Canteen Combo Prices and a Suspicious Fourth-Day Revenue
A college canteen sells three combo plates: A, B and C. Over three days the owner records only the number of each combo sold and the total revenue, because the individual prices were not printed. Day 1: 2 A, 3 B, 1 C, total ₹170. Day 2: 1 A, 2 B, 4 C, total ₹240. Day 3: 3 A, 1 B, 2 C, total ₹170. Let x, y and z be the prices in rupees of A, B and C. The owner wants to know whether these three days determine the prices uniquely, and a fourth day's sheet shows 4 A, 5 B and 6 C with total ₹430. The accountant suspects a copying error because the same prices should apply. The canteen committee also wants to know what happens if all prices are raised by ₹2, and whether the recorded totals would still be consistent with a unique solution. You are asked to set up the matrix system, solve it, and judge the fourth day's record.
1. Express the three days' records as a matrix equation AX = B and compute det A. What does the determinant tell you about uniqueness?
Read the worked answer
Let X = [x, y, z]^T and B = [170, 240, 170]^T. The equations are:
2x + 3y + z = 170 x + 2y + 4z = 240 3x + y + 2z = 170
So A = [[2,3,1],[1,2,4],[3,1,2]]. Expanding along the first row:
det A = 2(2×2 - 4×1) - 3(1×2 - 4×3) + 1(1×1 - 2×3) = 2(0) - 3(2 - 12) + 1(1 - 6) = 0 + 30 - 5 = 25.
Since det A ≠ 0, A is invertible and the system has a unique solution. The three days do determine the prices, provided the recorded quantities and totals are exact.
2. Solve for x, y and z, and check the solution against all three days.
Read the worked answer
Use row reduction on the augmented matrix. Swap R1 and R2:
[1 2 4 | 240] [2 3 1 | 170] [3 1 2 | 170]
R2 - 2R1 → [0, -1, -7 | -310], so y + 7z = 310. R3 - 3R1 → [0, -5, -10 | -550], so 5y + 10z = 550, or y + 2z = 110. Subtracting: 5z = 200, so z = 40. Then y = 110 - 80 = 30. Finally x = 240 - 2(30) - 4(40) = 20.
Check: Day 1: 2(20) + 3(30) + 40 = 40 + 90 + 40 = 170. Day 2: 20 + 2(30) + 4(40) = 20 + 60 + 160 = 240. Day 3: 3(20) + 30 + 2(40) = 60 + 30 + 80 = 170. Thus x = ₹20, y = ₹30, z = ₹40.
3. Evaluate the fourth day's record. If the total is ₹430, what can you conclude? If it should have been ₹470, does that remove the inconsistency? Distinguish a data-entry error from a model failure.
Read the worked answer
Using the prices just found, the expected revenue for 4 A, 5 B and 6 C is
4(20) + 5(30) + 6(40) = 80 + 150 + 240 = ₹470.
The recorded ₹430 is ₹40 short. Since det A = 25 ≠ 0, the first three days fix the prices uniquely; a fourth day at the same prices cannot legitimately total ₹430. The most reasonable conclusion is a data-entry error - a quantity or the total was copied incorrectly. If the total should be ₹470, the record is consistent and there is no contradiction.
This is not a model failure. A data error can be corrected while the linear pricing model remains intact. Model failure would require evidence that price depends on quantity, day or combo (for example, discounts), which would need a different model. If all prices rise by ₹2, each day's total rises by 2 times the number of combos sold: ₹12, ₹14 and ₹12 respectively. The coefficient matrix is unchanged, so the new system still has determinant 25 and still has a unique solution.