Inclusion and exclusion
1. There are 40 students in a class, of which 15 is in the math group, 18 is in the model airplane group, and 10 is in both groups. So how many people don't participate in both groups?
Solution: There are (15+18)-10 = 23 (people) in the two groups.
40-23= 17 (person) did not attend.
A: There are 17 people, and neither group will participate.
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There are forty-five students in a class who took the final exam. After the results were announced, 10 students got full marks in mathematics, 3 students got full marks in mathematics and Chinese, and 29 students got no full marks in both subjects. So how many people got full marks in Chinese?
Solution: 45-29- 10+3=9 (person)
A: Nine people got full marks in Chinese.
3.50 students stand in a row facing the teacher. The teacher asked everyone to press 1, 2,3, ..., 49,50 from left to right. Let the students who are calculated as multiples of 4 back off, and then let the students who are calculated as multiples of 6 back off. Q: How many students are facing the teacher now?
Solution: multiples of 4 have 50/4 quotients 12, multiples of 6 have 8 50/6 quotients, and multiples of 4 and 6 have 4 50/ 12 quotients.
Number of people turning back in multiples of 4 = 12, number of people turning back in multiples of 6 ***8, including 4 people turning back and 4 people turning back from behind.
Number of teachers =50- 12=38 (person)
A: There are still 38 students facing the teacher.
4. At the entertainment party, 100 students won lottery tickets with labels of 1 to 100 respectively. The rules for awarding prizes according to the tag number of lottery tickets are as follows: (1) If the tag number is a multiple of 2, issue 2 pencils; (2) If the tag number is a multiple of 3, 3 pencils will be awarded; (3) The tag number is not only a multiple of 2, but also a multiple of 3 to receive the prize repeatedly; (4) All other labels are awarded to 1 pencil. So how many prize pencils will the Recreation Club prepare for this activity?
Solution: 2+000/2 has 50 quotients, 3+ 100/3 has 33 quotients, and 2 and 3 people have 100/6 quotients.
* * * Preparation for receiving two branches (50- 16) * 2 = 68, * * Preparation for receiving three branches (33- 16) * 3 = 5 1, * * Preparation for repeating branches (2+).
* * * Need 68+5 1+80+33=232 (branch)
A: The club has prepared 232 prize pencils for this activity.
5. There is a rope with a length of 180 cm. Make a mark every 3 cm and 4 cm from one end, and then cut it at the marked place. How many ropes were cut?
Solution: 3 cm marker: 180/3=60, the last marker does not cross, 60- 1=59.
4cm marker: 180/4=45, 45- 1=44, repeated marker: 180/ 12= 15,15-/kloc-.
Cut it 89 times and it becomes 89+ 1=90 segments.
A: The rope was cut into 90 pieces.
6. There are many paintings on display in Donghe Primary School Art Exhibition, among which 16' s paintings are not in the sixth grade, and 15' s paintings are not in the fifth grade. Now we know that there are 25 paintings in Grade 5 and Grade 6, so how many paintings are there in other grades?
Solution: 1, 2,3,4,5 * * has 16, 1, 2,3,4,6 * * has15,5,6 * * has 25.
So * * has (16+ 15+25)/2=28 (frame), 1, 2,3,4 * * has 28-25=3 (frame).
A: There are three paintings in other grades.
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7. There are several cards, each with a number written on it, which is a multiple of 3 or 4. Among them, cards marked with multiples of 3 account for 2/3, cards marked with multiples of 4 account for 3/4 and cards marked with multiples of 12 account for 15. So, how many cards are there?
Solution: The multiple of 12 is 2/3+3/4- 1=5/ 12, 15/(5/ 12)=36 (sheets).
There are 36 cards of this kind.
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8. How many natural numbers from 1 to 1000 are divisible by neither 5 nor 7?
Solution: multiples of 5 have 200 quotients 1000/5, multiples of 7 have quotients 1000/7 142, and multiples of 5 and 7 have 28 quotients 1000/35. The multiple of 5 and 7 * * * has 200+ 142-28=3 14.
1000-3 14=686
A: There are 686 numbers that are neither divisible by 5 nor divisible by 7.
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9. Students in Class 3, Grade 5 participate in extracurricular interest groups, and each student participates in at least one item. Among them, 25 people participated in the nature interest group, 35 people participated in the art interest group, 27 people participated in the language interest group, 12 people participated in the language interest group, 8 people participated in the nature interest group, 9 people participated in the nature interest group, and 4 people participated in the language, art and nature interest groups. Ask how many students there are in this class.
Solution: 25+35+27-(8+ 12+9)+4=62 (person)
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 three circles A, B and C are all 30, the areas of overlapping parts of A and B, B and C, and 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 overlapping area of A, B and C =73+(6+8+5)-3*30=2.
Shadow area =73-(6+8+5)+2*2=58.
A: The shaded part is 58.
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-Author: abc
-Date of issue: 2004-12-1215: 45: 02
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Grade four 1 class 1 1 There are 46 students taking part in three extracurricular activities. Among them, 24 students from the math group and 20 students from the Chinese group participated. The number of people who participated in the art group was 3.5 times that of those who participated in both the math group and the art group, and 7 times that of those who participated in all three activities. The number of people who participated in both the literature and art group and the Chinese group was twice that of those who participated in all three activities, and the number of people who participated in both the math group and the Chinese group was 10. The number of people seeking to join the art troupe.
Solution: Let the number of people participating in the art group be x, 24+20+x-(x/305+2/7 * x+10)+x/7 = 46, and the solution is X=2 1.
A: The number of participants in the art group is 2 1.
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-Author: abc
-Date of issue: 2004-12-1215: 45: 43
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12. There are 100 books in the library. The borrower needs to sign the book. It is known that 33, 44 and 55 books in 100 have the signatures of A, B and C respectively, among which 29 books have the signatures of A and B, 25 books have the signatures of A and C, and 36 books have the signatures of B and C. How many of these books have not been borrowed by any of A, B and C?
Solution: The number of books read by three people is: A+B+C-(A+B+C+C)+A, B, C =33+44+55-(29+25+36)+ A, B, C =42+ A, B, C, A, C is the most.
Three people will always read 42+25=67 (books) at most, and at least 100-67=33 (books) have never been read.
A: At least 33 books in this batch have not been borrowed by any of A, B and C.
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-Author: abc
-Date of issue: 2004-12-1215: 46: 53
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13, as shown in Figure 8-2, five equal-length line segments form a pentagram. If exactly 1994 points on each line segment are dyed red, how many red dots are there on this five-pointed star?
Solution: There are 5* 1994=9970 red dots on the right side of the five elements. If you put a red dot on all the intersections, then at least there are red dots. These five lines have 10 intersections, so there are at least 9970- 10=9960 red dots.
A: There are at least 9960 red dots on this five-pointed star.