Recovering canteen prices from sales records
A composite school canteen in Bikaner sells only two items during the interval, samosas and kachoris, each at a fixed price. The secretary kept a day-wise record but lost the price list. On Monday, 45 samosas and 30 kachoris were sold for ₹780. On Tuesday, 30 samosas and 50 kachoris were sold for ₹760. On Wednesday a new volunteer wrote only 15 samosas and 10 kachoris - ₹260, and claimed that the price list could still be recovered from that entry. On Thursday the canteen introduced a combo pack of 2 samosas and 1 kachori priced at ₹30 to attract younger students, and sold 150 packs. The principal wants the original prices of the two items, an explanation of what the Wednesday record is worth, and a statement of how much the canteen lost or gained on Thursday compared with selling the same items at the regular prices.
1. Find the price of one samosa and one kachori using Monday's and Tuesday's records, and verify the solution.
Read the worked answer
Let a samosa cost ₹x and a kachori ₹y.
Monday: 45x + 30y = 780. Dividing by 15: 3x + 2y = 52 ... (i)
Tuesday: 30x + 50y = 760. Dividing by 10: 3x + 5y = 76 ... (ii)
Subtracting (i) from (ii): 3y = 24, so y = 8.
Substituting in (i): 3x + 16 = 52, so 3x = 36 and x = 12.
Verification: Monday 45(12) + 30(8) = 540 + 240 = ₹780 ✓; Tuesday 30(12) + 50(8) = 360 + 400 = ₹760 ✓.
The coefficients of x are equal in both reduced equations, which is why subtraction eliminates x cleanly instead of requiring substitution.
2. What is the worth of the Wednesday record - 15 samosas and 10 kachoris for ₹260? Is the volunteer right that the price list can be recovered from it?
Read the worked answer
Wednesday's equation is 15x + 10y = 260. Dividing throughout by 5: 3x + 2y = 52, which is exactly equation (i) from Monday.
The new record is therefore not a new equation - it is a multiple of an existing one. Geometrically the two lines coincide, so the pair has infinitely many solutions; the record is consistent with the known prices but adds no information.
Verdict: the volunteer is wrong. Wednesday cannot fix x and y, because for any x there is a y = (52 − 3x)/2 that satisfies it, e.g. x = 10, y = 11 also fits. To determine both prices you need a second equation whose coefficient ratio differs from 3 : 2, such as one involving unequal numbers of the two items. The Wednesday entry is useful only as a cross-check that the prices found from Monday and Tuesday are consistent.
3. Compare Thursday's combo pack of 2 samosas and 1 kachori at ₹30 with regular pricing, and state the gain or loss on 150 packs.
Read the worked answer
Regular price of one pack: 2x + y = 2(12) + 8 = ₹32.
Pack price charged: ₹30, so each pack is under-priced by 32 − 30 = ₹2.
For 150 packs: loss compared with regular sale = 150 × 2 = ₹300.
Revenue actually collected = 150 × 30 = ₹4,500, whereas regular pricing would have fetched 150 × 32 = ₹4,800. The canteen is not losing money on cost (we are not told the cost price), but it foregoes ₹300 of income. If the aim was a small promotional discount, ₹2 per pack is the exact size of the concession; if the canteen wanted no loss, it should have charged ₹32 or sold the pack only when the extra volume recovers ₹300.