The shopkeeper charged 12 rupees for a bunch of chocolate but I bargained to…

Concept: When a fixed amount of money (here, Rs 12) buys a changing number of identical items, the price per dozen scales as (total money ÷ quantity) × 12. A stated change in this per-dozen price between two purchase scenarios sets up an equation relating the two quantities, which solves as a quadratic.

Application: Set up the two price-per-dozen expressions and equate their difference to the stated Rs 1 drop.

1. Let the number of chocolates bought for Rs 12 before bargaining be x.

2. Price per dozen at this rate = (12/x) × 12 = 144/x rupees.

3. After bargaining, the same Rs 12 buys x + 2 chocolates, so the new price per dozen = 144/(x+2) rupees.

4. The new price per dozen is Rs 1 less than the old one: 144/x − 144/(x+2) = 1.

5. Multiplying through by x(x+2): 144(x+2) − 144x = x(x+2), which simplifies to x² + 2x − 288 = 0.

6. Solving the quadratic: x = [−2 ± √(4 + 1152)] / 2 = [−2 ± 34] / 2, so x = 16 (rejecting the negative root).

7. The question asks for the quantity received after bargaining, which is x + 2 = 18.

Cross-check: Before bargaining, 16 chocolates cost Rs 12, i.e., Rs 144/16 = Rs 9 per dozen; after bargaining, 18 chocolates cost Rs 12, i.e., Rs 144/18 = Rs 8 per dozen. The drop of Rs 9 − Rs 8 = Rs 1 matches the given condition, confirming the answer.