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: \(\boxed{200}\)

Answer 2

Approach: Before solving, count unknowns versus equations. With 4 items across 3 months there are 12 unknown monthly sales figures. List every fact given and see how many independent equations they produce.

Step 1: Let \(X_{Jan}, X_{Feb}, X_{Mar}\) denote item X's sales in each month. The facts translate to:
\[ A_{Jan}+A_{Feb}+A_{Mar} = 300, \qquad C_{Feb} = C_{Jan}+50, \qquad B_{Mar} = \tfrac{1}{2}A_{Feb}, \qquad \text{(sum of all 12 values)} = 1320. \]
That is only 4 independent equations relating 12 unknowns.

Step 2: Try two different sets of values, both obeying all four equations, and see if \(A_{Feb}\) changes.
Set 1: \(A_{Jan}=100, A_{Feb}=100, A_{Mar}=100\) (total 300); \(B_{Mar}=50\); pick \(C_{Jan}=40, C_{Feb}=90\); fill the remaining 6 unknowns so the grand total is 1320.
Set 2: \(A_{Jan}=50, A_{Feb}=200, A_{Mar}=50\) (total 300 too); \(B_{Mar}=100\); pick different C values; again fill the rest to reach 1320.
Both sets satisfy every stated condition, yet \(A_{Feb}\) is 100 in one and 200 in the other.

Final Answer: Feb sales of item A cannot be uniquely determined – the four given facts under-constrain the 12 unknowns, so the data is insufficient.