The school had a bake sale to raise money. A group of 11 students bought either chocolate chip cookis for $.40 cents each or brownies for $.50 cents each. The total amount they spent was $5.10. How many chocolate chip cookies and brownies did they buy?
I will assume that each of the 11 students bought either one cookie or one brownie.
let the number of cookies be c
then the number of brownies is 11-c
solve this equation:
40c + 50(11-c) = 510
(let me know if you don't get cookies =4
brownies = 7 ?)
To solve this problem, we can use a system of equations:
Let's assume:
x = number of chocolate chip cookies bought
y = number of brownies bought
From the problem, we can derive the following equations:
1. The total number of items bought equation: x + y = 11
2. The total cost equation: 0.40x + 0.50y = 5.10
To solve this system of equations, we can use the substitution or elimination method. Let's use the substitution method in this case.
We can rearrange equation 1 to solve for x:
x = 11 - y
Next, substitute this value of x into equation 2:
0.40(11 - y) + 0.50y = 5.10
Now, let's solve the equation step by step:
4.4 - 0.40y + 0.50y = 5.10
Combine like terms:
0.10y + 4.4 = 5.10
Subtract 4.4 from both sides:
0.10y = 0.70
Divide by 0.10 from both sides:
y = 7
So, they bought 7 brownies (y).
Now, substitute this value of y back into equation 1 to find x:
x = 11 - y
x = 11 - 7
x = 4
They bought 4 chocolate chip cookies (x).
Therefore, they bought 4 chocolate chip cookies and 7 brownies.