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Ask the master to help solve two Olympic math problems in the sixth grade of primary school
1.

Think of the distance as downhill. If the speed is the same, the distance up the mountain is twice as long as that down the mountain.

The distance up the mountain is two-thirds of the total distance.

Party A completed the whole journey, while Party B only completed 2/3+ 1/3* 1/2=5/6.

The speed of b is 5/6 of that of a.

Line A is 2/3 of the whole journey, while Line B is 400 meters shorter.

2/3 of Party A's whole journey and 2/3*5/6=5/9 of Party B's whole journey.

The whole journey is 400/(2/3-5/9)=3600 meters.

2/3*3600=2400 meters

2.

If Xiaoming's speed is 1, Xiaoguang's speed is 4/3 and motorcycle's speed is 16.

Design a plan:

Motorcycle and Xiaoguang go first, and Xiaoming walks.

After arriving at a certain place, the motorcycle put down the light and went back to pick up Xiao Ming. Xiaoguang is gone.

Finally, the motorcycle brought Xiaoming and Xiaoguang to school at the same time.

Suppose that the time for Xiaoguang to ride a motorcycle is x and the distance he travels is 16x.

During this time, Xiao Ming walked around for X.

When the motorcycle puts down the small light, the distance is Xiaoming 16x-x= 15x.

The motorcycle turns around to pick up Xiao Ming, and it takes time to pick up Xiao Ming:

15x/( 16+ 1)= 15/ 17x

During this time,

Xiao Ming's walking distance is 15/ 17x.

The walking distance of Xiaoguang is15/17x * 4/3 = 20/17x.

Two people apart:

15x+(20/ 17x- 15/ 17x)= 15x+5/ 17x

Then, take Xiaoming on a motorcycle and arrive at school with Xiaoguang at the same time. The required time is:

( 15x+5/ 17x)/( 16-4/3)= 195/ 187 x

Now, to sum up,

Xiao Ming's walking distance is: x+15/17x = 32/17x.

The walking distance of Xiaoguang is 20/17x+195/187x * 4/3 = 480/187x.

The walking distance between Xiaoming and Xiaoguang is:

32/ 17:480/ 187= 1 1: 15

Something like that I wonder if there is any mistake in the calculation. . .