Who knows where there is an explanation of the problem of cows grazing, non equations

1.The pasture has an evenly grown meadow that can feed 27 cows for 6 weeks, or 23 cows for 9 weeks, so how many weeks does it feed 21 cows? (This kind of problem was first asked by Newton, so it is also called the "Newtonian problem").

Analysis and Solution] 27 cows eat for 6 weeks, which is equivalent to 27 6 = 162 cows eat for 1 week, eat the original grass plus 6 weeks of new grass;

23 cows eat for 9 weeks, which is equivalent to 23 9 = 207 cows eat for 1 week, eat the original grass plus 9 weeks of new grass; So, there are 207-162 = 45 cows, and they eat 9-6 = 3 weeks of new grass So 45 3 = 15 cows can eat 1 week of new grass That is equivalent to giving 15 cows to eat the new grass So 27-15 = 12 cows eat the original grass in 6 weeks, and now there are 21 cows, minus 15 eating the grass, so 21-15 = 6 cows eat the original grass;

So it takes 12 6 6 = 12 (weeks), so 2l cows need to eat for 12 weeks

Comment: We find out how many years it takes to eat grass per unit of "1" area, so that the problem is reduced to a general engineering problem

General method: first find out how many cows are equivalent to the changing grass to eat: (the number of cattle heads A time A - the number of cattle heads B time B) (time A - time B);

Then perform the following calculations: (the number of cattle heads - the number of changing grass) when asking A (the number of cattle heads - the number of changing grass) = time C

Or: (The number of cattle heads in A - the number of heads of the grass in change) Time A Time C + the number of heads required in the number of heads of the grass in the change of the grass in the same number of heads

Example 3 In winter, 12 cows ate 720 kg of hay for 3 days. After taking away the 3 cows, there are 720 kilograms of hay to feed the remaining cattle for 4 days, is the hay enough? (Suitable for 5th grade).

1) How many kilograms of hay does 1 cow eat in 1 day?

720 12 3 = 20 (kg).

2) After taking 3 cows, how many cows are left?

12-3=9 (head).

3) How many kilograms of hay do 9 cows eat in 4 days?

20 9 4 = 720 (kg) comprehensive equation:

720 (kg).

A: 720 kg of hay is just enough.

Let the grassland grow x, the grass grows y every day, each cow eats z in one day, and it takes n days for 25 cows to eat up all the grass. Establish the equation according to the meaning of the topic: x+40y-10*40z=0 (1 formula) x+20y-15*20z=0 (2 formulas) x+ny-25nz=0 (3 formulas) (1 formula)-(2 formulas) get:

y=5z (4 formulas) 2*(2 formulas)-(1 formulas) get: x=200z (5 formulas) Bring (4 formulas) (5 formulas) into (3 formulas) to get: 200z+5nz-25nz=0 (6 formulas) (6 formulas) Approximate z to get:

200+5n-25n=0 Find n=10 (days) So 25 cows can be eaten in 10 days.

Remember.

Solve it with equations"Cows eat grass"The steps of the problem can be summarized in three steps:

1. Set the total amount of the original grass and the change of grass per unit time, generally set the original total amount as 1 and the change per unit time as x;

2. List **, which respectively represents the number of cattle, the total amount of time, the total amount of grass (the original total amount + the amount of change in a certain period of time), and the number of grasses per unit time of each cow.

3. Solve x according to the series of equations that the amount of grass eaten per unit time of each cow remains the same, so that the total amount of grass at any time can be found, and the number of grass eaten per unit time of each cow can also be found. Therefore, the unknown is set to y for the problem to be solved.

The following is an analysis of a few example questions:

Example: The grass on a pasture grows at a uniform rate every day. The grass can feed 27 cows for 6 weeks, or 23 cows for 9 weeks. So how many weeks is enough for 21 cows?

Solution: The first step: set the original grass volume of the pasture to 1, and the new grass x every week;

Step 2: Column ** is as follows:

Step 3: List the equations according to the fourth line equal to each other:

1-5x)/20*5 = 1-6x)/16*6 (1)

1-5x)/20*5 = 1-yx)/11y (2)

From (1) we get x = 1 30, and substituting (2) gives y = 8 (days).

Cows eat grass"Problems often appear in other forms such as intake and drainage, or queuing, and can also be solved by thinking about equations.

Consider how much grass each cow eats per day (i.e., unit consumption).

How much grass grows per day in the pasture (i.e., the amount of grass grown per unit).

How much grass was there in the pasture (the original base).

For example, 10 cows can eat for 10 days, and 9 cows can eat for 20 days, how many days can 5 cows eat?

Solution: Let the original grass be a, no cow eats x every day, and the grass grows y10*10*x=a+10y every day

9*20*x=a+20y

i.e. x = a 20 and y = 2a 5

Then 5 cattle consume a 4 per day, which is less than the growth so they can be eaten all the time.

What are the specific requirements? Anyway, there are 7 "stomachs" in the cow's body, isn't this just for eating grass, haha. Agree with it.

For those who know the equations, this problem is easy to solve.

Taking exercise 1 as an example, we have the following solution:

Suppose the original grass is x parts, the grass that grows every week is y parts, and each cow eats 1 part of grass per week.

Then you can list the equation:

x+6y=27×6

x+5y=30×5

The solution is x=90, y=12

If 42 cows are put and can be eaten in n weeks, then:

90+12n=42n

n=3 weeks.

Solution: Let the amount of grass eaten by each cow per day be 1, the grass grows x per day, and the grassland originally had grass yy+(6-1)x=24 6

y+(8-1)x=21×8

Solution: x=12, y=84

If 16 cattle are grazing, a day can eat the grass, 84 + 12 (a-1) = 16a solution a = 18 up to 12 cows, the grass can never be eaten Because the growth of grass is 12, a cow eats grass in a day 1

Solution: Let each cow eat 1 grass per day, and the grass grows (1 21 8 1 24 6) 8 6 12 original grass 1 21 8 12 8 72

72 16 12 18 (day).

To keep the grass from running out, put up to 12 cows. Eat only 12 long days a day

(21 8 24 6) 2 Find the amount of grass that grows per day The grass can never be eaten, so the total amount of grass that the cattle eat every day is not more than the amount of grass that grows per day, so the maximum is 12.

One tendon formula: the number of cattle minus the variable multiplied by the number of days is equal to the number of cattle minus the variable multiplied by the number of days, let the variable be x, grazing 16 cows, you can eat y amakusa! (24-x)*6=(21-x)*8=(16-x)*y, and the variable x=12 y=18 is solved

Each pump is set to pump 1 water per minute.

3 units, 40 minutes pumping: 3 40 = 120 servings.

6 units, 16 minutes pumping: 6 16 = 96 servings.

Difference: 120-96 = 24 parts.

These 24 parts are the water that seeps in 40-16 = 24 minutes.

Average water seepage per minute: 24 24 = 1 serving.

There was water in the well: 120-40 1=80 parts.

There are 9 good sock pumps, and it takes 80 (9-1) = 10 minutes to finish pumping.

90 minutes to drain, need:

80 90 + 1 2 units.

80-(2-1) 20=60 servings.

Also hungry: 60 (6-1)=12 minutes.

Are you fighting to take the civil service exam? The old elevator increases x per minute, then 5 (20 x) = 6 (15-x), and the solution is x = -10, that is, the elevator disappears 10 grids per minute in the process of walking, and the number of steps of the escalator is 6 * 25 150