How to find the least number of weigh trials when trying to get a decimal amount?

Question

Does it exist a way to obtain a quick answer to this problem instead of just guessing?

The problem is as follows:

A technician has only available a two pan scale to measure powdered sucrose. The only calibrated weighs available in the lab are one of $100\,g$. He is tasked to obtain $1.1\,kg$ of the sucrose from a bag which has only $2.5\,kg$ of the sugar. What would be the least number of weigh trials he can take to obtain the desired weight?.

The alternatives given are as follows:

$\begin{array}{ll} 1.&\textrm{4 trials}\\ 2.&\textrm{3 trials}\\ 3.&\textrm{2 trials}\\ 4.&\textrm{6 trials}\\ \end{array}$

Does it exist a way to solve this kind of problem without just guessing?. I'm stuck in this situation. Can someone help me here?.

If you split in two halves $2.5\,kg$ you can have $1.25\,kg$ in both sides. This would be in one trial.

In a second trial $1.25$ can be in one side and in the other $1.15+0.10$ using the other weigh. But I end up having one bag with $1.15\,kg$.

In the following weigh: (Using the other sack and returning to it the unused $100$ grams)

$1.25+0.10=1.35\,kg$

But here is where I'm stuck at. Can someone help me?. Does it exist an equation or something?. It would help me a lot to get a step-by-step explanation for this problem.

Answer 1

First, put 100g to the left pan. Then add powder to both left and right pan such that they balance. This way, You will have 100g weigh + 1.2kg powder on the left pan and 1.3kg powder on the right pan.

Remove the 1.3kg powder from the right pan, keep the 1.2kg powder from the left pan. Keep the 100g weigh on the left pan.

Now try to add powder from remaining 1.2kg powder to the right pan so that it balance the 100g weigh on the left pan. This way You will add 100g powder to the right pan, and what does that left You? 1.1kg powder left :)

Now the method (maybe not really):

$2.5kg=25\times 0.1kg$ and $1.1kg=11\times0.1kg$

$25=2^{4}+2^{3}+2^{0}$ and $11=2^{3}+2^{1}+2^{0}=2^{3}+2^{2}-2^{0}$

In one trial i can do one of the followings:

1. Substract $2^{0}\times0.1kg$ of powder. (this is what i did for the second trial)
2. Reduce all power of 2 by 1, and removing the $2^{0}$. (this is what i did for the first trial, $12=2^{4-1}+2^{3-1}$)
3. Reduce all power of 2 by 1, and maintaining the $2^{0}$. (this is if i keep the 1.3kg instead of 1.2kg on my first trial, $13=2^{4-1}+2^{3-1}+2^{0}$)

That being said, i want to move from $2^{4}+2^{3}+2^{0}$ to $2^{3}+2^{2}-2^{0}$ using aforementioned steps. Thus i did step 2 for the first trial, then step 1 for the second trial.

Maybe not really what You expect of a method.

Answer 2

The problem should really define what types of weighing are allowed. My solution:

Put the 100g on one side and split the sugar until it balances. You have 1.2kg with the weight and 1.3 kg on the other side

Move the weight to the other pan and remove sugar until it balances. There is still 1.2 kg of sugar on the other pan, so we have 1.1 kg on the pan with the weight

2 weighings

I am not aware of any general procedure for things like this. You are just supposed to be clever.