Difference between revisions of "2018 AMC 10A Problems/Problem 2"
(→Solution 1 (Easier)) |
MRENTHUSIASM (talk | contribs) m (→Solution 2) |
||
| Line 27: | Line 27: | ||
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> -> Liliane has <math>20\%</math> more soda. Our answer is <math>\boxed{\textbf{(A) } 20 \%}</math>. | 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> -> Liliane has <math>20\%</math> more soda. Our answer is <math>\boxed{\textbf{(A) } 20 \%}</math>. | ||
| − | + | ~lakecomo224 | |
== Video Solutions == | == Video Solutions == | ||
Revision as of 00:09, 23 August 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 1 (Easier)
WLOG, lets use 4 gallons instead of 1. When you work it out, you get 6 gallons and 5 gallons.
is of 
. Thus, we reach .
~Ezraft
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
