Decimal professional network bridge 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: (m) Passing time: (min) A: It takes 17. 1 min for this train to pass 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: (m) Train speed: (m) A: 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. Total distance: cave length: (meters) A: This cave is 60 meters long. Roi and his mother's age add up to 40 years old, and his mother's age is four times that of 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 the multiple of mother's age is: 4+ 1 = 5 (times) (2) the age of ROI is 40 ÷ 5 = 8 years old (3) the age of mother is 8× 4 = 32 years old. Comprehensive: 40 ÷ (4+65438+). 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 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 formula: 4. Two grain depots A and B originally stored 170 tons of grain, and then 30 tons were transported from warehouse A to warehouse B, and 10 tons were transported to warehouse B. At this time, the grain in warehouse A was twice as much as that in warehouse B. How many tons were originally stored in the two grain depots? 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 problems of the system of equations (1) 1. Tinplate can be made, and each tinplate can be made into 16 boxes or 43 box bottoms, and one box and two box bottoms can be made into a can. At present, there are 150 pieces of iron sheet. How many pieces of iron 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 as follows: number of boxes made+number of box bottoms = total number of boxes made of iron sheets b ×2= number of box bottoms: 86 pieces of iron sheets are used for the box and 64 pieces of iron sheets are used for the box bottoms. Odd and even numbers (1) In fact, students have come across many odd and even numbers in their daily life. 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, and the common ones are: the sum or difference of two even numbers with the property 1 is still even. 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 the heads of five cards can be turned down only when the total number of flop cards is odd. 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 spots in box A decreases 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. Special topic of the Olympic Games-the problem of weighing balls (for example 1) has four piles of balls with the same appearance, and each pile has four balls. 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 the two groups A and B on two plates of the balance, and then (1) if A=B, both A and B are genuine, then weigh 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 B > C. (2) If A > B, 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. Special topic of the Olympic Games-pigeon hole principle case 1 A group of * * * 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, make five drawers in five colors. According to the pigeon hole principle 1, as long as you take out six socks, there will always be two in a drawer, and these two can make a pair. If you take this pair, there are four pairs left. Add two more to make six, and then according to the principle of pigeon nest 1, you can make a pair and take it away. If you add two more pairs, you can get a third pair. So at least 6+2+2 = 10 pairs of socks will make three pairs. Thinking: 1. Can I use Pigeon Hole Principle 2 to get the result directly? 2. Change the requirements in the question to 3 pairs of socks with different colors. How many pairs of socks should I take out at least? How about changing the requirements in the question to three pairs of socks of the same color? There are 35 wooden balls of the same size in a cloth bag, including 10 white, yellow and red balls, 3 blue balls and 2 green balls. How many balls can be taken out at a time to ensure that at least 4 balls are of the same color? Analyze and solve the most unfavorable takeaway situation. The most unfavorable situation is that of the five balls taken out first, three are blue balls and two are green balls. Next, consider white, yellow and red as three drawers. Because these three colors of balls are equal to more than four, according to Pigeon Hole Principle 2, as long as the number of balls taken out is greater than (4- 1)×3=9, that is, at least 10 balls must be taken out, it can be guaranteed that at least four balls are in the same drawer (with the same color). So the total * * * must take out at least 10+5 = 15 balls to meet the requirements. Thinking: How about changing the requirements in the question to four different colors, or two colors with the same color? When we encounter the problem of "judging whether something is essential, at least several", think about it-pigeon hole principle, which is your way to "win". Special topic of the Olympic Games-reduction example 1 A person goes to the bank to withdraw money. The first time, he used 50 yuan to withdraw more than half of his deposit, and the second time, he used the remaining 100 yuan to withdraw more than half. At this point, there is 1250 yuan left in his passbook. What was his initial deposit? Analysis From the above-mentioned "repackaging" case, we should be inspired: if we want to restore, we have to do the opposite (backward). According to "the remaining half exceeds 100 yuan for the second time" and "the remaining half is less than 100 yuan" is 1250 yuan, so the remaining half is1250+100 =100. The comprehensive formula is [(1250+100) × 2+50 ]× 2 = 5500 (yuan). The general feature of reduction problem is that the results of four operations of a certain number are known, or the results of increasing or decreasing a certain number need the initial number (before the operation or before the increase or decrease). To solve the reduction problem, it is usually necessary to carry out the corresponding inverse operation in the opposite order of operation or increase or decrease. There are 26 bricks in Example 2. The two brothers scrambled to pick, and the younger brother scrambled to take the lead. Just as the brick was laid, my brother came. My brother saw that my brother picked too much, so he took half of it himself. The younger brother thought he could do it, so he took half from him. My brother wouldn't let me, so he had to give him 5 pieces, so my brother picked 2 pieces more than his brother. How many pieces was my brother going to choose at first? We have to figure out how many pieces our brothers will choose in the end. As long as we solve a sum and difference problem, we will know that my brother chooses "(26+2)÷2= 14" and my brother chooses "26- 14= 12". Tip: The corresponding "inverse operation" to solve the problem of reduction means: addition is reduction by subtraction, subtraction is addition, multiplication is division, and division is multiplication, which is originally addition (subtraction), and should be subtraction (addition), multiplication (division) and division (multiplication). For some complex reduction problems, we should learn to list and use tables to push backwards, which can not only clarify the quantitative relationship, but also facilitate checking. Olympic Special Topic-Chicken and Rabbit in the Same Cage: 1 Chicken and rabbit in the same cage, head ***46, feet *** 128. How many chickens and rabbits are there? [Analysis]: If there are all 46 rabbits, a * * should have 4×46= 184 feet, which is more than the known 65,438 feet 184- 128 = 56 feet. If the rabbit is replaced by a chicken, it will be reduced by 4-. Obviously, 56÷2=28, but 28 rabbits were replaced by 28 chickens. So the number of chickens is 28, and the number of rabbits is 46-28= 18. Solution: ① How many chickens are there? (4× 6-128) ÷ (4-2) = (184-128) ÷ 2 = 56 ÷ 2 = 28 (only) ② How many? 46-28= 18 (only) A: There are 28 chickens and 18 chickens. There are 100 chickens and rabbits, and chickens have 80 more feet than rabbits. How many chickens and rabbits are there? [Resolution]: This example is different from the previous example. What it gives is not the sum of their feet, but the difference of their feet. How to solve this? Assuming that 100 chickens are all chickens, the total number of feet is 2× 100=200 (only one). At this time, the number of rabbit feet is 0, and chicken feet are 200 more than rabbit feet, but in fact, chicken feet are 80 more than rabbit feet. So the difference between chicken feet and rabbit feet is much more than known (200-80) = 65430. The number of rabbit feet decreased by 4. Then, the difference between chicken feet and rabbit feet is increased by (2+4)=6 (only), so the number of chickens replacing rabbits is 120÷6=20 (only). There are chickens (100-20)=80 (only). Solution: (2× 100-80)÷(2+4)=20 (only). 100-20=80 (only). A: There are 80 chickens and 20 rabbits. There are three classes *** 135 students in the third grade of Hong Ying primary school. Class two has five more students than Class one, and Class three has seven fewer students than Class two. How many students are there in each class? [analysis 1] We assume that there are three classes with the same number of students, so it is easy to ask how many students there are in each class. Therefore, if there are three classes with the same number of students, can we analyze and solve them? Consider the following figure. If the number of people in Class Two and Class Three is the same as that in Class One, the number of people in Class Two will be 5 less than the actual number, and the number of people in Class Three will be 7-5=2 (people) more than the actual number. Then, please calculate, assuming that the number of people in Class 2 and Class 3 is the same as that in Class 1, what should be the total number of people in the three classes? Solution 1: the first category: [135-5+(7-5)] ÷ 3 =132 ÷ 3 = 44 (person) the second category: 44+5=49 (person) the third category. [Analysis 2] Assuming that there are as many people in Class 1 and Class 3 as in Class 2, there are 5 more people in Class/kloc-0 and 7 more people in Class 3. What is the total this time? Solution 2: (135+5+7) ÷ 3 =147 ÷ 3 = 49 (person) 49-5=44 (person), 49-7=42 (person) Answer: Class One, Class Two, Class Three, Grade Three. Example 4 Teacher Liu took 4 1 students to go boating in Beihai Park, and * * * rented a boat for 10. Each big boat takes six people and each small boat takes four people. How many boats did you rent? [Analysis] Let's consider it step by step: ① Assume that 10 charters are all large ships, and the ships have to take 6× 10 = 60 (people). ② Assume that the total number of people is 60-(41+1) =18 (people) more than the actual number. The reason for the increase is that it is assumed that all four people on board are six. (3) When a ship is a big ship, there are two more people, and the extra 18 people is 18÷2=9 (a ship) as a big ship. Solution: [6×10-(41+1) ÷ (6-4) =18 ÷ 2 = 9 (bars)10-9 =/kl.