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 $\boxed{\textbf{(A) } 20 \%}$

Solution 1 (Easier)

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

~Ezraft

Solution 2

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 $\boxed{\textbf{(A) } 20 \%}$ .

 ~lakecomo224

Video Solutions

https://youtu.be/vO-ELYmgRI8

https://youtu.be/jx9RnjX9g-Q

~savannahsolver

https://youtu.be/zMeYuDelX8E

Education, the Study of Everything

2018 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 1	Followed by Problem 3
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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

Page

Toolbox

Search