Difference between revisions of "2021 Fall AMC 12A Problems/Problem 5"
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== Solution 2== | == Solution 2== | ||
There are <math>40-1=40</math> spaces between the <math>41</math> telephone poles, so Elmer takes <math>44 \times 40 = 1760</math> total strides. Oscar takes <math>12 \times 40 = 480</math> total strides. Thus, <math>(5280 \div 480) - (5280 \div 1760) = 11-3=\boxed{\textbf{(B) }8}</math>. | There are <math>40-1=40</math> spaces between the <math>41</math> telephone poles, so Elmer takes <math>44 \times 40 = 1760</math> total strides. Oscar takes <math>12 \times 40 = 480</math> total strides. Thus, <math>(5280 \div 480) - (5280 \div 1760) = 11-3=\boxed{\textbf{(B) }8}</math>. | ||
| − | + | ||
| + | ~MrThinker | ||
==Video Solution by TheBeautyofMath== | ==Video Solution by TheBeautyofMath== | ||
Revision as of 14:35, 9 August 2022
- The following problem is from both the 2021 Fall AMC 10A #6 and 2021 Fall AMC 12A #5, so both problems redirect to this page.
Contents
Problem
Elmer the emu takes 
equal strides to walk between consecutive telephone poles on a rural road. Oscar the ostrich can cover the same distance in 
equal leaps. The telephone poles are evenly spaced, and the 
st pole along this road is exactly one mile (
feet) from the first pole. How much longer, in feet, is Oscar's leap than Elmer's stride?
Solution 1
There are 
gaps between the telephone poles, so the distance of each gap is 
feet.
Each of Oscar's leaps covers 
feet, and each of Elmer's strides covers 
feet.
Therefore, Oscar's leap is 
feet longer than Elmer's stride.
~MRENTHUSIASM
Solution 2
There are 
spaces between the telephone poles, so Elmer takes 
total strides. Oscar takes 
total strides. Thus, 
.
~MrThinker
Video Solution by TheBeautyofMath
for AMC 10: https://youtu.be/ycRZHCOKTVk
for AMC 12: https://youtu.be/jY-17W6dA3c?t=591
~IceMatrix
Video Solution by WhyMath
~savannahsolver
Video Solution by HS Competition Academy
~Charles3829
See Also
| 2021 Fall AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 4 |
Followed by Problem 6 |
| 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 AMC 12 Problems and Solutions | |
| 2021 Fall AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 5 |
Followed by Problem 7 | |
| 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 AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
