Question

A company placed orders with Suppliers A and B, buying at least eight units from each.
Supplier A's pricing: A fixed fee of \$120, with the first eight units at \$25 each, and \$7 for every additional unit.
Supplier B's pricing: A fixed fee of \$180, with the first eight units at \$18 each, and \$12 for each extra unit. After all purchases, including fixed costs, the average price per unit was \$23.
Select the number of units purchased from Supplier A and the number of units purchased from Supplier B that are jointly consistent with the given information. Make only two selections, one in each column.

Hint:

When solving Diophantine equations (\(ax+by=c\)), look for divisibility rules and properties like even/odd to quickly narrow down the possible integer values for one of the variables. This is much faster than random guessing.

Answer 1

Step 1: Formulate Cost Equations
Let \(a\) be the number of units from Supplier A (\(a \geq 8\)). Let \(b\) be the number of units from Supplier B (\(b \geq 8\)).

Cost from A (\(C_A\)): The cost consists of a fixed fee, the cost of the first 8 units, and the cost of the additional units (\(a-8\)). \[ C_A = 120 + (8 \times 25) + (a-8) \times 7 = 120 + 200 + 7a - 56 = 7a + 264 \]
Cost from B (\(C_B\)): Similarly, for Supplier B. \[ C_B = 180 + (8 \times 18) + (b-8) \times 12 = 180 + 144 + 12b - 96 = 12b + 228 \]
Step 2: Use the Average Price Information
The total number of units is \(a+b\). The total cost is \(C_{total} = C_A + C_B = (7a + 264) + (12b + 228) = 7a + 12b + 492\). The average price per unit is given as \$23. \[ \frac{\text{Total Cost}}{\text{Total Units}} = 23 \] \[ \frac{7a + 12b + 492}{a+b} = 23 \] \[ 7a + 12b + 492 = 23(a+b) = 23a + 23b \] Rearrange the terms to form a single linear Diophantine equation: \[ 492 = (23a - 7a) + (23b - 12b) \] \[ 16a + 11b = 492 \] Step 3: Solve the Diophantine Equation with Constraints
We need to find integer solutions for \(16a + 11b = 492\) where \(a, b \geq 8\) and \(a,b\) are from the set \{15, 17, 18, 19, 20, 22\}.

We can observe that 492 is even and \(16a\) is always even. This means \(11b\) must be even, which implies that \(b\) must be an even number.
From the given options, the possible even values for \(b\) are 18, 20, and 22. We can test these values.
Test b = 18: \[ 16a + 11(18) = 492 \implies 16a + 198 = 492 \implies 16a = 294 \] \(a = 294/16 = 18.375\). Not an integer.
Test b = 20: \[ 16a + 11(20) = 492 \implies 16a + 220 = 492 \implies 16a = 272 \] \(a = 272/16 = 17\). This is an integer solution.
Test b = 22: \[ 16a + 11(22) = 492 \implies 16a + 242 = 492 \implies 16a = 250 \] \(a = 250/16 = 15.625\). Not an integer.
Step 4: Final Answer
The only integer solution is \((a=17, b=20)\). Both 17 and 20 are present in the list of options. Therefore, the company purchased 17 units from Supplier A and 20 units from Supplier B.