Difference between revisions of "2018 AMC 10A Problems/Problem 2"
m (Fixed a small typo.) |
Alexyuan2017 (talk | contribs) (→Solution 3) |
||
(14 intermediate revisions by 5 users not shown) | |||
Line 12: | Line 12: | ||
<math>\textbf{(E) }</math> Liliane has <math>100\%</math> more soda than Alice. | <math>\textbf{(E) }</math> Liliane has <math>100\%</math> more soda than Alice. | ||
− | |||
== 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) } | + | 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 == | == 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> | + | 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>. |
+ | |||
+ | ~Ezraft | ||
+ | |||
+ | == 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+) | ||
+ | |||
+ | ==Video Solution (HOW TO THINK CREATIVELY!)== | ||
+ | https://youtu.be/zMeYuDelX8E | ||
+ | |||
+ | Education, the Study of Everything | ||
+ | |||
+ | |||
− | |||
== Video Solutions == | == Video Solutions == | ||
Line 28: | Line 40: | ||
~savannahsolver | ~savannahsolver | ||
− | |||
− | |||
− | |||
− | |||
== See Also == | == See Also == |
Latest revision as of 10:21, 18 August 2024
Contents
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 2
WLOG, lets use gallons instead of
. When you work it out, you get
gallons and
gallons. We have
is
of
. Thus, we reach
.
~Ezraft
Solution 3
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 (minor edits made by <B+)
Video Solution (HOW TO THINK CREATIVELY!)
Education, the Study of Everything
Video Solutions
~savannahsolver
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.