11k) or comparing capacities directly.
  • Taking (\text{lcm}(7,11)=77) as the answer without converting to bench count.
  • Forgetting adults and children counts are equal, not sections.

2020 AMC 10A Answer Key

QuestionAnswer
1E
2C
3A
4E
5C
6B
7C
8B
9B
10B
11C
12C
13B
14D
15E
16B
17E
18C
19E
20D
21C
22A
23A
24C
25A

2020 AMC 10A Topic Distribution

The 2020 AMC 10A featured 25 questions to be completed in 75 minutes, emphasizing advanced problem-solving and proof-based reasoning skills.

Detailed Module Analysis

ModuleQuestion NumbersWhat It Tests (Brief)
Algebra1, 3, 4, 5, 7, 8, 14, 17, 21Linear equations, ratios, functions
Number Theory6, 9, 22, 24Divisibility, GCD/LCM
Geometry10, 12, 19, 20, 23Triangles, solids
Combinatorics / Probability2, 11, 13, 15, 16, 18, 25Counting and probability

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