Difference between revisions of "2020 AMC 8 Problems/Problem 3"
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| − | The area of the garden is <math>6</math> ft <math>\times 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> | + | 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> |
~jmansuri | ~jmansuri | ||
Revision as of 03:31, 18 November 2020
Contents
Problem 3
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
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 
or 
.
Solution 2
The area of the garden is ft 
ft 
square feet. Since Carrie plants strawberry plants per square foot, it follows that she plants a total of 
strawberry plants. Since each strawberry plant produces on average 10 strawberries, it follows that she can expect to harvest 
strawberries 
.
~jmansuri
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. 
