Difference between revisions of "2023 AMC 10B Problems/Problem 1"

Revision as of 17:01, 15 November 2023

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 $\frac{1}{3}$ 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?

$\textbf{(A) }\frac{1}{12}\qquad\textbf{(B) }\frac{1}{4}\qquad\textbf{(C) }\frac{1}{6}\qquad\textbf{(D) }\frac{1}{8}\qquad\textbf{(E) }\frac{2}{9}$

Solution 1

We let $x$ denote how much juice we take from each of the first $3$ children and give to the $4$ th child.

We can write the following equation: $1-x=\dfrac13+3x$ , since each value represents how much juice each child (equally) has in the end. (Each of the first three children now have $1-x$ juice, and hte fourth child has $3x$ more juice on top of their initial $\dfrac13$ .)

Solving, we see that $x=\boxed{\textbf{(C) }\dfrac16}.$

~Technodoggo

Solution 2

We begin by assigning a variable $a$ to the capacity of one full glass. Quickly we see the four glasses are equivalent to $1a+1a+1a+\frac{a}{3}=\frac{10a}{3}$ . In order for all four glasses to have the same amount of orange juice, they have to each have $\frac{\frac{10a}{3}}{4}=\frac{5a}{6}$ . This means each full glass must contribute $1a-\frac{5a}{6} = \frac{1a}{6}$ where $a=1$ . $\boxed{\textbf{(C) }\frac{1}{6}}$

~vsinghminhas

Vide Solution 1 by SpreadTheMathLove

https://www.youtube.com/watch?v=SUnhwbA5_So

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.

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 ==See also==
-{{AMC10 box|year=2023|ab=B|before=First Problem|num-b=2}}
+{{AMC10 box|year=2023|ab=B|before=First Problem|num-a=2}}
-{{AMC12 box|year=2023|ab=B|before=First Problem|num-b=2}}
+{{AMC12 box|year=2023|ab=B|before=First Problem|num-a=2}}
 [[Category:Rate Problems]]
 {{MAA Notice}}

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