Difference between revisions of "2020 AMC 8 Problems/Problem 3"
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We have <cmath>\left(6 \ \cancel{\mathrm{ft}}\cdot8 \ \cancel{\mathrm{ft}}\right)\cdot\left(4 \ \frac{\bcancel{\mathrm{plants}}}{\cancel{\mathrm{ft}^2}}\right)\cdot\left(10 \ \frac{\mathrm{strawberries}}{\bcancel{\mathrm{plant}}}\right)=6\cdot8\cdot4\cdot10 \ \mathrm{strawberries}=\boxed{\textbf{(D) }1920} \ \mathrm{strawberries}.</cmath> | We have <cmath>\left(6 \ \cancel{\mathrm{ft}}\cdot8 \ \cancel{\mathrm{ft}}\right)\cdot\left(4 \ \frac{\bcancel{\mathrm{plants}}}{\cancel{\mathrm{ft}^2}}\right)\cdot\left(10 \ \frac{\mathrm{strawberries}}{\bcancel{\mathrm{plant}}}\right)=6\cdot8\cdot4\cdot10 \ \mathrm{strawberries}=\boxed{\textbf{(D) }1920} \ \mathrm{strawberries}.</cmath> | ||
~MRENTHUSIASM | ~MRENTHUSIASM | ||
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==Video Solution by WhyMath== | ==Video Solution by WhyMath== | ||
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~Interstigation | ~Interstigation | ||
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| + | ==Video Solution by North America Math Contest Go Go Go== | ||
| + | https://www.youtube.com/watch?v=_IjQnXnVKeU | ||
| + | (Please subscribe I Really need them!) | ||
==See also== | ==See also== | ||
{{AMC8 box|year=2020|num-b=2|num-a=4}} | {{AMC8 box|year=2020|num-b=2|num-a=4}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 06:42, 19 June 2021
Contents
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
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.
Solution 2
Looking at the units of each quantity, we observe that the answer will be the product of the number of square feet, the number of plants per square foot, and the number of strawberries per plant. This gives 
.
Solution 3 (One Sentence)
We have ![\[\left(6 \ \cancel{\mathrm{ft}}\cdot8 \ \cancel{\mathrm{ft}}\right)\cdot\left(4 \ \frac{\bcancel{\mathrm{plants}}}{\cancel{\mathrm{ft}^2}}\right)\cdot\left(10 \ \frac{\mathrm{strawberries}}{\bcancel{\mathrm{plant}}}\right)=6\cdot8\cdot4\cdot10 \ \mathrm{strawberries}=\boxed{\textbf{(D) }1920} \ \mathrm{strawberries}.\]](http://latex.artofproblemsolving.com/a/3/5/a35032cbeadd18f2f59023f740988662d6a53074.png)
~MRENTHUSIASM
Video Solution by WhyMath
~savannahsolver
Video Solution
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!)
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. 
