Set x boxes.
5X+4=6(X- 1)+3
5X+4=6X-3
X=7
2) 12 table tennis, one of which is of poor quality (I didn't tell you whether the bad one is light or heavy). I have given you a scale and can only weigh it three times. How can I know which one is of poor quality?
The premise is! ! I don't know whether the one with poor quality is heavy or light!
Divide into four groups: A, B, C and D, with 3 balls in each group.
The first weighing; There are two situations when Group A and Group B are placed on the balance:
(1) If A and B are not equal in weight, the bad balls are in one group, and C and D are good balls; ② If A and B are equal in weight, the bad ball is in group C and D, and A and B are good balls.
Second weighing: Group A and Group C are put on the balance. If the weights are not equal, if ① holds, A has a bad ball; if ② holds, C has a bad ball; On the other hand, if the weights of A and C are equal, if ① holds, B has a bad ball, and if ② holds, D has a bad ball. It is important that because one side of the balance is a known good ball, it is also known whether the bad ball is light or heavy.
The third weighing: randomly pick two balls from a group of bad balls to both ends of the balance. If their weights are not equal, the lighter (or heavier) one is a bad ball. If the weight is equal, the bad ball is the one left.
3)
Regarding the mathematics of table tennis single round robin, five athletes from A.B.C.D.E participated in the table tennis single round robin, and each game stipulated that the winner would get 2 points and the loser would get 0 points. The known results are as follows:
1.a and B tied for first place.
C and d tied for third place.
Find the score of e.
Because 1. A and b tied for first place. 2.c and D tied for third place, with the winner getting 2 points and the loser getting 0 points.
So A B won as many games, C D won as many games, and E ranked fifth.
Let's assume that A won four games and scored eight points, then B can only win three games, which is unreasonable.
So A won three games and B won three games. Suppose A loses to C, and then B loses to A, so B wins CDE.
Then C D won two games each, so D can only win C E, and C won A E, so E lost all.
Only o points.
X stands for winning and O stands for losing.
Look from left to right.
- A-B-C-D-E
A - x-o-x-x
B-o - x-x-x
C-x-o - o-x
Dioo x x
Oh, oh, oh, oh.
Because AB is tied for first place, D is at least third.
Look at the watch
such as
- A-B-C-D-E
A - x-o-x-x
A won, B D E lost to C.
4) 10 children take part in the table tennis single round robin (everyone has to play a game with others). Ask a * * * how many games?
1+2+3+4+5+6+7+8+9=45 (field)
5) Players play table tennis round robin. Every two people have to play a game. If there are n athletes, how many games will they play?
(N- 1+ 1)(N- 1)/2
6) How many ping-pong balls can be put in this box?
How many ping-pong balls with a diameter of 4cm can a cubic box (with a lid) with a side length of 30cm hold at most? If the cover is removed, how many ping-pong balls can you hold at most?
How many ping-pong balls can a basket with a diameter of 30cm (which can be considered as hemispherical without a cover) hold?
answer
This box can hold 7 balls on each side, so the bottom layer is 49, and the gap is 2*30(cm*cm), which is exactly the radius length of the balls. Therefore, in the box, if the balls are staggered and compact, the number of balls on each floor is the same. With each additional layer, the total height of the balls decreases by 4-2(√3) compared with that without staggered placement. Therefore, with the increase, the formula of the total height change is 4n-[4-2(√3)](n- 1)(n is a positive integer). You can find that there are at most 8 floors in the box. Therefore, with the cover, 49*8=392 balls can be accommodated at most.
Without a cover, the ball is placed in a cone with four bottoms. From top to bottom, the usual rules are 1 ball on the first floor, 4 balls on the second floor, followed by 9 balls, 16 balls, 25 balls, 36 balls and 49 balls, which meet the requirements of this question, so the total number of balls is 1+4.
If a hemispherical uncovered basket with a diameter of 30cm is used to hold small balls, there can be 37 balls on the surface of the basket (not protruding from the basket), with 1+4+7+ 19+37=68 above the water and 7+ 19=26 below the water, so the total number is 68+.