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

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<math>\textbf{(E) }</math> Liliane has <math>100\%</math> more soda than Alice.
 
<math>\textbf{(E) }</math> Liliane has <math>100\%</math> more soda than Alice.
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==Video Solution==
 
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https://youtu.be/zMeYuDelX8E
 
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Education, the Study of Everything
 
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==Video Solution==
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== Video Solutions ==
 
https://youtu.be/vO-ELYmgRI8
 
https://youtu.be/vO-ELYmgRI8
  
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~savannahsolver
 
~savannahsolver
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https://youtu.be/zMeYuDelX8E
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Education, the Study of Everything
  
 
== See Also ==
 
== See Also ==

Revision as of 17:51, 29 November 2020

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 Alica 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 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

See Also

2018 AMC 10A (Problems • Answer Key • Resources)
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Problem 3
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All AMC 10 Problems and Solutions

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