Difference between revisions of "2018 AMC 8 Problems/Problem 17"
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We know that Bella goes 2.5 feet per step and since Ella rides 5 times faster than Bella she must go 12.5 feet on her bike for every step of Bella's. For Bella, it takes 4,224 steps, and for Ella, it takes 1/5th those steps since Ella goes 5 times faster than Bella, taking her 844.8 steps. The number of steps where they meet therefore must be less than 844.8. The only answer choice less than it is <math>\boxed{\textbf{(A) }704}</math>. | We know that Bella goes 2.5 feet per step and since Ella rides 5 times faster than Bella she must go 12.5 feet on her bike for every step of Bella's. For Bella, it takes 4,224 steps, and for Ella, it takes 1/5th those steps since Ella goes 5 times faster than Bella, taking her 844.8 steps. The number of steps where they meet therefore must be less than 844.8. The only answer choice less than it is <math>\boxed{\textbf{(A) }704}</math>. | ||
| − | == Solution 3 == | + | == Solution 3 (Fractions) == |
We can turn <math>2 \tfrac{1}{2}</math> into an improper fraction. It will then become 5/2. Since Ella bikes 5 times faster, we multiply 5/2 by 5 to get 25/2. Then we add 5/2 to it in order to find the distance they walk and bike together in total. After adding, you should get 30/2 which is equal to 15. This means that after 15 times, they will meet. So you have to divide 10,560 by 15. The answer should be <math>\boxed{\textbf{(A) }704}</math>. | We can turn <math>2 \tfrac{1}{2}</math> into an improper fraction. It will then become 5/2. Since Ella bikes 5 times faster, we multiply 5/2 by 5 to get 25/2. Then we add 5/2 to it in order to find the distance they walk and bike together in total. After adding, you should get 30/2 which is equal to 15. This means that after 15 times, they will meet. So you have to divide 10,560 by 15. The answer should be <math>\boxed{\textbf{(A) }704}</math>. | ||
Revision as of 17:56, 20 January 2024
Contents
Problem
Bella begins to walk from her house toward her friend Ella's house. At the same time, Ella begins to ride her bicycle toward Bella's house. They each maintain a constant speed, and Ella rides 
times as fast as Bella walks. The distance between their houses is 
miles, which is 
feet, and Bella covers 
feet with each step. How many steps will Bella take by the time she meets Ella?
Solution 1 (Fast and Easy)
Every 10 feet Bella goes, Ella goes 50 feet, which means a total of 60 feet. They need to travel that 60 feet 
times to travel the entire 2 miles. Since Bella goes 10 feet 176 times, this means that she travels a total of 1760 feet. And since she walks 2.5 feet each step, 
~ alexdapog A-A
Solution 2 (Use Answer Choices to our advantage)
We know that Bella goes 2.5 feet per step and since Ella rides 5 times faster than Bella she must go 12.5 feet on her bike for every step of Bella's. For Bella, it takes 4,224 steps, and for Ella, it takes 1/5th those steps since Ella goes 5 times faster than Bella, taking her 844.8 steps. The number of steps where they meet therefore must be less than 844.8. The only answer choice less than it is 
.
Solution 3 (Fractions)
We can turn into an improper fraction. It will then become 5/2. Since Ella bikes 5 times faster, we multiply 5/2 by 5 to get 25/2. Then we add 5/2 to it in order to find the distance they walk and bike together in total. After adding, you should get 30/2 which is equal to 15. This means that after 15 times, they will meet. So you have to divide 10,560 by 15. The answer should be .
Video Solution (CREATIVE ANALYSIS!!!)
~Education, the Study of Everything
Video Solution by OmegaLearn
https://youtu.be/TkZvMa30Juo?t=1123
~ pi_is_3.14
Video Solution
~savannahsolver
https://www.youtube.com/watch?v=UczCIsRzAeo ~David
See Also
| 2018 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 16 |
Followed by Problem 18 | |
| 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 AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
