2024 AMC 10B Problems/Problem 1

Revision as of 11:45, 14 November 2024 by Kathan 17 (talk | contribs) (Solution 2)
The following problem is from both the 2024 AMC 10B #1 and 2024 AMC 12B #1, so both problems redirect to this page.

Problem

In a long line of people arranged left to right, the 1013th person from the left is also the 1010th person from the right. How many people are in line?

$\textbf{(A) } 2021 \qquad\textbf{(B) } 2022 \qquad\textbf{(C) } 2023 \qquad\textbf{(D) } 2024 \qquad\textbf{(E) } 2025$


Certain China Testpapers:

In a long line of people arranged left to right, the 1015th person from the left is also the 1010th person from the right. How many people are in line?

$\textbf{(A) } 2021 \qquad\textbf{(B) } 2022 \qquad\textbf{(C) } 2023 \qquad\textbf{(D) } 2024 \qquad\textbf{(E) } 2025$

Solution 1

If the person is the 1013th from the left, that means there is 1012 people to their left. If the person is the 1010th from the right, that means there is 1009 people to their right. Therefore, there are $1012 + 1 + 1009 = \boxed{\textbf{(B)} 2022}$ people in line.

Solution for certain China test papers:

If the person is the 1015th from the left, that means there is 1014 people to their left. If the person is the 1010th from the right, that means there is 1009 people to their right. Therefore, there are $1014 + 1 + 1009 = \boxed{\textbf{(D)} 2024}$ people in line.

~Aray10 (Main Solution) and RULE101 (Modifications for certain China test papers)

Solution 2

The person is 1013th person from the left is also the 1010th person from the right, so the same person is counted twice.

Therefore, there are 1013 + 1010 - 1 = (B) 2022 people in line.

~Kathan_17

2024 AMC 12B P1.png

Video Solution 1 (Fast and Easy ⚡🚀)

https://youtu.be/DIl3rLQQkQQ?feature=shared

~ Pi Academy

See also

2024 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
First Problem
Followed by
Problem 2
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 10 Problems and Solutions
2024 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
First Problem
Followed by
Problem 2
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

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