2019 AMC 10B Problem 1

P1-10~1 minPrint

Ratios, Percents & Averages Set up the equation/formula and compute; no special trick needed

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?

Show solution without answering

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 $\frac56 V_2$ :
$\frac{V_1}{V_2} = \frac{3}{4} \div \frac{5}{6} = \frac34 \cdot \frac65 = \frac{18}{20} = \frac{9}{10}.$

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

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)