Printable

2018 AMC 10A Problem 2

2018 AMC 10A Problem 2AMC 10 Step by Step

Name: Date: October 28, 2026

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

Liliane has $50\%$ more soda than Jacqueline, and Alice has $25\%$ more soda than Jacqueline. What is the relationship between the amounts of soda that Liliane and Alice have?

A)Liliane has $20\%$ more soda than Alice.
B)Liliane has $25\%$ more soda than Alice.
C)Liliane has $45\%$ more soda than Alice.
D)Liliane has $75\%$ more soda than Alice.
E)Liliane has $100\%$ more soda than Alice.

Answer

Key insight

Pick an easy amount for Jacqueline, like 100. Then Liliane has 150 and Alice has 125, and you just compare 150 with 125.

Solution

The question only compares amounts, so the actual size of Jacqueline's soda does not matter. Pick a number that makes percents easy.

1. Give Jacqueline $100$ ounces.
2. Liliane has $50\%$ more: $100 + 50 = 150$ ounces.
3. Alice has $25\%$ more: $100 + 25 = 125$ ounces.
4. Compare Liliane with Alice. Liliane has $150 - 125 = 25$ ounces more than Alice. As a percent of Alice's amount, that is
$\frac{25}{125} = \frac{1}{5} = 20\%.$

So Liliane has $20\%$ more soda than Alice.

The answer is $\boxed{\textbf{(A)}\ \text{Liliane has }20\%\text{ more soda than Alice.}}$

Why this works

"More than Alice" means the percent is measured against Alice's amount. The $25$ extra ounces are $25\%$ of Jacqueline's $100$ but only $20\%$ of Alice's $125$ . Whenever a percent problem gives no actual amounts, choose $100$ for the base amount: every answer is the same for any starting amount, and $100$ makes the arithmetic easy.

The trap

Subtracting the percents, 50% minus 25%, and answering 25%. The two percents are taken of Jacqueline's amount, not Alice's, so they cannot be subtracted.

Concise solution

Percent comparisons depend only on ratios, so pick a convenient amount for Jacqueline: $J = 100$ .

Then Liliane has $L = 1.5 \cdot 100 = 150$ and Alice has $A = 1.25 \cdot 100 = 125$ .

Compare Liliane to Alice:
$\frac{L}{A} = \frac{150}{125} = \frac{6}{5} = 1.2,$
so Liliane has $20\%$ more than Alice.

The answer is $\boxed{\textbf{(A)}\ \text{Liliane has }20\%\text{ more soda than Alice.}}$

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 · Test small/specific values or special cases to find or verify the answer