Question

The marching band is holding a fundraiser. The band is selling t-shirts for $15 and yearbooks for $26. The goal is to sell at least $2,800 in merchandise. Which of the following is a solution to this scenario?

45 t-shirts and 80 yearbooks
46 t-shirts and 82 yearbooks
47 t-shirts 80 yearbooks
48 t-shirts and 79 yearbooks

Answer 1

To solve the scenario, you need to set up an equation representing the total cost of the merchandise. By substituting the values from the options, you can determine the solution. The solution is 46 t-shirts and 82 yearbooks.

Step-by-step explanation:

To solve this problem, we need to set up an equation representing the total cost of the merchandise. Let's assign the number of t-shirts sold as 't' and the number of yearbooks sold as 'y'. The cost of T-shirts is $15 and the cost of yearbooks is $26. So the equation is 15t + 26y = 2800.

Now, let's substitute the values from the options and check which one satisfies the equation:

1. 45 t-shirts and 80 yearbooks: 15(45) + 26(80) = 675 + 2080 = 2755 (not equal to 2800)
2. 46 t-shirts and 82 yearbooks: 15(46) + 26(82) = 690 + 2132 = 2822 (greater than 2800)
3. 47 t-shirts and 80 yearbooks: 15(47) + 26(80) = 705 + 2080 = 2785 (not equal to 2800)
4. 48 t-shirts and 79 yearbooks: 15(48) + 26(79) = 720 + 2054 = 2774 (not equal to 2800)

Therefore, the solution to this scenario is 46 t-shirts and 82 yearbooks.