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2018 AMC 10A Problem 2

2018 AMC 10A Problem 2AMC 10 Step by Step

Name: Date: September 13, 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

Give Jacqueline a convenient amount such as 100; then Liliane has 150 and Alice 125, and 150/125 = 1.2 means 20% more.

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

Why this works

" $p\%$ more than" means "multiplied by $1 + \tfrac{p}{100}$ ," so both amounts are multiples of Jacqueline's. The comparison between Liliane and Alice is the quotient $\tfrac{1.5}{1.25}$ , which does not depend on Jacqueline's actual amount; that is why substituting $100$ is legitimate and fast. Percentages never subtract across different bases.

The trap

Subtracting the percentages (50% - 25% = 25%) instead of comparing the two amounts as a ratio.

Common mistakes

Subtracting the percentages (50% - 25% = 25%) instead of comparing the two amounts as a ratio.
Computing Alice relative to Liliane ( $125/150 \approx 83\%$ , i.e. about $17\%$ less) and then misreading the direction of the question.

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