I saw this puzzle and thought it was pretty hard. I still have no
idea how to formulate my equations, even though I have all the
diagrams & the general feel of it. Anyone care to push me in the
right direction ?

Several faucets were used to fill up six tanks. For one hour, all the faucets discharged water in a reservoir, which distributed it between four of these tanks: A, B, C and D. After that, for one hour, the faucets discharged water in a double funnel which directed half of the water to tanks E and F and the other half to the reservoir which, in turn, continued to distribute its water between tanks A, B, C and D. With this, tanks A, B, C and D were full. To fill tanks E and F up, it was necessary to use one faucet, which, for two hours, distributed its water between tanks E and F. After this all the six tanks were full. What was the number of faucets initially used? (Note: all the faucets had the same water flow rate and all the tanks had the same volume).

Originally posted by SoLiDUS
I saw this puzzle and thought it was pretty hard. I still have no
idea how to formulate my equations, even though I have all the
diagrams & the general feel of it. Anyone care to push me in the
right direction ?

Simply model everything mathematically. Name variables. Set up equations. Substitute. Solve.

For example:

F = flow rate
T = time in hours
X = number of original faucets
V = volume of each container.

Then, for F not 0, and T not 0, X = 8.

JMG.

Please be more specific, GodLied. After all, he said "I still have no
idea how to formulate my equations".

It is not very helpful to say "Set up equations. Solve."

Show him your solution.

Originally posted by James R
Please be more specific, GodLied. After all, he said "I still have no
idea how to formulate my equations".

It is not very helpful to say "Set up equations. Solve."

Show him your solution.

This is where being a persistent mensan comes in handy...

I did it, but you could say it's cheating:

I started by determining how much a faucet fills (in percentage)

By doing the diagram, we can see that before the last faucet is
used for 2 hours, the two tanks are already 25% full. So in 2 hrs
the faucet filled 150%, or, 75% for each tank in an hour.

After that, I decided to go by trial and error until 8, where I saw
that everything worked out like I wanted it to, being:

8 x 75% = 600% fill total.

Divide in 4 for each ABCD tank gets you 150 filled so far. After the
funnel divides, you get 300/4 = 75 + 150 each = 225 total. Then I
checked that the two last had 300/2 = 150 each. Then, 75 an hr
for both meaning 37.5x2 = 75 each, 150+75 = 225. All of them are
full at 225.

So really, I'm lacking the equations... I'm sure it was much simpler
and more logical than the way I did it...

Originally posted by James R
Please be more specific, GodLied. After all, he said "I still have no
idea how to formulate my equations".

It is not very helpful to say "Set up equations. Solve."

Show him your solution.

James, I should have given more than the method and the answer. The details are helpful. After a visit to the store I will come back and show the conversion of English to math for the purpose of solving this problem.

JMG.

I don't know how GodLied solved the problem (he/she is right anyway) but a double check is never a bad thing...Here is my solution:

Let

N=the number of initial faucets.
Qv=the water flow rate of faucets in [m³/s]
V=the volume of the tanks
Therefore the volume of water discharged by N faucets in an hour (3600 [s]) is N*Qv*3600.

1.For the tanks A,B,C and D we have:

4V=N*Qv*3600+(N/2)*Qv*3600 (1)

For an hour all those N faucets discharged water in tanks A,B,C,D ---> the volume of water in the before mentioned tanks is N*Qv*3600.Further we have the information that for another hour half of the water volume enters A,B,C,D (the number of faucets remain the same as initially N) that is (N/2)*Qv*3600 (the other half being directed toward the tank E and F).After that the tanks A,B,C,D are full.

2.For the tanks E and F we have:

2V=(N/2)*Qv*3600+Qv*7200 (2)

2 hours=2*3600 [s]=7200 [s]

As I've shown above half of the water is distributed toward tanks E and F [its volume=(N/2)*Qv*3600] but they are not full.To fill them it is needed a single faucet which to discharge water for another 2 hours (the volume of water dicharged=1*Qv*7200).

Solving (1)+(2) ---> N=8.

Quote: Several faucets were used to fill up six tanks. For one hour, all the faucets discharged water in a reservoir, which distributed it between four of these tanks: A, B, C and D. After that, for one hour, the faucets discharged water in a double funnel which directed half of the water to tanks E and F and the other half to the reservoir which, in turn, continued to distribute its water between tanks A, B, C and D. With this, tanks A, B, C and D were full. To fill tanks E and F up, it was necessary to use one faucet, which, for two hours, distributed its water between tanks E and F. After this all the six tanks were full. What was the number of faucets initially used? (Note: all the faucets had the same water flow rate and all the tanks had the same volume). :End quote.

The problem is: given six tanks of equal volume, how many faucets with equal flow rates does it take to fill the tanks when four tanks are filled by recieving flow from all faucets for a unit of time as well as half as much flow for another unit of time; and, the other two tanks are filled by half the flow from all faucets for one unit of time plus the flow from one faucet for two units of time.

Define:

X = number of original faucets
F = flow rate
T = unit of time
V = volume of any tank.

Then it follows that

4V = FTX + FTX/2

and

2V = FTX/2 + F(2T)(1)

which implies

4V = FTX + 4FT.

By substitution

FTX + FTX/2 = FTX + 4FT

Obviously, as long as F and T not zero,

X + X/2 = X + 4

Subtracting X from both sides we get

X/2 = 4

Solving for X we get

X = 8.

If you noticed FTX on both sides of the equality, you could have jumped to X/2 = 4. Your level of observation can reduce the number of documented steps taken. Part of mathematical incompetence lies in illiteracy in the form of inability to comprehend what is written. I made the solution as drawn out as possible for those who need it.

JMG.