The pool is equipped with five water pipes: A, B, C, D, and E

First, affirm that the problem is solvable, five unknowns, five equations;

According to the known conditions, it is not difficult to list the following five equations, assuming that the full once is m2a+2b=m; 6c+6d=m；10e+10a=m；3d+3e=m；15b+15c=m；

Simultaneous solution: a=11 60 m; b=19/60 m；c=-15/60 m；d=25/60 m；e=-5/60 m；

Therefore, the fastest water pipe is d, and it takes 60 25 = hours for a single pipe to be filled;

The highest water discharge efficiency is the C pipe, and the single pipe drainage needs 60 15 = hours to drain.

Fully open, that is, 2a+2b+2c+2d+2e=36, a+b+c+d+e=18......1

With the formula 1 and the known synopids a=0, b=2, c=13, d=-7, e=10

That is, d is to release water, and it will be released in 7 hours. A should be the fastest without water, with B.

Let the volume of the pool be, 6C+6D=1,10E+10A=1,3D+3E=1,15B+15C=1 and solve the equations to obtain: A=11 60, B=19 60, C=, D=5 12, E=-1 12. Therefore, the answers are respectively D pipe, and C pipe can be drained of water in an hour.

First of all, it is judged that the water injection pipe and the water discharge pipe, if A is the water discharge pipe, then it is the B water injection pipe, C is the water discharge pipe, D is the water injection pipe, E is the water discharge pipe, then A and E are the same water discharge pipes, therefore, A is the water injection pipe, B is the water discharge pipe, C is the water injection pipe, D is the water discharge pipe, E is the water injection pipe, but by A, B is full for 2 hours, E, A is full for 10 hours, it is also impossible, only A and B are water injection pipes, C is the water discharge pipe, D is the water injection pipe, E is the water discharge pipe, in the water injection pipe, by E, A is 10 hours, D and E are 3 hours, B and C are 15 hours, C and D are 6 hours to know the highest efficiency of D, in the discharge pipe, from C and D to 6 hours, D and E are 3 hours to know the highest efficiency of C, by A effect + B effect = 1 2, B effect + C effect = 1 15, A effect + E effect = 1 10 Know: C effect + E effect = 1 15 + 1 10-1 2=-1 3, and then by D effect + C effect = 1 6, D effect + E effect = 1 3 3 Know: D effect = (1 6 + 1 3 + 1 3) 2 = 5 12, C effect = 1 6-5 12=-1 4。

Therefore, the D water pipe with the highest water injection rate can fill the empty pool in 4 hours, and the C water pipe with the highest water discharge efficiency can fill the pool in 4 hours.

a+b=1 2 List similar equations and then it is easy to calculate, which is understood to be AB two pipes of water discharge 1 2 per hour, after the calculation, the positive number is the water intake, and the negative number is the water discharge.

If the volume of the reservoir is 1, the water will be filled for 1 A per hour and the water will be drained for 1 2 A per hour

Then 1 2 (1 a-1 2a) = a(h).

According to the inscription, the water inlet rate = 1 a, and the drainage rate is 1 2a, so (1 2) (1 a-1 2a).

1/2)÷(1/2a)

a (hours).

When are the inlet and drain pipes open at the same time?

It should be that the water left in the pool per hour accounts for 1 a-1 2a=1 2a of the whole pool, and the water in the whole pool is unit 1

Look at how many 1 2a are in 1, which is how many hours there are, so 1 (1 a-1 2a) = 1 1 2a = 2a (h).

The problem? When is the inlet and drain open at the same time, when is it full?

1 / 1/a) +1/2a ) 2a/3

According to the inscription, the water inlet velocity = 1 a, and the drainage velocity is 1 2a, so.

The time required to open the inlet and drain pipes at the same time = 1 2 (1 a-1 2a) = a h

The topic is A reservoir is filled with inlet pipes and drain pipes If the inlet pipe is opened separately, the pool can be filled with water: if the drain pipe is opened separately, 2ah can be drained of the full pond. Now that the reservoir has half a pool of water, and the inlet and drain pipes are opened at the same time for irrigation purposes, how long will it take to fill the reservoir?

Neither will I.

Let's get the topic clear first.

1 here is the total amount of water. 1 20 + 1 30 indicates the amount of water discharged in one minute if pipe B and pipe C are opened at the same time. The calculated 12 means that if the pool is full and pipe A is not open, it takes 12 minutes for B and C to cooperate, and we can see that B and C work together to release water at a rate of 1 12 per minute

However, the current situation is that pipe A is still flooding when B and C are discharging water, so the time for releasing the pool is extended, that is, it is extended by 18-12 = 6 minutes. Then 1 12*(18-12) 1 12*6 1 2That is, the water discharge rate per minute multiplied by the time when B and C cooperate, and the speed multiplied by the time is equal to the total amount.

The total amount here is the amount of water that A puts in 18 minutes1 2Then divide the total amount of 1 2 by the time it takes A to calculate 1 36 is the amount of water per minute of A (it can also be understood that if the outlet pipe is not working and only A is entering water, the filling time is 36 minutes).Now we can know that on one side it is pipe A that is entering the water and on the other side that pipe B is coming out.

Who's faster? Obviously, there is a gap between the inlet water of A and the water of B, and the difference is that the amount of water per minute (1 20-1 36), that is, the amount of water output per minute. Now we know the amount of water produced per minute (1 20-1 36) and the total amount of water is 1 (as I said at the beginning).

Then divide the total amount by the velocity, and you will get the time. I don't know if you can see it.

Set the slag content time of the pre-stove to be set in the unit of "1", and the amount of water entering 1 water pipe is 1, and open x water pipes first.

x*1*2/5+2x*1*3/5=8

x=5, then it's better to open 2x=10 water pipes.

Compare the first column with the third column to get the water injection rate: b e, compare the first column and the fifth column to get the water injection rate: a c, compare the second column and the fifth column to get the water injection rate: d b, compare the second column with the fourth column to get the water injection rate: e c, compare the third column and the fourth column to get the water injection speed: d a, from d b, b e, get d e, and because e c, get d c, from a c, d a, get d the fastest Answer: d

Assuming that the full tank volume is x and can be filled after n hours, then (x 18 + x 12-x 27) is the water that can be filled per hour.

From the meaning of the title, we can get n(x 18+x 12-x 27)=x, so let's do the math yourself.

The two brigades are compared to calculate the flow rate:

EA vs ED — Thana years — A >D

ea- ba ——e>b

bc - cd ——d>b

bc - ab ——c>a

DE - CD — Sensitive — C>E

It is concluded that the flow velocity of c is the largest, and the fastest can fill the pool.