Summary of Intellectual problems (4)

1. A city has a car hit people escape incident, the city only two colors of the car, blue 15% Green 85%, the incident when there is a person at the scene saw, he refers to the blue car, but according to the experts in the field analysis, then that condition can see the right probability is 80% then, the accident is the probability of the car is blue?

Answer:

Set to see the blue car for event a

See the green car for event B

See correct for event C

See Error for Event D

A occurs when the event C is asked to occur

P (c| A) =p (CA)/P (a) =p (a) p (c| A)/[p (a) P (c| A) +p (B) P (d| B)]

=15%*80%/(15%x80%+85%x20%)

2. One person had 240 kilograms of water, and he wanted to make money in the dry areas. He carries a maximum of 60 kilograms each time, and consumes 1 kilograms of water per kilometer (evenly water consumption). Assuming that the price of water at the starting point of 0, later, and transport distance is proportional to (that is, at 10 km is 10 yuan/kg, 20 km is 20 yuan/kg ...) And suppose that he must return safely, how much money can he make, please?

answer : Set travel x km

f (x) = (60-2x) *x, when X=15, has a maximum value of 450.

450x (240/60)

3. There are now a total of 100 horses with 100 stones, 3 species of horses, large horses, medium-sized horses and small horses. One horse can carry 3 pieces of stone at a time, the medium horse can carry 2 pieces, and the small horse 2 can carry a stone. Q. How many horses, medium and small horses are needed? (The crux of the problem is that it must be done with 100 horses)

Answer:6 kinds of results, respectively, set a large horse for x, medium y, small z-horse, so there are: 1.x+y+z=100; 2.3x+2y+z/2=100, can get 5x+3y=100, know y must be a multiple of 5, and x<20.

4. There is a sports competition containing M items, athletes a,b,c participate, in each project, first, second, third, respectively, X, Y, z, x, y, z are positive integers and x>y>z. Finally a score of 22 points, B and C are 9 points, B in the hundred-meter race to achieve the first. Ask for the value of M, and asked in the high jump who got second place.

Answer: X, Y, Z is a positive integer and x>y>z, so x+y+z≥6;b first and finally divided into 9, so x≤8. and M (x+y+z) =22+9+9 so the possible value of M is 2, 4, 5 m take 2 o'clock, a score 22>16≥2x, obviously impossible. M fetch 4 o'clock, the value of x, Y, Z is four cases 1:7,2,1 2:6,3,1 3:5,4,1 4:5,3,2 when and only when the second case can be satisfied with the first and eventually get 9 points 9=6+1+1+1 but not satisfied with C eventually also get 9 points so M is 5 at this time x, Y, Z There are two cases of the value of 1:5,2,1 2:4,3,1 the second type of value can not meet the B 9 points The first case B, 9=5+1+1+1+1 C, 9=2+2+2+2+1 A, 22=5+5+5+5+2 hundred meters race b first, A second, c third the rest of the second all C take the

5. The U2 choir has to get to the concert in 17 minutes, on the way must cross a bridge, four people from the same end of the bridge, you have to help them to reach the other end, it is very dark, and they have only one flashlight. At the same time can have up to two people to cross the bridge, and the bridge must hold a flashlight, so someone has to bring a flashlight to take, back and forth Bridge ends. Flashlights can not be lost in a way to pass. The walking speed of four people is different, if the two people walk with the slower speed of whichever. Bono takes 1 minutes to cross the bridge, the edge takes 2 minutes to cross the bridge, Adam takes 5 minutes to cross the bridge, and Larry takes 10 minutes to cross the bridge. How are they going to cross the bridge within 17 minutes?

Answer:

2+1 first 2.

Then 1 back to send the flashlight 1

5+10, 10 more.

2 Back to send flashlight 2

2+1 past 2

Total 2+1+10+2+2=17 minutes

6. A family has two children, one of which is a girl, asking the probability of another girl (assuming the same probability of having a male or female)

Answer:(the same as the puzzle summary (4) 1 questions)

Sample space (male) (female) (female) (male)

A= (one is known to be a girl) =) (female) (female) (male)

B= (another girl) = (female)

So P (b/a) =p (AB)/P (A) = (1/4)/(3/4) =1/3

7.12 Ball a balance, now know that only one and the other weight is different, ask how to call to use three times to find the ball. 13 of them? (Note that this question does not indicate that the weight of the ball is light or heavy)

Answer:

12 can find out whether it is heavy or light, 13 can only find out which ball, the weight does not know.

Make the ball into a ①②③④⑤⑥⑦⑧⑨⑩⑾⑿. (13 hours numbered ⒀)

First call: Put ①②③④ and ⑤⑥⑦⑧ on both sides of the balance,

The ㈠ is equal, indicating that the special ball is in the remaining 4 balls.

To weigh ①⑨ and ⑩⑾ for the second time,

⒈ such as equal, indicating ⑿ special, ① and ⑿ for the third weighing can be judged ⑿ is heavy or light

⒉ such as ①⑨<⑩⑾ Note that either the ⑩⑾ has a heavy, or ⑨ is light.

⑩ and ⑾ for the third time, such as equal description ⑨ light, can not find out who is the heavy ball.

⒊ such as ①⑨>⑩⑾ Note that either there is a light in the ⑩⑾, or the ⑨ is heavy.

⑩ and ⑾ for the third time, such as equal description of ⑨ weight, can not find out who is the light ball.

㈡ on the left side of the right side, indicating that there is a light or heavy

Weigh ①②⑤ and ③④⑥ for the second time.

⒈ such as equal, indicating that there is a heavy ⑦⑧, ① and ⑦ for the third weighing can be judged ⑦ and ⑧ who is the heavy ball

⒉ such as ①②⑤<③④⑥ Note that either there is a light in the ①②, or the ⑥ is heavy.

① and ② for the third time, such as equal description of ⑥ weight, can not find out who is the light ball.

⒊ such as ①②⑤>③④⑥ that either the ⑤ is heavy, or that one of the ③④ is light.

③ and ④ for the third time, such as equal description of ⑤ weight, can not find out who is the light ball.

㈢ to the right of the left and opposite ㈡.

When the 13 balls are ㈠, the next step is as follows.

To weigh ①⑨ and ⑩⑾ for the second time,

⒈, such as equal, indicating ⑿⒀ special, ① and ⑿ for the third weighing can be judged ⑿ or ⒀ special, but can not judge the severity.

See the ⒉⒊ of step ㈠ in the case of ⒉

8.100 people to answer five questions, 81 people answered the first question, 91 answers the second question correctly, 85 answer the third question correctly, 79 answer the fourth question correctly, 74 correct the fifth question correctly, the correct three questions or three questions above the person counts the pass, then, in this 100 people, at least () the person passes.

answer : Ask at least how many people have passed, that is to say that the number of failed to pass the fewest number of people. 100 people answer 5 questions, equivalent to do 500 questions, total number of questions answered: 81+91+85+79+74=410 (TAO), the number of errors are: 500-410=90 (road), wrong 3 or more failed, each wrong 3 the number of failed to pass the most, 90÷3=30 (people), The minimum number of passes is: 100-30=70 (person).

9.1,11,21,1211,111221, what's the next number?

answer : The next one is on the previous explanation so the new one should be 3 1 2 2 1 1:312,211

10. A total of three kinds of drugs, respectively, 1g,2g,3g, put in a number of bottles, can now be identified in each bottle only one of the drugs, and each bottle of pills enough, can only be called once to know each bottle is the type of medicine? What if there are 4 kinds of drugs? What about Class 5? N Class (n can be counted)? If there is a total of M bottles of N-type drugs (M,n is a positive integer, the quality of the drug varies but the quality of the various drugs known)? Can you just call it once and know what each bottle is? Note: There's a price, of course, that we don't have to call the pills.

Answer:

First look at the simplest case of a total of three kinds of drugs, respectively, heavy 1g,2g,3g.

We can give bottles numbered 1th, 2nd and 3rd, respectively.

Now take out 1 tablets from bottle 1th, 2nd bottles take out 10 tablets, 3rd bottles take out 100 tablets, weigh is a three-digit number.

One digit corresponds to the weight of the 1th bottle pill, 10 digits corresponds to the weight of the 2nd bottle pill, and the hundred numbers correspond to the weight of the 3rd bottle pill.

Problem solving.

Similarly, if there is a total of four kinds of medicine, the weight 1g,2g,3g,4g respectively.

We can give bottles numbered 1th, 2nd, 3rd and 4th, respectively.

Now take out 1 tablets from bottle 1th, 2nd bottles take out 10 tablets, 3rd bottles take out 100 pieces, 4th bottles take out 1000 tablets weighing is a four-digit weight.

One digit corresponds to the weight of the 1th bottle pill, the 10 digits correspond to the weight of the 2nd bottle pill, the hundred numbers correspond to the weight of the 3rd bottle pill, the thousand numbers correspond to the weight of the 4th bottle pill.

Problem solving.

The problem is solved, but too much waste of pills (though enough), there is no better way?

Then take three kinds of medicine, weigh 1g,2g,3g for example respectively.

We can give bottles numbered 1th, 2nd and 3rd, respectively.

Now take out 1 tablets from bottle 1th, 2nd bottle take out 4 tablets, 3rd bottle remove 16 pieces, weigh on the scale, and then the weight in four-digit notation.

Then the first digit on the right is the weight of the 1th-bottle pill, the second digit from the right is the weight of the 2nd-bottle pill, and the third digit from the right is the weight of the 3rd-bottle pill.

Problem solving.

Why is this?

The decimal 1 corresponds to the Quad 1, the decimal 4 corresponds to the Quad 10, and the decimal 16 corresponds to the four-binary 100

This is the same as taking 1, 10,100 pieces, and the last three digits of the decimal number to indicate no essential difference.

Why use the four-step system?

This is because the maximum weight of the pill is 3, and there is no one in the quad to move forward, so that each digit corresponds exactly to the weight of a bottle of pills.

Similarly, if there is a total of four kinds of medicine, the weight 1g,2g,3g,4g respectively.

We can give bottles numbered 1th, 2nd, 3rd and 4th, respectively.

Now take out 1 pieces from bottle 1th, 2nd bottle take out 5 tablets, 3rd bottle take out 25 pieces, 4th bottle Remove 125 pieces, weigh on the scale, and then the weight of five decimal numbers, you can read from right to left the weight of the pill 1 to 4th bottles.

Through the above discussion, the solution of this kind of problem can be obtained:

If there is a total of n drugs, placed in n bottles, and the heaviest pill weight m, all pills weight is a positive integer,

Then we can give the bottle numbers from 1th to N, respectively.

Now from 1th bottle out of 1 pieces, 2nd bottle take Out (m+1) tablets, 3rd bottle take Out (m+1) ^2 tablets, N bottle take Out (m+1) ^ (N-1) tablets, weigh the weight with (m+1) into the number, you can read from right to left the weight of the pills in the 1 to N bottles.

In fact, the result of this, the number of this (m+1), each digit is different, because the weight of each pill is different.

However, if the first digit on the right is the same as the second digit on the right, it means that bottle No. 1th is the same weight as the pill in bottle 2nd, and under the condition of the case, it is indicated that bottle 1th and bottle 2nd are the same medicine.

In other words, the above calculation method is suitable for the same pill in different bottles of the case.

So if there is a total of M bottles of N-type drugs, if the m>n is the above situation. (if M in summary, if a total of M bottles of N-type drugs, and the weight of the heaviest tablets in the N, all the weight of the pills are a positive integer,

Then we can give the bottle number 1th to M respectively.

Now remove 1 pieces from bottle 1th, 2nd bottle take Out (a+1) tablets, 3rd bottle take Out (a+1) ^2 tablets, n bottle out (a+1) ^ (m-1) tablets, on the scale, and then the weight with (a+1) into the number, you can read from right to left 1 to M bottle of the weight of the pill. Thus distinguishing the kinds of Chinese medicine in each bottle.

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Summary of Intellectual problems (4)