2023 AMC 12B Problems/Problem 1
- The following problem is from both the 2023 AMC 10B #1 and 2023 AMC 12B #1, so both problems redirect to this page.
Contents
Problem
Mrs. Jones is pouring orange juice into four identical glasses for her four sons. She fills the first three glasses completely but runs out of juice when the fourth glass is only full. What fraction of a glass must Mrs. Jones pour from each of the first three glasses into the fourth glass so that all four glasses will have the same amount of juice?
Solution 1
The first three glasses each have a full glass. Let's assume that each glass has "1 unit" of juice. It won't matter exactly how much juice everyone has because we're dealing with ratios, and that wouldn't affect our answer. The fourth glass has a glass that is one third. So the total amount of juice will be . If we divide the total amount of juice by 4, we get , which should be the amount of juice in each glass. This means that each of the first three glasses will have to contribute to the fourth glass.
~Sir Ian Seo the Great & lprado
Solution 2 (unnecessary numerical values)
Given that the first three glasses are full and the fourth is only full, let's represent their contents with a common denominator, which we'll set as 6. This makes the first three glasses full, and the fourth glass full.
To equalize the amounts, Mrs. Jones needs to pour juice from the first three glasses into the fourth. Pouring from each of the first three glasses will make them all full. Thus, all four glasses will have the same amount of juice. Therefore, the answer is
~Ishaan Garg
Solution 3
We let denote how much juice we take from each of the first children and give to the th child.
We can write the following equation: , since each value represents how much juice each child (equally) has in the end. (Each of the first three children now have juice, and the fourth child has more juice on top of their initial .)
Solving, we see that
~Technodoggo
Video Solution by Math-X (First understand the problem!!!!)
https://youtu.be/EuLkw8HFdk4?si=kYOwfMyrV1Wwtdev&t=62
~Math-X
Video Solution (Quick and Easy!)
~Education, the Study of Everything
Video Solution by SpreadTheMathLove
https://www.youtube.com/watch?v=SUnhwbA5_So
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution by Interstigation
https://youtu.be/gDnmvcOzxjg?si=cYB6uChy7Ue0UT4L
See also
2023 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 |
2023 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.