Math pool application problems, pool water discharge water inlet application questions how to do

If the volume of the pool is v, then A is injected with v20 water per minute, B is injected with v30 water per minute, and C is injected with v 15 minutes per minute.

Turn on B and C first, inject V 30 + V 15 water per minute, and inject a total of 5 minutes.

5*(v 30 + v 15)= v 2 water, then there is still v - v 2 = v 2 water to be injected.

Then by two ways of thinking:

1) It takes 20 minutes for A to fill a pool, so filling this half of the pool is 20 2 = 10 minutes.

2) Volume divided by velocity of A = time, then (v 2) (v 20) = 10 min.

Personally, I think the first method is easier to understand and faster, but it is only suitable for this question, and the second method can be used in a wide range.

I guess the original question asked how long it would take?

If you ask how long it takes to fill the entire pool, then the answer is correct

If the capacity of the pool is x, then A fills x 20 per unit of time (1 minute), in the same way B is x 30 and C x 15

According to the title, B and C are turned on for 5 minutes, and the water injection is x 30 * 5 + x 15 * 5 = x 6 + x 3 = x 2, that is, half of the water has been injected.

Then you can open the armor separately and inject the other half.

Yes (x 2) (x 20) = 10 minutes.

Hope it helps.

After 5 minutes of B and C, the whole pool was injected: 5 30 + 5 15 = 1 2

The remaining 1 2 of Note A takes 20 2 = 10 minutes.

Question: A pool has an inlet pipe and a drain pipe. If you open only the inlet pipe, you can fill the pool in 2 hours, and then turn off the inlet pipe and open only the drain pipe, and it will take 6 hours to empty the pool.

Q: How long does it take to fill the pool when I start with an empty pool and open the inlet and drain pipes at the same time?

Solution: If the volume of the pool is "1", the water inlet rate is 1 2 (1 hour into 1 2 pools of water), and the drainage rate is 1 6 (1 hour to drain 1 6 pool of water).

Time = Volume (Water Inlet Rate - Discharge Velocity) = 1 (1 2-1 6) = 3 (hours).

A: It takes 3 hours to fill up.

Solve the problem of water filling and discharging in the pool. It takes 8 hours for a pool to open the inlet switch to fill the water, and it takes 10 hours to open the sluice gate to fill the water. I listed an equation: 8x=10y=(x-y)z, solve the equation, and the middle "z" is the answer, right.

How to solve this equation. Solution: It takes x hours to fill the pool, according to the meaning of the topic and known conditions, there are:

1 8-1 10=1 x x=40 (hours) A: It takes 40 hours to fill the pool.

Use the amount of water incoming water minus the amount of water outgoing. Get the net inflow value.

Then divide the total volume by the net inflow value to get the time.

Think of a pool of water as 1.

Then the working efficiency of the inlet pipe is 1 pool per hour.

The working efficiency of the outlet pipe is to put 1 2 pools per hour.

So 2 hours can fill the pool.

1. The first load is not considered, only the self-weight, water pressure and earth pressure of the reservoir are considered. Open reservoir, load combination is full of water in the pool, no soil outside the pool; For closed pools, the load combination is that there is no water in the pool and soil outside the pool. When calculating, the bulk density of mortar masonry and concrete is taken as.

Underground pool, the backfill outside the pool wall is required to be tamped, and the bulk weight of the fill is taken when calculating the earth pressure, and the internal friction angle is taken as 30°.

2. The allowable bearing capacity of the foundation should be deduced according to the geological conditions, if the actual bearing capacity of the foundation does not meet the design requirements or the foundation will produce uneven subsidence, effective foundation treatment measures must be taken before the reservoir can be built. The foundation of the reservoir floor must have sufficient bearing capacity, flat and compact, otherwise it must be paved and compacted with gravel (or coarse sand).

3. The reservoir should be designed as standard as possible, or according to the relevant specifications of the five-level building. The bottom and side walls of the pool can be made of mortar masonry, plain concrete or reinforced concrete. Areas with an average temperature of 5 in the coldest month can also be brickwork, but cement mortar should be used to plaster.

When the bottom of the pool is made of slurry masonry, it should be built with slurry, and the mortar of the pool is not less than m10, and the thickness is not less than 25cm. When using concrete, the grade should not be lower than C15, and the thickness should not be less than 10cm. The soil foundation should be tamped with a depth of not less than 40cm.

The size of the pool wall should be calculated and determined according to the standard design or according to the requirements of the code.

4. The foundation of the reservoir is very important, especially in the collapsible loess area, if there is a slight leakage, it will endanger the safety of the project. Therefore, the use of monolithic reinforced concrete or plain concrete reservoirs should be given priority to the construction of reservoirs on collapsible loess. When the foundation soil is weak collapsible loess, the bottom of the pond should be tamped, and the tamping depth should not be less than 50cm; If the base soil is medium and strong collapsible loess, the tamping depth should be increased, and measures such as immersion and pre-sinking should be taken.

[1-(1 6-1 8)*40 60] 1 6 = 23 6 hours = 3 hours 50 minutes.

The pool will take another 3 hours and 50 minutes to fill up.

6 * 60 = 360 (points) 8 * 60 = 480 (points) 360 = 3 * 3 * 2 * 2 * 2 * 2 * 5 480 = 2 * 2 * 2 * 2 * 3 * 5

360 * 4 = 1440 (unit) 1440 60 6 = 4 (unit) 1440 60 8 = 3 (unit) 3 * 40 4 = 30 (points).

Only enter the pool when filling 1 5 pools with water in an hour.

Only when it is released for an hour 1 6 pools of water.

When it is opened at the same time, it is the water that enters the pool of 1 5-1 6 = 1 30.

So the time to fill up is 1 divided by 1 30 equals 30 hours.

The formula is 1 (1 5-1 6).

Divided into 2 parts.

Part I. Fill the pool with 1-4

i.e. A, state cover C with the same time, set this time to x minutes.

1/5+1/10）x=1/4

The explanation of the shouting trail answered:

x=5 6 Part II.

A, B, and C each open for one minute, and the water injection is.

So after A, B and C each opened for one minute, the second round was not completed.

After the completion of the first round, the pool is still 3 4-14 30 = 17 60 The second round, A opens one minute to complete 1 5

The pool is still 17 60-1 5 = 1 15

B completes 1 6 in one minute, (1 6) x = 1 15x = 2 5 (min).

All the time on Zheng Hui is the total time.

5 6 + 5 6 + 3 + 1 + 2 5 = 92 15 (minutes) 6 and 2 15 minutes.

In fact, don't be paralyzed by a few imitation slides when doing this type of problem. First of all, look at the first tube A can be filled in 15 minutes, indicating that the water that can be filled per minute is 1 15 of the pool water dissolution, the same is 1 10 for tube B, 1 9 for tube C, and 1 15 + 1 10 + 1 9 = 6 90 + 9 90 + 10 90 = 25 90 = 5 14

It is already known that three pipes per minute can be injected into the pool of 5 14, and how much time it takes to inject the pool of water should be very simple.

Answer: If A and B open the sparrow mold for x minutes at the same time, then A injects x 5 and B discharges x 6, and there are x 5-x 6=x 30 in the pool

A and C open x minutes at the same time: x 5 + x 10 = 3x 10 according to the meaning of the equation:

x/30+3x/10=1/4

Solution: x=3 4

The front took 2x=3 2=minutes.

3 4) (1 5-1 6 + 1 10) = (3 4) * (30 4) = 45 8<6 minutes.

Therefore: after A, B and C take turns once, then A and B, and C does not need to be injected with water for 1 minute for the last few times in the last year.

So: total time = 45 8+ minutes.

Let the velocities of A, C and B be xyz respectively

There is 5x=10y=6z. Seek the limbs to do and take y=;

Set the tube to open for m minutes.

Then there is mx-mz+mx+my=1 4*5*x to find m=minutes.

There are l calendar minutes for three minutes.

x+y-z）*l=3

5x ground so it takes minutes.

It is known that there are 1 6 pools of water, then there are 5 6 more to note.

Re-analysis, get:

A: You can fill 1 3 with water in one hour

B: One hour water injection: -1 4 (because it is a drainage pipe, it is a negative number) C: One hour water injection: 1 5

D: One hour water filling: -1 6 (because it is a drainage pipe, it is a negative number) Open the water pipe for 4 hours in the order of lifting.

It can be seen that water can be filled every 4 hours: 1 3 + (-1 4) + 1 5 + (-1 6) = 1 12 + 1 30 = (10 + 4) 120 = 7 60

Then the remaining 5 6 pools of water take time:

5 6) 7 60) = 50 7 = 7 + 1 7 4 hours.

However, after the 7th 4 hours, the remaining water in the pool is:

At this time, the A is filled with water, so the time for A to fill up the remaining water is:

1 60) 1 3) = 1 20 hours.

Plus the above 7*4=28 hours.

It can be seen that the time to fill the pool is 28+1 20=hours.

Suppose the pool volume is 1. If A's hourly water injection is x, then B's hourly water injection is (7 8-x) and the unary equation is listed.

40 60) x + (30 60) (7 8-x) = 1 2x = 3 8 (i.e. A has 3 8 per hour).

The amount of water injected per hour is .

7/8-x=4/8=1/2

Then it takes time for A to fill water alone:

1 x = 8 3 (hours).

B When filling water alone:

1 (1 2) = 2 (hours).

In fact, don't be paralyzed by a few tubes when doing this type of question. First of all, look at the first tube A can be filled in 15 minutes, indicating that the water that can be filled per minute is 1 15 of the pool water dissolution, the same is 1 10 for tube B, 1 9 for tube C, and 1 15 + 1 10 + 1 9 = 6 90 + 9 90 + 10 90 = 25 90 = 5 14

It is already known that three pipes per minute can be injected into the pool of 5 14, and it should be very simple to need 3 5 of the pool to fill the pool.