Everyone, please give me a few questions about the problem of cows grazing, and there is a solution,

I hope my offer and analysis will satisfy you.

1. A pasture of Xiaojun's family is overgrown with grass, and the grass grows at a uniform rate every day, and this pasture can be eaten by 10 cows for 20 days and 12 cows for 15 days. If Xiaojun's family has 24 cows, how many days can they eat?

Grass speed: (10 20 12 15) (20 15) = 4

Lao Cao (distance difference): According to: distance difference = speed difference and catch-up time.

10 4) 20=120 or (12 4) 15=120

Catch-up time = distance difference Speed difference: 120 (24 4) = 6 (days).

2. One ranch can feed 58 cows for 7 days, or 50 cows for 9 days. Assuming that the amount of grass grows equally every day, and the amount of grass eaten by each cow is also equal, how many cows can eat for 6 days?

Grass speed: (50 9 58 7) (9 7) = 22

Old grass (poor distance): 50 22) 9=252 or (58 22) 7=252

Finding a few cows is to find the speed of the cow, the speed of the cow = the distance difference and the time of catching up with the grass speed 252 6 22 = 64 (head).

3. There are 10 flood gates in a reservoir, and the water level of the existing reservoir has exceeded the safety line, and the upstream river water is still increasing at the same rate. In order to prevent floods, it is necessary to adjust the rate of flood discharge. Assuming that the velocity of each gate is the same, it is calculated that if one flood gate is opened, the water level will drop to a safe line in 30 hours; If the two floodgates are opened, the water level will drop to the safe line in 10 hours.

Now the flood command unit asks to bring the water level below the safety line within an hour, asking: at least how many gates should be opened at the same time?

Compared with the "cattle grazing problem", the amount of water originally injected into the reservoir is equivalent to the "original amount of grass", the amount of water injected when the flood gate is opened is equivalent to the amount of "new grass", and the amount of water injected per hour is equivalent to "the amount of new grass growing every day". In this way, we can answer the problem according to the idea and method of solving the problem of "cows eat grass".

Solution: 1. The amount of water newly injected per hour is.

1 30 2 10) (30 10) = pcs).

2. The original amount of water before the flood discharge is:

1 hour).

3. The total water volume of the reservoir per hour is.

15+ hours).

The number of sluice gates that need to be opened at least at the same time to bring the water level in the reservoir below the safety line in 5 hours is.

A: In order to bring the water level of the reservoir below the safe line within an hour, at least 4 gates need to be opened at the same time.

The key to answering these questions is to find the amount of water that is being injected per hour and the amount of water that is already being injected.

1.If there is a meadow growing at a uniform rate on the pasture that can feed 27 cows for 6 weeks, or 23 cows for 9 weeks, how many cows can it feed for 18 weeks?

2.If there's a pasture that grows at a uniform rate and can feed 12 cows for 25 days, or 24 cows for 10 days, how many cows can it feed for 20 days?

3.As the weather gets colder, the grass on the pasture not only does not grow, but decreases at a fixed rate It is known that the grass on a certain meadow can feed 20 cows for 5 days, or 15 cows for 6 days According to this calculation, how many cows can be fed for 10 days?

4.As the weather gradually gets colder, the grass on the pasture decreases at an even rate every day It is calculated that the grass on the pasture can feed 20 cows for 5 days, or 16 cows for 6 days So, how many days can it feed 11 cows?

5.A piece of grass that grows at a uniform rate can feed 16 cows for 20 days or 100 sheep for 12 days If a cow eats grass in a day equal to the amount of grass that 5 sheep eat in a day, how many days can this grass be eaten by 10 cows and 75 sheep together?

Give a few first, and try it yourself.

The formula for the cow grazing problem is:1) The growth rate of the grass = (the corresponding number of cows, the number of days eaten more, the corresponding number of cowheads, the number of days eaten) (the number of days eaten more, the number of days eaten less).

2) The amount of original grass = the number of cows, the number of days eaten, the growth rate of grass, the number of days eaten.

3) Number of days eaten = amount of original grass (number of cows, growth rate of grass).

4) The number of cattle heads = the number of days eaten by the original grass fiber + the growth rate of the grass.

For example, the elimination of the erection of the grandson:

At a certain station, queues start a few minutes before the ticket is checked, and the number of passengers comes every minute is the same. It takes 30 minutes to open 4 ticket gates at the same time and 20 minutes to open 5 chain ticket gates at the same time from the start of ticket inspection to the disappearance of the queue waiting for ticket checking. How many minutes does it take to open 7 wickets at the same time?

With this kind of theme, the ticket gate can be regarded as a cow and the passenger as a grass. Direct set of formulas - passenger speed per minute = (4 30-5 20) (30-20) = 2.

Passengers in line = (5-2) 20=60.

60=（7-2）t。

t=12。

45+5 2 cars transport 50 carts 5 per day

45+9 4 trolleys transport 54 trolleys.

2 large cars and 4 small cars 2*5+4* a day

3 days 45+3 = 16*3

It can be fully stocked in 3 days.

I teach you. Five days full is one-fifth of the bicycle, and nine days full of clouds is one-ninth of the bicycle.

You know how to do it.

3 days. The key to this problem is to find the parameters.

Junior high school questions? When I was in elementary school, I didn't know how to solve it.

The basic idea: assuming that the rate of grazing for each cow is "1", according to the two different eating methods, find the difference in the total amount of grass; By finding out the reason for this difference, we can determine the growth rate of the grass and the total amount of grass.

Basic characteristics: the amount of original grass and the growth rate of new grass are constant;

Key question: Determine two invariant quantities.

The basic formula: growth = (longer time number of cattle heads for a long time - number of cattle heads for a short period of time) (long time - short time);

Total grass amount = long time number of cattle heads for a long time - growth for a long time;

There are four basic formulas commonly used in the problem of cattle grazing:

The problem of cows grazing, also known as the problem of growth and decline, was proposed by the great British scientist Newton in the 17th century. The condition of a typical cow grazing problem is to find how many cows can eat the same grass by assuming that the growth rate of grass is fixed, and the number of days it takes for different numbers of cattle to eat the same grass is different. Because the number of days eaten is different, and the grass grows every day, the stock of grass constantly changes with the number of days eaten.

There are four basic formulas commonly used to solve the problem of cattle grazing, which are:

1) The growth rate of the grass = (the corresponding number of cows, the number of days eaten more, the corresponding number of cowheads, the number of days eaten) (the number of days eaten more, the number of days eaten less).

2) The amount of original grass = the number of cows, the number of days eaten, the growth rate of grass, the number of days eaten. `

3) Number of days eaten = amount of original grass (number of cows, growth rate of grass).

4) The number of cattle heads, the amount of grass, the number of days eaten, and the growth rate of grass.

These four formulas are the basis for solving the problem of growth and decline.

Since the grass is constantly growing in the process of grazing in the process of cattle grazing, the key to solving the problem of growth and decline is to find ways to find invariants from change. The original grass on the pasture is unchanged, and although the new grass is changing, the amount of new grass growing every day should be the same because it grows at a uniform rate. It is because of this invariant that the above four basic formulas can be derived.

See what I know.

Look at the library again.

The problem of cattle grazing is Newton's Bai problem, which was raised by Newton.

du got its name.

"A pile of grass can be eaten by 10 cows for 3 days, and 6 cows for a few days? "It's a very simple question, use.

Genus 3*10 6=5 (days). If you replace "a pile of grass" with "a growing meadow", the problem is not so simple. Because grass grows every day, the amount of grass is constantly changing.

The problem with this type of irregular (uniformly varied) total amount of work is the problem of cattle grazing.

The key to solving this type of question is to find a way to find the constant quantity from change. The original grass on the pasture is unchanged, and although the new grass is changing, because it grows at a uniform rate, the new grass grows every day. Correctly calculate the amount of grass that is already in the meadow and the new grass that grows every day, and the problem will be easily solved.

For example, a meadow grows grass at a uniform rate every day. If it can be eaten for 24 cows for 6 days, 20 cows can be eaten in 10 days. So, how many days can it feed 19 cows?

Answer: (20*10-24*6) (10-6)=14 (copies).

24*6-14*6=60 (portions) 60 (19-14)=12 (days).

1. More per day: 20-15=5

If there are more than 12 days: 12*5=60

Well depth: 60 (12 10-1) = 300 (decimeter)2, 60 more per minute 20 (27-24) = 9 levels If 3 minutes are more than 3 * 9 = 27 levels.

The elevator has 27 (3 2-1) = 54 steps.

180 6 = 30 (only for the ship).

Suppose there are x passengers per minute at the time of ticket check-in, and there are z passengers lined up a few minutes before the ticket check-in, and Y passengers are checked in each ticket gate per minute during the check-in, and assuming that it takes t minutes for the queue to disappear when opening 7 ticket gates, then there are:

4 wickets: z+30x=4*30y....Equation 1;

5 ticket gates: z+20x=5*20y....Equation 2;

7 wickets: z+tx =7*ty....Equation 3;

Subtract the left and right sides of Equation 1 and Equation 2 to obtain: x=2y....Equation 4;

Bringing Equation 4 into either Equation 1 or Equation 2 yields: z=60y....Equation 5;

From Equation 3: t=z (7y-x).Equation 6;

Bringing equations 4 and 5 into 6, we get: t=12

Solution: Suppose a cow eats one serving of grass every day.

So: 21 cows eat 168 grasses in 8 days.

24 cows ate 144 grasses for 6 days.

168-144=24 (pieces).

Growing grass per day: 24 divided by the number of days is (8-6 equals 2) equals 12 (grains per day) 8 times 12 equals 96 herbs, 168 minus 96 = 72, can also be tested separately.

72 is the original grass.

96 divided by (16-12) = 28

Answer: (1) You can eat the grass in 28 days.

Answer: (2) If you grow 12 pieces of grass every day, you can only raise 12 cows permanently!

1) The growth rate of the grass = (the corresponding number of cows, the number of days eaten more, the corresponding number of cowheads, the number of days eaten) (the number of days eaten more, the number of days eaten less).

2) The amount of original grass = the number of cows, the number of days eaten, the growth rate of grass, the number of days eaten. 3) Number of days eaten = amount of original grass (number of cows, growth rate of grass).

4) The number of cattle heads, the amount of grass, the number of days eaten, and the growth rate of grass.