Quantitative relationship questions, quantitative relations are answered

Remember mine for the correct answer, thanks!

Idea: Let the first to the last name of the character be a b c d e f g h i j, find f

1: The first place A and the second place B in the game have not lost a game, then.

1st place is definitely tied with 2nd place.

Then the first place score is: 17 points (up to 8 games to win, one game to tie with second place).

The second place score is: 16 points (a maximum of 7 games to win, one game to the first place, and at least the other team to the other).

2: The top two have 20 more points than the third place finisher.

Then the third place has 13 points.

3: The score of the fourth place is equal to the score of the last four: and d c 13

Can be pushed: 4th place d 12 points.

Then the sum of the bottom four must be d 12 points.

4:10 students have different scores.

Then there is a maximum of 0 points. And the maximum score is 17 points.

5: When there is no 0 point appearing, there will be a :

j≥1 i≥2

h≥3 g≥4

f 5 d = f + g + h + i + j 15 contradicts d 12 points.

Therefore, a score of 0 must appear.

When 6:0 appears, then the penultimate place i has at least 2 points (at least wins the penultimate J), and the third-to-last H has at least 4 points (at least wins the penultimate first, lower and penultimate I, because when i wins one point, h will have at least one more point than i, and if i wins i, then h will directly get 2 points). The same penultimate will have at least 6 points.

That is: d = f + g + h + i + j 0 + 3 + 4 + 6 = 113 and d 12, contradictory.

Or: d = f + g + h + i + j 0 + 2 + 4 + 6 = 12 and d 12

Then d = 12

So: the 1-3 equal sign of ideas is established.

i.e. a=17, win, 8 and 1

b = 16 wins 6 and 2

c = 13 wins 5 and 3 (push: if you win 4, then and 5, a total of 10 people and a and b draw, so the number is not enough.) i.e. only 5 and 3 can be won).

d = 12g = 6 wins 2 and 2

h=4 wins, 2 losses, 0

i=2 wins, 1 losses, 8

J=0 loses 9 and now there is only e,f unknown. It can be found based on the number of wins, losses, draws, and total games.

I'm going to get off work, so you can watch and beg slowly. I'm sorry.

The above reasoning is basically correct, but b is 7 wins and 2 draws, and in the end it is still not possible to reason, there are four unknowns!

Let's assume that the workload of project A is A and the workload of project B is B.

It is known that the velocity of A is a 13 and b 7, and the velocity of B is a 11 and b 9

In fact, the shortest way to cooperate is that A and B first cooperate on project A, and then cooperate with project B.

A 13)+(A 11)=24 143a Then do project A first, 5 and 23 24 days to complete, B 7 + B 9 = 16 63b, the next 1 24 days to start project B, complete 2 189b, the remaining 187 189b

After 3 days of doing project B, 144 189b was completed, leaving the last day of project workload of 43 189b, A and B needed 43 189 (16 63) = 43 48 days.

Suppose it takes T days to complete two projects.

1/13+1/7+1/11+1/9)t=1

The solution is t=9009 3794 days, i.e. about 2 days and 9 hours. Therefore, on the last day, the two teams had to work together for about 9 hours to complete the task.

There are p(5,4)=120 different arrangement methods, choose a

Analysis: For the sake of good talk, we call the five stations ABCDE Zhan, first arrange 2 two cars in A Zhan, a total of 10 arrangements, and win the three stations of 3 two cars, a total of 6 arrangements, so under this arrangement, a total of 5

6 10 kinds of arrangement, the remaining three stations arrange 2 two cars and A station arranges two cars in the same situation, so there are a total of 60 4 = 240 kinds of arrangement. See figure for details.

This can be calculated using the permutations and combinations knowledge learned in high school.

The more numbers, the better.

I am a retired employee who loves to play numbers games with my children.

Let the velocities of A and B be x and y, respectively.

3x-3y=15

Solve the equation to get B = kilometers of hours.

Let the velocities be A and B respectively.

A + B = 15

3A-3B = 15

Solve the equation to get B = kilometers of hours.

Assuming that the first time is divided into groups x and the second time is divided into groups y, then:

7x+4=5y+2 (i.e., the number of party members in the two groups is equal), 3x=2y (i.e., the number of party activists in the two groups is equal), and the equation is solved to obtain x=4, y=6

There are more party members than party activists (7 4 + 4) - (3 4) = 20 (people).

A has an advantage in doing B, B has an advantage in doing A, let A finish B in 7 days, at this time, B still has 4 11 A, everyone does the rest of A together

4 11 divided by (1 13 + 1 11) = 2 and 1 6 That is, in the end, you only need to do 1 6 days to complete the task.