Difference between revisions of "2018 AMC 8 Problems/Problem 17"
m (→Solution) |
(→Video Solution) |
||
(24 intermediate revisions by 16 users not shown) | |||
Line 1: | Line 1: | ||
− | ==Problem | + | ==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 5 times as fast as Bella walks. The distance between their houses is <math>2</math> miles, which is <math>10,560</math> feet, and Bella covers <math>2 \tfrac{1}{2}</math> feet with each step. How many steps will Bella take by the time she meets Ella? | + | 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 <math>5</math> times as fast as Bella walks. The distance between their houses is <math>2</math> miles, which is <math>10,560</math> feet, and Bella covers <math>2 \tfrac{1}{2}</math> feet with each step. How many steps will Bella take by the time she meets Ella? |
<math>\textbf{(A) }704\qquad\textbf{(B) }845\qquad\textbf{(C) }1056\qquad\textbf{(D) }1760\qquad \textbf{(E) }3520</math> | <math>\textbf{(A) }704\qquad\textbf{(B) }845\qquad\textbf{(C) }1056\qquad\textbf{(D) }1760\qquad \textbf{(E) }3520</math> | ||
− | ==Solution== | + | ==Solution 1== |
− | |||
− | == | + | Since Ella rides 5 times faster than Bella, the ratio of their speeds is 5:1. For Bella, we have D/R = t and for Ella, we have D/5r = t, however we know times for both girls must be the same, and so that means in D/5R = t, numerator becomes 5D(Ella travels 5 times the distance that Bella does). 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 <math>\boxed{\textbf{(A) }704}</math>. |
− | |||
− | + | ||
+ | ==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. For Bella it takes 4224 steps and for Ella it would take 1/5th those steps since Ella goes 5 times faster than Bella so it takes 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>. | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/ycZ381n_1bQ | ||
+ | |||
+ | ~savannahsolver | ||
+ | |||
+ | https://www.youtube.com/watch?v=UczCIsRzAeo | ||
==See Also== | ==See Also== |
Latest revision as of 23:29, 16 April 2022
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
Since Ella rides 5 times faster than Bella, the ratio of their speeds is 5:1. For Bella, we have D/R = t and for Ella, we have D/5r = t, however we know times for both girls must be the same, and so that means in D/5R = t, numerator becomes 5D(Ella travels 5 times the distance that Bella does). 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 .
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. For Bella it takes 4224 steps and for Ella it would take 1/5th those steps since Ella goes 5 times faster than Bella so it takes 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 .
Video Solution
~savannahsolver
https://www.youtube.com/watch?v=UczCIsRzAeo
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.