The Puzzle There are ABC three inlet taps on a pool

1/a+1/b=1/3

1/a+1/c=1/4

a/b+1/c=1/5

Add the three formulas to get.

2(1/a+1/b+1/c)=47/60

1/a+1/b+1/c=47/120

So the time it takes to fill the pool when the three taps are turned on is 120 47 hours.

a+b=1/3

a+c=1/4

b+c=1/5

a+b+c=(1 3+1 4+1 5) 2=47 120 The time required to open the three faucets is to fill the pool is:

1 (47 120) = 120 47 (hours).

It can be considered that the flow rate of each faucet is A, B, and C respectively, and the total volume of the pool is D.

3（a+b)=d

4(a+c)=d

5(b+c)=d

Convert: a+b=1 3d

a+c=1/4d

b+c=1/5d

The two sides of the equation are added to each other.

2（a+b+c)=47/60d

can be obtained. 120/47(a+b+c)=d

So the time it takes to open the three faucets to fill the pool is 120 47 hours.

Assumptions can be made as follows:

a, b, c per hour irrigation rate is x, y, z respectively, set the total water volume of the pool is 1 There is an equation as follows:

3x+3y=1

4x+4z=1

5y+5z=1

Solving the above equation gives x=23 120, y=17 120, z=7 120, so the time it takes to open the three faucets is 1 (23 120 + 17 120 + 7 120) = 120 47 hours.

It takes xyz hours to set ABC to open separately.

1/x+1/y=1/3

1/x+1/z=1/4

1/y+1/z=1/5

Add the three formulas to 2 x+2 y+2 x=47 601 x+1 y+1 z=47 120

ABC together with 47 120 per hour to complete, that is, it takes 2 and 26 47 hours.

It takes 3 + 4 + 5 = 12 hours to fill three pools of water, and it takes 4 hours to fill one pool, so it only takes 8 to 3 hours to fill one pool.

Think of filling a pool of water as '1'. Then a faucet is filled with water per hour 1 4 = 1 4, B faucet is injected with water per hour 1 6 = 1 6, A and B two faucets are opened at the same time, and water can be injected per hour 1 4 + 1 6 = 5 12.

A and B two faucets are turned on at the same time, and it takes time to fill a pool of water

The equivalence relation of this question is that the total water storage capacity of the pool is constant.

The equivalence relation of this problem is that the amount of irrigation is constant, and it is solved according to the series of equations of this level.

Set A and B faucets to fill up the water for x hours at the same time, and the equation is listed

1/4+1/6=1/x

Solution x=4 6 (4+6)=hours, that is, A and B faucets are turned on at the same time to fill the pool.

Example 3, there is a pool with water inlet pipe, water outlet pipe, the water inlet and outlet of the water pipe in each unit time is certain, set from a certain time, only water inlet and no water out within 4h, no water inlet and only water out in the subsequent time, the relationship between the time x (h) and the water amount y (m3) obtained (as shown in the figure).

1) How much water is inlet into the water inlet pipe for 4 hours? How much water is ingress per hour?

2) When 0 x 4, what is the relationship between y and x?

3) What is the amount of water in the pool when x=9?

4) If after 4 hours, only the water is discharged without water, how many hours can the water in the pool be discharged?

Analysis: In this problem, the meaning of the abscissa is the time of water inflow and outflow, the ordinate represents the amount of water in the pool, and when 0 x 4 is seen from the image, y is the proportional function of x; When x>4, y is a primary function of x.

Solution: (1) According to the image, a total of 20m3 of water is inlet for 4h, so the water inflow per hour is 5m3

2) y is a proportional function, let y=kx, because its image passes through the point (4,20), so 20=4k, k=5, i.e. y=5x(0 x 4)

3) From the image, it can be seen that when x=9, y=10, that is, the amount of water in the pool is 10m3

4) Since x 4, the image is a straight line, so y and x conform to a one-time functional relationship, let y=kx b, from the image, the straight line passes through the point (4,20), (9,10).

Let y=0, then 2x 28=0, x=14

14 4=10, so after 4 hours, only the water is released without water, and the water in the pool can be discharged in 10 hours.

The discharge rate of each outlet pipe per minute is 1 part.

2x8=16 (portions).

3x5=15 (parts) water intake is:

The original amount of water is:

16-8x1 3=40 3 or:

15-5x1/3=40/3

40 3) (1 3) = 40 (minutes).

Solution 2: The amount of water discharged by 2 outlet pipes in 1 minute is 1 8+ inlet water.

The water volume discharged by 3 outlet pipes in 1 minute is 1 5 + the amount of water discharged in 1 minute by 1 outlet pipe is 1 5-1 8 = 3 401 The water inlet of 1 minute is 3 40x2-1 8 = 1 40 or:

3/40x3-1/5=1/40

1 (1 40) = 40 (minutes).

(2×8-3×5)÷（8-5）=1/3

40 3 1 3 = 40 (minutes) The outlet pipe is opened 40 minutes later than the inlet pipe.

Let a water inlet pipe have a water inlet of x in one minute, a water outlet pipe that has a water inlet of y in one minute, and the water that has been put into the pool is z.

z+8*x=2*8*y

z+5*x=3*5*y

From the above two formulas, it can be seen that y=3x, z=40x, that is, the amount of water that has been put into the pool is the amount of water that has been inlet into a water inlet pipe for 40 minutes, that is.

The inlet pipe is opened for 40 minutes before the outlet pipe is opened, and the outlet pipe is opened 40 minutes later than the inlet pipe.

Test point: Cattle grazing problem Analysis: The water discharged from the outlet pipe can be divided into two parts:

One part is the amount of water that was in the water outlet before it was opened, and the other part is the water that was put into the inlet pipe during the period between the start of draining and the emptying of the water Since the original amount of water is unchanged, it is possible to compare the time and amount of water discharged between the two drains

Set the water discharged from the pool by the outlet pipe per minute is 1 part, then the water discharged by the 2 outlet pipes in 8 minutes is 2 8 = 16 (parts), and the water discharged by the 3 outlet pipes in 5 minutes is 3 5 = 15 (parts), and the water discharged in these two times includes the original water volume and the water inflow from the beginning of the discharge to the emptying period The subtraction of the two is the amount of water put in in 8-5 = 3 (minutes), so the water inflow per minute is (16-15) 3 = 13

Part) Let's say let's 1

3 outlet pipes are specially discharged to the new water inlet pipe, the two phases are offset, and the rest of the outlet pipes discharge the original water, and the amount of original water can be found as: (2-1

3 (servings) or (3-1.)

3 (copies) Answer: Solution: Set the water discharged by the outlet pipe for 1 part per minute The water inflow per minute is:

2 8-3 5) = 1 3 (portions);

The water inlet pipe was opened in advance:

2-1 3) 8 1 3 ,5 3 8 3,40 (min).

Answer: The outlet pipe is opened 40 minutes later than the inlet pipe Comments: Although there is no "cow eating grass" on the surface, because the total amount of water is changing evenly, "water" is equivalent to "grass", and the water inlet pipe is in.

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.

4 hours! Don't know the process! The answer is written like this. We're in for a long time! I just sat down to this question.

15／7。Let a be a water outlet x per hour, and set b ... y, let c...

z。Set up a pool to hold water. Then (1)2x+2z=k.

2）3y+3z=k。(3)7z+x+y=k (1)+(2)=2*(3) gives (4)y=9z. Substituting (4) into (2) gives (5)30z=k.

Substituting (4) and (5) into (3) gives x= filling the pool with a time of 30z 14z = 15 7

Test Focus: Engineering Problems

Analysis: It is required to open A separately, and the pool can be filled in a few hours, and the ergonomics of A are required first; Considering the total amount of work as the unit "1", the ergonomics of C can be X, then the ergonomics of A is X+12, and the ergonomics of B is X+13;From the meaning of the title: 1-7x+(x+12+x+13-x)=1, after obtaining the ergonomics of c, the ergonomics of a can be obtained, and then according to the "total amount of work, ergonomics = working time", substitute the numerical calculation

Answer: Solution: Let the ergonomics of c be x, then the ergonomics of a is x+12, and the ergonomics of b is x+13, which can be obtained from the question: 1-7x+(x+12+x+13-x)=1,116-6x=1,x=536;

1 (536+12)=3623 (hours);

Answer: Open a alone, 3623 hours can fill the pool

Comments: This problem is a complex engineering application problem, the key to solving this problem is to first set a quantity as an unknown number, and other quantities are also expressed by unknown numbers, and then according to the question column, the solution can be

1/a+1/b=1/3

1/a+1/c=1/4

1/b+1/c=1/5

Add the three formulas to get.

2(1/a+1/b+1/c)=47/60

1/a+1/b+1/c=47/120

So the time it takes to fill the pool when the three taps are turned on is 120 47 hours.

As you can see from the table: ABA

CBC adds the left and right sides of the above three equations to get ABC

60, so, 1a

BC120, the time it takes to fill the pool when the three faucets are turned on is:

hours); A: Well, the time it takes to fill the pool when the three taps are turned on is 226 hours