2018 AMC 8 Problems/Problem 17

Revision as of 17:36, 1 November 2020 by Nezha33 (talk | contribs) (Solution)

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

Since Ella rides 5 times as fast as Bella, Ella rides at a rate of $\frac{25}{2}$ or $12 \tfrac{1}{2}$. Together, they move $15$ feet towards each other every unit. Dividing $10560$ by $15$ to find the number of steps Bella takes results in the answer of $\boxed{\textbf{(A) }704}$

Solution 2

Since Ella rides 5 times faster than Bella, the ratio of their speeds is 5:1. This means that Bella travels 1/6 of the way, and 1/6 of 10560 feet is 1760 feet. Bella also walks 2.5 feet in a step, and 1760 divided by 2.5 is $\boxed{\textbf{(A) }704}$.

Solution 3

We know that Ella's speed is $12.5$ st/s (steps per second) and Bella's speed is $2.5$ st/s. Let the speeds in which they are walking and riding at be the slope of two equations, one for Bella and one for Ella. We make $s$ the amount of steps she takes and $d$ distance from their houses. Since we are finding where they meet, we need to solve for the intersections of the two equations for Ella and Bella. The equations we get are:

\[d = -12.5s + 10560\] \[d = 2.5s\]

Subtracting both equations from each other gives us:

\[-15s + 10560 = 0\]

Solving that equation grants us:

\[s = 704\]

Since we are finding the number of steps they take until they meet each other, we can stop there. The solution is

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

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

Invalid username
Login to AoPS