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

Line 16: Line 16:
 
==Solution 4==  
 
==Solution 4==  
 
  The answer is <math>6 \times 8 \times 10 \times 4 = 1920</math> or <math>\boxed{\textbf{(D) }1920}</math>.
 
  The answer is <math>6 \times 8 \times 10 \times 4 = 1920</math> or <math>\boxed{\textbf{(D) }1920}</math>.
 +
-oceanxia
  
 
-franzliszt
 
-franzliszt

Revision as of 17:57, 19 November 2020

Problem 3

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

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 $6 \times 8 \times 10 \times 4 = 1920$ or $\boxed{\textbf{(D) }1920}$.

Solution 2

The area of the garden is $6$ ft $\times$ $8$ ft $= 48$ square feet. Since Carrie plants $4$ strawberry plants per square foot, it follows that she plants a total of $48 \times 4=192$ strawberry plants. Since each strawberry plant produces on average 10 strawberries, it follows that she can expect to harvest $192 \times 10=1920$ strawberries $\implies\boxed{\textbf{(D) }1920}$.
~junaidmansuri

Solution 3

Note that $6\cdot 8 = 48$, so Carrie has $4\cdot 48 = 192$ strawberry plants. Each plant produces $10$ strawberries, so the final answer is $192\cdot 10 = \textbf{(D)}\ 1920$.'

Solution 4

The answer is $6 \times 8 \times 10 \times 4 = 1920$ or $\boxed{\textbf{(D) }1920}$.

-oceanxia

-franzliszt

Video Solution

https://youtu.be/7S0wAZMy2ZQ

~savannahsolver

See also

2020 AMC 8 (ProblemsAnswer KeyResources)
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