Primary School Mathematics Olympiad Problem 3 , the process should be analyzed in detail. 20

In the first encounter, Xiao Zhang walked a kilometer, and in the second encounter, Xiao Zhang walked a distance of +3 kilometers, and the distance traveled by Xiao Zhang was 3 times that of the first encounter, that is, a kilometer.

Distance between A and B = km.

Xiao Zhang speed: Xiao Wang speed = :(: 8

Each time they met, they walked twice as far as they met, and twice as far as they did when they first met.

During the fourth encounter, Xiao Zhang walked 7 times the distance of the first encounter, that is, kilometers, at this time, Xiao Zhang was 2 kilometers away from place B, a kilometer away from place A.

Solution: When Xiao Zhang met for the first time, he traveled for a kilometer, and when he met for the second time, Xiao Zhang should travel another kilometer, then he passed through B and was 3 kilometers away from B, so the whole journey was a kilometer, and when he met for the fourth time, Xiao Zhang traveled a kilometer, that is, he traveled 3 whole journeys + 2 kilometers (2 kilometers from B and B), and a kilometer away from A place.

Two athletes, Xiao Zhang and Xiao Wang, train in race walking, with Xiao Zhang starting from A and Xiao Wang starting from B at the same time, walking back and forth between the two places (returning immediately after arriving at the other place). They met for the first time a kilometer from A and a second time when Zhang left B and was 3,000 meters away from B. How many kilometers away from Adi when they meet for the fourth time?

Solution: The total weight of the luggage of 3 people exceeds the weight of 35 yuan for one person, and it is divided into 3 people, and the amount of money that exceeds the weight = 3 + 5 + 7 = 15 (yuan) so if the free weight of 1 person is counted as overweight, the amount of money required = (35-15) 2 = 10 (yuan).

So the ratio of the baggage weight of 3 people = (10 + 3) :( 10 + 5) :(10 + 7) = 13:15:17, so A's baggage weight = 26 kg.

B's baggage weight = 30 kg.

C's baggage weight = 34 kg.

1. Solution: If one person carries it, you need to pay a baggage fee of 35 yuan.

And divided into 3 people, the amount of money exceeding the weight = 3 + 5 + 7 = 15 (yuan).

Therefore, if the free weight of 1 person is counted as overweight weight, the amount of money required = (35-15) 2 = 10 (yuan).

Therefore, if the luggage is not free, it costs 35 + 10 = 45 yuan, which is 45 90 = yuan (yuan per kilogram of luggage).

So A's baggage: (10+3) kg.

B's baggage: (10+5) kg.

C's baggage: (10+7) kg.

2. Solution: If one person carries it, you need to pay a baggage fee of 35 yuan.

And divided into 3 people, the amount of money exceeding the weight = 3 + 5 + 7 = 15 (yuan).

Therefore, if the free weight of 1 person is counted as overweight weight, the amount of money required = (35-15) 2 = 10 (yuan).

Therefore, the ratio of the baggage weight of 3 people = (10 + 3) :(10 + 5) :(10 + 7) = 13:15:17, that is, 90 is divided into 13 + 15 + 17 = 45 parts, so A's baggage weight = 90 * (13 45) = 26 kg.

B's baggage weight = 90 * (15 45) = 30 kg.

C's baggage weight = 90 * (17 45) = 34 kg.

Each person can carry a free weight of x kg, and for each additional 1 kg, it costs y yuan for 90-3x = (3+5+7) y

90-x = 35/y

So x= 20

y=So the baggage weight of belt C is 20+7 =34(kg).

The baggage fee for the free weight portion per person is (35-3-5-7) 2=10 yuan.

90 (35+10)=2 kg.

C's baggage weight (10 + 7) * 2 = 34 kg.

Solution: The weight of A is the weight of one free baggage plus the excess baggage weight of 3 yuan.

The weight of B and C is one free baggage weight plus the excess baggage weight of 5 yuan, and one free baggage weight plus the excess baggage weight of 7 yuan.

Then the total weight of the three people is 3 free baggage weight plus the excess baggage weight of 15 yuan, according to the known, the total weight is 90kg, that is, the weight of 3 free baggage plus the excess baggage weight of 15 yuan = 90kg, then it can be easily obtained: the total weight of 1 free baggage plus 5 yuan of baggage fee is 30kg, and according to the weight of three people, the baggage fee to be paid is 35 yuan, which can be understood as a free baggage weight plus the total weight of the excess baggage weight of 35 yuan =90kg

The above comparison shows that the free baggage weight can be exceeded by (90-30) (35-5) = 2kg for every dollar paid for baggage

Then the free baggage weight is 20kg

The weight of the baggage with belt C is 20 + 7 * 2 = 34kg

270 + 58 + 58 = 386 people did not decide on newspapers and periodicals.

510 + 330 + 120 = 960 people subscribed.

Among them, 270 people were ordered twice, 58 people were ordered three times, 960-270-58*2=574 people.

So the number of subscribers was 574.

Undecided 960-574 = 386 people.

Solution: Let a {students who subscribe to the "Essay Newspaper"}, b = {students who subscribe to the "Math Newspaper"}, c = {students who subscribe to the "Science and Technology Newspaper"}, i = a b c - a b - a c - b c a b c = 510 + 330 + 120-270 + 58 = 748 (people).

Students who do not subscribe to any newspapers are: i- a b c = 960-748 = 212 (person).

Applying the "repulsion principle" formula, a b c is a student who subscribes to at least one newspaper) Answer: There are 212 students in the school who do not subscribe to any newspaper.

A total of 510 + 330 + 120 = 960 copies were subscribed by the title.

The number of people who set a report is 960-58*3-270*2=246 people.

The number of unsubscribed people is 960-58-270-246 = 386.

270x2 + 58x3 + one type of subscription = 510 + 330 + 120

Derives a = 246

Not ordered = 960-246-270-58 = 386

270x2 + 58x3 + one subscription = 510 + 330 + 120 one subscription = 246 no subscription = 960-246-270-58 = 386

A total of 960 copies were subscribed, minus 270 because 270 people had subscribed to an extra newspaper, and minus 2*58 because 58 people had ordered two more newspapers.

574 ordered newspapers and 386 did not order.

Since the speed ratio of the hour and minute hands is 5:60=1:12, this clock is only (61+4 11) (12-1) 12=minutes per hour, so this clock is slower than standard time.

Slow (minutes) per day.

Because the speed ratio of the hour and minute hands is 5:60 = 1:12

61+4 11) (12-1) 12=minutes (minutes.)

The principle of addition, we take the remaining 5 rounds and give it a number.

In these 5 games, the first team only needs to win 2 more games to end, so the number of games he can win is:

1 and 5 (four kinds).

2 and 5 (three) 3 and 5, (two) 4 and 5 (one).

Every winning situation is different. But he was guaranteed to win, as Yidu only won three games.

There are a total of 4+3+2+1 combinations of 4+3+2+1 combinations: there is a feature that two elements are selected from 5 elements, so after taking any element, there are still four elements left to combine with it.

There are four scenarios. As.

1 and 2, 1 and 3, 1 and 4, 1 and 5.

2 and 1, 2 and 3, 2 and 4, 2 and 3.

3 and 1, 3 and 2, 3 and 4, 3 and 5

4 and 1, 4 and 2, 4 and 3, 4 and 5

5 and 1, 5 and 2, 5 and 3, 5 and 4

There are five elements, each of which has four combinations. There are 5 (5-1)=20 species.

But among the 20 types, 1 and 2 are the same as 2 and 1, and although the order is different, the result is the same. So half of these 20 are the same.

So the combinatorial principle calculation is: 5 (5-1) 2

Personal understanding - 1: To win the fourth game, the third game must be won, there is only one case; 2: To win the fifth game, then the first.

There are three or four games and only one game wins, 2 takes 1 for 2; 3: To the sixth game of victory, then the first.

3, 4 and 5 rounds take 1 as 3; 4: To win the seventh game, then the first.

The three, four, five, and six rounds of 4 take 1 as 4. Total 4+3+2+1 10.

In fact, it is a wrong way to use the principle of addition or multiplication!

This kind of practical problem generally needs to consider the meaning of existence: for this question, if the game is over after 4 games, (that is, Party A wins 4:0), then the next few games do not need to be played, that is to say, the next few games have no meaning of existence!

Therefore, the idea of solving this problem is: classification discussion (how many rounds of the game is completed as the classification standard) 4 rounds: 1 kind.

5 rounds: 2 types.

6 rounds: 3 types.

7 rounds: 4 types.

To sum up: there are a total of 10 possibilities.

How to explain when adding.

If the third game is won, then one game can be won, there are 4 types if the fourth game is won, then one game can be won, and there are 3 types if the fifth game is won, then one game can be won, and there are 2 types.

If you win the sixth game, then 7 wins are enough, and there is 1 kind.

A total of 1+2+3+4=10 kinds.

When multiplying, how do you explain 5 and (5-1) respectively?

The first win can be selected in a game, there are 5 kinds of second wins in the remaining 4 rounds to choose one, so there are 5 * 4 = 20 kinds but the first time to choose 3, the second time to choose 4 and the first time to choose 4, the second time to choose 3 The situation is actually the same, that is, it is double-counted, so it is necessary to divide by 2, that is, 5 * 4 2 = 10 kinds.

The most straightforward way to do this is to combine them yourself, and the remaining 5 games are represented by 1, 2, 3, 4, and 5, and if you want to get two wins, there are.

1,2), (1,3), (1,4), (1,5), (2,3), (2,4), (2,5), (3,4), (3,5), (4,5) These possibilities are a total of 10 situations.

Addition: 3rd game A wins, A wins 4 possibilities, 3rd game A loses 4th game, there are 3 possibilities, 3rd and 4th game loses, 5th game wins, there are 2 possibilities, 3rd, 4th, 5th game wins, 6th game wins, there are 1 possibility, (4+3+2+1=10) Multiplication: 5 games win 2 games, 5 games A wins one game, there are 5 possibilities, win another game, to win one game in the remaining 4 games, there are (5-1) = 4 possibilities considering that the 2 games win and the order has nothing to do with dividing by 2, 5 (5-1) 2,

9。Answer: Suppose A did day a, then B did day 16-a.

The quantity is fixed, assumed to be 1

Then the velocity of A is 1 12 per day, and the velocity of B is 1 18 per day, which can be columnar: 1 12*x+ 1 18*(16-x)=1, which gives 1 48*x=2 18

x=6 (days).

10。Answer: Suppose "Fifth Grade" is book A, and "Sixth Grade" is (120-a) book (120-a)*5-a*

This gives a=50

50 books for 5th grade and 70 books for 6th grade.

9.Solution: The armor has been dry for x days.

-x+--16-x）=1

x = 410 solution: set five years x book.

x=50

If you first press 15 per box, then the last 7 boxes should be added to each box 2 more, we can first convert this sentence into "if you first press 15 per box, then 15-2 * 7 = 1 (pcs)".

If there are 12 in each box, there will be 11 more, which translates into "if there are 12 in each box, 12-11 less = 1 (pcs)".

So now, one less for each box, so we can solve it like this:

The least common multiple is 180, and it is less than 1, and then 180-1 = 179, but 179 is not "more than three hundred", so it is multiplied, 180 * 2-1 = 359 (one) (note to multiply first and then subtract) to meet the requirements of "more than three hundred", so the answer is 359.

If you add 1, you are a multiple of 180

The minimum is 180-1=179.

179 + 180 = 359.

Theoretically, this is the difference between theoretical mathematics and life mathematics. In other words, the raw material of 1 rice can be cut into 2 rice and 1 rice. You can analyze it slowly in this way.

Rice: If you buy 50 sticks, you will have 100 sticks of rice and 50 sticks of rice.

Buy another 25 and you will have 50 sticks of rice, 50 sticks of rice.