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2006 AMC 10A Problem 3

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2006 AMC 10A Problem 3AMC 10 Step by Step

Name: Date: October 28, 2026

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

The ratio of Mary's age to Alice's age is $3:5$ . Alice is $30$ years old. How old is Mary?

A) $15$
B) $18$
C) $20$
D) $24$
E) $50$

Answer

Key insight

Think of the ages as 3 equal parts and 5 equal parts. Alice's 5 parts are 30 years, so one part is 6 years.

Solution

A ratio of $3:5$ means Mary's age is $3$ equal parts and Alice's age is $5$ of the same parts.

Alice's $5$ parts make $30$ years, so one part is $30 \div 5 = 6$ years.
Mary has $3$ parts: $3 \times 6 = 18$ years.
Check: $18 : 30$ simplifies (divide both by $6$ ) to $3 : 5$ . Correct.

The answer is $\boxed{\textbf{(B)}\ 18}$ .

Why this works

A ratio tells you how many equal "parts" each quantity has. Once you know one quantity, you know the size of one part, and then everything else follows. A quick sense check helps: Mary's ratio number $3$ is smaller than Alice's $5$ , so Mary must be younger than $30$ .

The trap

Flipping the ratio and computing 30 times 5/3 = 50, the wrong choice (E). Mary's number in the ratio is smaller, so she must be younger than 30.

Concise solution

Think of the ages as $3$ parts and $5$ parts of the same size. Alice's $5$ parts equal $30$ years, so one part is $30 \div 5 = 6$ years.

Mary has $3$ parts: $3 \cdot 6 = 18$ years.

Equivalently, $\dfrac{\text{Mary}}{30} = \dfrac{3}{5}$ gives $\text{Mary} = \dfrac{3 \cdot 30}{5} = 18$ .

The answer is $\boxed{\textbf{(B)}\ 18}$ .

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