Printable

2019 AMC 10B Problem 1

2019 AMC 10B Problem 1AMC 10 Step by Step

Name: Date: October 28, 2026

Ratios, Percents & Averages · difficulty 1 of 5 · about 1 min · position P1-10

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?

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

Answer

Key insight

The same water fills 5/6 of the first container and 3/4 of the second. Pick an amount of water that makes both containers whole numbers.

Solution

The water does not change when it is poured. So one amount of water is $\frac56$ of the first container and also $\frac34$ of the second. Choosing a friendly amount of water makes everything concrete.

Pick an amount of water that divides nicely by both $5$ and $3$ , say $15$ liters.
$15$ liters is $\frac56$ of the first container. One sixth of it is $15 \div 5 = 3$ liters, so the whole container is $6 \cdot 3 = 18$ liters.
$15$ liters is $\frac34$ of the second container. One quarter of it is $15 \div 3 = 5$ liters, so the whole container is $4 \cdot 5 = 20$ liters.
The ratio of the first volume to the second is $18 : 20 = 9 : 10$ .

The amount $15$ was only a convenient choice. Any other amount scales both containers by the same factor, so the ratio stays $\frac{9}{10}$ .

The answer is $\boxed{\textbf{(D)}\ \frac{9}{10}}$ .

Why this works

One thing (the water) is described as a fraction of two different wholes. Giving that thing a concrete size lets you rebuild each whole, and then comparing them is easy. As a check, the first container is only $\frac56$ full but the second is $\frac34$ full with the same water, so the first container must be the smaller one, and the ratio is less than $1$ .

Alternative approach

Call the volumes $V_1$ and $V_2$ . The water gives the equation $\frac56 V_1 = \frac34 V_2$ . Divide both sides by $\frac56 V_2$ :
$\frac{V_1}{V_2} = \frac{3}{4} \div \frac{5}{6} = \frac34 \cdot \frac65 = \frac{18}{20} = \frac{9}{10}.$

The trap

Flipping the division and getting 10/9, or subtracting the two fractions instead of comparing them.

Concise solution

Let the containers have volumes $V_1$ and $V_2$ . The water is the same before and after pouring, so
$\frac{5}{6}V_1 = \frac{3}{4}V_2.$
Solve for the requested ratio:
$\frac{V_1}{V_2} = \frac{3/4}{5/6} = \frac{3}{4}\cdot\frac{6}{5} = \frac{18}{20} = \frac{9}{10}.$

The answer is $\boxed{\textbf{(D)}\ \frac{9}{10}}$ .

Review in the AoPS books

Introduction to Algebra, Chapter 6: Ratios and Percents
Introduction to Algebra, Chapter 7: Proportion (§7.1, §7.2, §7.3)
Prealgebra, Chapter 13: Data and Statistics (§13.1)

Techniques

Set up the equation/formula and compute; no special trick needed