AMC 10 Step by Step

2018 AMC 10A Problem 2

P1-10~1 minPrint

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

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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 100100 ounces.
2. Liliane has 50%50\% more: 100+50=150100 + 50 = 150 ounces.
3. Alice has 25%25\% more: 100+25=125100 + 25 = 125 ounces.
4. Compare Liliane with Alice. Liliane has 150−125=25150 - 125 = 25 ounces more than Alice. As a percent of Alice's amount, that is
25125=15=20%.\frac{25}{125} = \frac{1}{5} = 20\%.

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

The answer is (A) Liliane has 20% more soda than Alice.\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 2525 extra ounces are 25%25\% of Jacqueline's 100100 but only 20%20\% of Alice's 125125. Whenever a percent problem gives no actual amounts, choose 100100 for the base amount: every answer is the same for any starting amount, and 100100 makes the arithmetic easy.

Concise solution

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

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

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

Compare Liliane to Alice:
LA=150125=65=1.2,\frac{L}{A} = \frac{150}{125} = \frac{6}{5} = 1.2,
so Liliane has 20%20\% more than Alice.

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

Review in the AoPS books

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Problem © Mathematical Association of America (MAA), American Mathematics Competitions. Reproduced for non-commercial educational use. Solution and commentary are original to this site.