2018 AMC 8 Problems/Problem 17


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 $5$ times as fast as Bella walks. The distance between their houses is $2$ miles, which is $10,560$ feet, and Bella covers $2 \tfrac{1}{2}$ feet with each step. How many steps will Bella take by the time she meets Ella?

$\textbf{(A) }704\qquad\textbf{(B) }845\qquad\textbf{(C) }1056\qquad\textbf{(D) }1760\qquad \textbf{(E) }3520$

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 $10560\div60=176$ 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, $1760\div2.5=\boxed{\textbf{(A) }704}$

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 $\boxed{\textbf{(A) }704}$.

Solution 3 (Fractions)

We can turn $2 \tfrac{1}{2}$ into an improper fraction. It will then become $\frac{5}{2}$. Since Ella bikes 5 times faster, we multiply $\frac{5}{2}\cdot 5=\frac{25}{2}$. Then we add $\frac{5}{2}+\frac{25}{2}$ to find the distance they walk and bike together for each step of Bella's: $\frac{30}{2} = 15$. This means that they travel 15 ft. of distance for each step that Bella takes. Divide 10,560 by 15 to find that Bella takes $\boxed{\textbf{(A) }704}$ steps.

Video Solution (CREATIVE ANALYSIS!!!)


~Education, the Study of Everything

Video Solution by OmegaLearn


~ pi_is_3.14

Video Solution



https://www.youtube.com/watch?v=UczCIsRzAeo ~David

See Also

2018 AMC 8 (ProblemsAnswer KeyResources)
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. AMC logo.png