## Casks of Wine

A man has an 8 gallon cask of wine and wants to get half of it in a 5 gallon cask and he only has a 3 gallon measure; how does he do it?

*Answers can be submitted below in the ‘comments’ section if you can figure it out. To give everyone a fair chance, however, will keep submissions under wraps for about a day. Thanks*

Joe| Guest June 18th, 2007 - 10:15 pmFill the 5 gallon cask, then fill the 3 gallon cask with the 5 gallon cask. You now have 3 gallons in the 8 gallon cask, 2 gallons in the 5 gallon cask, and 3 gallons in the 3 gallon cask.

Pour the 3 gallons from the 3 gallon cask into the 8 gallon cask, then fill the 3 gallon cask from the 5 gallon cask. You now have 6 gallons in the 8 gallon cask and 2 gallons in the 3 gallon cask.

Fill up the 5 gallon cask with the 8 gallon cask. You now have 5 gallons in the 5 gallon cask, 2 gallons in the 3 gallon cask, and 1 gallon in the 8 gallon cask.

Finally, fill the 3 gallon cask with the 5 gallon cask and you’re left with 4 gallons in the 5 gallon cask.

Mario Berges| Guest June 18th, 2007 - 11:09 pmFirst he fills the three gallon cask and pours the contents into the five gallon one.

He repeats this process again, but only until the five gallon cask is full.

This will leave him with 1 gallon on the three gallon cask.

Then he pours the contents of the five gallon cask back into the eight gallon one.

Lastly, he pours the 1 gallon content of the three gallon cask into the now empty five gallon one, and fills/pours the three gallon one again to make up 4 gallons.

biobot| Guest June 19th, 2007 - 2:33 amHe fills the 5 gallon barrel. Pours it into the 3 gallon barrel until that’s full. That’s 2 gallons left in the 5 gallon barrel. Repeat.

Conny| Guest June 19th, 2007 - 4:13 amThe following sequence shows the respective volume of whine in

each of the three cascs.

More precisely, the three-tuples denote the volume of whine in the …

[8 gallon casc, 5 gallon casc, 3 gallon measure].

[8,0,0],

[3,5,0],

[3,2,3],

[6,2,0],

[6,0,2],

[1,5,2],

[1,4,3]

First you fill 5 gallons into the 5 gallon casc. Then you fill 3 gallons from the 5 gallon task. And so on.

jason| Guest June 19th, 2007 - 6:47 ambuy a four gallon bucket!! let A=8. B=5, and C=3

Pour A into C, then C into B.

Pour A into C, then C into B again(1 gallon remainder in C)

Pour B into A, then C into B.

Pour A into C again, then into B, giving you 4 gallons

miws| Guest June 19th, 2007 - 9:03 amfirst fill up the 5 gallon cask (5gc) and from this fill up the 3 gallon cask, so you have 3g in the 8gc, 2g in the 5gc and 3g in the 3gc. put the 3g from the 3gc into the 8gc and the 2g from the 5gc to the 3gc. Now you have 6g in the 8gc, 0g in the 5gc and 2g in the 3gc. Now fill up the 5gc from the 8gc and fill up the 3gc with the 5gc. Since the 3gc already had 2g there is only room for 1g, which leaves the 5gc with 4g – half of the original content from the 8gc.

Mark M.| Guest June 19th, 2007 - 9:32 amFill the 3 gallon cask and dump it into the 5 gallon, refill the 3 gallon and top off the 5 gallon, leaving 1 gallon of wine in the 3 gallon cask. Empty the 5 gallon cask into the 8 gallon. Place the 1 gallon into the 5 gallon container then refill the 3 gallon and add to the 5 gallon cask leavig 4 gallons in the 5 gallon cask.

MosieGT| Guest June 19th, 2007 - 10:59 amTilt the cask 45 degrees, so that only half of the wine can spill out into the 5 gallon cask.

Jordan| Guest June 19th, 2007 - 11:01 amassuming the 8 gallon cask is full, he could tip the cask and pour the wine into the 5 gallon cask until the top surface of the liquid forms a line between the open pouring edge of the cask and the line where the bottom of the cask intersects with the circular wall of the cask, thus creating a forty five degree angle, indicating one half.

behnam| Guest June 19th, 2007 - 11:48 am8 3 5

——

8 0 0

3 0 5

3 3 2

6 0 2

6 2 0

1 2 5

1 3 4

Alexey| Guest June 19th, 2007 - 12:36 pmAt the start we have

8 gallons in 8 gallon cask – 0 in 5 gallon cask – 0 in 3 gallon cask

The sequence is as follows:

3-5-0

3-2-3

6-2-0

6-0-2

1-5-2

And, finally 1-

4-3Shawn| Guest June 19th, 2007 - 1:34 pmCask A – 8 gallon capacity, contains 8 gallons (A-8)

Cask B – 5 gallon capacity, contains 0 gallons (B-0)

Cask C – 3 gallon capacity, contains 0 gallons (C-0)

1. Initial State: A-8, B-0, C-0

2. Fill Cask B from Cask A: A-3, B-5, C-0

3. Fill Cask C from Cask B: A-3, B-2, C-3

4. Empty Cask C into Cask A: A-6, B-2, C-0

5. Empty Cask B into Cask B: A-6, B-0, C-2

6. Fill Cask B from Cask A: A-1, B-5, C-2

7. Fill Cask C from Cask B: A-1, B-4, C-3

The 5-gallon container, Cask B, now contains 4 gallons of wine, exactly half of the original 8 gallons.

Ian| Guest June 19th, 2007 - 2:54 pmhe pours 3 gallons into the 5 gallon cask. he takes another three and fills it to the top, with out spilling a drop. he will next pour all of the wine in the 5 gallon cask back into the 8 gallon cask. he pours the remainding amount of wine in the 3 gallon measure into the 5 gallon cask. he gets another 3 gallons from the 8 gallon cask and pours it into the 5 gallon cask.

problem solved.

Tom Darby| Guest June 19th, 2007 - 2:56 pmDUH! Pour 3 gallons into the 3-gallon container, set it aside, pour the remainder in 5 gallon bucket. Big deal.

Dusty| Guest June 19th, 2007 - 7:00 pm1.empties 3 gals into measure which is transfered to 5 gal cask.

2.empties 3 gals into measure leaving 2 gals in original cask.

3.empties 2 gals from measure into 5 gal cask leaving 1 gal in measure.

4.empties 5 gals from 5 gal cask into original cask which now contains 7 gals

5.empties 1 gal from measure into 5 gal cask

6.empties 3 gals from original cask into measure which is transferred to 5 gal cask leaving 4 gals in each cask.

hope explanation is correct and not too long winded.

Abhijit| Guest June 19th, 2007 - 8:10 pmIt requires the following steps :

1.Prestart position : cask A contains 8 litres , cask B and C are empty

2.Distribute this 8 litres from A to B & C . Now A is empty and B contains 5 litres and C contains 3 litres .

3.Empty the cask C by transferring its content to cask A. Now A contains 3 litres B contains 5 litres and C is empty .

4.Tansfer 3 litres from cask B to cask C . Now A contains 3 litres , B contains 2 litres and C contains 3 litres ( i.e, it is full )

5.Transfer the content of cask C to cask A . Now , cask A contains 6 litres B contains 2 litres and the C is empty .

6. Transfer the content ( ie. 2 litres) of cask B to cask C . Now A contains 6 litres B is empty and C contains 2 litres.

7. Transfer 5 litres from A to B . Now A contains 1 litre , B contains 5 litres and C 2 litres .

8. Fill up the empty portion i.e, 1 litre of C from B . Now A contains 1 litre C contains 3 litres .

9.Now transferring the content of C to A gives half of original content of the cask A i.e, 4 litres.

I think there must be some shorter way . But i have failed to find out . I am looking forward to better responses from others .

Azri| Guest June 19th, 2007 - 9:22 pm(8) (5) (3)

8 0 0

3 5 0

3 2 3

6 2 0

6 0 2

1 5 2

1 4 3

4 4 0

RLP| Profile June 19th, 2007 - 10:08 pmLots of right answers and good explanations;

the 45 degree angle thing caught me by surprise

Mark| Guest June 20th, 2007 - 3:55 amhmm… Whether the 45 degree tilt works depends very much on the shape of the cask. If its height is equal to its width (among other requirements, such as symmetry), it could work. Otherwise, if it’s taller than it is wide you would pour too little at 45 degrees tilted. If it’s shorter than it is wide you would pour too much.

Leslie| Guest June 20th, 2007 - 2:30 pmFirst column is the 8 gl cask, Second Column is the 3 gl measure and the third is the 5 gl cask into which we will end up with 4 gls.

Step 1. The 8 gl cask is full and the 3 gl measure is empty as is the 5 gl

Represented by 8 – 0 – 0

Step 2. Pour out 3 gl to the measure 5 – 3 – 0

Step 3. Pour the 3 gl from measure into the 5 gl cask 5 – 0 – 3

Step 4. Pour 3 gl from the 5 remaining gls from the

original 8 gl cask into the measure 2 – 3 – 3

Step 5. Fill the 5 gl cask to capacity from 3 gl measure 2 – 1 – 5

(and by reason of capacity you are left with

1 gal left in the measure because only 2 gls can

fit into the 5 gl cask)

Step 6. Pour the 5 gls back into 1st cask with 8 gl 7 – 1 – 0

capacity

Step 7. Pour the one gl from the measure into the 7 – 0 – 1

5 gl capacity

Step 8. Pour out 3 gls into the measure from the 7 gls 4 – 3 -1 Step9. Finally pour the 3 gls into the final cask and you end up with

4 – 0 – 4

Ian| Guest June 20th, 2007 - 7:16 pmtom, that would be 5 gallons in the 5 gallon cask, not 4.

MosieGT| Guest June 21st, 2007 - 1:18 pmMark that is a good point however casks are most usualy round, as shown in the image.

Assuming that the cask is of the usual shape, tilt the cask so that for example the bottom left “edge” is horizontal to the top right “edge” then only half of the wine shall spill out.

Ok i know my use of the word “edge” isn’t quite correct as there is no “edge” as such around the circumference of a cylindrical item. but i think you could still get my point.

Ian| Guest June 21st, 2007 - 2:33 pmhay! this was the 100th brainteaser post! I didn’t notice that before.

RLP| Profile June 21st, 2007 - 5:54 pmIan- thanks for pointing out

if I realized it myself, probably would’ve pulled out this extra special puzzle I’ve had in waiting…will try to save for the 200th

remember this ‘comment’ here 60-90 days from now

actually, for the past 2 months, I’ve been trying to get this neat brain teaser made. Was hoping to have it ready for the site’s 1 year anniversary last month, but it’s taken longer than expected. Hoping to have it up sometime over the next couple days.

Mark W| Guest June 21st, 2007 - 6:25 pmMosieGT indeed it’s a smart plan. We could calculate the angle assuming the appropriate symmetry:

angle = 90 – arctan(width/height)

marcus| Guest June 27th, 2007 - 1:32 amempty 3 gallons into the measure, transfering it to the 5 gallon cask.

take 3 gallons from the 8 gallon and pour 2 gallons of that into the 5 gallon filling it.

what you should now have is

2 gallons in the 8 gallon cask.

5 gallons in the 5 gallon cask.

1 gallon in the 3 gallon cask.

now empty the 5 gallons bck into the 8 gallon cask making 7 gallons, pour the 1 gallon into the 5 gallons, refill the container and empty a further 3 gallons, you know have 4 gallons of shaken up wine.

ann| Guest June 27th, 2007 - 9:38 amI KNOW YOU FILL THE 5 GALLON CASK AND THEN TAKE THAT AND FILL THE 3 GALLON CASK AND YOU HAVE 2 LEFT BUT I CANT GET PAST THERE BUT I AM WORKING ON IT

nick| Guest June 30th, 2007 - 5:17 pmvolumes in 8 gallon cask A, 5 gallon cask B and 3 gallon cask C respectively:

8,0,0

5,0,3 (3 from a to c)

5,3,0 (3 from c to b)

2,3,3 (3 from a to c)

2,5,1 (2 from c to b)

7,0,1 (5 from b to a)

7,1,0 (1 from c to b)

4,1,3 (3 from a to c)

4,4,0 (3 from c to b)

azris method is better with one step

carolyn d| Guest August 10th, 2007 - 6:12 pmpours some of it into the 3 gallon, then pours the rest into the 5

Sanjay| Guest September 4th, 2007 - 9:10 pmThis question was in the movie Die Hard 3 (or Die Hard with a Vengence) The bad guy gave the good guys this question, and had to solve this in time to stop the bad guys or something.

Good question, and good movie!