Newton problem (grazing problem)

1:

The pasture is covered with grass and grows at a constant speed every day. This pasture can feed 10 cows for 20 days, 15 cows 10 days. How many days can 25 cows eat?

Answer:

(1) 10 How many cows can eat grass for 20 days?

10×20=200 (head)

(2) 15 cows 10 How many cows can you eat in a day?

15× 10= 150 (head)

(3)(20- 10) How many cows can eat on the grass for one day?

200- 150=50 (head)

(4) How many cows can long grass feed in a day?

50÷ 10=5 (head)

(5) How many cows can the pasture grass feed?

200-5×20= 100 (head)

Or150-5×10 =100 (head)

(6) Can feed 25 cows for x days?

100+5X = 25 times

Original grass +X day grass =25 cows eat up.

X=5

2:

A pond with underground spring water, if 50 people carry water, it will take 20 hours to finish the pond water; If 80 people carry water, it can be finished in 8 hours. Q 100 people carry water, how many hours does it take to finish the pond water? If the pond water is not used up, how many people should be restricted to carry it?

Answer:

(1)50 people carry water for 20 hours. How many people can carry it?

50×20= 100 (person)

(2)80 people carry water for 8 hours. How many people can carry water for one hour?

80×8=640 people

(3)(20-8) How many people can pick an hour for the rising water?

1000-640=360 (person)

(4) How many people can pick up the rising water in an hour?

360÷ 12=30 (person)

(5) There is water in the pond. How many people can pick one hour?

1000-30×20=400 (person)

Or 640-30×8=400 (person)

(6) 100 people carry water and finish pond water in x hours?

400+30X= 100X

X equals five and five sevenths.

Question 2:

Because 1 hour flood is enough for 30 people to carry, if you add 1 person to carry the water in the original pond, you have to carry the water in the pond sooner or later, so the number of people should be controlled within 30, that is, only carry the flood, without reducing the original 400.

Special topics on Olympic mathematics

The chicken and the rabbit are in the same cage.

A cage contains chickens and nine-headed birds (the mythical nine-headed birds). If the total number of heads is 60 and the total number of feet is 40, how many chickens and nine-headed birds are there in the cage?

Solution 1:

(1) The total number of feet is 40, and the number of chickens and nine-headed birds can be found.

Namely: 40÷2=20 chickens and 20 birds with nine heads.

(2) Assume that all 20 chickens are chickens.

There should be 1×20=20 20 heads.

(3) But there are actually 60 heads,

60-20 = 40, missing 40 heads.

(4) Omitted is that each bird with nine heads is short of 9- 1=8 heads.

(5) Use 40÷8=5

You can find out the number of nine-headed birds.

Namely: 5 birds with nine heads and 0/5 chickens/kloc.

Solution 2:

(1) Because a chicken with nine heads and a bird with nine heads have two feet, we can find out the number of chickens with nine heads and birds with nine heads.

40÷2=20

(2) Assuming that all 20 birds are nine-headed,

There should be 9×20= 180 heads.

(3) There are actually only 60 heads,

180-60= 120 extra heads.

(4) The extra 120 head means too many chickens.

9- 1= 8 chickens.

(5) The number of chickens can be found out.

120÷8= 15

Then 20- 15=5 birds.

Analysis of classic good questions in primary school mathematics review ......

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