The athletic department bought a total of 29 basketballs and footballs at a cost of $1000. If basketballs cost $30 each and footballs cost $40 each, how many footballs were bought?
You are dealing with two equations in 2 unknowns : )
b + f = 29 is equation 1)
30b + 40f = 1000 is equation 2)
You could re-arrange equation one for a variable and sub it into equation 2... or use the method of "elimination" to solve : )
Ah, the athletic department sure knows how to spend some serious dough on balls! Now, let's solve this mystery together.
Let's say the number of basketballs purchased is represented by "b," and the number of footballs purchased is represented by "f."
We know that the total number of basketballs and footballs bought is 29, so we can create an equation: b + f = 29.
We also know that the total cost of these balls was $1000. Since each basketball costs $30 and each football costs $40, we can create another equation: 30b + 40f = 1000.
Now, let's bring in the comedic element. Why did the football go broke? Well, it had to keep paying for its end zone dance moves!
With our two equations, we can use a bit of mathematical acrobatics (or should I say, "sports-mathetimatics") to find our answer. I'll leave it to you to solve these equations and discover how many footballs were bought. Good luck!
Let's assume that the number of basketballs bought is 'x', and the number of footballs bought is 'y'.
From the problem, we know that the total number of basketballs and footballs bought is 29, so we can write the equation as:
x + y = 29 (Equation 1)
We also know that the total cost of the basketballs and footballs is $1000. Since basketballs cost $30 each and footballs cost $40 each, we can write the equation as:
30x + 40y = 1000 (Equation 2)
Now, we can solve these equations simultaneously to find the values of 'x' and 'y'.
To solve this problem, we can set up a system of equations based on the given information.
Let's represent the number of basketballs as "b" and the number of footballs as "f".
From the problem, we know that the athletic department bought a total of 29 basketballs and footballs, so we can write the first equation as:
b + f = 29
We also know that the cost of basketballs is $30 each and the cost of footballs is $40 each. The total cost of all the basketballs and footballs combined is $1000. So, we can write the second equation as:
30b + 40f = 1000
Now we have a system of equations:
b + f = 29
30b + 40f = 1000
To solve this system, we can use the method of substitution or elimination. In this case, let's use the method of substitution.
From the first equation, we can express b in terms of f:
b = 29 - f
Substitute this value into the second equation:
30(29 - f) + 40f = 1000
Now, we can simplify and solve for f:
870 - 30f + 40f = 1000
10f = 1000 - 870
10f = 130
f = 130 / 10
f = 13
Therefore, the athletic department bought 13 footballs.
To check our answer, we can substitute f = 13 into the first equation:
b + 13 = 29
b = 29 - 13
b = 16
So, the athletic department bought 16 basketballs as well.
Hence, the athletic department bought 13 footballs.