Printable View
You have 12 coins. They are completely identical except six of them weigh 24g and the other six weigh 25g. You have only a balance scale to sort them out. What is the minimum number of weighings which guarantees all the coins to be sorted?
Multiple answers possible with this one. https://forums.windrivers.com/images.../2005/03/1.gif
__________
Find possible answers below.
10.Quote:
Originally Posted by a d e p t
6Quote:
Originally Posted by a d e p t
If you were lucky enough to weigh each of the 6 coins of one type, you would then know that the 6 left were all of the other wight. :thumbs2:
You could do it in 5 if you were lucky.Quote:
Originally Posted by Camaro80z
No, sorry.Quote:
Originally Posted by edball
And there seems to be something misssing...
..oh yeah - no explanation given!
If I'm wrong, why should I explain it ? This question is making my brain hurt anyway !Quote:
Originally Posted by a d e p t
11 assuming the first 2 coins were different and each coin weighed against one of the first 2 went into a different pile until you got to the last coin which would go into the pile with only 5 coins.Quote:
Originally Posted by a d e p t
Steel is denser than air, therefore it would take a higher volume of air to displace the weight of the steel in the boat over the surface of the water when the boat was upright.Quote:
Originally Posted by a d e p t
I like Ed's initial answer of 10. You pick one coin at random and place it on one side of the scale. You have 11 coins remaining. After measuring 10 of the coins against the initial coin and separating them based on your results, you will know what the weight of the last coin must be so there's no need to measure it. You will also know which pile the initial coin goes in so no need to weigh that one.Quote:
Originally Posted by a d e p t
You could get lucky and do it in fewer tries but that sounds like the smallest number of measurements to guarantee they are all sorted.
The answer is C
Just had to post in a thread that had been going on for so long
Hrm, I now think the answer is eight. You place six coins on each side and note the level of each side of the scale. Swap one coin from each side to the other side. Worst case, five measurements each result in the two coins being identical so you now have six piles, each consisting of two identical coins.
Two more weighings of two of those piles against each other will result in one of three scenarios
1-two piles of four, each of which is known to contain heavy or light coins
2-two piles of four, neither of which is known to be heavy or light
3-one pile of four identical coins and two piles of two coins known to be heavy or light
Scenario 1, one more weighing of any weighed two coin pile against either unweighed pile will sort them all out
Scenario 2, same method as scenario 1
Scenario 3, the two unweighed piles must be identical. weigh either of them against either of the two separate weighed two coin piles that are known to be heavy or light.
So worst case with this method is five weighings plus two weighings plus one weighing equals eight.
I say eight also, but I used a different weighing method.Quote:
Originally Posted by a d e p t
Simple logic can eliminate some of these steps depending on the situation, but this is given worst case scenario.
Step 1: Split coins up into four groups of three, labeled A, B, C and D.
Step 2: Weigh groups A and B (1), this will determine either even or one side is heavy.
Step 3: Weigh 2 coins from group A (2), then weigh one of those and the third from group A (3). This should determine light and heavy for the three coins in group A, using logic on the three weighing results. Three down, 9 to go.
Step 4: Weigh 2 coins from group B (4), the weigh one of those and the third from group B (5). This should determine light and heavy for the three coins in group B, using the same logic as in step 3. 6 down, 6 to go.
Step 5: Weigh group C and group D (6), they will be either even or heavy.
Step 6: Weigh 2 coins from group C (7).
Step 7: At this point you will either weigh the third coin from group C against one of the other two, or you will weigh two coins from group D (8). Anyway it goes, you should be able to determine placement with a little logic.
I have used this method repeatedly and have come up with 5 to 8 weighings, depending on the situation. I'm not sure I've explained this properly, but it's difficult given the various possible situations. Using this method, however, 8 is the maximum number of weighings that guarantees separation.
Extra point goes to you.Quote:
Originally Posted by edball
3 points.Quote:
Originally Posted by cisco2
3 points here too.Quote:
Originally Posted by Deity