Difference between revisions of "2018 AMC 10A Problems/Problem 2"
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== Problem == | == Problem == | ||
| − | 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 | + | 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. | ||
| − | == Solution == | + | == Solution 1 == |
| − | 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) } 20 \%}</math> | + | 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>. |
| + | |||
| + | == Solution 2 == | ||
| + | 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>. | ||
| + | |||
| + | ~Ezraft | ||
| + | |||
| + | == Solution 3 == | ||
| + | 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>. | ||
| + | |||
| + | ~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 == | == See Also == | ||
Latest revision as of 09:21, 18 August 2024
Contents
Problem
Liliane has 
more soda than Jacqueline, and Alice has 
more soda than Jacqueline. What is the relationship between the amounts of soda that Liliane and Alice have?
Liliane has 
more soda than Alice.
Liliane has more soda than Alice.
Liliane has 
more soda than Alice.
Liliane has 
more soda than Alice.
Liliane has 
more soda than Alice.
Solution 1
Let's assume that Jacqueline has 
gallon(s) of soda. Then Alice has 
gallons and Liliane has 
gallons. Doing division, we find out that 
, which means that Liliane has more soda. Therefore, the answer is 
.
Solution 2
WLOG, lets use 
gallons instead of . When you work it out, you get 
gallons and 
gallons. We have 
is of . Thus, we reach .
~Ezraft
Solution 3
If Jacqueline has 
gallons of soda, Alice has 
gallons, and Liliane has 
gallons. Thus, the answer is 
Liliane has more soda. Our answer is .
~lakecomo224 (minor edits made by <B+)
Video Solution (HOW TO THINK CREATIVELY!)
Education, the Study of Everything
Video Solutions
~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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
