Question

The members of the boosters organization at your high school bought new balls for the school. They spent $30.00 per basketball and $24.00 per football, spending a total of $840.00. They bought 8 more footballs than basketballs. How many of each type of ball did they buy?

Answer 1

The boosters organization bought 12 basketballs and 20 footballs.

Step-by-step explanation:

Let's use algebra to solve this problem.

Let's assume the number of basketballs purchased is x.

Since they bought 8 more footballs than basketballs, the number of footballs purchased would be x+8.

According to the given information, the cost of each basketball is $30 and the cost of each football is $24. The total cost of all the balls is $840.

• Cost of basketballs: 30x
• Cost of footballs: 24(x+8)

From the given information, the total cost of both types of balls is $840:

30x + 24(x+8) = 840

Simplify the equation:

30x + 24x + 192 = 840

Combine like terms:

54x + 192 = 840

Subtract 192 from both sides:

54x = 648

Divide both sides by 54:

x = 12

The number of basketballs purchased is 12.

The number of footballs purchased is 12 + 8 = 20.