Grazing problem

Assume that each ticket gate can pass 1 ticket per minute.

1 mouth, 20 minutes can pass: 1×20=20 copies.

2 ports, which can pass in 8 minutes: 2×8= 16 copies.

Poor: 20- 16=4 copies.

These four copies are people who come in line within 20-8= 12 minutes.

The average number of people coming every minute is: 4÷ 12= 1/3.

Originally: 20- 1/3×20=40/3 copies.

Open three mouths, and you can pass three copies per minute.

Except for 1/3 copies queued every minute, those already queued can pass 3- 1/3=8/3.

It takes 40/3÷8/3=5 minutes to pass all.

2.

The length of the rope remains the same.

It is equal to 2×2=4 times the circumference of trunk 1× 2 = 2m.

It is also equal to 3× 1=3 times1.5× 3 = 4.5m trunk circumference.

Trunk circumference: (4.5-2) ÷ (4-3) = 2.5m.

Rope length: 2.5× 4+2 = 12m.

3.

A set of purchase price: 9 10( 1+30%)= 700 yuan.

Another set of purchase price: 910 ÷ (1-30%) =1300 yuan.

Total purchase price: 700+ 1300=2000 yuan.

Total selling price: 9 10×2= 1820 yuan.

1820