# 2019 AMC 10B Problems/Problem 1

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, $\frac{72}{80} \Rightarrow \frac{9}{10}$.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 