Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2018 AMC 10A Problems/Problem 2"

m (→‎Solution)
(→‎Solution)
Line 15: Line 15:
 
== 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 <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 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>
 +
 +
== Solution 2 ==
 +
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>.
 +
 +
[i]Answer by lakecomo224[\i]
  
 
== See Also ==
 
== See Also ==

Revision as of 11:11, 13 February 2018

Problem

Liliane has $50\%$ more soda than Jacqueline, and Alice has $25\%$ more soda than Jacqueline. What is the relationship between the amounts of soda that Liliane and Alica have?

$\textbf{(A) }$ Liliane has $20\%$ more soda than Alice.

$\textbf{(B) }$ Liliane has $25\%$ more soda than Alice.

$\textbf{(C) }$ Liliane has $45\%$ more soda than Alice.

$\textbf{(D) }$ Liliane has $75\%$ more soda than Alice.

$\textbf{(E) }$ Liliane has $100\%$ more soda than Alice.

Solution

Let's assume that Jacqueline has $1$ gallon of soda. Then Alice has $1.25$ gallons and Liliane has $1.5$ gallons. Doing division, we find out that $\frac{1.5}{1.25}=1.2$, which means that Liliane has $20\%$ more soda. Therefore, the answer is $\boxed{\textbf{(A) } 20 \%}$

Solution 2

If Jacqueline has $x$ gallons of soda, Alice has $1.25x$ gallons, and Liliane has $1.5x$ gallons. Thus, the answer is $\frac{1.5}{1.25}=1.2$ -> Liliane has $20\%$ more soda. Our answer is $\boxed{\textbf{(A) } 20 \%}$.

[i]Answer by lakecomo224[\i]

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10A_Problems/Problem_2&oldid=91170"