Difference between revisions of "2024 AMC 10B Problems/Problem 1"
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| + | {{duplicate|[[2023 AMC 10B Problems/Problem 1|2023 AMC 10B #1]] and [[2023 AMC 12B Problems/Problem 1|2023 AMC 12B #1]]}} | ||
| + | ==Problem== | ||
| + | In a long line of people arranged left to right, the 1013th person from the left is also the 101th person from the right. How many people are in line? | ||
| + | |||
| + | <math>\textbf{(A) } 2021 \qquad\textbf{(B) } 2022 \qquad\textbf{(C) } 2023 \qquad\textbf{(D) } 2024 \qquad\textbf{(E) } 2025</math> | ||
| + | |||
| + | ==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 <math>1012 + 1 + 1009 = \boxed{\textbf{(B) } 2022}</math> | ||
| + | ==See also== | ||
| + | {{AMC10 box|year=2024|ab=B|before=First Problem|num-a=2}} | ||
| + | {{AMC12 box|year=2024|ab=B|before=First Problem|num-a=2}} | ||
| + | |||
| + | {{MAA Notice}} | ||
Revision as of 00:08, 14 November 2024
- The following problem is from both the 2023 AMC 10B #1 and 2023 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 101th person from the right. How many people are in line?
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 
See also
| 2024 AMC 10B (Problems • Answer Key • Resources) | ||
| 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 (Problems • Answer Key • Resources) | |
| 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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
