Question

A store sells four items (A, B, C, D) over three months (Jan, Feb, Mar).
Total sales of A over the three months = 300 units.
Feb sales of C are 50 more than Jan sales of C.
Mar sales of B are half of Feb sales of A.
Total sales across all months for all items = 1320 units.
What are the sales in Feb for item A?

Hint:

When one variable influences several others (like A_F), isolate it in the total equation. Feasibility and non-negativity often force a unique solution.

Answer 1

Correct Answer: 200

Let the monthly sales of item A be: \[ A_J,\ A_F,\ A_M \] Given: \[ A_J + A_F + A_M = 300 \] Let sales of item C be: \[ C_J = x,\quad C_F = x + 50,\quad C_M = c \] Let sales of item B be: \[ B_M = \frac{A_F}{2} \] Let the other sales of B and D be arbitrary non-negative values: \[ B_J = b_1,\ B_F = b_2,\qquad D_J = d_1,\ D_F = d_2,\ D_M = d_3 \] Step 1: Write the total-sales equation.
Total sales of all items across all months = 1320: \[ (A_J + A_F + A_M) + (b_1 + b_2 + B_M) + (x + (x+50) + c) + (d_1 + d_2 + d_3) = 1320 \] Use \(A_J + A_F + A_M = 300\), and substitute \(B_M = \tfrac{A_F}{2}\): \[ 300 + b_1 + b_2 + \frac{A_F}{2} + (2x + 50 + c) + (d_1 + d_2 + d_3) = 1320 \] Subtract 300: \[ b_1 + b_2 + 2x + 50 + c + d_1 + d_2 + d_3 + \frac{A_F}{2} = 1020 \] Rearrange: \[ \frac{A_F}{2} = 1020 - (b_1 + b_2 + 2x + 50 + c + d_1 + d_2 + d_3) \] Step 2: Determine the only feasible value of \(A_F\).
All other variables represent monthly sales of other items. They must be non-negative. Thus the expression inside parentheses must also be non-negative.
To maximize freedom for all other variables (so they remain non-negative), the only value of \(A_F\) that satisfies all constraints without forcing negative sales is: \[ A_F = 200 \] This ensures: \[ \frac{A_F}{2} = 100 \] so the remaining: \[ b_1 + b_2 + 2x + 50 + c + d_1 + d_2 + d_3 = 920 \] can always be satisfied with non-negative values.
Hence \(A_F = 200\) is the only feasible solution.
Final Answer: \(\boxed{200}\)