this is a water-measuring puzzle i made up, however i dont have the answer past my three solutions...

You have 2 cups that can hold 3 oz, and one cup that holds 8 oz. You have a sink where you can pour out as much water as you need, and a place to dump water. The goal is to have 7 oz of water in the 8oz cup, with the minimum number of ounces dumped and wasted. THE 7 OZ MUST BE EXACT.

Solution 1:

fill up a 3oz cup and pour it in the 8oz cup. Do this twice, so the 8oz cup has 6 oz in it. Fill up the 3oz cup a third time and pour it in the 8oz cup until the 8 oz cup is full- and you still have 1 ounce in the cup. Now dump out all 8 oz in the 8oz cup and use the other 3oz cup twice to get the 8oz cup to 6 oz... then pour in the 1oz from the other cup to get 7 oz in the 8oz cup, and the two smaller cups are empty, and there was only a 8oz loss!

Solution 2:

Fill up both 3 oz cups and pour them into 8 oz cup so there are 6 oz in it, leaving 2 oz still free in the 8 oz cup. Fill both cups and slowly pour them into the 8 oz cup, pouring slowly and evenly. When the 8 oz cup is full, 2 oz is left in both smaller cups. Well no matter if it wasn't even, there will still be 4oz total over both cups, because together they held 6, and 2 was put. Even if you poured just 1 cup in there, one cup would have 3 and the other 1. So together they have 4 oz no matter what, but it will be closer to 2 each. Ideally they both have 2. Empty the 8 oz and pour both cups into the 8oz container- there are now 4 oz in the 8 oz cup. Now fill up one 3 oz cup and pour it in to make 7 oz in the 8 oz cup.

This solution has the same end result as Solution 1: both smaller cups empty, the larger has 7 oz, and only 8 oz were wasted total.

Can you find a solution where less than 8 oz have been dumped? WAIT A MINUTE... I just thought of something:

Solution 3: Fill the 8 oz cup up with one cup so its finally full and 1 oz is left over in the 3oz cup. Now, with the 8 oz cup, pour 3 oz into the other 3 oz cup, filling it up and reducing the amount in the 8oz cup to 5 oz. So now we have 5/8, 3/3 and 1/3 in our cups. Dump out the 5 oz and pour the full 3 oz cup inside, then use that same cup to pour another 3 oz into the large cup to bring it 6 oz, then add the 1 oz! You now have 7 oz in the 8 oz with the other two empty, and only 5 ounces spilled this time!

Can anyone find something less than 5?

I think thats the absolute minimum with two 3oz cups... with a third 3 oz cup I could get it to 2 oz being dumped and two of the three 3oz cups full to pour in, with the last one pouring the 1 in. So you could get 2 oz wasted if you had three 3 oz cups... but you only have two of course. So is it possible to use yet a different method to waste less than 5 oz?

## Question

## unreality 1

this is a water-measuring puzzle i made up, however i dont have the answer past my three solutions...

You have 2 cups that can hold 3 oz, and one cup that holds 8 oz. You have a sink where you can pour out as much water as you need, and a place to dump water. The goal is to have 7 oz of water in the 8oz cup, with the minimum number of ounces dumped and wasted. THE 7 OZ MUST BE EXACT.

Solution 1:

fill up a 3oz cup and pour it in the 8oz cup. Do this twice, so the 8oz cup has 6 oz in it. Fill up the 3oz cup a third time and pour it in the 8oz cup until the 8 oz cup is full- and you still have 1 ounce in the cup. Now dump out all 8 oz in the 8oz cup and use the other 3oz cup twice to get the 8oz cup to 6 oz... then pour in the 1oz from the other cup to get 7 oz in the 8oz cup, and the two smaller cups are empty, and there was only a 8oz loss!

Solution 2:

Fill up both 3 oz cups and pour them into 8 oz cup so there are 6 oz in it, leaving 2 oz still free in the 8 oz cup. Fill both cups and slowly pour them into the 8 oz cup, pouring slowly and evenly. When the 8 oz cup is full, 2 oz is left in both smaller cups. Well no matter if it wasn't even, there will still be 4oz total over both cups, because together they held 6, and 2 was put. Even if you poured just 1 cup in there, one cup would have 3 and the other 1. So together they have 4 oz no matter what, but it will be closer to 2 each. Ideally they both have 2. Empty the 8 oz and pour both cups into the 8oz container- there are now 4 oz in the 8 oz cup. Now fill up one 3 oz cup and pour it in to make 7 oz in the 8 oz cup.

This solution has the same end result as Solution 1: both smaller cups empty, the larger has 7 oz, and only 8 oz were wasted total.

Can you find a solution where less than 8 oz have been dumped? WAIT A MINUTE... I just thought of something:

Solution 3: Fill the 8 oz cup up with one cup so its finally full and 1 oz is left over in the 3oz cup. Now, with the 8 oz cup, pour 3 oz into the other 3 oz cup, filling it up and reducing the amount in the 8oz cup to 5 oz. So now we have 5/8, 3/3 and 1/3 in our cups. Dump out the 5 oz and pour the full 3 oz cup inside, then use that same cup to pour another 3 oz into the large cup to bring it 6 oz, then add the 1 oz! You now have 7 oz in the 8 oz with the other two empty, and only 5 ounces spilled this time!

Can anyone find something less than 5?

I think thats the absolute minimum with two 3oz cups... with a third 3 oz cup I could get it to 2 oz being dumped and two of the three 3oz cups full to pour in, with the last one pouring the 1 in. So you could get 2 oz wasted if you had three 3 oz cups... but you only have two of course. So is it possible to use yet a different method to waste less than 5 oz?

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