Difference between revisions of "2021 Fall AMC 12A Problems/Problem 5"
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<math>\textbf{(A) }6\qquad\textbf{(B) }8\qquad\textbf{(C) }10\qquad\textbf{(D) }11\qquad\textbf{(E) }15</math> | <math>\textbf{(A) }6\qquad\textbf{(B) }8\qquad\textbf{(C) }10\qquad\textbf{(D) }11\qquad\textbf{(E) }15</math> | ||
| − | == Solution == | + | == Solution 1 == |
There are <math>41-1=40</math> gaps between the <math>41</math> telephone poles, so the distance of each gap is <math>5280\div40=132</math> feet. | There are <math>41-1=40</math> gaps between the <math>41</math> telephone poles, so the distance of each gap is <math>5280\div40=132</math> feet. | ||
| − | Each of Oscar's | + | Each of Oscar's leaps covers <math>132\div12=11</math> feet, and each of Elmer's strides covers <math>132\div44=3</math> feet. |
Therefore, Oscar's leap is <math>11-3=\boxed{\textbf{(B) }8}</math> feet longer than Elmer's stride. | Therefore, Oscar's leap is <math>11-3=\boxed{\textbf{(B) }8}</math> feet longer than Elmer's stride. | ||
~MRENTHUSIASM | ~MRENTHUSIASM | ||
| + | |||
| + | == Solution 2== | ||
| + | There are <math>41-1=40</math> gaps between the <math>41</math> telephone poles, so Elmer takes <math>44 \cdot 40 = 1760</math> strides in total, and Oscar takes <math>12 \cdot 40 = 480</math> leaps in total. Therefore, the answer is <math>(5280 \div 480) - (5280 \div 1760) = 11-3=\boxed{\textbf{(B) }8}</math>. | ||
| + | |||
| + | ~MrThinker | ||
| + | |||
| + | ==Video Solution (Simple and Quick)== | ||
| + | https://youtu.be/895dqtet8MA | ||
| + | |||
| + | ~Education, the Study of Everything | ||
| + | |||
| + | |||
| + | ==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== | ||
| + | https://youtu.be/AEmEKsHe2i0 | ||
| + | |||
| + | ~savannahsolver | ||
| + | |||
| + | ==Video Solution by HS Competition Academy== | ||
| + | https://youtu.be/0HjK-M6BJ8E | ||
| + | |||
| + | ~Charles3829 | ||
| + | |||
| + | ==Video Solution== | ||
| + | https://youtu.be/TjWjrApq-OQ | ||
| + | |||
| + | ~Lucas | ||
==See Also== | ==See Also== | ||
Latest revision as of 13:28, 13 June 2023
- 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 gaps between the telephone poles, so Elmer takes 
strides in total, and Oscar takes 
leaps in total. Therefore, the answer is 
.
~MrThinker
Video Solution (Simple and Quick)
~Education, the Study of Everything
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
Video Solution
~Lucas
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. 
