The marching band is holding a fundraiser. The band is selling t-shirts for $22 and yearbooks for $23. The goal is to sell at least $2,400 in merchandise. Which of the following is a solution to this scenario?
Option A: Sell 50 t-shirts and 50 yearbooks
50 t-shirts would bring in 50 x $22 = $1,100
50 yearbooks would bring in 50 x $23 = $1,150
Total sales = $1,100 + $1,150 = $2,250
This is not a solution as it falls short of the goal of selling at least $2,400 in merchandise.
Option B: Sell 70 t-shirts and 60 yearbooks
70 t-shirts would bring in 70 x $22 = $1,540
60 yearbooks would bring in 60 x $23 = $1,380
Total sales = $1,540 + $1,380 = $2,920
This is a solution as it exceeds the goal of selling at least $2,400 in merchandise.
To find a solution to this scenario, we need to determine the combination of t-shirts and yearbooks that will allow the marching band to reach their goal of selling at least $2,400 in merchandise.
Let's assume the marching band sells x t-shirts and y yearbooks.
The cost of x t-shirts is $22x.
The cost of y yearbooks is $23y.
According to the scenario, the goal is to sell at least $2,400 in merchandise, so we have the equation:
$22x + $23y ≥ $2,400
This inequality ensures that the total sales value is equal to or greater than $2,400.
Now, we need to find the values of x and y that satisfy this inequality. Since we don't have any other information or conditions in the scenario, there are multiple possible solutions. Without additional constraints, we can't determine a specific solution.
Possible solutions (x, y) could be:
(100, 100) - selling 100 t-shirts and 100 yearbooks
(200, 0) - selling 200 t-shirts and no yearbooks
(0, 100) - not selling any t-shirts and selling 100 yearbooks
These are just a few examples, and there are infinitely many other combinations that would satisfy the inequality.