One, the "two people" mentioned in the question of two people can also be two groups, two teams and so on. Example 1 A can finish the work in 9 days, and B can finish the work in 6 days. Now A has done it for three days, and B continues to finish the rest. B How many days does it take to finish all the work? Scheme 1: Take this job as 1, where A can complete one ninth of this job every day, and three days is 1/3. B can finish one-sixth of the work every day, (1-1/3) ÷1/6 = 4 (days) A: B needs to finish all the work within 4 days. Scheme 2: The least common multiple of 9 and 6 is 18. Let the total workload be 65438+. B finish 3 copies every day. B The time required to complete the remaining work is (18- 2 × 3)÷ 3= 4 (days). Scheme 3: The working efficiency ratio of A and B is 6∶ 9= 2∶ 3. A has done 3 days, which is equivalent to 2 days for B. The time required for B to complete the remaining work is 6-2=4 (. Solution: * * * did it for 6 days, but A did it for 24 days and B did it for 24 days. Now, A does 0 days and B does 40=(24+ 16) days. This shows that A has worked for 24 days, and B can be used instead of 16 days. So, if B does it alone, it will take time. The required time is 75 days. Answer: It takes 75 days for Party A or 50 days for Party B to do it alone. A project can be completed by Party A for 63 days, and then by Party B for 28 days. If both parties cooperate, it will take 48 days to complete. Now Party A does it alone for 42 days, and then Party B does it alone. How many more days does Party B need to do? Solution: The comparison is as follows: A does 63 days, B does 28 days; A does it for 48 days and B does it for 48 days. It is known that A does 63-48= 15 (days) less and B does 48-28=20 (days) more. It is concluded that A worked alone for 42 days, 63-42=2 1 (day) less than 63 days, which is equivalent. B It will take 28+28= 56 days. A: B It will take another 56 days. Example 4 A project is completed by team A alone 10 days, and team B alone for 30 days. Now the two teams cooperate, during which team A has a rest for 2 days and team B has a rest for 8 days (no rest on that day). Q from beginning to end * * Scheme 1: Team A will work alone for 8 days and Team B will work alone for 2 days. * * * The remaining workload is the cooperation of two teams, and the number of days required is 2+8+ 1= 1 1 (days). A: It cost 65438+ from beginning to end. Party B shall complete 1 copy every day. Team A has done it alone for 8 days, and after team B has done it alone for 2 days, it still needs the cooperation of the two teams (30-3× 8-/kloc-0 /× 2) ÷ (3+1) =1(days). Option 3: Team A will do 1. There is also the workload of (Team A) 10-8= 2 (days), which is equivalent to the workload of Team B 2×3=6 (days). After Team B worked alone for 2 days, there is still the workload of (Team B) 6-2=4 (days). 4=3+ 1, of which 3 days can be used. So the two teams just need to cooperate again 1 day. Solution 4: Break up and think together (The topic says that Team A and Team B don't rest together, so we assume that they rest together. ) The daily workload of Team A is110, and that of Team B is 1/30, because Team A has rested for 2 days and Team B has rested for 8 days. 2, so let's assume that A has a rest for two days and B also has a rest. Then Party A starts to work, and Party B takes a rest: 8-2=6 (days). During these six days, Party A independently completed110× 6 = 6/10, and the remaining workload was 1-6/ 10 = 4/60. 4/ 10 of the engineering quantity needs the cooperation of both parties: (4/10) ÷ (110+1/30) = 3 days. So it needs * * * 8+3= 1 1 (days) ● Example 5 A project, team A completed 20 days alone, and team B completed 30 days alone. Now they are working together, during which Team A has a rest for 3 days and Team B has a rest for a few days. It took 65433 days from start to finish. Scheme 1: If both teams don't rest for 16 days, the workload that can be completed is (1÷ 20) ×16+(1÷ 30 )×16 = 4/3, because 20× 3 =11/The rest day of Team B is11/60 ÷ (1/30) =1/2 A: B. B finish 2 copies every day. The workload that the two teams did not do during the break was (3+2)× 16- 60= 20 (copies). Therefore, the rest days of B are (20- 3 × 3)÷ 2= 5.5 (days). Option 3: Team A does it for 2 days, which is equivalent to Team B doing it for 3 days. Team A is equivalent to 4.5 days' rest for Team B. If Team A doesn't rest for 0/6 days, Team A will only work for 4 days, which is equivalent to 6 days' work for Team B. The rest days for Team B are 16-4.5 = 5.5 (days). There are two jobs in Example 6, and it takes 65,438+00 days for Zhang to complete the job alone. It takes 8 days for Li to complete work A and 20 days for Li to complete work B. If two people can cooperate in each job, how many days will it take to complete these two jobs? Solution: Obviously, Li's work efficiency as A is high, and Zhang's work efficiency as B is also high. Therefore, let Li do A first and Zhang do B first. Let B's workload be 60 copies (/kloc-the least common multiple of 0/5 and 20), with Zhang completing 4 copies every day and Li completing 3 copies every day for 8 days. Finish Work A. At this time, Zhang still has (60-4×8) copies of Work B. The cooperation between Zhang and Li needs (60-4×8)÷(4+3)=4 (days). 8+4= 12 (days). A: At least these two tasks have been completed. Solution: We assume that the workload of this project is 30 copies, with Party A completing 3 copies every day and Party B completing 2 copies every day. With the cooperation of two people, * * * completes 3× 0.8+2 × 0.9= 4.2 (copies). Because two people should work together as little as possible, and the one who works alone should be the one with high efficiency. Because it will be completed within 8 days, the number of days for two people to cooperate is (30-3×8)÷(4.2-3)=5 (days). Obviously, it finally became a problem of "chickens and rabbits in the same cage". If Party A and Party B cooperate to finish a job, Party A's work efficiency will be faster than when it is done alone. If this work is always done by Party A alone, how many hours will it take? Solution: The work done by one person in 6 hours is 6 hours for two people to cooperate, and the work done by one person in A is 6 hours. A The time required to complete this work alone is A:A It takes 33 hours to complete this work alone. Most examples in this section have been "rounded". However, "integer" does not make the calculation of all engineering problems simple. This is the case in Example 8. Example 8 can also be an integer. It's a little convenient to find B, but it doesn't do much good. There is no need to reinvent the wheel. Second, there are at least three people in the multi-person project. Of course, the multi-person problem is more complicated than the two-person problem, but the basic idea of solving the problem is still similar. Example 9 A and B worked together for 36 days. Solution: Suppose the workload of this job is 1. The cooperation between Party A, Party B and Party C is completed every day, minus the workload of Party B and Party C. Answer: It takes 90 days for Party A to do it alone. Example 9 can also be rounded off. Assuming that the total workload is 180 copies, Party A and Party B complete 5 copies every day, and Party B and Party C complete 4 copies every day. ● Example 10 For a job, Party A needs to do it alone 12 days, Party B needs to do it alone 18 days and Party C needs to do it alone for 24 days. Party A worked for a few days, then Party B worked for a few days, and Party B worked for three times as many days as Party A, then Party C worked for a few days, and finally completed it. Solution: A does 1 day, B does 3 days, and C does 3×2=6 (days). It means that A does 2 days, B does 2×3=6 (days), C does 2×6= 12 (days), and three people do 2+6+. 18,24 has an easily found least common multiple of 72. It can be assumed that the total workload is 72. Every day Party A completes 6, Party B completes 4 and Party C completes 3. A project takes 1 1 in total, and it takes 13 days for Party A, Party B and Party C to cooperate. Solution: Two days' work of Party C is equivalent to four days' work of Party B, and the work efficiency of Party C is 4÷2=2 (times) that of Party B. The cooperation between Party A and Party B is 1 day, which is the same as that of Party B's four days. That is to say, Party A works 1 day, which is equivalent to Party B's working for 3 days. Party A's working efficiency is 3 times that of Party B, and they *. Party A needs to answer: It takes 26 days for Party A to work alone. In fact, when we calculate the working efficiency ratio of Party A, Party B and Party C as 3∶2∶ 1, we know that Party A works 1 day, which is equivalent to the cooperation of Party B and Party C 1 day. It takes 13 days for three people to cooperate, of which Party B and Party C will complete it. It can be converted into 13 days to complete. Example 12 A job can be completed by three people in group A in eight days, and by four people in group B in seven days. How long will it take two people in group A and seven people in group B to finish the work? Scheme 1: Suppose the workload of this work is 1. Everyone in group a can finish it every day. Everyone in group b can finish it every day. A: The work can be finished in three days. Scheme 2: Three people in Group A can finish it in eight days, then two people can finish it in 12 days. Four people in group B can finish it in seven days, so seven people can finish it in four days. Now, regardless of the number of people, the problem becomes: Group A can work alone 12 days, and Group B can work alone for 4 days. How many days will the cooperation be completed? Elementary school arithmetic should make full use of the particularity of given data. Scheme 2 is a typical example of flexible use of proportion. If you do your mental arithmetic well, you will get the answer soon. ● Example 13 It takes 10 days for Workshop A to make a batch of parts. If Workshop A and Workshop B do it together, it only takes 6 days. Workshop b and workshop c work together, which takes 8 days to complete. Now the three workshops work together, and when they are completed. Scheme 1: We still assume that the total workload is 1. Party A completes more parts than Party B every day, so the total number of parts in this batch is the number of parts manufactured in workshop C. The least common multiple of Answer 2: 10 and 6 is 30. Let's assume that the total workload of manufacturing parts is 30. Party A completes 3 parts every day, and Party A and Party B jointly complete 5 parts every day. From this, it is concluded that B completes 2 copies every day. B and c are finished together in 8 days. B completes 8×2= 16 (copies), and C completes 30- 16= 14 (copies). It is known that the efficiency ratio of Party B and Party C is 16: 14 = 8: 7. It is known that the efficiency ratio of Party A and Party B is 3: 2 = 12: 8. Taken together, the efficiency ratio of Party A, Party B and Party C is 12: 8: 7. The number of parts produced by C is 2400÷( 12- 8) × 7= 4200 (pieces). ● Example 14 It takes 10 hour for Party A to move goods in the warehouse, 12 hour for Party B and 15 hour for Party C.. Yes Solution: Suppose the workload of moving goods in a warehouse is 1. Now it is equivalent to three people * * * completing workload 2, and the required time is 3 hours for a: C to help A and 5 hours for B.. The key to solve this problem is to calculate the time for three people to move two warehouses together. Of course, the calculation of this problem can also be rounded off. Suppose the total workload of moving a warehouse is 60. One every hour B delivers 5 per hour, and C delivers 4 per hour. It takes 60 × 2÷ (6+ 5+ 4)= 8 (hours) for three people to act together. A needs C to help carry (60- 6× 8)÷ 4= 3 (hours). B needs C to help carry (60-5× 4) water pipes. The problem is the same as the engineering problem. Water injection or drainage in a pool is equivalent to a project, and water injection or drainage is the workload. The water injection quantity or displacement per unit time is the working efficiency. As for injection and drainage, there are problems, and the workload will only increase. Therefore, the idea of solving the water pipe problem is basically the same as that of solving the engineering problem. For example, 15, when the pipes A and B are opened at the same time, the pool can be filled in 9 minutes. Now open tube A first, 10 minutes later open tube B. After 3 minutes, the pool was filled with water. It is known that pipeline A injects 0.6 cubic meters more water per minute than pipeline B. What is the volume of this pool? Solution: The water injection per minute of A is: (1-1/9× 3) ÷10 =115. The water injection rate per minute of B is: 1/9- 1/. 45)=27 (cubic meters) Answer: The volume of the pool is 27 cubic meters. There are some water pipes, and they inject the same amount of water every minute. Now, open some of them, and after a predetermined time of 1/3, double the opened water pipes, and the pool can be filled at a predetermined time. If it starts, open the 10 water pipe. Analysis: After adding water pipes, the water pipes are twice as many as before, and the water injection time is 1- 1/3=2/3, and 2/3 is twice as much as 1/3, so the water injection during this period after adding water pipes is four times as much as that in the previous period. Assuming that the capacity of the pool is 1 and the water injection ratio in the first two periods is 1: 4, the water injection amount in the predetermined period is1(1+4) =1/5. 10 water pipe is opened at the same time, and water can be injected as scheduled. The water injection quantity of each water pipe is110, which is 1/3 of the scheduled time, and the water injection quantity of each water official is110×1/3 =. Water pipe 1/5÷ 1/30=6 (root) solution: the water injection ratio in the first two phases is1:[(1-1/3) ÷1/3. 5 1/3 The water injection quantity of each water pipe at the scheduled time is:1÷10×1/3 =1/30. The number of water pipes to be opened at first is 1/5 ÷ 1/30. Example 17 reservoir has two water inlets A and C and two water drains B and D. To fill a pool, it takes 3 hours to open the A pipe, 5 hours to open the C pipe, 4 hours to open the B pipe and 6 hours to open the D pipe. Now there is one-sixth of the water in the pool, such as pressing A, B, C, D and A. Later (20 hours), the problem of water in the pool was similar to the widely circulated "frog climbing well": a frog falling into a dry well had to climb 30 feet to reach the wellhead, and it always climbed 3 feet and slipped 2 feet every hour. How many hours does it take a frog to climb to the wellhead? It seems that it only climbs 3- 2= 1 (feet) per hour, but after 27 hours, it climbs 1 hour and climbs 3 feet to the wellhead. So the answer is 28 hours, not 30 hours. Example: 18 A reservoir has 4 cubic meters of water flowing in every minute. If five faucets are turned on, 2. Solution: First, calculate the water output per minute of 1 faucet. Two and a half hours is 60 minutes more than 1 half hour, and the water inflow is 4 × 60= 240 (cubic meter). Time is in minutes, 1 tap water output per minute is 240 ÷ (5× 150). The water output of 8 faucets 1.5 hours is 8 × 8 × 90, of which the water inflow in 90 minutes is 4 × 90, so the raw water of the pool is 8 × 8 × 90-4 × 90= 5400 (m3). Turn on 13 faucet, and 8× 13 water can be discharged every minute. The rest will release raw water, and it takes 5400 ÷(8 × 13- 4)=54 (minutes) to empty the original 5400. Answer: It takes 54 minutes to turn on 13 faucet and empty the pool. There are two parts of water in the pool, raw water and fresh water, so it is needed. The key to solve this problem is to find out the original water in the pool. This is implied in the question. For example, 19, in a pool, groundwater seeps into the pool from four walls, and the amount of water infiltrated per hour is fixed. Open tube A for 8 hours, and open tube C 12 hours. If the pipes A and B are opened, the water can be discharged in 4 hours. Solution: The water filled in the pool is 1. One pipe is lined up every hour for 4 hours. Therefore, B and C run in parallel, and the displacement per hour is B and C, and the time required to discharge the water full of the pool is A:B and C, it takes 4 hours and 48 minutes to discharge the water full of the pool. This issue should also be considered separately. Both the original water quantity and the seepage water quantity of the pool (full pool) are unknown, just like the specific workload in engineering problems. Here, the two kinds of water are set to "1" respectively, but confusion should be avoided. In fact, it can also be rounded off. Raw water can be set as the least common multiple of 8 and 12, which was written by Newton, a great British scientist in the 24th century. The book puts forward a problem of "cows eat grass", which is an interesting arithmetic problem. Essentially similar to the examples 18 and 19. The topic involves three quantities: original grass, new grass and grass eaten by cattle. This is almost the same as the original water quantity, seepage water quantity and water discharged from water pipes. There are three examples in Example 20. 2 1 A cow ate the grass in the second pasture in 9 weeks. /kloc-how many cows can the grass in the third pasture eat in 0/8 weeks? Solution: the total amount of grass eaten = the amount of grass eaten by a cow per week × the number of cows × the number of weeks. According to this formula, the amount of grass eaten by a cow in a week can be set as the measurement unit of grass. Original grass +4 weeks new grass = 12×4. Original grass +9 weeks new grass =7×9. The new grass grows every week is (7×9- 12×4)÷(9-4)=3. Then the original grass is 7×9-3×9=36 (or 12×4-3×4). For the third pasture, the total amount of original grass and new grass growing in 18 weeks is that these grasses can feed 90×7.2÷ 18=36 cows 18 weeks. A: 36 cows can eat the grass in the third pasture within 18 weeks. Solution of Example 20 and Example 19 In fact, if Example 19 has another condition, such as "Open the B tube, and the water in the pool can be emptied in 10 hour", the quantitative relationship between "new length" and "original length" can be found, but this is only Example 65438+. The problem of "cattle eating grass" can appear in various forms. Due to the limitation of space, we will only give one more example. For example, the 2 1 exhibition started at 9: 00, but there were still people queuing to enter. Since the first audience arrived, the number of visitors per minute has been the same. If you open three entrances, no one will line up at 9: 09; if you open five entrances, no one will line up at 9: 05. Solution: We assume that an entrance can enter 1 calculation unit per minute. Admission from 9: 00 to 9: 09 is 3x9, and admission from 9: 00 to 9: 05 is 5x5. Because the audience is more than 9-5=4 (minutes), the audience coming every minute is (3×9-5×5)÷(9-5)=0.5. 9, and the audience who came before 5× 5-0.5 = 22.5. It takes 22.5÷0.5=45 minutes for these audiences to arrive. A: What is the arrival time of the first audience? Team A digs for three days, while Team B digs for one day, which can reach 3/ 10 of this canal. How many days does it take for each team to dig separately? Analysis: After the cooperation between Party A and Party B 1 days, A did it for another 2 days * * * 3/10-/6 = 4/302 ÷ (3/10-/6) = 2. Answer: It takes 15 days for Party A to do it alone, and 10 days for Party B to do it alone ... If Party A does it alone, then Party A can finish it 2 days in advance and Party B will finish it 3 days later. Now Party A and Party B will cooperate for two days, and the rest will be done by Party B alone, within the stipulated date. If both parties cooperate, how long will it take to finish the work? Solution: The specified time is X days. (A needed X-2 days alone, B needed X+3 days alone, and A * * * did it for 2 days. Otsuichi worked for x days)1(x-2) × 2+x/(x+3) =1x =12. The required completion is 1/, and it takes 12 days. = 1÷( 1/6) =6 days A: It takes 6 days for two people to cooperate. Example: A project is completed by Party A alone for 63 days, then by Party B for 28 days, and both parties cooperate for 48 days. A does it for 42 days first, and B does it for a few days? Answer: Let A's work efficiency be X, and B's work efficiency be Y 63x+28y = 1 48x+48y =1x =1/84y =12 b Need to do (/kloc-0)