Ticket pricing with a rounding trap
A small cinema sells two ticket classes. Adult tickets cost 9.50 euros and student tickets cost 6.25 euros. On a given evening the box office reports 143 tickets sold and total revenue of 1147.75 euros. The cashier also notes that 12 tickets were sold as part of a promotion where each ticket was rounded down to the nearest whole euro at the moment of sale, but the revenue total was recorded from the rounded amounts. A manager wants to reconstruct how many adult and student tickets were sold. She sets up the naive system 9.5a + 6.25s = 1147.75 and a + s = 143, then worries the promotion rounding makes the system inconsistent, since 1147.75 is not a multiple of 0.25 in the way she expects after rounding.
1. Solve the naive system ignoring the promotion. Then determine whether the solution is a valid pair of nonnegative integers, and explain what that tells you about the rounding concern.
Read the worked answer
From a + s = 143 we get s = 143 - a. Substitute: 9.5a + 6.25(143 - a) = 1147.75.
Expand: 9.5a + 893.75 - 6.25a = 1147.75, so 3.25a = 254, giving a = 78.1538..., not an integer.
So the naive system has no integer solution. That is informative: the revenue figure cannot come from unrounded prices with 143 tickets. The rounding story is plausible, but we cannot simply solve the exact system; we must model the rounding.
Note also that 1147.75 is a multiple of 0.25, consistent with prices in quarter-euros, so the non-integrality is a real signal, not a decimal artefact.
2. Model the promotion precisely: exactly 12 of the tickets had their price rounded down to the nearest whole euro before being added to revenue. Let a and s be the true counts. Write the correct equation and find all integer solutions with a + s = 143.
Read the worked answer
Each promoted adult ticket contributes 9 instead of 9.50 (a loss of 0.50); each promoted student ticket contributes 6 instead of 6.25 (a loss of 0.25). If k of the 12 promoted tickets were adult, revenue is
9.5a + 6.25s - 0.5k - 0.25(12 - k) = 1147.75.
With s = 143 - a the unrounded revenue is 893.75 + 3.25a, so
893.75 + 3.25a - 0.25(12 + k) = 1147.75, i.e. 3.25a = 257 + 0.25k.
Multiply by 4: 13a = 1028 + k. Since 0 <= k <= 12 and a integer, k must satisfy 1028 + k divisible by 13. 1379 = 1027, so k = 12 gives 1040 = 1380, hence a = 80, s = 63, and all 12 promoted tickets were adult.
Check: 80(9.5) + 63(6.25) - 12(0.5) = 760 + 393.75 - 6 = 1147.75. Consistent.
3. Suppose the manager instead only knows the total revenue is 1147.75 and that between 10 and 14 tickets were rounded down, with no record of class. Explain why the answer may not be unique and give the range of possible adult counts.
Read the worked answer
Generalise: with m promoted tickets and k adult among them, 13a = 1028 + k where 0 <= k <= m. For each m in 10..14, k ranges over 0..m, so a = (1028 + k)/13 requires 1028 + k ≡ 0 mod 13, i.e. k ≡ -1028 ≡ -1 ≡ 12 mod 13.
Within 0..14 the only such k is 12. So a = 1040/13 = 80 whenever m >= 12; if m = 10 or 11, k = 12 is impossible, so no solution exists.
Hence the data force a = 80, s = 63, with exactly 12 promoted adult tickets, provided m >= 12. If m <= 11 the reported revenue is impossible. The answer is unique in the feasible cases, which is the useful conclusion: the congruence, not the exact m, pins the counts.