A ship sailed from Port A to Port B, and when sailing on this route, it walked 3/5 of the whole journey at normal speed. After that, slow down 10 knots and keep driving at this speed to port B. In this way, the time of this voyage after deceleration is the same as that without deceleration. What is the normal speed of this ship on this route?

The normal speed of the third ship on this route is X.

3/5÷x =( 1-3/5)÷(x- 10)

3(x- 10)=2x

x=30

The normal speed of this ship on this route is 30 knots.

The distance between Party A and Party B is 125km. From Party A to Party B, some people ride cars and others ride bicycles. Bikes leave 4 hours earlier than cars and arrive 1/2 hours later. It is known that the ratio of cycling speed to cycling speed is 2: 5. What are the speeds of bicycles and cars?

Let the self-help speed be x km/h, then the riding speed is 5x/2km/h.

Then the bus ride time is:125 ÷ 5x/2 = 50/x.

Then there is the equation: 125/x-50/x=4.5 (according to the time difference between riding and riding).

The solution is x = 50/3 km/h.

Then the speed is: 5/2 * 50/3 =125/3 km/h.

A motorcade plans to transport m tons of goods in t days. If n tons have been transported, (n is less than m), the number of days of remaining goods transportation is t 1=__, and the average tonnage of goods shipped every day is a=____.

Daily freight volume: metric tons

Then the number of days required to transport the surplus goods is: (m-n) ÷ m/t = (m-n) * t/m.

A = metric tons

It takes the same time for a ship to sail 80 kilometers downstream and 60 kilometers upstream. Given the speed of the current of 3 km/h, find the speed of the ship in still water.

Let the speed of the ship in still water be x,

Then the downstream speed is x+3.

The upstream speed is x-3.

Then it is: 80/(x+3)=60/(x-3)

Solve the equation, x = 21km/h.