Milk problem?

What sort of scale is it? One that is used by people to measure their own weight? Analogue or digital?

I would insert the coin and then quickly place the cartons on the scale one by one, see which one tips the scale by less then 1 kilogram.

Welcome to A2K by the way

gee...thanx for the welcome...

well, the type of the scale is not the subject here...the most important thing here is, how to know which is the wrong carton by one time use of the scale...

got to do with a bit calculus....

try again, good luck!

There is a contradiction here. It says he has to deliver 10 cartons, but at the same time it's written s that one of them should not be sent. It's not written how many cartons he has in the car. If he only has 10 cartons total, he would have to drive back to get a new carton anyway, in order to satisfy all the customers (or maybe there is one single customer?).

I don't know all the details I would like to know, so I have to make presumptions.

1. He puts 5 boxes on the weight. Presuming that the bottles and the carton doesn't have any weight (very unrealistic). If the weight is less than 5 kilos he can safely deliver the other 5 cartons.
a)With the money he got for the milk he delivered (Again presuming that he is delivering to several customers AND that they are paying with cash) he can use another one of those coin-insert-scales at the street (I have never seen those).
b) If he doesn't get paid cash, he can ask his customers if they have a scale he can use.

2. He simply opens the boxes and compares the bottles. I presume that it's not important that the boxes stay unopened, if they were closed to begin with. Maybe he has to open the cartons anyway, because he is delivering individual bottles of milk. It should be easy to see that one box has bottles with less milk (presuming they are see-trough). In this case there is not even a point in using the scale; unless he just wants to make sure he found the right carton.

He will mark all bottles in each carton with a number from 1 to 10 and take one bottle from the first carton, two from the second and so on up to the 10th and weighs the whole lot of these massed bottles of milk.

In an ideal world where all bottles are 100g, the weight of the amassed bottles of milk will be 5500 grams (10*11/2*100 grams). If the first carton is light the massed bottles will be 5 gram light---the second carton 10 gram light and so on until if it the 10th carton it will be 50 grams light. The suspect carton will be determined by how light the massed bottles weigh (n*5 grams where n is the suspect carton).

Rap

Ah... I didn't understand that the bottles were individually selectable.
Raprap provides an elegant and very correct solution. Thank you, raprap

yup...raprap got it right !!! gee...was it simple like that ??