There are 13 difficult Olympiad questions, math is in a hurry, math masters must enter Oh, I only ha

Question 1: (1200+6*80) 6

Question 2: Let the commander be x

530+x）/40=（380+x）/30

Question 3: Set the line for x seconds.

182+1034+18x=20x

Question 4: Set the train length x meters.

1200+x+18x）/75=x/15

Question 5: ((51-1)*40) 2=1000(m-min) — Calculate and change units by yourself.

one''''（1200-6*80）／6=120

II. II. II''''Set train captain x. （530+x）／40=（380+x）／30 。x=70

Three''''Set x seconds. 20x-182=1034+18x。 x=608

Four''''Set the full length x. （x+1200）／75=x／15。 x=300

Five''''(51-1)*40 2=1000m min=60km h

Six''''Express: (15-10)*12=60m Slow: (15-10)*9=45m

Seven''''Set the speed x'Length: 310+y=30xx=13m, s y=80m

Eight''''Set the speed x'Length (100+y) x=20 x=20 y=300

Nine''''（2*10+90）／10=11m／s

Ten''''1034／（20-18）=517s

In the first semester, there will be x male students and y female students.

It can be obtained from the title.

x+y=1450

1+(1 25))x+(1-(1 35))y=1460.

x=750y=700

In the finals of the Multiple Intelligences Competition, there are only three questions. It is known that: (1) 25 students from a certain school participated in the competition, and each student solved at least one problem; (2) Among all students who did not solve the first problem, the number of students who solved the second problem was twice as large as the number of students who solved the third problem

3) The number of students who only solved the first problem was 1 more than the number of students who solved the first problem; (4) Half of the students who solved only one problem did not solve the first problem, so what is the number of students who solved only the second problem?

Classmates, do you want to go to high school? I saw a few and wouldn't do it at all...

I hope to ask for advice, and I will send it to you, thank you!

Out of 1000 people, at most a few people know each other or a few don't know each other.

The natural number a, if you multiply a by 3, you get a number b,

Olympiad How to do the concentric method to make odd and even magic squares?

12 balls, one of which is not the same quality!

Tool: Balance.

Requirements:"Weigh it three times and find out the difference!

Based on the time of meeting and the time of arrival, it can be assumed that the meeting will be in the BC segment.

Let ab=x, bc=y, ca=z, then (x 20) + (y 15) + (z 30) = 1, z 15) + (y 30) + (x 20) = substitute the elimination element to obtain: y = z + 6

If BC segment A goes A, then (x 20) + (a 15) = 28 60 (z 15) + (y-a) 30 = 28 60 first eliminate a, and then substitute y=z+6 into x=20-4z and x=20-4z and y=z+6 into (x 20)+(y 15)+(z 30)=1, and get z=,x=4

According to this method, it is assumed that the encounter is in the X and Z segments, and there is no solution.

So in the BC segment, ab=4