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The distance between Party A and Party B is 490km. A slow train starts from station a and travels 80km per hour. An express train starts from station B and travels 100km per hour. It's urgent! If the slow train runs for 30 minutes first and the two trains run in opposite directions, how many hours does the fast train run into the slow train? Use the equation to solve the problem
Set the fast train to meet the local train after X hours
80×30/60＋（80＋100）X＝490
180X＝490－40
180X＝450
X＝450÷180
X＝2.5
A rides a bike and B walks from a and B. after meeting, a arrives at B in 15 minutes and B arrives at a in 1 hour. Find out the speed ratio between a and B
Let the meeting time be t
After a journey of 15 minutes and B journey of T minutes, the speed ratio of a and B is t: 15
If B travels for 60 minutes and a travels for t minutes, the speed ratio of a and B is 60: T
So there is: T: 15 = 60: T
T = 30
Therefore, the speed ratio of a and B is 30:15 = 2:1
Party A and Party B ride bicycles to each other from a and B. Party A starts for an hour before Party B starts. After another four hours, they meet at C. after meeting each other, they can see each other
A continues to move forward at the original speed, and B also moves forward at the original speed after 20 minutes' rest. As a result, a arrives at B 40 minutes later than B arrives at A. It is known that B travels 4 kilometers more per hour than a, and the speed of a and B is calculated
Suppose the speed of a is x km / h, the speed of B is (x + 4) km / h, the distance between a and B is x + 4 (x + X + 4) = 9x + 16 km, AC = 5x km, BC = 4 (x + 4) = 4x + 16 km. It takes one hour more for a to take BC than B to take AC
（4X+16）/X=5X/（X+4）+1
The solution is: X1 = 16, X2 = - 2 (rounding off)
Therefore, the speed of a is 16 km / h and that of B is 20 km / h
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