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## Water From 3 Jugs Into Two Pools

Logic puzzles require you to think. You will have to be logical in your reasoning.

 Puzzle ID: #19031 Fun: (2.1) Difficulty: (2.86) Category: Logic Submitted By: nenad Corrected By: MarcM1098

You have:
- two 10-liter jugs, full of water
- one 9-liter jug, empty
- two large pools, empty

Your objective is to end up having 4 liters of water in each pool, and one liter of water in the 9-liter jug.

You have nothing else at your disposal and you can not measure using eye or shape of jugs. You can not pour water from pools, only into them.

Label the 10-liter jugs as A and B, the 9-liter jug as C, and the two pools as Pa and Pb. The problem can be solved in 33 steps, as shown below. Each step is shown in one row, starting with the step number, followed by the jug we poured water from, then the quantity of water, then the jug/pool we poured water into... i.e. A ->( 1)-> Pb means we poured 1 liter of water from jug A (being 1st 10-liter jug) to pool Pb (being 2nd pool). Finally in the row we show how much water we have in each jug/pool after performing this step.

------------------------------------
#: From -> Qty -> To = A B C Pa Pb
------------------------------------
0: _____________ = 10 10 0 0 0
1: A ->( 9)-> C = 1 10 9 0 0
2: A ->( 1)-> Pa = 0 10 9 1 0
3: C ->( 9)-> A = 9 10 0 1 0
4: B ->( 9)-> C = 9 1 9 1 0
5: B ->( 1)-> Pa = 9 0 9 2 0
6: C ->( 1)-> A = 10 0 8 2 0
7: C ->( 8)-> B = 10 8 0 2 0
8: A ->( 9)-> C = 1 8 9 2 0
9: A ->( 1)-> Pa = 0 8 9 3 0
10: C ->( 9)-> A = 9 8 0 3 0
11: A ->( 2)-> B = 7 10 0 3 0
12: B ->( 9)-> C = 7 1 9 3 0
13: B ->( 1)-> Pa = 7 0 9 4 0
14: C ->( 3)-> A = 10 0 6 4 0
15: C ->( 6)-> B = 10 6 0 4 0
16: A ->( 9)-> C = 1 6 9 4 0
17: A ->( 1)-> Pb = 0 6 9 4 1
18: C ->( 9)-> A = 9 6 0 4 1
19: A ->( 4)-> B = 5 10 0 4 1
20: B ->( 9)-> C = 5 1 9 4 1
21: B ->( 1)-> Pb = 5 0 9 4 2
22: C ->( 5)-> A = 10 0 4 4 2
23: C ->( 4)-> B = 10 4 0 4 2
24: A ->( 9)-> C = 1 4 9 4 2
25: A ->( 1)-> Pb = 0 4 9 4 3
26: C ->( 9)-> A = 9 4 0 4 3
27: A ->( 6)-> B = 3 10 0 4 3
28: B ->( 9)-> C = 3 1 9 4 3
29: B ->( 1)-> Pb = 3 0 9 4 4
30: C ->( 7)-> A = 10 0 2 4 4
31: C ->( 2)-> B = 10 2 0 4 4
32: A ->( 9)-> C = 1 2 9 4 4
33: C ->( 8)-> B = 1 10 1 4 4

After the 33rd step we end up with 4 liters in each pool (Pa and Pb), and 1 liter in the 9-liter jug (C), as requested in the problem.

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