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

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

Revision as of 16:59, 8 February 2018

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

Let's assume that Jacqueline has $1$ gallon 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 $\boxed{\textbf{(A)}\text{Liliane has$ (Error compiling LaTeX. Unknown error_msg)20\%$more soda than Alice.}}$ (Error compiling LaTeX. Unknown error_msg)

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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