Small topics in mathematics

Bridge crossing problem (1)

1. A train passes the 6700-meter-long Nanjing Yangtze River Bridge. The train conductor140m, and the train runs 400m per minute. How many minutes does it take for this train to cross the Yangtze River Bridge?

Analysis: This question is about transit time. According to the quantitative relationship, we know that if we want to find the transit time, we must know the distance and speed. Distance is the length of bridge plus the length of car. The speed of the train is a known condition.

Total distance: 6700 meters

Passing time: (6700+140) ÷ 400 =17.1(minutes)

A: It takes 17. 1 minute for this train to cross the Yangtze River Bridge.

A train is 200 meters long, and it takes 30 seconds for the whole train to cross a 700-meter-long bridge. How many meters does this train travel per second?

Analysis and solution: this is a bridge-crossing problem for speed. We know that if we want to find speed, we need to know the distance and the elapsed time. The distance can be calculated by using the known conditions of bridge length and vehicle length, and the transit time is also a known condition, so the vehicle speed can be calculated conveniently.

Total distance: 700 meters.

Train speed: 200+700=900 meters, 900÷30=30 meters.

This train runs 30 meters per second.

A train is 240 meters long. This train runs 15 meters per second. It takes 20 seconds from the front of the train to the whole car leaving the cave. How long is this cave?

Analysis and solution: the train crossing the cave is the same as the train crossing the bridge. When the locomotive enters the cave, it is equivalent to the locomotive getting on the bridge; The whole car out of the hole is equivalent to getting off the bridge at the rear. Finding the length of the cave in this problem is equivalent to finding the length of the bridge. We must know the total distance and length of the car. The length of the car is a known condition, so we must use the speed and transit time given in the question to calculate the total distance.

This cave is x meters long.

(X+240)/ 15=20

X+240=300

X = 60

This cave is 60 meters long.

And folding problems.

1.Roi is 40 years old with his mother, who is four times as old as Roi. How old are Roi and his mother?

Let's take the age of Roi as 1 times, and "the age of mother is four times that of Roi", then the sum of the age of Roi and mother is equivalent to five times that of Roi, that is, (4+ 1) times, and it can also be understood that five copies are 40 years old. So how many times is 1, and how many times is it four times?

The sum of (1)Roi and his mother's age multiple is: 4+ 1 = 5 (times).

(2) Age of ROI: 40 ÷ 5 = 8 years old.

(3) Mother's age: 8× 4 = 32 years old.

Comprehensive: 40 ÷ (4+ 1) = 8 years old and 8× 4 = 32 years old.

To ensure the correctness of this question, please verify that

(1) 8+32 = 40 years old (2) 32 ÷ 8 = 4 (times)

The calculation results meet the requirements, so the problem is correct.

2. Two planes A and B fly in the opposite direction from the airport at the same time, flying 3600 kilometers in three hours. The speed of A is twice that of B. What are their speeds respectively?

Knowing that two planes fly 3600 kilometers in three hours, we can find out the flight distance of two planes per hour, that is, the speed sum of two planes. As can be seen from the figure, the sum of this speed is equivalent to three times the speed of plane B, so that the speed of plane B can be calculated, and then the speed of plane A can be calculated according to the speed of plane B.

Planes A and B travel at speeds of 800 kilometers and 400 kilometers per hour respectively.

3. My brother has 20 extracurricular books and my brother has 25 extracurricular books. How many extra-curricular books did my brother give him? His extra-curricular books are twice as many as his brother's?

Thinking: (1) What is the same number of topics before and after my brother gives my brother extra-curricular books?

(2) What conditions do I need to know if I want to ask my brother how many extracurricular books to give my brother?

(3) If the extracurricular books left by my brother are regarded as 1 time, how many times can my brother's extracurricular books be regarded as the extracurricular books left by my brother?

On the basis of thinking about the above problems, ask my brother how many extracurricular books he should give his brother. First check how many extracurricular books my brother has left according to the conditions. If we regard my brother's extracurricular books as 1 time, then my brother's extracurricular books can be regarded as twice as many as my brother's extracurricular books, that is to say, some multiples of the two brothers are equivalent to three times that of my brother's extracurricular books, and the total number of extracurricular books of the two brothers is always the same.

(1) The number of extracurricular books owned by the two brothers is 20+25 = 45.

(2) After the elder brother gave his younger brother several extracurricular books, some multiples of the two brothers were 2+ 1 = 3.

(3) The number of extracurricular books left by my brother is 45 ÷ 3 = 15.

(4) The number of extracurricular books given by elder brother to younger brother is 25- 15 = 10.

Try to list the comprehensive formulas:

4. The two grain depots A and B originally stored 170 tons of grain, and then transported 30 tons from the depot A to the depot B 10 tons. At this time, the stored grain of A is twice that of B. How many tons did the two grain depots originally store?

According to the two grain depots A and B, the original grain storage was 170 tons, and then 30 tons were transported from the depot A to the depot B 10 tons. At this time, how many tons of grain were stored in the two depots * * *. According to "At this time, A's grain storage is twice that of B", if B's grain storage is 1 times, then A and B's grain storage is equivalent to 3 times that of B. So find out how many tons of grain storage B has at this time, and then find out how many tons of grain storage B has. Finally, we can find out how many tons of grain warehouse A originally stored.

Warehouse A originally stored 130 tons of grain, and warehouse B originally stored 40 tons of grain.

Solve the application problem of equations (1)

1. tin can be made, and each tin can be made into 16 boxes or 43 boxes. A box of two boxes can be made into a can. There are 150 pieces of tin at present. How many pieces of tin can be used to make the box body and the bottom just match?

According to the meaning of the question, this question has two unknowns, one is the number of iron pieces in the box and the other is the number of iron pieces at the bottom of the box, so it can be expressed by two unknowns. To ask for these two unknowns, we must find two equal relationships from the problem, list two equations, and combine them to form an equation.

The equivalent relationship between them is: the number of sheets in a box+the number of sheets at the bottom of a box = the total number of iron sheets.

B number of boxes manufactured ×2= number of boxes manufactured.

86 pieces of tinplates are used as the box body and 64 pieces of tinplates are used as the box bottom.

Odd and even numbers (1)

In fact, in daily life, students have been exposed to many odd numbers and even numbers.

Any number divisible by 2 is called even number, and even numbers greater than zero are also called even numbers; All numbers that are not divisible by 2 are called odd numbers, and odd numbers greater than zero are also called odd numbers.

Because even numbers are multiples of 2, this formula is usually used to represent even numbers (in this case, integers). Because any odd number divided by 2 is 1, odd numbers (here, integers) are usually expressed by formulas.

Odd and even numbers have many properties, the common ones are:

The sum or difference of two even numbers of the attribute 1 is still an even number.

For example: 8+4= 12, 8-4=4, etc.

The sum or difference of two odd numbers is also an even number.

For example: 9+3= 12, 9-3=6, etc.

The sum or difference between odd and even numbers is odd.

For example: 9+4= 13, 9-4=5, etc.

Is odd sum odd, odd sum even, even sum even.

Property 2 The product of odd number and odd number is odd number.

The product of even number and integer is even number.

Attribute 3 Any odd number cannot be equal to any even number.

1. There are 5 playing cards, and the picture is up. Xiao Ming turns four cards at a time. So, can he turn all five cards down after a few times?

Students can try. Only by flipping the card an odd number of times can its image change from top to bottom. If you want all five cards to face down, you must flip each card odd times.

The sum of five odd numbers is odd, so only when the total number of flop is odd can all five cards face down. Xiao Ming turns four pages at a time, no matter how many times, the total number of pages is even.

So no matter how many times he flips, it is impossible to make all five cards face down.

2. Box A contains 180 white chess pieces and 18 1 black chess pieces, and Box B contains 18 1 white chess pieces. Li Ping randomly draws two pieces from Box A at a time. If the two pieces are of the same color, he takes out an albino from Box B. If the two pieces are different colors, he puts the sunspot back in the armor box. So after how much he took, there was only one piece left in the armor box. What color is this one?

No matter what kind of chess pieces Li Ping takes out of the armor box, he always puts a chess piece in the armor box. So every time he takes it, the number of pieces in box A decreases by one, so after he takes180+181-1= 360 times, there is only one piece left in box A.

If he takes out two sunspots, the number of sunspots in box A will be reduced by two. Otherwise, the number of sunspots in box A remains unchanged. In other words, every time Li Ping takes out a box, the number of sunspots is even. Since 18 1 is odd, odd minus even equals odd. So the number of sunspots left in the armor box should be odd, and the odd number not greater than 1 is only 1, so the one left in the armor box should be sunspots.

Olympic Special Topic-Weighing Ball

Example 1 There are 4 piles of balls with the same appearance, 4 balls in each pile. It is known that three piles are genuine and one pile is defective. Each quality ball weighs 10g, and each quality ball weighs11g.. Please weigh it with a balance and find out the defective pile.

Solution: Take 1, 2, 3 and 4 balls from the first, second, third and fourth piles in turn. Put this 10 ball on the balance and weigh it together. The total weight is a few grams more than100g, and the first pile is defective balls.

There are 27 balls with the same appearance, and only one ball is defective, which is lighter than the genuine one. Please weigh it with the balance only three times (no weight) to find out the defective ball.

Solution: the first time: divide 27 balls into three piles, each pile has 9 balls, and take two of them and put them on two plates of the balance. If the balance is unbalanced, you can find a lighter pile; If the balance is balanced, then the remaining pile must be lighter, and the defective products must be in the lighter pile.

Second time: Divide the pile judged to be lighter for the first time into three piles, each with three balls, weigh two piles according to the above method, and find out the pile with lighter defective products.

Third pass: Take out two of the three lighter balls found in the second pass and weigh them once. If the balance is unbalanced, the lighter ball is defective. If the balance is balanced, the remaining one that is not weighed is defective.

Example 3 Take 10 balls with the same appearance, and only one ball is defective. Please weigh it three times with the balance to find out the defective products.

Solution: Divide 10 balls into three, three and 1 4 groups, and express the four groups of balls and their weights as A, B, C and D respectively. Weigh Group A and Group B on two plates of the balance, and then

(1) If A=B, both A and B are genuine, and then called B and C. If B=C, it is obvious that the ball in D is defective; If B > C, the defective product is in c, and the defective product is lighter than the genuine product. Then take out two balls in C and weigh them, and you can draw a conclusion. If b < c, we can also draw a conclusion by imitating the situation of b > C.

(2) if A > B, then both c and d are credible. If b and c are called again, there can be no B=C or B < C (B > C). Why? If B=C, the defective product is in A, and the defective product is heavier than the genuine product. Then take out two balls in A and weigh them, and you can draw a conclusion. If b < c, you can also draw a conclusion before imitation.

(3) if a < b, similar to the case of a > b, we can draw a conclusion through analysis.

Olympic Special Topic-Pigeon Cage Principle

Example 1 A group has 13 students, at least two of whom have birthdays in the same month. Why?

Analysis shows that there are 12 months in a year, and anyone's birthday must be in one of these months. If this 12 month is regarded as 12 drawers, the birthdays of 13 students are regarded as 13 apples, and 13 apples are put into 12 drawers, then there must be at least one drawer.

Example 2 Any four natural numbers, in which the difference between at least two numbers is a multiple of 3. Why is this?

Analysis and solution must first understand a law. If the remainder of two natural numbers divided by 3 is the same, then the difference between the two natural numbers is a multiple of 3. The remainder of any natural number divided by 3 is either 0, 1, or 2. According to these three situations, natural numbers can be divided into three categories, which are the three "drawers" we want to make. We regard four numbers as "apples". According to the pigeon hole principle, there must be at least two numbers in a drawer. In other words, four natural numbers are divided into three categories, at least two of which are the same category. Because they belong to the same category, the remainder of these two numbers divided by 3 must be the same. Therefore, the difference between any four natural numbers and at least two natural numbers is a multiple of 3.

There are 15 pairs of socks of the same size and five colors mixed in the box. How many socks can you take out of the box at least to ensure three pairs of socks (socks are left and right)?

Analysis and solution Imagine taking six or nine socks out of the box and making three pairs of socks. The answer is no.

Engineering problems

1, 1. Party A and Party B cooperate to complete a work. Because of good cooperation, Party A's work efficiency is improved by one tenth, Party B's work efficiency is improved by one fifth, and Party A and Party B cooperate for four hours to complete two fifths of all the work. The next day, B worked alone for another 4 hours, and there was still 13/30 unfinished. How many hours does it take to finish the work alone?

Solution: When Party B works alone for 4 hours1-2/5-13/30 = 3/5-13/30 =1/6.

Party B's work efficiency =( 1/6)/4== 1/24.

B One person needs to do it 1/( 1/24)=24 hours.

The work efficiency of B is improved 1/5, and then it is (1/24) x (1+1/5) =1/20.

The sum of the improved work efficiency of both parties = (2/5)/4 =110.

Then the improved work efficiency a =110-1/20 =1/20.

A original work efficiency = (1/20)/(1+110) =1/22.

A it takes one person to do it 1/( 1/22)=22 hours.

2. A project can be completed by two people, A and B, in six days. If A works for 3 days first and B works for 7 days, it can be finished. B How many days does it take for a person to finish this project?

AB cooperation can be completed every day 1/6.

A does it for 3 days first, B does it for 7 days,

It can be seen as AB cooperation for 3 days, and B alone 7-3=4 days.

AB cooperation can be completed in 3 days: 1/6×3= 1/2.

B did it alone for 4 days and finished 1- 1/2= 1/2.

B Do it alone and finish it every day: 1/2÷4= 1/8.

B It takes 1÷ 1/8=8 days to complete it alone.

3. Both Party A and Party B start to climb the mountain from the foot of the mountain at the same time, and go down immediately after reaching the top of the mountain. Both of them went down the mountain at twice the speed of going up the mountain. When A reached the top of the mountain, B was 400 meters away. When A returns to the foot of the mountain, B just goes halfway down the mountain to find the distance from the foot of the mountain to the top of the mountain.

Solution: The speed of going downhill is twice that of going uphill, so suppose,

Downhill is also regarded as uphill, and its length is 1/2 of uphill.

Speed is the speed of going up the mountain.

Then, the original distance up the mountain accounts for 2/3 of the total distance.

Downhill distance accounts for 1/3 of the total distance.

A return to the foot of the mountain, Otsuichi * * * the whole process:

2/3+ 1/3× 1/2=5/6

The speed of b is 5/6 of that of a.

When A reached the top of the mountain, he walked two-thirds of the way.

B should have completed the course: 2/3×5/6=5/9.

In fact, B did 2/3 of the whole journey minus 400 meters.

So the whole journey is: 400÷(2/3-5/9)=3600 meters.

The distance from the foot of the mountain to the top of the mountain is 3600×2/3=2400 meters.

4. A project is contracted by Party A and Party B, which can be completed in 2.4 days, and requires payment of 1.800 yuan, contracted by Party B and Party C for 3 days and 3/4 days, and contracted by Party A and Party C for 2 days and 6/7 days, which requires payment of 1.600 yuan, which is guaranteed.

Total ergonomics of Party A and Party B: 1/(2 plus 2/5) = 5/ 12.

Ergonomic sum of ethylene and propylene: 1/(3 3/4) = 4/ 15.

A-C Ergonomics Sum: 1/(2 6/7) = 7/20.

Ergonomics sum of A, B and C: (5/12+4/15+7/20)/2 = 31/60.

A work efficiency: 31/60-4/15 =1/4.

Ergonomics: 3 1/60-7/20= 1/6.

Work efficiency: 31/60-5/12 =110.

What can be done in a week is a and B.

Daily project payment of both parties: 1800/(2 2/5) = 750 yuan.

Daily project cost of Party B and Party C: 1500/(3 3/4) = 400 yuan.

Daily project cost of Party A and Party C:1600/(2 7) = 560 yuan.

Daily project cost of Party A, Party B and Party C: (750+400+560)/2=855 yuan.