Who will do this question, the question of the first year of junior high school.

Let the number of rooms be x, and the number of people who will not be able to stay in the first type of room is y, which can be obtained.

x-1)*4+y=3x+2

Simplification. x=6-y

From the meaning of the title. 1≤y＜4

So. When y=1.

x=5 The number of boys is 17.

When y=2.

x = 4 for 14 boys.

When y=3.

x = 3 for 11 boys.

If the number of rooms is x, there are (3x+2) male students in the class.

4 (x-1) + 1 "3 x + 2 " 4 (x-1) + 33 " x "5 so x = 3, 4, 5

There were 11, 14, and 17 boys in the class.

Because A and B have the same number of rooms, then for each room, A has one more person than B, and if A happens to be full, the extra number of people should be exactly the number of rooms. In fact, A has a room that is not full, that is to say, the last room may live in 1, 2, 3 people, three cases, and B room is full, and there are two more, which can be seen as the same number of rooms as A, but there are 5 more people, and the people in our room that are not full are also regarded as extra people, at this time, 5-1, 5-2, 5-3 is not the number of rooms that A is full? In this way, it can be calculated that when there are three rooms in each room, there are 11 boys, 14 in each of the four rooms, and 17 in each of the five rooms.

Set up the rest of the bureau to go to the flat road x kilometers, uphill y kilometers, tong rent rolling downhill z kilometers.

Yes. x+y+z=142

x/30+y/28+z/35=

x/30+y/35+z/28=

Solving the equation system type is obtained, x=30, y=42, z=70, that is, 30 kilometers on a flat road, 42 kilometers uphill and 70 kilometers downhill when going.

The probability that no one wins the lottery = the probability that 5 people will not win = 7 10 * 6 9 * 5 8 * 4 7 * 3 6 = 1 12

Pick D. Agree.

If you pass the Olympiad in elementary school, you can easily do the combination of numbers.

You look at ** understand it, don't understand it and ask me.

3.1 2 times 2 events 2 3 times 4 events 3 4 times 4 events 4 times 4 events.

Solution: If there is a problem with a problem, an easy problem with b, and a problem c with only two people solving it, then a+b+c=100

So 2a+2b+2c=200

a+3b+2c=60*3=180

So a-b=20

i.e. a=20+b

Answer: There are more difficult questions than easy questions, 20 more questions.

Let A and B coincide as a, A and C coincide b, and B and C coexist c A, B, and C coincide i

The puzzle is easy for i.

60+60+60-2a-2b-2c-3i

and because a+b+c+2i=180-100

It's easy. 180-2（80-2i）-3i=20+i

So there are many easy questions, 20 more questions.

Solution: Let A and B coincide as a, A and C coincide b, B and C coincide c A, B, and C coincide i

60+60+60-2a-2b-2c-3i

And because a+b+c+2i=180-100

It's easy. 180-2（80-2i）-3i=20

There are many easy questions, 20 more questions.

There are only easy questions and difficult questions:

A represents the number of problems solved by A, B represents B, C represents C, and ABC represents the number of problems solved by all three, A+B+C-ABC=100, ABC=80

That is, there are 80 easy questions, 20 difficult questions, and 60 more easy questions than difficult questions.

Because everyone has solved 60 questions, three people should solve 180 questions, and A, B and C have solved a total of 100 questions, indicating that some problems have been solved by three people, some problems have only been solved by two people, and some problems have only been solved by one person, and the number of problems solved by two people and three people is 180-100=80, and if there are x easy questions, the number of problems solved by two people is 80-2x, and the number of problems is 100-(80-2x)-x=20+x, so the number of problems - the number of easy problems = (20+x)-x=20, so, there are many problems, 20 more.

Personal calculations.

1.There are no easy questions.

There are at least 20 questions, and one person solves 100*2-60*3=20, that is, the problem, and vice versa, up to 80 questions, medium class, 60*3-100=802There is no medium class.

Up to 40 questions have been solved (60*3-100) 2=40, i.e. easy questions, otherwise, up to 60 difficult questions.

3.There are no problems.

If it is not true, 100*2-60*3=20 then there will be at least 20 difficult questions to summarize: set up difficult x questions, medium class y lanes, easy questions z questions.

20<=x<=60 0<=y<=80 0<=z<=40x+y+z=100 1.

x+2y+3z=180 2 formula.

Find the positive and negative of x-z.

Formula 1 * 2 -&2 = x-z = 20

60 3 100 80 (puzzle) ......There are many problems.

100 80 20 (easy).

There are many easy questions, and twenty more questions than hard. First of all, there are three situations for this 100 questions, difficult, easy, and medium (two people answer correctly), three people answer a total of 180 questions, and 80 questions are repeated. "Medium" is repeated once, and "Difficult" is repeated twice.

That is, medium + difficult + difficult = 80, and easy + medium + difficult = 100, subtract the two formulas to obtain: easy - difficult = 20

Your -2 is counted as 2, which is of course wrong.

It just so happens that I'm in my first year of junior high school, but I won't