一道五年级奥数题!急!!!

2024年11月18日 19:53
有5个网友回答
网友(1):

用方程解可以吗:
设甲的速度是X千米/小时,则乙的速度是X-12,甲乙两地的距离是4.5X千米,
甲乙相遇时,共走了时间是4.5+31.5/X 小时,相遇时乙走了4.5X-31.5千米
因此可以写出方程:(X-12)(4.5+31.5/X)=4.5X-31.5
得出:X=42

网友(2):

设甲每小时行驶X千米,乙为Y千米/小时
两个方程式:X=Y+10
4.5X=(31.5*Y/X)+4.5y+31.5

计算结果:X=17.5 Y=7.5

网友(3):

12*4.5=54
54-31.5=22.5
31.5-22.5=9
9/12=3/4
31.5/(3/4)=42

网友(4):

提问不完整

网友(5):

甲车每小时比乙车多行12千米,即甲车行驶四个半小时到达西站后甲车比乙车多行驶了12*4.5=54千米;
再没有停留,立即原路返回,在距离西站31.5千米的地方和乙车相遇,即此时乙车又行驶了54-31.5=22.5千米;
在之前行驶过四个半小时后甲车原路返回的这段时间内甲车比乙车多行驶了31.5-22.5=9千米;
由于甲车每小时比乙车多行12千米,所以在行驶过四个半小时后需要用时9/12=3/4=0.75H方能在距离西站31.5千米的地方和乙车相遇。
甲车通过0.75H从西站行驶到31.5千米的地方,该车的速度为:31.5/0.75=42千米/H

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