2018 AMC 10A Problems/Problem 2: Difference between revisions
m moved a video solution down |
m Fixed a small typo. |
||
| Line 1: | Line 1: | ||
== 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 20:17, 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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing
