# Profit/loss word problem.

Problem statement

A dishonest shopkeeper uses a false weight of 900gm instead of 1kg, if he promises to sell his goods at cost price, what is his overall profit?

I approached this problem like this,

when a customer buys 1 kg, he gets 0.9 kg.(profit of 0.1kg for shopkeeper)

when a customer buys 2 kg, he gets 1.8 kg.(profit of 0.2kg for shopkeeper)

So, if the shopkeeper has x kg originally, he will be able to sell (10x/9)kg.

It gives me the correct answer but I still can't visualize this.

For e.g. Let's say the shopkeeper has 9 kg originally, and sells all of it, in this case he will sell 8.1 kg(there is 900gm left). So, in total he will sell, 9.9kg(9kg+900gm leftover). The earlier equation gives 10kg.

What is wrong with reasoning?

• If the shopkeeper sells all of $9$ kg, he will have sold $9$ kg for the price of $10$ kg. Jun 6 at 11:43
• @N.F.Taussig Can you point out what is wrong with my reasoning, which gives me 9.9.
– Max
Jun 6 at 11:54
• The last comment was a bit loose so I deleted it. The last $900$ gm will be treated by the customer as $1$ kg, so the customer will pay for it as if it's worth $1$ kg. That means even though, from the shopkeeper's point of view, the last $900$ gm is actually $900$ gm in weight, what he earns out of it is what $1$ kg is worth. So the leftover $900$ gm is to be treated as $1$ kg when sold. That's what gives $10$kg instead of $9.9$ at the end. Jun 6 at 12:02
• Other (more difficult) way to think about it: now shopkeeper has 900g left, he sells "900g", but actually he sells 810g, and has 90g left. Repeating it, he will sell total of 9kg + 900g + 90g + 9g + ... = 10kg. Jun 6 at 12:34
• @mihaild I considered it, but the problem with this reasoning is that "he uses 900gm weight instead of 1kg, but it is possible that for weights below 1kg, he has the correct weights." What is wrong with my analysis? The question mention only one crooked weight, right? 1kg => 900gm
– Max
Jun 6 at 13:09

$$\begin{array}{c|c|c|c} \text{Math notation} & \text{Relative amount} & \text{Customer} & \text{Shopkeeper} \\\hline x & \text{smaller} & \text{he gets} & \text{has originally} \\\hline 10x/9 & \text{bigger} & \text{buys} & \text{sell} \\\hline \end{array}$$ In your last example, you use different logic when you say