A school planned to buy x calculators for a total cost $216. Supplies agreed to offer a discount $.80 per calculator. the school was then able to get three extra calculators for the same amount of money. (a) orginal price of each calculator; (b) price of each calculator after the discount.
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With a cost of $216, $.80 (80 cents) is not a significant discount to even be considered.
To find the original price of each calculator, we'll first set up an equation based on the given information.
Let's say the original price of each calculator is "p". The school planned to buy "x" calculators, so the total cost would be "x * p".
According to the problem, the total cost of x calculators is $216. So, we can write the equation:
x * p = 216 (Equation 1)
Now, let's assume that after the discount, the price of each calculator becomes "d". According to the discount offer, the school saves $0.80 on each calculator. Therefore, the new price after the discount can be written as "(p - d)".
The school is able to get three extra calculators for the same amount of money, so the total cost for (x+3) calculators is also $216. Using the new price after the discount, we can write the equation:
(x + 3) * (p - d) = 216 (Equation 2)
We have a system of two equations (Equation 1 and Equation 2) with two variables (p and d). Let's solve this system to find the answers to the given questions.
Step 1: Solve Equation 1 for p
Divide both sides of Equation 1 by x:
p = 216/x
Step 2: Substitute the value of p into Equation 2
(x + 3) * (216/x - d) = 216
Step 3: Simplify Equation 2
Expand the left side of the equation:
216 - dx + 648/x - 3d = 216
Step 4: Rewrite the equation and collect like terms:
- dx + 648/x - 3d = 0
Step 5: Multiply through by x, to get rid of the fraction:
- dx^2 + 648 - 3dx = 0
Step 6: Combine the quadratic terms:
dx^2 - 3dx + 648 = 0
Step 7: Factorize or use the quadratic formula to solve the equation for x.
The quadratic equation can be factored as:
(x - 12)(x - 54) = 0
Setting each factor equal to zero gives us two possible values for x:
x - 12 = 0 --> x = 12
x - 54 = 0 --> x = 54
Step 8: Substitute the values of x back into Equation 1 to find the original price of each calculator.
For x = 12:
p = 216/12 = 18
For x = 54:
p = 216/54 = 4
Therefore, there are two possible answers to the first part of the question:
(a) The original price of each calculator could be $18 if the school planned to buy 12 calculators.
OR
(a) The original price of each calculator could be $4 if the school planned to buy 54 calculators.
To find the price of each calculator after the discount, we need to find the value of "d" in Equation 2. Let's use one of our found values for p and x to solve for d.
Using the value p = 18 and x = 12:
(12 + 3)(18 - d) = 216
15(18 - d) = 216
270 - 15d = 216
- 15d = 216 - 270
- 15d = - 54
d = -54 / -15
d = 3.6
Therefore, the price of each calculator after the discount is $3.60 if the original price is $18.
(b) The price of each calculator after the discount is $3.60.