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# my sink puzzle

5 replies to this topic

### #1 unreality

unreality

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Posted 26 July 2007 - 05:27 PM

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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### #2 mdsl

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Posted 26 July 2007 - 08:37 PM

Not sure if this is 9 "wasted" or 3 wasted.

0-0-8 Fill the 8 oz
3-3-2 Overspill both 3 oz
0-3-2 Dump 3 oz
2-3-0 Transfer from the 8 to the 3
2-3-8 Refill the 8 oz
3-3-7 Overspill 1 oz from the 8 to the 3

Target 7 oz in 8 oz cup, 3 oz dumped, 6 oz in other cups.
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### #3 unreality

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Posted 26 July 2007 - 10:54 PM

in ur drawing, the first two are 3 oz cups, the last is the 8 oz cup, so how did you come up with this:

2-3-8 Refill the 8 oz
7-3-3 Overspill 1 oz from the 8 to the 3

the 3oz cup cant hold 7 oz...?
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### #4 mdsl

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Posted 27 July 2007 - 12:33 PM

right, on my paper the 8 oz came first. I changed it because in the problem the 3 oz cups came first, but then I didn't pay attention to what I was typing. simple mistake, sorry if it confused you so much.
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### #5 normdeplume

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Posted 27 July 2007 - 01:28 PM

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.

Using this method, could allow you to waste 0 water, depending on acuracy. pouring 2 cups slowly together and leaving 2oz in each cup. This would allow you to pour 1 oz from the full 8 oz cup leaving
7 3 2, with no spillages. Of course accuracy would be the problem.
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### #6 unreality

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Posted 27 July 2007 - 08:48 PM

exactly. But notice my OP- it said "THE 7 MUST BE EXACT". It can't be guesswork. my Solution #2 wasn't guesswork. It didn't have to be even double pouring to work. It could've been all outta one cup and none outta the other and it would have still worked. Remember, it must be exactly 7.

guys, dont fret yourselves to hell though, I'm not even sure there's a solution better than my #3 one.
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