2020 AMC 8 Problems/Problem 3
Contents
- 1 Problem
- 2 Solution 1
- 3 Solution 2
- 4 Video Solution by NiuniuMaths (Easy to understand!)
- 5 Video Solution (🚀 Just 1 min 🚀)
- 6 Video Solution by WhyMath
- 7 Video Solution by The Learning Royal
- 8 Video Solution by Interstigation
- 9 Video Solution by North America Math Contest Go Go Go
- 10 Video Solution by STEMbreezy
- 11 See also
Problem
Carrie has a rectangular garden that measures feet by feet. She plants the entire garden with strawberry plants. Carrie is able to plant strawberry plants per square foot, and she harvests an average of strawberries per plant. How many strawberries can she expect to harvest?
Solution 1
Note that the unit of the answer is strawberries, which is the product of
- square feet
- plants per square foot
- strawberries per plant
By conversion factors, we have
~MRENTHUSIASM ~Bobthegod78
Solution 2
The area of the garden is square feet. Since Carrie plants strawberry plants per square foot, there are a total of strawberry plants, each of which produces strawberries on average. Accordingly, she can expect to harvest strawberries.
Video Solution by NiuniuMaths (Easy to understand!)
https://www.youtube.com/watch?v=8hgK6rESdek&t=9s
~NiuniuMaths
Video Solution (🚀 Just 1 min 🚀)
~Education, the Study of Everything
Video Solution by WhyMath
~savannahsolver
Video Solution by The Learning Royal
~The Learning Royal
Video Solution by Interstigation
https://youtu.be/YnwkBZTv5Fw?t=92
~Interstigation
Video Solution by North America Math Contest Go Go Go
https://www.youtube.com/watch?v=_IjQnXnVKeU (Please subscribe I Really need them!)
~North America Math Contest Go Go Go
Video Solution by STEMbreezy
https://youtu.be/L_vDc-i585o?list=PLFcinOE4FNL0TkI-_yKVEYyA_QCS9mBNS&t=89
~STEMbreezy
See also
2020 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
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.