A certain river has two piers, the upper and lower piers, 90km apart. Every day, two passenger ships with the same speed, A and B, depart from the two piers at the same time and travel towards each other. One day, ship A dropped an object when it set off from the upstream pier. The object floated on the water and drifted down the water. It met with ship A 1km after 2 minutes. How many hours after ship B's departure is it expected to meet the object?

The basic formula for the encounter problem: encounter time = encounter distance ÷ (speed of floating object + speed of boat B). The speed of floating object here is the speed of water flow, so encounter time = encounter distance ÷ [speed + (speed of B) Ship speed - water speed)] = encounter distance ÷ ship B’s speed.

The distance traveled by the float and boat B is the distance between the two docks of 90km. Therefore, just by finding out the speed of boat B, you can find out the time when they met. We also know that the speed of ship A = the speed of ship B, so just find the speed of ship A. And because the floating object and ship A are traveling along the current and in the same direction, the speed of ship A = the quotient of the distance between the two and the traveling time.

Speed ??of ship A: 1000÷2=500 (m/min)

Encounter time: 90000÷500=180 (min)=3 (hour)

Or the speed of ship A is: 1÷(2÷60)=1÷2×60=30 (km/h)

That is, the speed of ship B is 30 kilometers/h

p>

The encounter time is: 90/30=3 (hours)

Answer: The floating object meets boat B after three hours

Mathematical Olympiad question for fifth grade primary school—— Quick calculations and clever calculations

In daily life and when solving mathematical problems, we often need to perform calculations. In mathematics class, we have learned some simple calculation methods, but if we are good at observation and diligent in thinking, there are still many problems in calculation. Being able to find more clever calculation methods will not only enable you to calculate well and quickly, but also make you smarter and more alert.

Example 1: Calculation: 9.996+29.98+169.9+3999.5

Solution: It seems that the addition in the formula cannot be calculated using the simple arithmetic method learned in mathematics class. However, this Just add a little to each of several numbers to become a whole ten, a whole hundred or a whole thousand. After these numbers are "rounded up", it will be easy to calculate. Of course, you have to remember that any increase you make when "rounding up" must be subtracted back.

9.996+29.98+169.9+3999.5

=13174000-(0.004+0.02+0.1+0.5)

=4210-0.624

=4209.376

Example 2: Calculation: 1+0.99-0.98-0.97+0.96+0.95-0.94-0.93+…+0.04+0.03-0.02-0.01

Solution: The numbers in the formula start from 1 and decrease by 0.01 until the last number is 0.01. Therefore, there are 100 numbers in the formula and the operations in the formula are to add two numbers and then subtract two. number, then add two more numbers, then subtract two more numbers...arranged in this order.

Since the arrangement of numbers and the arrangement of operations are very regular, according to the rules, we can consider adding brackets to each group of four numbers. Does the operation result of each group of numbers also have certain rules? It can be seen that the first number in each group of numbers is subtracted from the third number, and the second number is subtracted from the fourth number. Each result is 0.02, and the total is 0.04. Then, the operation of each group of numbers (that is, each bracket) The results are all 0.04. The 100 numbers in the entire calculation are divided into 25 groups, and the result is the sum of 25 0.04s.

1+0.99-0.98-0.97+0.96+0.95-0.94-0.93+…+0.04+0.03-0.02-0.01

=（1+0.99-0.98-0.97 )＋（0.96＋0.95-0.94-0.93）＋…＋（0.04＋0.03-0.02-0.01）

=0.04×25

=1

< p>If you can flexibly apply the rules of number exchange, you can also group and add brackets to calculate according to the following method:

1+0.99-0.98-0.97+0.96+0.95-0.94-0.93+…+0 .04+0.03-0.02-0.01

=1+(0.99-0.98-0.97+0.96)+(0.95-0.94-0.93+0.92)+…+(0.03-0.02-0.01)

=1

Example 3: Calculation: 0.1+0.2+0.3+…+0.8+0.9+0.10.11+0.12+…+0.19+0.20

Solution: The numbers in this formula are arranged like an arithmetic sequence, but if you look closely, you will see that it actually consists of two arithmetic sequences. 0.1+0.2+0.3+…+0.8+0.9 is the first arithmetic sequence, followed by Each number is 0.1 more than the previous number, and 0.10.11+0.12+...+0.19+0.20 is the second arithmetic sequence. Each subsequent number is 0.01 more than the previous number, so it should be divided into two Segments are calculated by summing arithmetic sequences.

0.1+0.2+0.3+…+0.8+0.9+0.10.11+0.12+…+0.19+0.20

= (0.1+0.9)×9÷2+( 0.10＋0.20）×11÷2

=4.5+1.65

=6.15

Example 4: Calculation: 9.9×9.9+1.99

p>

Solution: One of the two factors of 9.9×9.9 in the formula is expanded by 10 times, and the other factor is reduced by 10 times. The product remains unchanged, that is, this multiplication can become 99×0.99; 1.99 can be divided into 0.99+1 and, after this change, the calculation is relatively simple.

9.9×9.9＋1.99

=99×0.99＋0.99+1

=（99＋1）×0.99+1

=100

Example 5: Calculation: 2.437×36.54+243.7×0.6346

Solution: Although the two multiplication calculations in the formula do not have the same factors, the 2.437 of the previous multiplication and the following The two numbers 243.7 of a multiplication have the same digits, but the positions of the decimal points are different. If the decimal points of the two factors of a multiplication are moved by the same number of places in opposite directions, so that the two numbers become the same, then multiplication can be used The distributive law is simplified.

2.437×36.54+243.7×0.6346

=2.437×36.54+2.437×63.46

=2.437×(36.54+63.46)

=243.7

*Example 6: Calculation: 1.1×1.2×1.3×1.4×1.5

Solution: Although several numbers in the formula are an arithmetic sequence, But the calculation formula is not a summation, and the result of this calculation cannot be calculated using the method of summing arithmetic sequences.

Students who usually pay attention to accumulating calculation experience may notice that the product of the three numbers 7, 11 and 13 is 1001, and when multiplying a three-digit number by 1001, just write the three-digit number continuously. Two times is their product, for example, 578×1001=578578. This question can be calculated according to this method, and the correct number can be calculated skillfully.

1.1×1.2×1.3×1.4×1.5

=1.1×1.3×0.7×2×1.2×1.5

=1.001×3.6

=3.6036

Calculate the following questions and write down the simplified calculation process:

1.5.467+3.814+7.533+4.186

2.6.25 ×1.25×6.4

3.3.997+19.96+1.9998+199.7

4.0.1+0.3+…+0.9+0.11+0.13+0.15+…+0.97+0.99

5.199.9×19.98-199.8×19.97

6.23.75×3.987+6.013×92.07+6.832×39.87

*7.20042005× 20052004－20042004×20052005

*8． (1+0.12+0.23)×(0.12+0.23+0.34)-(1+0.12+0.23+0.34)×(0.12+0.23)

Calculate the following questions and write down the simplified calculation process ：

1.6.734-1.536+3.266-4.464

2.0.8÷0.125

3.89.1+90.3+88.6+92.1+88.9+90. 8

4.4.83×0.59+0.41×1.59-0.324×5.9

5.37.5×21.5×0.112+35.5×12.5×0.112

Fifth Grade Mathematical Olympiad Questions in the Second Volume

Name Class Score

Use a simple method to calculate the following questions.

20.36-7.98-5.02-4.36 117.8÷2.3-4.88÷023

9.56×4.18-7.34×4.18-0.26×4.18

1. There are 123 Divide the children into groups of 12 or 7 so that there is no remainder. It is also known that the total number of groups is about 15.

So, how many groups of 12 people are there? How many groups of 7 are there?

2. Zhang Ni’s average score in the five exams is 88.5 points, and the full score for each exam is 100 points. In order to make the average score reach over 92 points as soon as possible, how many more times will Zhang Ni take the exam to get a full score?

3. The sum of the ages of the father and his three sons is 108 years. If another 6 years pass, the father's age will be exactly equal to the sum of the ages of his three sons. Ask your father how old he is now?

4. Process a batch of parts. The original plan was to process 80 parts per day, and the task was completed on schedule. Due to the improved production technology, 100 pieces were actually processed every day. In this way, not only the processing task was completed 4 days ahead of schedule, but 100 more pieces were processed. How many parts do they actually machine?

5. A pool can hold 8 tons of water. The pool is equipped with an inlet pipe and an outlet pipe. When both pipes are opened, a pool of water can be drained in 20 minutes. It is known that the water inlet pipe brings in 0.8 tons of water into the pool every minute. How many tons of water does the outlet pipe discharge per minute?

6. Cut a wire into 15 sections. One section is 8 meters long, and the other section is 5 meters long. A total length of 8 meters is 3 meters longer than a total length of 5 meters. How many meters long is this wire?

7. Divide a big fish into three parts: head, body and tail. The tail weighs 4 kilograms. The weight of the head is equal to the weight of the tail plus half the weight of the body, and the body The weight is equal to the weight of the fish head plus the weight of the tail. How many kilograms does this big fish weigh?

8. The sports room needs to pay 287 yuan to buy 5 footballs and 4 basketballs, and 154 yuan to buy 2 footballs and 3 basketballs. So how much does it cost to buy a football and a basketball?

9. There are 14 pieces of RMB of 5 yuan and 10 yuan, and the price is 100 yuan. How many 5 yuan coins and 10 yuan coins are there?

10. Someone walked 30.5 kilometers from village A to village B over the top of the mountain. It took 7 hours. He walked 4 kilometers per hour up the mountain and 5 kilometers per hour down the mountain. . If the speed of going up and down the mountain remains unchanged, how long will it take to return to Village A along the original road from Village B?

11. Two people, A and B, are walking toward each other from places A and B at the same time. A rides a bicycle at 16 kilometers per hour, and B rides a motorcycle at 65 kilometers per hour. A meets B 62.4 kilometers away from the starting point. How many kilometers are there between AB and AB?

12. The tortoise and the hare raced. The hare ran 35 kilometers per minute, and the tortoise climbed 10 meters per minute. The hare took a nap on the way, and when he woke up, he found that the tortoise was already 50 meters ahead of him. How long will it take for the hare to catch up with the tortoise?

13. On a 600-meter-long circular track, a brother and a sister ran clockwise at the same starting point at the same time, meeting each other every 12 minutes. If the speed of the two people remains unchanged and they still start from the original starting point at the same time, and the brother changes to running in a counterclockwise direction, they will meet every 4 minutes. How many minutes does it take for each of them to run a lap?

14. In still water, the speeds of two ships, A and B, are 20 kilometers per hour and 16 kilometers per hour respectively. The two ships set off from a certain port along the current. B departed 2 hours earlier than A. If the water speed is Traveling 4 kilometers per hour, how many hours does it take for A to catch up with B?

15. It takes 40 seconds for a train to cross a 440-meter bridge and 30 seconds to cross a 310-meter tunnel at the same speed. What is the speed and length of the train?

16. A bookshelf is divided into upper and lower floors. The number of books on the upper floor is 4 times that of the lower floor. After taking 5 books from the lower layer and placing them on the upper layer, the number of books on the upper layer is exactly 5 times that of the lower layer. How many books are there on the lower floor?

17. There are 1,800 kilograms of goods, which are loaded on three trucks A, B, and C. It is known that the number of kilograms loaded in car A is exactly twice that of car B. Car B carries 200 kilograms more than car C. Cars A, B and C each

Inclusion and exclusion

1. There are 40 students in a certain class, 15 of whom participate in the math group and 18 in the aircraft model group. Both groups of 10 people participated. So how many people don’t participate in either group?

Explanation: The two groups *** have (15+18)-10=23 (people),

The number of people who did not participate is 40-23=17 (people)

p>

Answer: 17 people did not participate in either group.

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2. 45 students in a class took the final exam. After the results were announced, 10 students got full marks in mathematics, and 3 students got full marks in both mathematics and Chinese. , 29 people did not get full marks in these two subjects. So how many people have perfect scores in Chinese?

Answer: 45-29-13=9 (persons)

Answer: There are 9 people who got full marks in Chinese.

3. 50 students stood in a line facing the teacher. The teacher first asked everyone to count 1, 2, 3,..., 49, 50 from left to right; then asked the students who reported the number to be a multiple of 4 to turn back, and then asked the students to report the number to be a multiple of 6. Turn backwards. Question: How many students are there currently facing the teacher?

Solution: There are 12 50/4 quotients for multiples of 4, 8 50/6 quotients for multiples of 6, and 4 50/12 quotients for multiples of both 4 and 6.

The number of people who turned backward in multiples of 4 = 12, and the number of people who turned backward in multiples of 6 = 8 people, of which 4 people turned backward and 4 people turned back.

Number of students facing teachers = 50-12 = 38 (people)

Answer: There are still 38 students facing teachers.

4. At the entertainment party, 100 students drew lottery tickets with labels ranging from 1 to 100. The rules for awarding prizes based on lottery ticket tag numbers are as follows: (1) If the tag number is a multiple of 2, 2 pencils will be awarded; (2) If the tag number is a multiple of 3, 3 pencils will be awarded; (3) If the tag number is a multiple of 2, Prizes can be claimed repeatedly in multiples of 3; (4) All other tag numbers will be awarded 1 pencil. So how many pencils are there as prizes prepared by the fair for this event?

Solution: There are 50 100/2 quotients for multiples of 2, 33 100/3 quotients for multiples of 3, and 16 100/6 quotients for multiples of 2 and 3 people.

The *** preparation for receiving 2 branches (50-16)*2=68, the *** preparation for receiving 3 branches (33-16)*3=51, the *** preparation for repeated collection 16*(2+3)=80, prepare the rest 100-(533-16)*1=33

***Need 68+51+833=232 (supports)

*** p>

Answer: There are 232 prize pencils prepared by the entertainment club for this event.

5. There is a 180 cm long rope. Make a mark every 3 cm from one end and every 4 cm. Then cut the marked place. Ask how many pieces the rope was cut into?

Explanation: The mark of 3 centimeters: 180/3=60. In the end, without marking, 60-1=59 pieces

The mark of 4 centimeters: 180/4=45, 45 -1=44, repeated marks: 180/12=15, 15-1=14, so there are actually 59+44-14=89 marks in the middle of the rope.

After cutting 89 times, it becomes 89+1=90 segments

Answer: The rope was cut into 90 segments.

6. There were many paintings on display at the Donghe Primary School art exhibition, 16 of which were not from the sixth grade, and 15 of which were not from the fifth grade. Now we know that there are 25 paintings by *** in the fifth and sixth grades. How many paintings by *** in other grades?

Explanation: There are 16 in Grades 1, 2, 3, 4, and 5, there are 15 in Grades 1, 2, 3, 4, and 6, and there are 25 in Grades 5 and 6.

So the total number is (16+15+25)/2=28 (frames), and the grades 1, 2, 3, and 4 have 28-25=3 (frames)

p>

Answer: There are 3 paintings from other grades.

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7. There are a number of cards. Each card has a number written on it. It is a multiple of 3 or a multiple of 4. Among them, the card marked with a multiple of 3 There are 2/3, cards marked with multiples of 4 account for 3/4, and there are 15 cards marked with multiples of 12. So, how many cards are there in a day?

Answer: The multiples of 12 are 2/3+3/4-1=5/12, 15/(5/12)=36 (pieces)

Answer: These cards There are 36 cards in a ***.

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8. Among the natural numbers from 1 to 1000, they are neither divisible by 5 nor divisible by 7 How many are there?

Solution: There are 200 1000/5 quotients for multiples of 5, 142 1000/7 quotients for multiples of 7, and 28 1000/35 quotients for both 5 and 7. There are 20142-28=314 multiples of 5 and 7.

1000-314=686

Answer: There are 686 numbers that are neither divisible by 5 nor divisible by 7.

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9. Students in Class 3 of fifth grade participate in extracurricular interest groups, and each student participates in at least one activity. Among them, 25 people participated in the nature interest group, 35 people participated in the art interest group, 27 people participated in the Chinese interest group, 12 people participated in the Chinese and art interest groups, and 8 people participated in the nature and art interest groups. Naturally, 9 people also participated in Chinese interest groups, and 4 people participated in all three subject interest groups: Chinese, art, and science. Find the number of students in this class.

Solution: 25+35+27-(8+12+9)+4=62 (persons)

Answer: The number of students in this class is 62.

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10. As shown in Figure 8-1, it is known that the areas of the three circles A, B, and C are all 30. A and B, B and C, The areas of the overlapping parts of A and C are 6, 8, and 5 respectively, and the total area covered by the three circles is 73. Find the area of ??the shaded part.

Solution: The area of ??the overlapping parts of A, B, and C = 73+ (6+8+5)-3*30=2

The area of ??the shaded part = 73-(6 +8+5)+2*2=58

Answer: The area of ??the shaded part is 58.

________________________________________

-- Author: abc

-- Release time: 2004-12-12 15:45:02

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11. There are 46 students in a fourth grade class participating in 3 extracurricular activities. Among them, 24 people participated in the math group and 20 people participated in the Chinese group. The number of people participating in the art group was 3.5 times the number of people who participated in both the math group and the art group, and 7 times the number of people who participated in all three activities. The number of people who also participate in the Chinese group is twice the number of people who participate in all three groups. There are 10 people who participate in both the math group and the Chinese group. Find the number of people participating in the literary and art group.

Explanation: Suppose the number of people participating in the literary group is X, 24+2X-(X/305+2/7*X+10)+X/7=46, the solution is /p>

Answer: The number of people participating in the literary group is 21.

________________________________________

-- Author: abc

-- Release time: 2004-12-12 15:45:43

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12. There are 100 books in the library, and those who borrow books need to sign on the books. It is known that among the 100 books, there are 33, 44 and 55 books signed by A, B and C respectively. Among them, 29 books are signed by both A and B, and 25 books are signed by both A and C. There are 36 books signed by B and C. Ask how many books in this batch of books are at least not borrowed by anyone from A, B, or C?

Explanation: The number of books that three people have read together is: A + B + C - (A, B + A, C + B, C) + A, B, C = 33 + 44 + 55 - (29 +25+36) + A, B, and C = 42 + A, B, and C. When A, B, and C are the largest, the three of them have read the most books, because A and C have only read 25 books together, which is more than A, B, and B and C* **have read very few of them, so A, B, and C have both read at most 25 books together.

The three of them have read at most 42+25=67 (books), and they have read at least 100-67=33 (books).

Answer : There are at least 33 books in this batch that have not been borrowed by any of A, B, and C.

________________________________________

-- Author: abc

-- Release time: 2004-12-12 15:46:53

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13. As shown in Figure 8-2, five equally long line segments form a five-pointed star. If exactly 1994 points on each line segment are dyed red, then what is the minimum number of red points on this five-pointed star?

Solution: There are 5*1994=9970 red dots on the right side of the five lines. If a red dot is placed on all intersections, the red dots will be the least. These five lines have 10 intersections, so There are at least 9970-10=9960 red dots

Answer: There are at least 9960 red dots on this five-pointed star.

Related pictures for this topic are as follows:

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-- Author: abc

-- Release time: 2004-12-12 15:47:12

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14. A, B, and C water 100 pots of flowers at the same time. It is known that A has watered 78 pots, B has watered 68 pots, and C has watered 58 pots. How many pots of flowers have been watered by all three people?

Explanation: A and B must have 78+68-100=46 pots that have been watered at the same time, and C has 100-58=42 that have not been watered, so all three people have watered at least 46 -42=4 (pots)

Answer: There are at least 4 pots of flowers that have been watered by 3 people.

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-- Author: abc

-- Release time: 2004-12-12 15:52:54

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15. A, B, and C are all reading the same story book. There are 100 stories in the book. Everyone starts with a certain story and reads forward in order. It is known that A has read 75 stories, B has read 60 stories, and C has read 52 stories. So what is the minimum number of stories that A, B, and C have read together?

Explanation: B and C*** have read at least 652-100=12 stories together. A must read these 12 stories no matter where he starts.

Answer: A, B, and C*** have read at least 12 stories together.

________________________________________

-- Author: abc

-- Release time: 2004-12-12 15:53:43

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15. A, B, and C are all reading the same story book. There are 100 stories in the book. Everyone starts with a certain story and reads forward in order. It is known that A has read 75 stories, B has read 60 stories, and C has read 52 stories. So what is the minimum number of stories that A, B, and C have read together?

Explanation: B and C*** have read at least 652-100=12 stories together. A must read these 12 stories no matter where he starts.

Answer: A, B, and C*** have read at least 12 stories together.

________________________________________

-- Author: cxcbz

-- Release time: 2004-12-13 21:53:23

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The following is a quote from abc’s speech at 2004-12-12 15:42:17:

8. Among the natural numbers from 1 to 1000, they cannot be divided by 5. , how many numbers are there that are not divisible by 7?

Solution: There are 200 quotients of 1000/5 for multiples of 5, 142 quotients of 1000/7 for multiples of 7, and 28 quotients of 1000/35 for multiples of both 5 and 7. There are 20142-28=314 multiples of 5 and 7.

1000-314=686

Answer: There are 686 numbers that are neither divisible by 5 nor divisible by 7.

The division in the question should be integer division.

________________________________________

-- Author: cxcbz

-- Release time: 2004 -12-13 21:56:00

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The following is a quote from abc’s speech at 2004-12-12 15:45:02:

11. There are 46 students in a fourth grade class participating in 3 extracurricular activities. Among them, 24 people participated in the math group and 20 people participated in the Chinese group. The number of people participating in the art group was 3.5 times the number of people who participated in both the math group and the art group, and 7 times the number of people who participated in all three activities. The number of people who also participate in the Chinese group is twice the number of people who participate in all three groups. There are 10 people who participate in both the math group and the Chinese group. Find the number of people participating in the literary and art group.

Explanation: Suppose the number of people participating in the literary group is X, 24+2X-(X/305+2/7*X+10)+X/7=46, the solution is /p>

Answer: The number of people participating in the literary and art group is 21.

1. In the third class of fourth grade, 19 people subscribed to "Youth Digest", 24 people subscribed to "Learn and Play", and 13 people subscribed to both. How many people subscribe to "Youth Digest" or "Learn and Play"?

2. There are 58 people learning piano in the kindergarten, 43 people learning painting, and 37 people learning both piano and painting. How many of them only learn piano and only learn painting

< p>People?

3. Among the natural numbers from 1 to 100:

(1) How many numbers are there that are multiples of 2 and multiples of 3?

(2) How many numbers are multiples of 2 or 3?

(3) How many numbers are there that are multiples of 2 but not multiples of 3?

4. The statistics of the mid-term exam results of a certain class of mathematics and English are as follows: 12 people scored 100 points in English, 10 people scored 100 points in mathematics, two subjects

There were 3 people who got 100 points in both subjects, and 26 people who didn't get 100 points in both subjects. How many students are there in this class?

5. There are 50 people in the class, 32 people can ride bicycles, 21 people can roller skate, and 8 people can do both. How many people can’t do both?

6. There are 42 students in a class, 30 students participate in the sports team, and 25 students participate in the art team, and each student participates in at least one team.

How many people are participating in both teams in this class?

Answers to the test questions

1. There are 19 people in the third class of fourth grade who subscribe to "Youth Digest", 24 people who subscribe to "Learn and Play", and 13 people who subscribe to both. people. How many people subscribe to "Youth Digest"

or "Learn and Play"?

19 + 24—13 = 30 (people)

Answer: There are 30 people who subscribe to "Youth Digest" or "Learn and Play".

2. There are 58 people learning piano in the kindergarten, 43 people learning painting, and 37 people learning both piano and painting. How many of them only learn piano and only learn painting

< p>People?

Number of people who only learn piano: 58—37 = 21 (people)

Number of people who only learn painting: 43—37 = 6 (people)

3. Among the natural numbers from 1 to 100:

(1) How many numbers are there that are multiples of 2 and multiples of 3?

If it is a multiple of 3 and a multiple of 2, it must be a multiple of 6

100÷6 = 16...4

Therefore, it is a multiple of 2 There are 16 multiples of 3

(2) How many numbers are multiples of 2 or 3?

100÷2 = 50, 100÷3 = 33...1

50 + 33—16 = 67 (pieces)

Therefore, it is 2 There are 67 numbers that are multiples or multiples of 3.

(3) How many numbers are there that are multiples of 2 but not multiples of 3?

50—16 = 34 (numbers)

Answer: There are 34 numbers that are multiples of 2 but not multiples of 3.

4. The statistics of the mid-term exam results of a certain class of mathematics and English are as follows: 12 people scored 100 points in English, 10 people scored 100 points in mathematics, two subjects

There were 3 people who got 100 points in both subjects, and 26 people who didn't get 100 points in both subjects. How many students are there in this class?

12 + 10—3 + 26 = 45 (people)

Answer: There are 45 students in this class.

5. There are 50 people in the class, 32 people can ride bicycles, 21 people can roller skate, and 8 people can do both. How many people can’t do both?

50—(30 + 21—8) = 7 (people)

Answer: There are 7 people who can’t do both.

6. There are 42 students in a class, 30 students participate in the sports team, and 25 students participate in the art team, and each student participates in at least one team.

How many people are participating in both teams in this class?

30 + 25—42 = 13 (people)

Answer: There are 13 people participating in both teams in this class.

A certain class of students took the entrance exam, and the number of students who got full marks was as follows: 20 people in mathematics, 20 people in Chinese, 20 people in English, 8 people got full marks in mathematics and English, and 7 people got full marks in mathematics and Chinese. Among people, there are 9 people who got perfect marks in Chinese and English, and 3 people who didn’t get perfect marks in three subjects. How many people can this class have at most? What is the minimum number of people?

The analysis and solution are shown in Figure 6. Students with perfect scores in mathematics, Chinese, and English are included in this class. Suppose there are y people in this class, represented by a rectangle. A, B, and C represent mathematics, People who get full marks in Chinese and English are known to have A∩C=8, A∩B=7, B∩C=9. A∩B∩C=X.

From the inclusion-exclusion principle, we have

Y=A+B+c-A∩B-A∩C-B∩C+A∩B∩C+3

That is, y=2220-7-8-9+x+3=39+x.

Below we examine how to find the maximum and minimum values ??of y.

It can be seen from y=39+x that when x takes the maximum value, y also takes the maximum value; when x takes the minimum value, y also takes the minimum value. The number of people who got full marks, so the number of them must not exceed the number of people who got full marks in two subjects, that is, x ≤ 7, x ≤ 8 and x ≤ 9, from which we get x ≤ 7. On the other hand, the students who got full marks in mathematics are Maybe no one got full marks in Chinese, which means there are no students who got full marks in all three subjects, so x≥0, so 0≤x≤7.

When x takes the maximum value 7, y has the maximum value 39+7=46. When x takes the minimum value 0, y has the minimum value 39+0=39.

Answer: This class has a maximum of 46 people and a minimum of 39 people.

Question 1. The salesperson exchanged a RMB of 5 yuan and a RMB of 50 cents into 28 RMB with a face value of 1 yuan and 1 jiao. How much are the two kinds of RMB exchanged? open?

Question 2. There are 50 pieces of RMB of one yuan, two yuan and five yuan. The total face value is 116 yuan. It is known that there are 2 more yuan of one yuan than two yuan. Ask about the three denominations of RMB. How many of each?

Question 3. There are 400 movie tickets of 3 yuan, 5 yuan and 7 yuan, each worth 1920 yuan. Among them, the number of 7 yuan and 5 yuan tickets is equal. The three prices of movie tickets are How many of each?

Question 4. Use large and small cars to transport goods. Each large car holds 18 boxes and each small car holds 12 boxes. Now there are 18 trucks worth 3024 yuan. If each box is cheaper 2 yuan, then the value of this batch of goods is 2520 yuan. Question: How many large and small cars are there?

Question 5. A truck transports ore, which can be transported 20 times a day on sunny days and 12 times a day on rainy days. It has been transported 112 times a day, with an average of 14 times a day. During these days, How many days will it rain?

Question 6. A batch of watermelons has been shipped and is to be sold in two categories. The large ones are 0.4 yuan per kilogram and the small ones are 0.3 yuan per kilogram. The maximum value of this batch of watermelons is 290 yuan. If each The price of kilograms of watermelons has been reduced by 0.05 yuan. This batch of watermelons can only be sold for 250 yuan. Question: How many kilograms of large watermelons are there?

Question 7. Two people A and B are throwing darts. It is stipulated that every time they hit the target, they will be awarded 10 points, and every time they miss the target, they will be deducted 6 points. They each throw 10 times, and *** will get 152 points, of which A scored 16 more points than B. Question: How many times did each of them score?

Question 8. There are 20 questions in a certain mathematics competition. Each correct answer is worth 5 points. A wrong answer will not only give you no points, but also deduct 2 points. In this competition, Xiao Ming Scored 86 points. Question: How many questions did he answer correctly?

1. Solution: Suppose there are x pieces of 1 yuan and (28-x) pieces of 1 dime

x+0.1(28-x)=5.5

< p>0.9x=2.7

x=3

28-x=25

Answer: There are 3 cards for one dollar and 25 cards for one dime.

2. Solution: Suppose there are 2)+5(52-2x)=116

x+2x-4+260-10x=116

7x=140

x=20 < /p>

x-2=18

52-2x=12

Answer: There are 20 cards for 1 yuan, 18 cards for 2 yuan, and 12 cards for 5 yuan.

3. Solution: Suppose there are x pieces of 7 yuan and 5 yuan each, and (400-2x) pieces of 3 yuan

7x+5x+3(400-2x)=1920

12x+1200-6x=1920

6x=720

x=120

400-2x=160

Answer: There are 160 cards worth 3 yuan, 120 cards worth 7 yuan and 5 yuan each.

4. Solution: Total number of goods: (3024-2520) ÷ 2 = 252 (boxes)

There are x large cars and (18-x) small cars

p>

18x+12(18-x)=252

18x+216-12x=252

6x=36

x=6 < /p>

18-x=12

Answer: There are 6 large cars and 12 small cars.

5. Solution: Number of days = 112÷14=8 days

Suppose x day is a rainy day

20(8-x)+12x=112 < /p>

160-20x+12x=112

8x=48

x=6

Answer: There are 6 rainy days.

6. Solution: Number of watermelons: (290-250) ÷ 0.05 = 800 kilograms

Suppose large watermelons x kilograms

0.4x+0.3(800 -x)=290

0.4x+240-0.3x=290

0.1x=50

x=500

Answer : There is a large watermelon of 500 kilograms.

7. Solution: A’s score: (152+16)÷2=84 points

B: 152-84=68 points

Suppose A wins x Times

10x-6(10-x)=84

10x-66x=84

16x=144

x =9

Suppose B hits y times

10y-6(10-y)=68

16y=128

y= 8

Answer: A scored 9 times and B scored 8 times.

8. Solution: Suppose he answers x questions correctly

5x-2(20-x)=86

5x-42x=86

p>

7x=126

x=18

Answer: He answered 18 questions correctly.