Difference between revisions of "2018 AMC 8 Problems/Problem 17"

(Solution)
(Solution)
Line 6: Line 6:
 
==Solution==
 
==Solution==
 
Since Ella rides 5 times as fast as Bella, Ella rides at a rate of <math>\frac{25}{2}</math> or <math>12 \tfrac{1}{2}</math>. Together, they move <math>15</math> feet towards each other every unit. Dividing <math>10560</math> by <math>15</math> to find the number of steps Ella takes results in the answer of <math>\boxed{\textbf{(A) }704}</math>
 
Since Ella rides 5 times as fast as Bella, Ella rides at a rate of <math>\frac{25}{2}</math> or <math>12 \tfrac{1}{2}</math>. Together, they move <math>15</math> feet towards each other every unit. Dividing <math>10560</math> by <math>15</math> to find the number of steps Ella takes results in the answer of <math>\boxed{\textbf{(A) }704}</math>
 +
 
Solution 2
 
Solution 2
  

Revision as of 20:42, 11 April 2019

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

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 Ella 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 704.

-UnstoppableGoddess

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