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"

(→‎Solution)
(Added Problem header, resized Solution header, and LaTeXeD the 20%)
Line 1: Line 1:
 +
== 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 Alica have?
 
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 Alica have?
  
 +
<math>\textbf{(A) }</math> Liliane has <math>20\%</math> more soda than Alice.
 +
 +
<math>\textbf{(B) }</math> Liliane has <math>25\%</math> more soda than Alice.
 +
 +
<math>\textbf{(C) }</math> Liliane has <math>45\%</math> more soda than Alice.
 +
 +
<math>\textbf{(D) }</math> Liliane has <math>75\%</math> more soda than Alice.
  
−
<math>\textbf{(A) }</math>Liliane has <math>20\%</math> more soda than Alice.
+
<math>\textbf{(E) }</math> Liliane has <math>100\%</math> more soda than Alice.
−
<math>\textbf{(B) }</math>Liliane has <math>25\%</math> more soda than Alice.
 
−
<math>\textbf{(C) }</math>Liliane has <math>45\%</math> more soda than Alice.
 
−
<math>\textbf{(D) }</math>Liliane has <math>75\%</math> more soda than Alice.
 
−
<math>\textbf{(E) }</math>Liliane has <math>100\%</math> more soda than Alice.
 
  
−
===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)}}</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)}}</math>
  
 
== See Also ==
 
== See Also ==

Revision as of 20:52, 8 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 $\boxed{\textbf{(A)}}$

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=90696"