Ama's tomato crate: deal pricing, break-even and profit per pile
Ama sells tomatoes at Lilongwe market. Each Monday she buys one crate holding 60 piles of tomatoes for MK 48 000 and pays a driver MK 6 000 to bring it from the farm. She sells one pile for MK 1 200, and she also offers a deal: any 3 piles for MK 3 000. On Monday morning a customer asks for exactly 7 piles. By midday Ama has sold 15 of the three-pile deals and no single piles. She still has 15 piles in the crate and expects them to be soft by Tuesday morning, so she plans to sell the rest at the single price only. Ama is not sure whether she has covered her costs yet, and she does not want to give away profit without noticing. Her friend Grace tells her to sell the whole crate through the three-pile deal next week because 'the deal brings customers faster'.
1. The customer wants exactly 7 piles. Work out the cheapest way for Ama to charge for them, and state how much the customer saves compared with paying the single price for every pile.
Read the worked answer
Option 1 – all single: 7 × MK 1 200 = MK 8 400.
Option 2 – use the deal: a deal covers exactly 3 piles at MK 3 000, i.e. MK 1 000 per pile, but it cannot be used for part of a group. With 7 piles she can make 2 full deals (6 piles) plus 1 single pile: 2 × MK 3 000 + MK 1 200 = MK 6 000 + MK 1 200 = MK 7 200. Three deals would need 9 piles, which is more than the customer asked for.
Saving: MK 8 400 − MK 7 200 = MK 1 200.
Common error: charging 7 × MK 1 000 = MK 7 000. The MK 1 000 per pile figure is only the deal rate when piles are bought in threes; the seventh pile must be charged at the single price.
2. From her Monday midday position (15 deals sold, 15 piles left), find the smallest number of the remaining piles she must sell at the single price to cover the total cost of the crate, and say whether she can do it.
Read the worked answer
Total cost = MK 48 000 + MK 6 000 = MK 54 000, so each pile costs 54 000 ÷ 60 = MK 900.
Money already taken: 15 deals × MK 3 000 = MK 45 000. Still to recover: MK 54 000 − MK 45 000 = MK 9 000. Each single pile brings MK 1 200, so 9 000 ÷ 1 200 = 7.5 piles.
7 piles give 7 × 1 200 = MK 8 400, which is still MK 600 short, so she must round up to 8 piles (8 × 1 200 = MK 9 600).
She has 15 piles left, so yes - easily. If all 15 sell at MK 1 200 she takes MK 45 000 + MK 18 000 = MK 63 000, a profit of MK 9 000.
Judgement: the deals have already recovered about 83% of the cost (45 000 ÷ 54 000), so almost all the profit sits in the last 15 piles. They must be sold before they spoil, or the crate ends near break-even.
3. Evaluate Grace's advice to sell the whole crate through the three-pile deal. Use profit per pile and total profit for the crate to support your judgement.
Read the worked answer
Cost per pile = MK 900.
- Deal: MK 3 000 ÷ 3 = MK 1 000 per pile, so profit = MK 100 per pile.
- Single: MK 1 200 per pile, so profit = MK 300 per pile.
Whole crate by deal: 60 × MK 1 000 = MK 60 000, profit = 60 000 − 54 000 = MK 6 000. Whole crate at single price: 60 × MK 1 200 = MK 72 000, profit = MK 18 000.
Grace is right that deals move stock faster, but each pile sold in a deal gives up MK 200 of profit; over 60 piles that is MK 12 000 lost compared with single sales. At MK 100 profit per deal pile, one spoiled pile (MK 900 lost) wipes out the profit on 9 deal piles.
Judgement: do not sell the whole crate through the deal. The deal price is above cost, so it is useful late in the day for piles that would otherwise spoil, but the single price should stay for the rest. Any price below MK 900 per pile is a loss, however fast it sells.