A 6th grade math problem. Rapid drops!!

Assuming it's all cars, then there should be wheels:

20 4 80 pcs.

There are actually a total of 50 wheels, assuming that there are more wheels than the actual one:

80 50 30 pcs.

This is due to the fact that a car has 2 more wheels than a two-wheeled motorcycle, so it means that there are a total of two-wheeled motorcycles

30 2 15 (cars).

There are cars: 20 15 5 (cars).

They are all regarded as two-wheeled vehicles, and there should be wheels 2*20=40 (pcs), which is 50-40=10 (pcs) less than the existing wheels

If each four-wheeler is less than 2 wheels, then there are four-wheelers: 10 2 = 5 (vehicles) and two-wheelers: 20-5 = 15 (vehicles).

Chickens and rabbits haven't learned from the same cage, right?

There are 5 sedans and 15 motorcycles.

Those who have learned to solve equations directly with equations.

I haven't learned it, and I can solve it like this.

First of all, all 20 cars are counted as 2 wheels, 20 cars are 40 wheels, that is, each car has a few two wheels, and the total number of wheels is 50, and the other 10 wheels are not counted by cars, that is, there are 5 cars.

A car has 4 wheels, 10 cars have 40 wheels, a motorcycle has 2 wheels, 5 cars have 10 wheels, and 40 plus 10 has 50.

Ahem ! I'm in my first year of junior high school. Maybe you still don't understand.

Set the two-wheeled motorcycle as XFour-wheeled cars are YThen:

x+y=20

2x+4y=50 (omitted).

The result is: x=15 y=5

There are x two-wheeled motorcycles and Y four-wheeled cars.

x+y=20

2x+4y=50

Solution x=15y=5

If they were all cars, they would have 80 wheels, 30 more, so there should be 15 two-wheeled motorcycles and 5 four-wheeled cars.

If we remove two wheels from each car, 20 cars will have 40 wheels, and the extra 10 wheels will belong to the car, 2 for each car, so there will be 5 cars. The remaining 15 are motorcycles!

A car 1 10 each time

B car 1 15 each time

A and B combined transport each time 1 10 + 1 15 = 1 6

Required: 1 1 6 = 6 times.

What happened to the rope again.

7+7) (2-1)=14 m Well surface depth 14 + 7 = 21 m Rope length.

If the rope is doubly fed, both strands are 7 meters shorter, it should be like this (7+7x2) (2-1)=21 meters, and the well depth is 21+7=28 meters.

You didn't make the latter case clear, I guess it's a double-strand analysis: suppose the rope is 2x long and the well is y

Then there is a question to know: when the single strand of the rope is still 7m: 2x-7=y, when the double strand is missing, the rope is missing 7m: x+7=y

Two equations with two unknowns, add the two equations to give 3x=2y, and bring any one of them to get y=21, x=14

If it is not tied into two strands, it should be noted that the rope length cannot be set to 2x, but if it is a few strands, it should be set to several x, and the solution is the same).

The rope is 28 meters, and the wellhead is 21 meters above the water.