# 2018 AMC 10A Problems/Problem 2

## Problem

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

## Solution 1

Let's assume that Jacqueline has $1$ gallon(s) of soda. Then Alice has $1.25$ gallons and Liliane has $1.5$ gallons. Doing division, we find out that $\frac{1.5}{1.25}=1.2$, which means that Liliane has $20\%$ more soda. Therefore, the answer is $\boxed{\textbf{(A)}}$.

## Solution 2

WLOG, lets use $4$ gallons instead of $1$. When you work it out, you get $6$ gallons and $5$ gallons. We have $6 - 5 = 1$ is $20\%$ of $5$. Thus, we reach $\boxed{\textbf{(A)}}$.

~Ezraft

## Solution 3

If Jacqueline has $x$ gallons of soda, Alice has $1.25x$ gallons, and Liliane has $1.5x$ gallons. Thus, the answer is $\frac{1.5}{1.25}=1.2$ -> Liliane has $20\%$ more soda. Our answer is $\boxed{\textbf{(A)}}$.

~lakecomo224

## Video Solution (HOW TO THINK CREATIVELY!)

Education, the Study of Everything

## Video Solutions

~savannahsolver

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 