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. |
Revision as of 21:17, 26 January 2021
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
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
Video Solutions
~savannahsolver
Education, the Study of Everything
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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