2018 AMC 10A Problems/Problem 2: Difference between revisions

Latest revision as of 09:21, 18 August 2024

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

Solution 2

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

~Ezraft

Solution 3

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

~lakecomo224 (minor edits made by <B+)

Video Solution (HOW TO THINK CREATIVELY!)

https://youtu.be/zMeYuDelX8E

Education, the Study of Everything

Video Solutions

https://youtu.be/vO-ELYmgRI8

https://youtu.be/jx9RnjX9g-Q

~savannahsolver

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

@@ Line 13: / Line 13: @@
 <math>\textbf{(E) }</math> Liliane has <math>100\%</math> more soda than Alice.
+== 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)}}</math>.
+== Solution 2 ==
+WLOG, lets use <math>4</math> gallons instead of <math>1</math>. When you work it out, you get <math>6</math> gallons and <math>5</math> gallons. We have <math>6 - 5 = 1</math> is <math>20\%</math> of <math>5</math>. Thus, we reach <math>\boxed{\textbf{(A)}}</math>.
-== Solution 1 ==
+~Ezraft
-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 3 ==
+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> <math>=></math> Liliane has <math>20\%</math> more soda. Our answer is <math>\boxed{\textbf{(A)}}</math>.
+~lakecomo224 (minor edits made by <B+)
-== Solution 1 (Easier)==
+==Video Solution (HOW TO THINK CREATIVELY!)==
+https://youtu.be/zMeYuDelX8E
+Education, the Study of Everything
-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}.
-== 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>.
-  ~lakecomo224
 == Video Solutions ==
@@ Line 33: / Line 40: @@
 ~savannahsolver
-https://youtu.be/zMeYuDelX8E
-Education, the Study of Everything
 == See Also ==

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