4 bags of coins:
2 of them contain 100 10-ounce coins, the other 2 contain 100 12-ounce coins.
You are allowed to use a regular weigh scale (one that tells the weight) once,
and a pan scale once.
How can you determine the coin weights in each bag?
NOTE: I want the solution that uses the least number of coins (Nerd)
Same as the other one? I just changed a little the procedures for this one.
Take 7 coins, 1 from the first, 2 from the second and 4 from the third.
7 coins make 70 ounces.
So, if you weigh 72 ounces, the bags containing the 12 ounce coins are 1st and 4th.
74 -> 2nd and 4th
76 -> 1st and 2nd
78 -> 3rd and 4th
80 -> 1st and 3rd
82 -> 2nd and 3rd
That's a nice solution; but it can be done using 3 coins (and BOTH scales!).
By the way, the lighter coins weigh 10 ounces: so your 700 should be 70. (not important!)
Woops, I was messed up the weighs, sorry (Surprised)
I'll edit this part.
Okay, I think that if you take 6 coins, then:
1 from 1st.
2 from 2nd.
3 from 3rd.
62 -> 1st and 4th
64 -> 2nd and 4th
66 -> 1st and 2nd OR 3rd and 4th
68 -> 1st and 3rd
70 -> 2nd and 3rd
Now, we use the pan balance, take the 1 and 2 from the 1st and 2nd bags on one side, and the others (ie 3 from 3rd bag) on the other side. The one being the heavier will be the heavier coins.
Yes.....BUT that's 6 coins; how do you do it with 3 coins only?
Ok, I think that's easier and I didn't thought about it earlier XD
Ok, take any coin, one from any two bags.
Put each on one side of the pan balance.
If equal, then weigh one coin with the weigh scale. If reading 10, then both coins from the bags are 10 ounces coins. Same otherwise.
If unequal, the heavier is the 12 ounce coin (obviously). Then take a third coin from any other 2 bags and weigh on the weigh scale. Reading gives the weigh of the coin.