the athletic department ordered a total of 29 basketballs and footballs for $1000. if the basketballs cost $30 each and the footballs cost $40 each, how many footballs were purchased?
To answer this question, we can set up a system of equations based on the given information.
Let's say the number of basketballs ordered is represented by "b" and the number of footballs ordered is represented by "f".
According to the problem, the total number of basketballs and footballs combined is 29, so we have the equation:
b + f = 29
Additionally, we know that the cost of each basketball is $30 and the cost of each football is $40. So, the total cost of all the basketballs and footballs together is $1000, giving us the equation:
30b + 40f = 1000
Now we have a system of two equations:
Equation 1: b + f = 29
Equation 2: 30b + 40f = 1000
To solve this system of equations, we can use one of the common methods such as substitution or elimination.
Let's solve it using the elimination method.
Multiply Equation 1 by 30 to make the coefficients of "b" in both equations the same:
30(b + f) = 30(29)
30b + 30f = 870
Now, we have the following system:
Equation 3: 30b + 30f = 870
Equation 2: 30b + 40f = 1000
Now, subtract Equation 3 from Equation 2 to eliminate the term "b":
(30b + 40f) - (30b + 30f) = 1000 - 870
10f = 130
Divide both sides of the equation by 10 to solve for "f":
f = 130 / 10
f = 13
Therefore, 13 footballs were purchased by the athletic department.
If it wasn't urgent at 1:55, how come it was at 1:59?
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