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

m (moved a video solution down)
(Solution 1 (Easier))
(8 intermediate revisions by 2 users not shown)
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 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 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.
Line 16: Line 16:
 
== Solution 1 ==
 
== Solution 1 ==
 
Let's assume that Jacqueline has <math>1</math> gallon(s) 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(s) 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 1 (Easier)==
 +
 +
WLOG, lets use 4 gallons instead of 1. When you work it out, you get 6 gallons and 5 gallons.<math>6 - 5 = 1</math> is <math>20\%</math> of <math>5</math>. Thus, we reach <math>\boxed{\textbf{(A) } 20 \%}</math>.
 +
 +
~Ezraft
  
 
== Solution 2 ==
 
== Solution 2 ==

Revision as of 21:43, 1 February 2021

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 Alice 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 1

Let's assume that Jacqueline has $1$ gallon(s) 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 1 (Easier)

WLOG, lets use 4 gallons instead of 1. When you work it out, you get 6 gallons and 5 gallons.$6 - 5 = 1$ is $20\%$ of $5$. Thus, we reach $\boxed{\textbf{(A) } 20 \%}$.

~Ezraft

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 $\boxed{\textbf{(A) } 20 \%}$.

 ~lakecomo224

Video Solutions

https://youtu.be/vO-ELYmgRI8

https://youtu.be/jx9RnjX9g-Q

~savannahsolver

https://youtu.be/zMeYuDelX8E

Education, the Study of Everything

See Also

2018 AMC 10A (ProblemsAnswer KeyResources)
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