Question

十万火急!!!！！！！！2道六年级奥数题!!!

用方程解:
1、一条猎狗追30米外的一只狐狸。狗跳一次为2米，狐狸跳一次为1米，狐狸跳3次的时间狗只能跳2次，猎狗跑？米才能追上狐狸？
2、某工厂第一车间人数比第二车间人数的五分之二少30人，如果从第二车间调10人到第一车间，则第一车间人数就是第二车间人数的四分之三，两车间原来各有？人？
回答得好+多10分或以上！！！
一定要用方程，越详细越好！！！！
请各位不要发飙= =
谢谢合作
对啊,第二题貌似不对啊...我算出的结果不是整数
1能用方程吗?

Answer 1

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Answer 2

1----
设猎狗狐狸分别跑xy 则有
x=y+30
y/x=(3*1)/2*2
得x=120
y=90
2------
设第一 第二车间各有人xy
x=2/5y-30
x+10=(y-10)3/4
你自己解吧

Answer 3

第一道设狐狸跑x米.方程为x/2=(x+30)/(2*2)最后答案120米，抓住它们所用的时间相同，第二道，设第二车间为x人方程为0.2x-30=(x-10)*0.75-10

Answer 4

1.解：狐狸跳3次的时间狗只能跳2次 狐狸跳3米狗却跳了4米，狗跑4米追上了狐狸1米 的距离，要想缩短30米的距离，猎狗跑了米数：30/1/4=120米 即猎狗跑120米才能追上狐狸。

Answer 5

(1)
在相同时间内，狐狸跳3次可以跳3米，而够跳2次可以跳4米，所以要跳60次才可以追上，则一共跳了60*2=120米
(2)
设第一第二车间原来各有x,y人
2/5*y-x=30
x+10=3/4(y-10)
得出结果负，题目有错