Difference between revisions of "2018 AMC 10A Problems/Problem 2"
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===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 20% more soda. Therefore, the answer is <math>\boxed{\textbf{(A)}}</math> | ||
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+ | == See Also == | ||
+ | |||
+ | {{AMC10 box|year=2018|ab=A|before=1|num-a=3}} | ||
+ | {{MAA Notice}} |
Revision as of 14:54, 8 February 2018
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 Alica 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
Let's assume that Jacqueline has gallon of soda. Then Alice has gallons and Liliane has gallons. Doing division, we find out that , which means that Liliane has 20% more soda. Therefore, the answer is
See Also
2018 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by 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 |
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