75 logic reasoning questions

[1] Suppose there is a pond with an infinite amount of water in it. There are two empty kettles with 5-litre and 6-litre capacities respectively. The question is how to get 3 litre water from the pond with only the two kettles.

From full 6 to empty 5 inverted, 1 litre left, put this 1 litre down 5 miles, then 6 left full, pour 5 inside, because 5 contains 1 litre of water, therefore, 6 can only pour 4 liters of water to 5, then the remaining 2 liters of water to 6, into the empty 5, then fill 6 to 5 inverted 3 liters, the remaining 3 liters.

[2] Zhou Wen's mother is a member of Yulin cement plant. One day, Zhou Wen came to the laboratory to do his work. After that, I want to go out and play. "Wait, mom will take a test of your question." She went on to say, "You see these 6 glasses are used only for testing. The first three are filled with water, and the last three are empty. Can you move only one glass cup to separate the cup filled with water from the empty one? "Zhou Wen, who loves his brains, is a famous" clever "in the school. She just wanted to do it for a while. Please think about how "Xiao smart" works?

Set the cup number to abcdef, ABC to full, and Def to empty. Just pour the water in B into E.

[3] three young men fell in love with a girl at the same time. They decided to fight with a pistol to decide who would marry the girl. Xiao Li's hit rate is 30%, Xiao Huang is better than him, the hit rate is 50%, and the best gunner is Xiao Lin. He never makes a mistake and the hit rate is 100%. Because of this obvious fact, for the sake of fairness, they decided to follow the order: Xiao Li shot the gun first, Xiao Huang second, and Xiao Lin finally. Then it repeats until they have only one person left. Who has the greatest chance to survive these three individuals? What policies should they adopt?

Kobayashi will kill porn when it's his turn and Xiao Huang is not dead. Then he will make a single pick with cainiao Li.

Therefore, Huang will beat the forest if the forest is not dead, or else he will die.

After computation and comparison (the process is omitted), Mr. Li will decide to beat Mr. Lin first.

As a result, Xiao Li has the vitality of 873/2600 ≈ 33.6%;

Xiao Huang has the vitality of 109/260 ≈ 41.9%;

Kobayashi has 24.5% of its vitality.

Oh, in this way, the first shot of Xiao Li will go to the ground. Then, of course, he will beat the enemy and who will beat the enemy;

As always, Xiao Huang beat the forest first, but Xiao Lin killed Huang first, and his family was narrow!

Finally, Li, Huang and Lin had a survival rate of about 38: 27: 35;

Cainiao have a high chance of returning their beauty to their lives.

Li Xianzhi shot an empty gun (if the partner works for zhonglin, he will suffer the most). Huang will select Lin to take a shot (if Lin is not hit, he must have finished playing first) lin will select Huang to take a shot (after all, it has a high hit rate) Li Huang showdown 0.3: 0.280.4 possibility Li Lin showdown 0.3: 0.60.6 probability success rate 0.73

Li and Huang beat Lin and Li Huang showdown 0.3: 0.40.7*0.4 possibility Li Lin showdown 0.3: 0.7*0.6*0.70.7*0.6 probability success rate 0.64

[4] Two prisoners are in the same room. Every day, the prison will provide a pot of soup for the room so that the two prisoners can score points by themselves. At first, these two people often have disputes because they always think that the other party has more soup than their own. Later, they found a perfect solution: let another person select first. So the dispute is settled. However, now there is another new prisoner in this cell, and now there are three people here. You must find a new way to maintain peace between them. What should I do? Press: psychological problems, not logical problems

It is to let a split the soup. After the split, Party B and Party C pick the soup for themselves in any order, and leave the remaining bowl to Party. In this way, the total number of people B and C must be the biggest one they can get. Then mix the soups of the two and then try again.

[5]] Place n round coins of the same size on the desktop of a rectangle. Some of these coins may be incomplete on the desktop, or some may overlap with each other. When another coin is placed and its center is inside the desktop, the new coin must overlap with some of the original coins. Prove that the entire desktop can be completely covered with 4 n coins.

To prevent the new coin from overlapping the original coin, the center distance between the two coins must be greater than the diameter. That is to say, for any point on the desktop, the distance to the nearest center is less than 2. Therefore, the entire desktop can be covered with n coins with a radius of 2.

Scale down the desktop and coin by a factor. Then, the length and width of each small desktop half of the original desktop can be covered with n coins with a radius of 1. Then, divide the original table into four equal small tables, so each small table can be covered with n coins with a radius of 1. Therefore, the entire desktop can be covered with 4 n coins with a radius of 1.

[6] A ball and a straight ruler with a length of about 2/3 in the diameter of the ball. How do you measure the ball's radius? There are many ways to see who is more clever

[7]] Five Yuan coins of the same size. What should we do if we need to contact each other?

Put 1 at the bottom, 2 3 on the top of 1, and 4 5 on the top of 1.

【8】 guess Card problem Mr. S, Mr. P, Mr. Q they know there are 16 cards in the drawer of the table: red peach A, Q, 4 black peach J, 8, 4, 2, 7, 3 K, Q, 5, 4, 6 blocks A, 5. Professor John picked a card from the 16 cards and told Mr. P about the number of cards and the color of the card to Mr. Q. Professor John asked Mr. P and Mr. Q: Can you tell from the known points or colors what the card is? So Mr. S heard the following conversation: Mr. P: I don't know this card. Mr. Q: I know you don't know this card. Mr. P: Now I know this card. Mr. Q: I know. After hearing the above conversation, Mr. s thought about it and correctly launched the card. Excuse me: What is this card?

Box 5

[9] A professor of logic has three students, and all three of them are very smart! One day, the professor gave them a question. The professor pasted a piece of paper on each person's head and told them that each person's paper had a positive integer, and the sum of two numbers is equal to the third one! (Each person can see the other two numbers, but not his/her own.) The professor asked the first student: Can you guess your own number? Answer: No. Ask the second question. No. The third question. No. ask the first question. No. The second question. No. The third question. I guess it is 144! The professor smiled with satisfaction. Can you guess the number of the other two people?

After the first round, it indicates that any two numbers are different. In the second round, the first two people did not guess, indicating that none of them is twice that of others. Now we have the following conditions: 1. Each number is greater than 02. It ranges from two to two. 3. Any number is not twice the other number. Each number may be the sum or difference of the other two. The third person can guess 144, and one of the three conditions must be excluded. Assume that the difference between two numbers is x-y = 144. At this time, 1 (X, Y> 0) and 2 (X! = Y). Therefore, to deny x + y, we must make 3 unsatisfied, that is, x + y = 2y, and obtain x = y, not true (otherwise the first round can be guessed), so it is not the difference between two numbers. Therefore, it is the sum of two numbers, that is, x + y = 144. Similarly, if both 1 and 2 are satisfied, 3 is not satisfied, that is, x-y = 2y. If the two equations are the same, x = 108, y = 36.

The order of the two rounds of guesses is as follows: first round (1, 2), second round (3, 1, 2 ). In this way, the information obtained at the end of each round is the same (that is, the preceding three conditions ).

Let's assume we are C. Let's take a look at how C is made: C sees 36 of A and 108 of B. Because of the condition, the sum of two numbers is the third, then you can either 72 or 144 (guess this is because 108 is the sum of 36 and 72, and 144 is the sum of 108 and 36. Raise your hand if you cannot understand this sentence ):

If your (c) is 72, then B can see it in the second round. below is the idea of C being 72 and B: in this case, B sees 36 of A and 72 of C, so he can guess himself, 36 or 108 (guess this is because 36, if 36 and 36 are equal to 72,108, the sum of 36 and 108 is ):

If your (B) header is 36, then C can be seen in the first round. below is the idea of B's 36 and C: in this case, C sees 36 of A and 36 of B, so he can guess himself, 72 or 0 (this is not explained ):

If your (c) header is 0, then a can see it in the first round. The following figure shows how C is 0 and A: in this case, A sees B's 36 and C's 0, so he can guess himself, is 36 or 36 (this is not explained), then he can report his 36. (Then, reverse push and reverse push) Now, A does not report its own 36 in the first round. C (in B's imagination) can know that it is not 0 on its head, if other ideas are the same as those of B (that is, B's head is 36), C can report its own 72 in the first round. Now C does not report his 36 in the first round. B (in C's imagination) can know that his head is not 36, if other ideas are the same as those of C (that is, C's head is 72), B can report its 108 in the second round. Now, if B does not report his 108 in the second round, C can know that his head is not 72, and the only thing on C is 144.

[10] A car crashed and escaped from a city. There were only two color cars in the city, blue 15% green 85%. When the incident happened, a person saw the car at the scene. He testified that it was a blue car, however, according to the expert's on-site analysis, the possibility of correct conditions at that time is 80%. What is the probability of a car being a blue car?

15% * 80%/(85% * 20% + 15% * 80%)

[11]A person who has 240 kilograms of Water wants to transport to a dry area to make money. He can carry up to 60 kilograms each time, and each forward one kilometer must consume 1 kg of water (even water consumption ). Assume that the price of water is 0 at the place of departure, and later is proportional to the transportation distance (that is, 10 yuan/kg at 10 kilometers, 20 RMB/kg at a distance of 20 kilometers ......), assuming that he must return safely, how much can he earn at most?

F (x) = (60-2x) * X. When x = 15, the maximum value is 450.

450 × 4. In addition, it must be proved that 60 kg of water is the optimal solution for each operation.

[12] There are now 100 horses and 100 stones. There are 3 types of horses, large horses, and medium-sized horses and small horses. A horse can carry three stones at a time, and a medium-sized horse can carry two, while a small horse can carry two stones at a time. How many horses, medium-sized horses and small horses are required? (The key to the problem is that you must use up 100 horses)

6 RESULTS: X horses, y horses, and Z horses, respectively. For example, 1.x+ y + z = 100; 2.3x + 2y + z/2 = 100, we can obtain 5x + 3y = 100, y must be a multiple of 5, and x <20.

[13] 1 = 5, 2 = 15, 3 = 215, 4 = 2145 so 5 =?

Because 1 = 5, 5 = 1.

[14] There are 2n people queuing into the cinema, the fare is 50 cents. Among the 2n, N are only 50 cents, and n are $1 (paper ). When a stupid cinema starts selling tickets, there is no 1 cent. Q: How many queuing methods are there to make a 50 cent change to a cinema every time you buy a ticket for $1?

Note: One dollar = 100 cents for a person who owns one dollar, has a banknote, and cannot be broken into two 50 cents.

This question can be implemented using recursive algorithms, but the time complexity is the Npower of 2. You can also use the dynamic programming method. The time complexity is the square of N, which is much simpler to implement, but the most convenient thing is to directly use the formula: the number of queues = (2n )! /[N! (N + 1)!].

If you don't consider whether the cinema can find money, there will be a total of (2n )! /[N! N!] Queuing method (that is, extracting the number of combinations of n persons from 2n persons). For each queuing method, if the cinema cannot find money, it is called unqualified, this queuing method is (2n )! /[(N-1 )! (N + 1)!] (N-1-1 combination of individuals from 2n), so the number of qualified queuing types is (2n )! /[N! N!] -(2n )! /[(N-1 )! (N + 1)!] = (2n )! /[N! (N + 1)!]. Why is the unqualified number (2n )! /[(N-1 )! (N + 1)!], It's too complicated to say, so I won't talk about it here.

[15] A person bought a chicken for 8 yuan and sold it for 9 yuan. Then he thought it was not worthwhile. He bought it again for 10 yuan and sold it to another person for 11 yuan. Ask him how much he earned?

2 RMB