1. ## Interesting puzzle

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).

2. ## Re: Interesting puzzle

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.

3. ## Re: Re: Interesting puzzle

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."

4. ## Re: Re: Re: Interesting puzzle

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."

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...

5. ## Re: Re: Re: Interesting puzzle

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."

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.

6. 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 [m<sup>3</sup>/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.

7. ## solution

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.

8. GodLied is correct. Here's an easier explanation (well, at least for those who don't get algebraic equations easily): I posted 2 versions so if you still don't understand version 1, read version 2 too.
given X is the number of faucets

Version 1
X faucets poured water to 4 tanks A B C D in 1 hour (first hour).
In the next hour, X faucets poured half to the 4 tanks A B C D and half to the 2 tanks E F which means (X/2 faucets poured water to 4 tanks and the other X/2 faucets to the 2 tanks in this hour).

***X faucets pouring water to 4 tanks in 1 hour is like X/2 faucets pouring water to 2 tanks in 1 hour (both equations say that X/4 faucets poured the same volume of water to each tank for 1 hour), so, we can leave the first hour and the X/2 faucets pouring water to the 2 tanks E F in the 2nd hour aside.
***We can get X based on X/2 filling up 4 tanks A B C D in the 2nd hour and the 1 faucet filling up the 2 tanks E F for 2 hours in the 3rd and 4th hour.
***Based on these, we can say that---------->
X/2 faucets filling up the 4 tanks in 1 hour (or X faucets filling up 8 tanks in 1 hour) is the same as 1 faucet filling up the 2 tanks in 2 hours (or 2 faucets filling up 2 tanks in 1 hour). When I say "like" or "the same", I mean in proportion.

***The proportions for the two equations are X:8 (X faucets is to 8 tanks) for the first equation, and 2:2 (2 faucets is to 2 tanks) for the 2nd equation.
***based on the 2nd equation with the proportion 2:2, we can say that the proportion is 1 is to 1.
so---------->
in X:8, X is equal to 8 (because 1:1 equals 8:8)

Version 2:
After the First hour:
A B C D each has water from X/4 faucets
After the Second hour:
A B C D each has water from X/4+X/8 faucets and is full and
E F each has water from X/4 faucets
After the Third and Fourth hour (2 hours):
E F each has the water from X/4 faucets (from the second hour) and water from half of the 1 faucet running for the last 2 hours and is full
***After filling up the six tanks with X/4 water each, what made them full is the:
water from X/2 faucets in the second hour (for A B C D) and water from 1 faucet running for the last 2 hours (for E F)----->which means that the X/2 faucets running for 1 hour puts the same amount of water as from the 1 faucet running for 2 hours, to each of the 6 tanks. It means that these two equations have the same proportion of "water to tank" per hour (X/2:4 = 2:2) or (X:8 = 2:2). From that, we can get X=8.

It's hard for me to form sentences that will make this easier to understand but I already did my best so I hope you understand why this post is so long.

9. If you can understand algebraic expressions/equation btw, GodLied showed the best and simplest solution.

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