2018 AMC 10A Problems/Problem 2: Difference between revisions
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===Solution=== | ===Solution=== | ||
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) | 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> | ||
== See Also == | == See Also == | ||
Revision as of 15:59, 8 February 2018
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 Alica 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
Let's assume that Jacqueline has 
gallon of soda. Then Alice has 
gallons and Liliane has 
gallons. Doing division, we find out that 
, which means that Liliane has 20% more soda. Therefore, the answer is 
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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