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Difference between revisions of "2018 AMC 10A Problems/Problem 2"

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<math>\textbf{(D) }</math> Liliane has <math>75\%</math> more soda than Alice.
 
<math>\textbf{(D) }</math> Liliane has <math>75\%</math> more soda than Alice.
  
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<math>\textbf{(E) }</math> Liliane has <math>100\%</math> more soda than Alice.
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<math>\textbf{(E) }</math> Liliane has farted on Alice and Alice poured her soda on Liliane so Alice has infinity% more than Alice.
  
 
== Solution 1 ==
 
== Solution 1 ==

Revision as of 17:55, 1 December 2023

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 farted on Alice and Alice poured her soda on Liliane so Alice has infinity% more 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!)

https://youtu.be/zMeYuDelX8E

Education, the Study of Everything



Video Solutions

https://youtu.be/vO-ELYmgRI8

https://youtu.be/jx9RnjX9g-Q

~savannahsolver

See Also

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

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