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Difference between revisions of "2020 AMC 8 Problems/Problem 3"

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==Problem 3==
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==Problem==
 
Carrie has a rectangular garden that measures <math>6</math> feet by <math>8</math> feet. She plants the entire garden with strawberry plants. Carrie is able to plant <math>4</math> strawberry plants per square foot, and she harvests an average of <math>10</math> strawberries per plant. How many strawberries can she expect to harvest?
 
Carrie has a rectangular garden that measures <math>6</math> feet by <math>8</math> feet. She plants the entire garden with strawberry plants. Carrie is able to plant <math>4</math> strawberry plants per square foot, and she harvests an average of <math>10</math> strawberries per plant. How many strawberries can she expect to harvest?
  
 
<math>\textbf{(A) }560 \qquad \textbf{(B) }960 \qquad \textbf{(C) }1120 \qquad \textbf{(D) }1920 \qquad \textbf{(E) }3840</math>
 
<math>\textbf{(A) }560 \qquad \textbf{(B) }960 \qquad \textbf{(C) }1120 \qquad \textbf{(D) }1920 \qquad \textbf{(E) }3840</math>
  
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==Solution==
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==Solution 1==
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The answer is the product of the area of the field, the amount of strawberries per plant, and the amount of plants in one square feet. The answer is <math>6 \times 8 \times 10 \times 4 = 1920</math> or <math>\boxed{\textbf{D }1920}</math>.
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Note that the unit of the answer is <b>strawberries</b>, which is the product of
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 +
* square feet
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 +
* plants per square foot
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 +
* strawberries per plant
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By conversion factors, we have <cmath>\left(6 \ \color{red}\cancel{\mathrm{ft}}\color{black}\cdot8 \ \color{red}\cancel{\mathrm{ft}}\color{black}\right)\cdot\left(4 \ \frac{\color{green}\cancel{\mathrm{plants}}}{\color{red}\cancel{\mathrm{ft}^2}}\right)\cdot\left(10 \ \frac{\mathrm{strawberries}}{\color{green}\cancel{\mathrm{plant}}}\right)=6\cdot8\cdot4\cdot10 \ \mathrm{strawberries}=\boxed{\textbf{(D) }1920} \ \mathrm{strawberries}.</cmath>
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~MRENTHUSIASM ~Bobthegod78
  
 
==Solution 2==
 
==Solution 2==
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The area of the garden is <math>6</math> ft <math>\times</math> <math>8</math> ft <math>= 48</math> square feet. Since Carrie plants <math>4</math> strawberry plants per square foot, it follows that she plants a total of <math>48 \times 4=192</math> strawberry plants. Since each strawberry plant produces on average 10 strawberries, it follows that she can expect to harvest <math>192 \times 10=1920</math> strawberries <math>\implies\boxed{\textbf{(D) }1920}</math>.<br>
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The area of the garden is <math>6 \cdot 8 = 48</math> square feet. Since Carrie plants <math>4</math> strawberry plants per square foot, there are a total of <math>48 \cdot 4=192</math> strawberry plants, each of which produces <math>10</math> strawberries on average. Accordingly, she can expect to harvest <math>192 \cdot 10 = \boxed{\textbf{(D) }1920}</math> strawberries.
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~jmansuri
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 +
==Video Solution by NiuniuMaths (Easy to understand!)==
 +
https://www.youtube.com/watch?v=8hgK6rESdek&t=9s
 +
 
 +
~NiuniuMaths
 +
 
 +
==Video Solution (🚀 Just 1 min 🚀)==
 +
https://youtu.be/8jL971T8scM
 +
 
 +
~Education, the Study of Everything
 +
 
 +
==Video Solution by WhyMath==
 +
https://youtu.be/7S0wAZMy2ZQ
 +
 
 +
~savannahsolver
 +
 
 +
==Video Solution by The Learning Royal==
 +
https://youtu.be/eSxzI8P9_h8
 +
 
 +
~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==  
 
==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}}

Latest revision as of 06:13, 24 January 2024

Problem

Carrie has a rectangular garden that measures $6$ feet by $8$ feet. She plants the entire garden with strawberry plants. Carrie is able to plant $4$ strawberry plants per square foot, and she harvests an average of $10$ strawberries per plant. How many strawberries can she expect to harvest?

$\textbf{(A) }560 \qquad \textbf{(B) }960 \qquad \textbf{(C) }1120 \qquad \textbf{(D) }1920 \qquad \textbf{(E) }3840$

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 \[\left(6 \ \color{red}\cancel{\mathrm{ft}}\color{black}\cdot8 \ \color{red}\cancel{\mathrm{ft}}\color{black}\right)\cdot\left(4 \ \frac{\color{green}\cancel{\mathrm{plants}}}{\color{red}\cancel{\mathrm{ft}^2}}\right)\cdot\left(10 \ \frac{\mathrm{strawberries}}{\color{green}\cancel{\mathrm{plant}}}\right)=6\cdot8\cdot4\cdot10 \ \mathrm{strawberries}=\boxed{\textbf{(D) }1920} \ \mathrm{strawberries}.\]

~MRENTHUSIASM ~Bobthegod78

Solution 2

The area of the garden is $6 \cdot 8 = 48$ square feet. Since Carrie plants $4$ strawberry plants per square foot, there are a total of $48 \cdot 4=192$ strawberry plants, each of which produces $10$ strawberries on average. Accordingly, she can expect to harvest $192 \cdot 10 = \boxed{\textbf{(D) }1920}$ strawberries.


Video Solution by NiuniuMaths (Easy to understand!)

https://www.youtube.com/watch?v=8hgK6rESdek&t=9s

~NiuniuMaths

Video Solution (🚀 Just 1 min 🚀)

https://youtu.be/8jL971T8scM

~Education, the Study of Everything

Video Solution by WhyMath

https://youtu.be/7S0wAZMy2ZQ

~savannahsolver

Video Solution by The Learning Royal

https://youtu.be/eSxzI8P9_h8

~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. AMC logo.png

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