# Does there exist a way to find the least number of trials when a certain weight is requested?

I'm not very sure how to tackle this sort of problem:

It is as follows:

At a groceries store there is a two pan scale and two weighs which are of $$250\,g$$ and the other of $$500\,g$$. It is known that the store has only $$5\,kg$$ of sugar on sale. Find the least number of weigh trials that must be made in order to fill an order of two sacks containing $$2.625\,kg$$ of sugar and the other $$2.375\,kg$$.

The alternatives given in my book are as follows:

$$\begin{array}{ll} 1.&\textrm{1 trial}\\ 2.&\textrm{2 trials}\\ 3.&\textrm{3 trials}\\ 4.&\textrm{4 trials}\\ 4.&\textrm{5 trials}\\ \end{array}$$

In this problem I'm sort of lost. Typically what I've attempted to do was that I can balance out half of the sack which has 5\,kg of the sugar hence in one sack I end up with $$2500\,g$$ and the other $$2500\,g$$.

I've attempted to add the two weighs given $$500+250=750$$ but this doesn't exactly produce a result which I could use. How exactly should I proceed in these kinds of problems. Can someone help me?.

Since I'm a slow learner, I'd like to get a very detailed explanation and a strategy on how to solve this sorts of problems.

First,

Put the 250g weight in one bowl and split the 5kg so that the scales balance.

..and, well, you are done.

As for the strategy for finding the solution, look for integer multiples of the weights (and their sums and differences) in the amounts of sugar (and their sums and differences). If you find none, try halving some of the numbers, which is easy to do on balance scales. Here you get two "direct hits" right off the bat, so you'd expect the solution to be very simple:

$$2.625\text{kg} + 2.375\text{kg} = 5\text{kg}$$
$$2.625\text{kg} - 2.375\text{kg} = 250\text{g}$$

Then you need to figure out how to exploit those coincidences in the most optimal way; I don't think there's any shortcut to that other than trying all the possibilities. Once you have found some way of getting the desired result, you can ignore any method that takes longer than that, so this isn't such a tedious task it seems.

• Please, don't use "most" with "optimal" :) Mar 15, 2020 at 12:45
• @Bass Wait. Does this means it will be achieved in one trial right?. It seems that I have to try out for the two weighs first and see if splitting the sugar would result into the answer. Mar 15, 2020 at 21:12
• @Bass So does this means that it is just in one trial?. I don't see any other sentence belonging to the solution so I'm assuming that's it. Mar 16, 2020 at 21:23

In:

3 weighings.

How?

Make two sacks of $$2500g$$. Take $$250g$$ out of one. Make two piles of $$125g$$. Add one pile to each sack.

• The problem requested in least trials but did not indicate to use all the weighs. I don't see any need to make those two additional piles of $125g$ unless using what you mentioned on two halfs of the sack. Mar 16, 2020 at 21:21

Put on one side of the scale the 250 gr weight then put on the other side the 500 gr weight. Put the 5000gr bag on one side and an identical empty bag on the other side. Start taking sugar from the full bag and put it on the other till the scale is balanced. Then you have 5750gr divided by 2 equals to 2875gr if you take the weights off the scale then you have 2875-5oo=2375 and 2875-250=2625 which are the weights you want. So you have 1 weighing.

• Apparently it was just required one trial. Mar 15, 2020 at 21:10
• @Chris Steinbeck Bell. If you say it takes 1 weighing you know better so that might be. Mar 15, 2020 at 23:47
• As suggested by one of the answerers there is required only one trial but you say it is two. I think this is not correct. Mar 16, 2020 at 21:19
• @Chris Steinbeck Bell. Now you the answer you want. Mar 17, 2020 at 16:07