How can you get 13 pounds of coffee by using all three weights each trial?

Question

The puzzle is as follows [with minor copy edits for grammar]:

A merchant has a kiosk nearby our market. He has a sack of 52 pounds of coffee to sell. Assuming that he only uses a two pan scale and three weights of 7 pounds, 4 pounds and 1 pound, how many weighing trials at minimum does he have to do in order to get us 13 pounds of coffee from his sack? Assume that in all those weighing trials he will always use the three weights.

The choices given are:

1. 1 trial
2. 2 trials
3. 3 trials
4. 4 trials

This problem seems to be an adaptation from a reprinted copy of an intelligence APA exam of the mid 1980s on psychometry for IQ which is based on Thurstone-Catell's exams of the 1960s.

I'm not sure exactly how to tackle this puzzle. The thing is that the condition which forces to solve this problem by using the three weights make the problem difficult.

If it allowed to use any of the weights, you could split 52 in two halves of 26 each in one trial, and in a second and final trial we end with the 13 pounds requested.

I'm stuck on how to arrange those weights to get the necessary weight to answer. Can someone help me here?

Answer 1

Using all three weights in each trial is an interesting requirement, I haven't seen such a requirement before.

It's not clear what exactly counts as a trial, in particular whether a trial may consist of moving coffee from one side to the other until the scale is balanced, but assuming that's valid, here's a way to do it in

two

trials, using all three weights in each trial:

First, put the 7 and 1 weights on the left side and the 4 weight on the right side. Pour all the coffee on both pans such that the scale is balanced. There will be 24 pounds of coffee on the left side and 28 on the right side.

Next, put the 7 weight on the left side and the 4 and 1 weights on the right side. Remove the 28 pounds of coffee and use all 24 remaining pounds to balance the scale again. There will be 11 pounds on the left side and the desired 13 pounds on the right side.

Alternative solution:

First, put the 7 weight on the left side and the 4 and 1 weights on the right side. Pour all the coffee on both pans such that the scale is balanced. There will be 25 pounds of coffee on the left side and 27 on the right side.

Now remove the 27 pounds of coffee from the right side and put all weights onto the right side. If you remove coffee from the left side until the scale is balanced again, you will have removed exactly 13 pounds.

Answer 2

This started from Magma's answer, and I wanted to prove it. I start with a simplification of the problem and analyze the cases.

I finish by offering a completely different approach.

First, Replace the weights with single weight equivalents. These are 12, 10, 4, 2. with the provision that we must use one and only one.
Secondly, The weight always goes on the left otherwise we just have a symmetrical cases.

L + W = R // balance
L + R = C // Total Coffee
L + L + W = C (Sub Balance in Total Coffee)
L = (C-W)/2
R = C - L

For our first trial, C = 52 or 0, and only one of these is significant.
C W L R Config
52 12 20 32 7+4+1
52 10 21 31 7+4-1
52 4 24 28 7+1-4
52 2 25 27 7-1-4

The L and R columns are all the possible measurements we can get from one trial. If we were to put the weight on the right side, simply walk around the scales and it becomes left.

Assume the 13 + weight pounds will end up in L (Symmetry again)
13 = (C-W)/2
26 = (C-W)
By inspection, all the values in the L column are too small satisfy this constraint.
Also, since 26 is even, and all Ws are even, we are left with 32 and 28 as candidates.

Using 32, we can get C-W as 20,22,28,30 -- None of these are 26
Using 28, we can get C-W as 16,18,24,26

So, For our first weighing, we get 24 lbs of coffee and 7+1 pounds of weight in the left tray, and 28 pounds of coffee and 4 lbs of weigh in the right.
Put the 24 lbs of coffee back in the bag.

For our second trial, We have 13 coffee + 7 w in the left and 15 coffee and (1 + 4) w in the right.

Alternately, if we want 13 pounds in the right,
13 = C - (C - W)/2
26 = 2C - C + W
26 = C + W Similarly to above, we end up using 24 pounds of Coffee and 2 pounds of weight i.e. 11 + 7 = 13 + (4 + 1)

Then I checked Magma's alternate, and it is the only solution of this form.

But, I came up with a different approach that may be interesting,

Directly weigh out 12 lbs of coffee. Put it in the weightless sales bag.
In the left tray, put the 12 lb bag of coffee, and 7 lbs of weight.
In the right tray put the remaining 5 lbs of weight. Now add all the coffee from the bulk bag (40 lbs) to balance the scales.
You will end up with 13 loose, 12 bagged and 7 weight = 32lbs on the left. There will be 27 loose + 5 weight = 32 on the right.
Now, dump the sales bag into the bulk bag. Put the 13 pounds of coffee from the left tray in the sales bag and close the deal!

Answer 3

Two

Here's how to get it done.

Step 1:
Put 7 pounds on LHS, 4+1 pounds on RHS; we get 2 pounds of coffee out of the bag

Step 2:
Split the 2 pound coffee equally to both scales. This makes the scale balanced with 1 pound on each side. Now put 7+4+1 on left hand side, pour the remaining coffee on right hand side.

When the scale is balanced, we will have 13 pounds of coffee on right hand side

If we need to ensure all coffee is back in the bag, pour the left hand side coffee back into the bag. That will ensure only 13 pounds from right hand side is outside.

Answer 4

I have a bit of a different way to answer this one than the existing questions.

Number of trials:

two

Method:

First, add all weights to the left side of the scale totaling 12 pounds of weight. Add coffee to the right side of the scale until balanced, finishing trial one.
Leave the coffee on the right, and add the 1 pound weight to the right hand side. Add coffee to the left hand side until balanced leaving you with 12 pounds on the right and 1 pound on the left, finishing trial two.
Remove all the coffee measured - this is your thirteen pounds.

Edit: my solution is incorrect, many apologies. Can’t fix this method without breaking the rules of the puzzle but leaving it up anyway