各位高手,帮帮忙!!一道小学数学题!!不要太复杂的解,有详细的解释,最好是方程,还要有思路!!谢谢

2024-12-21 10:35:01
推荐回答(5个)
回答1:

设距离为y,初始车速为x,y的单位看做千米,x的单位看做千米/分钟(很怪异,但是为了解题更简单)
可以看出,规定时间是y/x,将此看做一个已经存在的定值,不去管它。
根据条件1:车速变为1.25倍,可以提前24分钟到,由此列出公式:y/1.25x=y/x-24。
根据条件2:先开80千米,在提速,提前10分钟,像上一个公式一样写出来:80/x(前80千米以规定速度行驶需要的时间)+(y-80)/(4/3x)(后面的路程提速后需要的时间)=y/x-10(总时间)
大约的解题思路就是这样,可能算起来会麻烦点,不过既然是奥赛题,那还是自己算出结果比较好,懂得思路后在做解答会比直接得出数字更好。

回答2:

路程一定,速度和时间成反比,车速提高25%,即现车速是原车速的1+25%=1.25倍,那么规定到达时间就是现到达时间的1.25倍,现在到达时间为:24÷(1.25-1)=96(分钟),规定到达时间为:96+24=120(分钟)。将速度提高3分之一,那么可以提前10分钟到达,即120-10=110(分钟)到达。在80千米以外的路程里,速度是原速度的(1+1/3)=1又1/3倍,原时间是现时间的
1又1/3倍,现时间为:10÷(1又1/3 -1)=30(分钟),原时间是:30+10=40(分钟)或
30×1又1/3=40(分钟)。那80千米所用时间为:120-40=80(分钟),速度一定,时间和路程成正比,所以:80千米以外的路程为x,则:40:80=x:80(前一个80分钟,后一个80千米)
x=40×80÷80=40(千米)
甲乙两地相距:80+40=120(千米)

回答3:

速度提升25%,提升后速度和原速的比为5:4,提升后所需时间比和原速所需时间的比为4:5。差24分钟,就可求出全用原速走,就需120分钟。
再看,速度提升了3分之一,就为原来的一又3分之一,也就是3分之4,原速跑完80千米之后,时间就应该减少到原来的4分之3。
减少了跑完80千米后时间的4分之1,这个差为10分钟,也就是说,用跑完80千米后提升的速度跑,还需30分钟。如果这时不提速,用原速继续跑,还需40分钟,这种行程方法全用原速,所需时间就应该为上面说的120分钟。
120分钟减去40分钟等于80分钟,这个求的是用原速跑80千米所需的时间。可以很快的求出:这辆车的速度为80除以80等于1千米\时。
最后,1*120=120千米。这就是路程。

回答4:

小学能不能设置两个未知数?
设定原来的速度为x,总路程为y,列式子
y/(x * 125%) + 24 = y / x
80/x + (y - 80) * 3 / x * 4 + 10 = y/x
如果不能用未知数的话就把x倒出来一个全是已知数的式子就是答案了

回答5:

车速提高25%,即为原来的5/4倍,则时间为原来的4/5,提前1/5,则所用时间为
24/(1/5)=120
将速度提高3分之一,即为原来的4/3倍,则时间为原来的3/4,提前1/4,则
假设没有提速原来需要时间10/(1/4)=40分钟 提高速度后实际为30分钟
原来的80千米用了80分钟,
所以甲乙两地相距120千米

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