The two ports of AB are 300 kilometers apart, and if Ship A sails from A to B along the water

Let the speed of B be x and the speed of the water be y, when the first ship is downstream, the meeting time is h1, and when the second boat is downstream, the meeting time is h2

Rule. ac=(27+y)*h1,cb=(x-y)*h1,ab=(27+x)*h1=300

ad=(x+y)*h2,db=(27-y)*h2,ab=(27+x)*h2=300

It can be seen that H1 = H2 = H, and the calculation of Cd is divided into two cases:

1.When AC is less than AD, cd=ab-ac-db=300-(27+y)*h-(27-y)*h=300-(27+y+27-y)*h

300-54h, and it is known that cd=30, then h=(300-30) 54=270 54=5

Because (27+x)*h=300, then x=300 5-27=33(km h).

2.When AC is greater than AD, cd=ab-ad-cb=300-(x+y)*h-(x-y)*h=300-(x+y+x-y)*h

300-2xh, and it is known that cd=30, then xh=135, because (27+x)*h=300, then 27h+xh=300, that is, 27h=300-135=165, and h=165 27=55 9, then x=135 (55 9)=243 11, about 22km h).

If the speed of the ship A is the speed of the ship B.

Let the speed of ship B be x, and the meeting time is t;

27t+xt=300

xt-27t=30

From the above equation, it can be solved: x=33 , t=5, that is, the speed of ship B is 33km h.

If the speed of the ship A is the speed of the ship B.

27t+xt=300

27t-xt=30

Get x= t=, which should be fractions, both approximately equal.

Upstairs only one answer is calculated, if ship A is faster than B, then d is closer to A, and V B = 243 11km h can be calculated in a similar way

According to the inscription, the speed of the ship A is v A=450 18=25 (km h) v B = 450 15 = 30 (km h).

If t hours have elapsed from the departure bench of ship B to ship B catching up with ship A, then 2*25+25*t=30*t gets t=10 (h) At this time, ship B has traveled a total of 10*30=300 (km), 1,2 1 18 450 [(1 15-18) 450] = 10 hours.

10 1 18 450 = 300 km, 1,450 2 18 (450 15-450 18) = 10 hours.

450 15 10 = 300 meters, 1, A and B two ships to go from port A to port B, it is known that the two ports of AB are 450 kilometers apart, it takes 18 hours for ship A to complete the journey, and it takes 15 hours for ship B to complete the journey, and after 2 hours after ship A leaves for friends to go in advance, ship B goes to chase ship A, how many kilometers does ship B have to travel to catch up with ship A?

Downstream velocity 600 30 = 20 km h.

The speed against the water is 600 40 = 15 km h.

Water velocity (20-15) 2=km/h.

Ship speed in still water kilometers per hour.

Solution: From B to port A, it takes 3 2=6 hours less for ship A than for ship B.

The time taken by ship A is 1 6 5 = 5 6 of ship B

The time taken by ship B is 6 (1-5 6) = 36 hours.

The time taken by ship A is 36 5 6 = 30 hours.

The speed of the first ship is 540 30 = 18 km/h.

The speed of the B ship is 540 36 = 15 km/h.

A: The speed of ship A is 18 kilometers per hour, and the speed of ship B is 15 kilometers per hour.

If the speed of the ship A is the speed of the ship B.

Let the speed of ship B be x, and the meeting time is t;

27t+xt=300

xt-27t=30

From the above equation, it can be solved: x=33 , t=5, that is, the speed of ship B is 33km h.

If the speed of the ship A is the speed of the ship B.

27t+xt=300

27t-xt=30

Get x= t=, which should be fractions, both approximately equal.

How do you feel incomplete, I don't understand what the hell are you asking for!!

Isn't this a math problem?,What to ask?,Write it clearly.

Let's finish the questions first......(Wipe sweat).

Let be: B's speed x km/h, it takes n hours for two ships to meet, 27+x)n=300

x-27)n=30

x = 33 kmh.

Ship A returns immediately to port B, and meets B head-on after a few hours, and the distance of the encounter is: kilometers);

When ship A arrives at port B, the distance traveled by ship B is: 200-40=160 (km);

The speed ratio of the two ships A and B sailing along the river is: 200:160=5:4;

When the velocity of the water is x km;

45+x）：（35+x）=5：4，5（35+x）=4（45+x），175+5x=180+4x，x=5．

A: The water flow speed is 5 kilometers per hour

So the answer is: 5

According to the inscription, the speed of the ship A is v A=450 18=25 (km h) v B = 450 15 = 30 (km h).

If t hours have elapsed from the departure of ship B to the time when ship B catches up with ship A, then 2*25+25*t=30*t get t=10 (h) At this time, ship B has traveled a total of 10*30=300 (km).