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Revision as of 17:00, 16 February 2019

The following problem is from both the 2019 AMC 10B #1 and 2019 AMC 12B #1, so both problems redirect to this page.

Problem

Alicia had two containers. The first was $\tfrac{5}{6}$ full of water and the second was empty. She poured all the water from the first container into the second container, at which point the second container was $\tfrac{3}{4}$ full of water. What is the ratio of the volume of the first container to the volume of the second container?

$\textbf{(A) } \frac{5}{8} \qquad \textbf{(B) } \frac{4}{5} \qquad \textbf{(C) } \frac{7}{8} \qquad \textbf{(D) } \frac{9}{10} \qquad \textbf{(E) } \frac{11}{12}$

Solution

Let the first jar's volume be $A$ and the second's be $B$. It is given that $\frac{3}{4}A=\frac{5}{6}B$. We find that $\frac{B}{A}=\frac{3/4}{5/6}=\boxed{\frac{9}{10}}.$

We already know that this is the ratio of smaller to larger volume because it is less than $1.$

Solution 2

We can set up a ratio to solve this problem. If x is the volume of the first container, and y is the volume of the second container, then: \[\frac{5}{6}x = \frac{3}{4}y\]

Cross Multiplying allows us to get $\frac{x}{y} = \frac{3}{4} \cdot \frac{6}{5} \Rightarrow \frac{18}{20} \Rightarrow \frac{9}{10}$. Thus the ratio of the volume of the first container to the second container is $\boxed{(\text{D})\frac{9}{10}}$

An alternate solution is to plug in some maximum volume for the first container - let's say $72$, so there was a volume of 60 in the first container, and then the second container also has a volume of $60$, so you get $60 \cdot \frac{4}{3} \Rightarrow 80$. Thus, $\frac{72}{80} \Rightarrow \frac{9}{10}$.

See Also

2019 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
2019 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. AMC logo.png

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