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Xiaoming walks from place a to place B, and Xiaoqiang walks from place B to place a at the same time. When they reach the destination, both sides immediately return. During the walking process, their respective speeds remain unchanged. The first time they meet is 40 kilometers away from place a, and the second time they meet is 15 kilometers away from place B. then the distance between a and B is______ Kilometers
Suppose the distance between a and B is s kilometers, then: the speed of a is V1, the speed of B is V2, the time of the first meeting of a and B is T1, the time of the second meeting is T2, the journey of two people is as shown in the figure below: from the meaning of the question: v1t1 + v2t1 = sv1t1 = 40v1t2 + v2t2 = 3sv2t2 = 2s-15, we can get: S = 105 (kilometers), so the answer is 105
Xiaoming's distance is one fourth more than Xiaohua's, but Xiaohua's time is one third more than Xiaoming's. what's Xiaohua's speed?
How to write this problem? Hurry up, thank you
If Xiaohua's distance is x and Xiaoming's time is t, then Xiaoming's distance is 5x / 4 and Xiaohua's time is 4T / 3
Xiaohua's speed is faster than Xiaoming's speed V1: V2 = x / (4T / 3): 5x / 4T = 3:5
So Xiaohua express is 3 / 5 of Xiaoming's
After a period of running training, Xiao Ming ran 1200 meters, the time was shortened from 5 points to 4 points, and the speed was increased by ()%
Original speed = 1200 △ 5 = 240
Current speed = 1200 △ 4 = 300
300÷240=1.25
25% faster
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