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Several super difficult math problems
1. Suppose Xiaoming is now X years old and Grandpa should be 6x years old. (6x+a)/(x+a)=5 and x=4a.

(6x+b)/(x+b)=4 x = 1.5b。

X is an integer, so X should be a multiple of 3 and also a multiple of 4.

When x= 12, a=3, b=8, grandpa is 72 years old, which fits the question.

When x=24, a=6 and b= 16, grandpa is too old to be realistic.

So my grandfather is 72 years old.

2. Set the meeting time t (minutes).

(300+240)*t=2400a(a is an integer, indicating how many times my brother has met.

When a= 1, t=40/9, the distance from a is (40/9)*300=4000/3.

When a=2, t=80/9, the distance from A is 2400*2-(80/9)*300=6400/3.

When a=3, t=40/3, the distance from A is 4800-(40/3)*300=800.

When a=4, t= 160/9, and the distance from a is (160/9) * 300-4800 =1600/3.

When a=5, t=200/9, and the distance from a is 5600/3.

When a=6, t=80/3, and the distance from a is 800.

When a=7 and t=280/9, the ground distance from A is 800/3.

A if it is older, it will take more than 35 seconds, so the shortest distance should be 800/3, the seventh meeting.

3, x people decide, do y days.

(x-5)(y+ 1)=xy。

(x-5-8)(y+ 1+2)=xy

Solve the equation X=65, y= 12.

That is,/kloc-no one did it for 0/2 days.

These three questions are really troublesome! ! !