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

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== Problem ==
 
== Problem ==
  
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Liliane has <math>50\%</math> more soda than Jacqueline, and Alice has <math>25\%</math> more soda than Jacqueline. What is the relationship between the amounts of soda that Liliane and Alica have?
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Liliane has <math>50\%</math> more soda than Jacqueline, and Alice has <math>25\%</math> more soda than Jacqueline. What is the relationship between the amounts of soda that Liliane and Alice have?
  
 
<math>\textbf{(A) }</math> Liliane has <math>20\%</math> more soda than Alice.
 
<math>\textbf{(A) }</math> Liliane has <math>20\%</math> more soda than Alice.
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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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== Solution ==
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== Solution 1 ==
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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 <math>20\%</math> 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(s) 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 <math>20\%</math> more soda. Therefore, the answer is <math>\boxed{\textbf{(A)}}</math>.
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== Solution 2 ==
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WLOG, lets use <math>4</math> gallons instead of <math>1</math>. When you work it out, you get <math>6</math> gallons and <math>5</math> gallons. We have <math>6 - 5 = 1</math> is <math>20\%</math> of <math>5</math>. Thus, we reach <math>\boxed{\textbf{(A)}}</math>.
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~Ezraft
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== Solution 3 ==
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If Jacqueline has <math>x</math> gallons of soda, Alice has <math>1.25x</math> gallons, and Liliane has <math>1.5x</math> gallons. Thus, the answer is <math>\frac{1.5}{1.25}=1.2</math> <math>=></math> Liliane has <math>20\%</math> more soda. Our answer is <math>\boxed{\textbf{(A)}}</math>.
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~lakecomo224 (minor edits made by <B+)
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==Video Solution (HOW TO THINK CREATIVELY!)==
 +
https://youtu.be/zMeYuDelX8E
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 +
Education, the Study of Everything
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== Video Solutions ==
 +
https://youtu.be/vO-ELYmgRI8
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https://youtu.be/jx9RnjX9g-Q
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~savannahsolver
  
 
== See Also ==
 
== See Also ==

Latest revision as of 09:21, 18 August 2024

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 (minor edits made by <B+)

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)
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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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