A university bookstore recently sold a wire-bound graph-paper notebook for $2.64, and a college-ruled notebook for $3.61. At the start of spring semester, a combination of 50 of these notebooks were sold for a total of $168.86. How many of each type were sold?
Write one equation for the total number sold and another for the total money received. Each equation will have two unknowns, the number sold of each type.
Then solve the two equations in two unknowns.
There comes a time when you have to attempt these yourself. There is no time like the present.
Here is what I got.
3.61 * 38 = 137.18
2.64 * 12 = 31.68
137.18 + 31.68 = 168.86
wire bound graph paper notebook @ 2.64
bought 12
college ruled notebook @ 3.61 bought 38
did I do this right?
To solve this problem, we will use a system of equations. Let's assume that x represents the number of wire-bound graph-paper notebooks sold and y represents the number of college-ruled notebooks sold.
We can set up two equations based on the given information:
1) The total number of notebooks sold is 50:
x + y = 50
2) The total cost of the notebooks sold is $168.86:
2.64x + 3.61y = 168.86
Now we have a system of equations we can solve to find the values of x and y.
First, let's solve the first equation for x:
x = 50 - y
Next, substitute this expression for x in the second equation:
2.64(50 - y) + 3.61y = 168.86
Now we can simplify and solve for y:
132 - 2.64y + 3.61y = 168.86
0.97y = 36.86
y ≈ 38.04
Since we can't have a fraction of a notebook, let's round y down to 38 and calculate x using the first equation:
x = 50 - y
x = 50 - 38
x = 12
Therefore, 12 wire-bound graph-paper notebooks were sold, and 38 college-ruled notebooks were sold.