Statement: You have a 11 litre and a 19 litre water container, each container has no markings except for that which gives you it's total volume. You also have a running tap. You must use the containers and the tap in such away as to exactly measure out 6 litres of water. How is this done? Can you generalize the solution?
Need your assistance to solve this question.
But I had solved similar question as shown below:
Statement:
You have a 3 and a 5 litre water container, each container has no markings except for that which gives you it's total volume. You also have a running tap. You must use the containers and the tap in such away as to exactly measure out 4 litres of water. How is this done? Can you generalise the form of your answer?
Answer:
There are two ways to solve this, maybe the question could be modified to say the 5 litre can doesn't fit under the tap...
Number 1: $\cdot$ Fill the 5 litre can from the tap
$\cdot$ Empty the 5 litre can into the 3 litre can - leaving 2 litres in the 5 litre can.
$\cdot$ Pour away the contents of the 3 litre can.
$\cdot$ Fill the 3 litre can with the 2 litres from the 5 litre can - leaving 2 litres in the 3 litre can.
$\cdot$ Fill the 5 litre can from the tap.
$\cdot$ Fill the remaining 1 litre space in the 3 litre can from the 5 litre can.
$\cdot$ Leaving 4 litres in the 5 litre can.
Number 2:
$\cdot $ Fill the 3 litre can from the tap.
$\cdot$ Empty the contents of the 3 litre can into the 5 litre can.
$\cdot$ Fill the 3 litre can from the tap.
$\cdot$ Empty the contents of the 3 litre can into the 5 litre can. - Leaving the 5 litre can full and 1 litre in the 3 litre can.
$\cdot$ Pour away the contents of the 5 litre can
$\cdot$ Pour the 1 litre from the 3 litre can into the 5 litre can.
$\cdot$ Fill the 3 litre can from the tap.
$\cdot$ Empty the contents of the 3 litre can into the 5 litre can.
$\cdot$ Leaving 4 litres in the 5 litre can.
Generalised Form:
We have two solutions above which both give the same answer of 3 litres. It turns out that if you make it algebraic the answer does not have the same form and it's sort of coincidence. Lets call the smaller container A and the lager B and work it through.
Number 1: Empty the 5 litre can into the 3 litre can - leaving 2 litres in the 5 litre can. ie $B - A$ in $B$ \ Fill the 3 litre can with the 2 litres from the 5 litre can - leaving 2 litres in the 3 litre can. As in space in A of $A - (B - A) = 2A - B$ Fill the remaining 1 litre space in the 3 litre can from the 5 litre can. Leaving 4 litres in the 5 litre can. So the amount in $B$ is $B - (2A - B)$ or $2B - 2A$
So the generalised form of solution 1 is $2B - 2A$ in $B$.
Number 2: Fill the 3 litre can from the tap. Empty the contents of the 3 litre can into the 5 litre can. Which gives us $B - A$ space in $B$. Fill the 3 litre can from the tap. Empty the contents of the 3 litre can into the 5 litre can. - Leaving the 5 litre can full and 1 litre in the 3 litre can. ie. removing $(B - A)$ from $A$ leaves $A - (B - A)$ or $2A - B$ in $A$ Pour away the contents of the 5 litre can. Pour the 1 litre from the 3 litre can into the 5 litre can. ie $2A - B$ in $B$ Fill the 3 litre can from the tap. Empty the contents of the 3 litre can into the 5 litre can. Leaving 4 litres in the 5 litre can. ie $A + (2A - B)$ or $3A - B$ in $B$.
So the generalised form of solution 2 is $3A - B$ in $B$.
There are some restrictions, such as in solution 2; $2A < B$ ($2A$ must be less than $B$,) but other wise we can make up other puzzles like 5 Litres and 9 Litres to get 6 Litres. or 8 Litres. All exactly the same form.
Hint: Some (really just some?) of the achievable filling states are the ones marked below plus by extending the arrow accordingly