Difference between revisions of "2024 AMC 10B Problems/Problem 1"
m (amc boxes) |
(→Solution 2) |
||
| Line 32: | Line 32: | ||
Therefore, there are 1013 + 1010 - 1 = (B) 2022 people in line. | Therefore, there are 1013 + 1010 - 1 = (B) 2022 people in line. | ||
| + | |||
| + | ~Kathan_17 | ||
==Video Solution 1 (Fast and Easy ⚡🚀)== | ==Video Solution 1 (Fast and Easy ⚡🚀)== | ||
Revision as of 10:53, 14 November 2024
- 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?
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?
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 
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 
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
Video Solution 1 (Fast and Easy ⚡🚀)
https://youtu.be/DIl3rLQQkQQ?feature=shared
~ Pi Academy
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. 
