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When car a finishes 120 kilometers in 1 hour and 30 minutes, it just finishes one third of the whole journey. At this time, car B's journey is 60 kilometers more than two fifths of the whole journey. After a few hours, two cars meet each other. The best way to solve the problem is to write it down!)
120*3=360
120/1.5=80
(360*2/5+60)/1.5=136
136*x+80*x=360
x=360/216=1.67
A. B is 450 km away from each other. A and B set out from each other and run in opposite directions. It is known that a's speed is 120 km / h and B's speed is 80 km / h. After how long, the two vehicles are 50 km apart
First of all, there are two answers to this question
At the beginning of approach, there is a time point, 50 km apart, (450-50) / (120 + 80) = 2 hours
When they meet, they continue to move forward, away from each other, another time point, 50 kilometers apart, (450 + 50) / (120 + 80) = 2.5 hours
A and B start from city a to city B at the same time. Car a travels 45 kilometers per hour and car B 40 kilometers per hour. After 30 minutes, car a returns to city a at the same speed
After a delay of 30 minutes in city a, he went to city B at the same speed, but he arrived 30 minutes later than city B. to find the distance between the two cities, he had to solve the equation in five minutes,
x/45+1=x/40
x=360
If you go to the railway station by motorcycle from home, if you walk 30 kilometers per hour, it will be 15 minutes earlier than the driving time. If you walk 18 kilometers per hour, it will be 15 minutes later than the driving time. Now you plan to arrive at the railway station 10 minutes earlier than the driving time, then the speed of the motorcycle should be ()
A. 25km / h B. 26km / h C. 27km / h D. 28km / h
Suppose the distance from motorcycle to railway station is s kilometers, and the time of driving to railway station is t hours. If you walk 30 kilometers per hour, it is 15 minutes earlier than the driving time, then t-s30 = 14. If you walk 18 kilometers per hour, it is 15 minutes later than the driving time, then s18-t = 14
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