Engineering Problems are Mathematics in Primary and Secondary SchoolsApplied questionThe key point in teaching is the extension and supplement of fraction word problems and the cultivation of studentsLogical thinking abilityImportant tools.It isfunctionOne-to-one correspondenceThe powerful infiltration of ideas in word problems.Engineering problems are also difficult points in textbooks.
Engineering problems are word problems that regard the total amount of work as a unit "1". They are abstract and difficult for students to recognize.
In teaching, how to let students establish correct concepts is the key to mathematical word problems.This lesson runs through the concepts of engineering problems from beginning to end, aiming to enable students to understand and master the concepts.
Introduce it through actual conversation.The concept of suspension and penetration is introduced.The purpose is to let students review and understand the concepts of total workload, working time and working efficiency and the quantitative relationship between them.Preliminary review reinforces the concept of engineering problems.
Through comparison, establish the concept.Give full play to students' dominant position in teaching, and use students' existing knowledge "including and excluding" to solve cooperation problems.
Rationally apply the concept of reinforcement.On the basis of perception, students have initially formed the concept representation in their minds and have the concept prototype.Some students just accepted the concept, but have not yet fully digested it.Therefore, exercises have been prepared to help students understand, understand and digest concepts through students' application, so that students can find solutions to engineering problems more skillfully.After a lot of practice, the students will draw out a number of work questions, and let the students find the answers to the questions themselves.
Example
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In daily life, doing a certain thing, manufacturing a certain product, completing a certain task, completing a certain project, etc., all involve the three quantities of total workload, work efficiency, and work time. The basic quantitative relationship between them is - work efficiency × time=total workload.
In primary school mathematics, we all call them "engineering problems" when we discuss the relationship between these three quantities.
Let's take a simple example: A can finish a job in 15 days, and B can finish it in 10 days. How many days can the two people work together?
A job is regarded as a whole, so the work quantity can be calculated as 1. The so-called work efficiency is the amount of work completed in a unit time. The time unit we use is "day", and one day is a unit,
According to the basic quantity relation
Workload ÷ work efficiency=work time
1÷（1/15+1/10）
=6 (days)
A: It takes six days for the two to cooperate
This is the most basic problem in engineering problems. Many examples introduced in this lecture are developed from this problem.To calculateintegerChange (use integer to calculate as much as possible), such as the method used in Example 3 and Example 8 in Lecture 3, to set more shares of the workload.It's the same as the previous question, 10 and 15Least common multipleIt's 30.If the total workload is 30, then Party A will complete 2 copies every day, and Party B will complete 3 copies every day. The number of days required for cooperation between the two is:
30 ÷ (2+3)=6 (days)
It is more convenient to calculate by numbers.
3: 2. Or "the workload is fixed, and the work efficiency is inversely proportional to the time". The ratio of work efficiency of A and B is 10:15=2:3
Method summary
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1： Basic quantity relation
1. WorkEfficiency × time=total workload 2. Work efficiency=total workload ÷ work time 3. Work time=total workload ÷ work efficiency.
2： Basic characteristics
Assume that the total amount of work is "1", and the efficiency=1/time.
Ten:Cattle grazing: 1. New grass quantity, 2. Original grass quantity, 3. Problem solving.
Example analysis
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When we know the ratio of work efficiency between the two, we can also flexibly answer the question from the perspective of proportion.
Therefore, in the following examples, we do not fully adopt the practice of "setting the workload as a whole 1" in common textbooks, but focus on "integer" or "from the perspective of proportion", which may make our solution ideas more flexible.
1、 Two person question
The "two people" in the title can also be two groups, two teams, etc
●Example 1A job can be completed in 9 days by Party A and 6 days by Party B.Now Party A has done the work for three days, and the rest of the work will be completed by Party B. How many days will Party B need to do to complete all the work?
Solution 1: Think of this work as 1. A can finish one ninth of this work every day and one third of it in three days.
B can complete one sixth of this work every day, (1-1/3) ÷ 1/6=4 (days)
Answer: B needs 4 days to complete all the work
Solution 2: 9 and 6Least common multiple18. Assume that the total workload is 18 copies. Party A completes 2 copies every day, and Party B completes 3 copies every day. The time required for Party B to complete the remaining work is
(18 - 2 × 3) ÷ 3=4 (days)
Solution 3: The ratio of work efficiency between A and B is
6∶ 9= 2∶ 3.
Three days for Party A is equivalent to two days for Party B. The time required for Party B to complete the remaining work is 6-2=4 (days)
●Example 2A job can be completed within 30 days through cooperation between Party A and Party B. After 6 days of joint work, Party A leaves and Party B continues to work for 40 days. How many days will it take for Party A or Party B to complete the job alone?
Solution: After six days,
It turns out that Party A does it for 24 days, and Party B does it for 24 days,
Now, Party A does 0 days, and Party B does 40=(24+16) days
This shows that the work that A did in 24 days can be replaced by the work that B did in 16 days. Therefore, the work efficiency of A is that of B
(times)
Six days for Party A is equivalent to six days for Party B
(days),
If Party B does it alone, the required time is 6+4+40=50 days.
If Party A does it alone, the time required is
day
Answer: The time required for Party A or Party B to do it alone is 75 days and 50 days respectively
●Example 3A project can be completed after 63 days by Party A and 28 days by Party B;In case of cooperation between Party A and Party B, it will take 48 days to complete.Now, Party A will do it alone for 42 days, and then Party B will do it alone. How many more days does Party B need to do?
Solution: First compare as follows:
63 days for Party A and 28 days for Party B;
It takes 48 days for Party A and 48 days for Party B
It is known that Party A needs to do 63-48=15 (days) less, and Party B needs to do 48-28=20 (days) more, so it can be concluded that Party A's work efficiency
B is efficient
(times)
First, Party A did it alone for 42 days, 63-42=21 (days) less than 63 days,
It is equivalent to what Party B needs to do
(days)
Therefore, B still needs to do
28+28=56 (days)
Answer: B still needs 56 days.
●Example 4For a project, team A will work alone for 10 days, and team B will work alone for 30 days. Now the two teams work together, and team A has two days off, while team B has eight days off (there is no such thing as two teams taking the same day off). How many days from the beginning to the completion?
Solution 1: Team A will work alone for 8 days and Team B will work alone for 2 days to complete the workload
The remaining workload is the cooperation of the two teams, and the number of days required is
2+8+1=11 (days)
A: It took 11 days from the beginning to the completion
Solution 2: Set the total workload as 30 copies. Party A completes 3 copies every day, and Party B completes 1 copy every day. After 8 days in Team A alone and 2 days in Team B alone, the two teams need to cooperate
(30 - 3 × 8 - 1 × 2) ÷ (3+1)=1 (day)
Solution 3: One day for Team A is equivalent to three days for Team B
After 8 days in Team A alone, the remaining workload (Team A) is 10-8=2 (days), which is equivalent to 2 × 3=6 (days) in Team B. After 2 days in Team B alone, the remaining workload (Team B) is 6-2=4 (days)
4=3+1，
Three days can be completed by Team A in one day, so the two teams only need to cooperate for another day
Solution 4：
Method: Split break and think together (The question said that the two teams did not have a rest together, so we assumed they had a rest together.)
The daily workload of Team A is 1/10 and that of Team B is 1/30. Because Team A has rested for 2 days and Team B has rested for 8 days, we assume that when Team A has rested for 2 days, Party B is also resting.When Party A starts to work, Party B will have a rest: 8-2=6 (days), then Party A has completed 1/10 × 6=6/10 of the project alone in these 6 days, and the remaining workload is 1-6/10=4/10, and the remaining 4/10 is the work volume completed by both Party A and Party B together, so 4/10 of the work volume needs the cooperation of Party A and Party B: (4/10) ÷ (1/10+1/30)=3 days.Therefore, from the beginning to the completion: 8+3=11 (days)
●Example 5For a project, Team A will work alone for 20 days and Team B will work alone for 30 days. Now they are working together, and Team A has three days off, while Team B has several days off. From the beginning to the completion, it takes 16 days. How many days has Team B rested?
Solution 1: If the two teams do not rest for 16 days, the workload that can be completed is (1 ÷ 20) × 16+(1 ÷ 30) × 16=4/3
The workload that the two teams did not do during the rest period is 4/3-1=1/3
The unfinished workload of Team B during the rest period is 1/3-1/20 × 3=11/60
The rest days of Team B are 11/60 ÷ (1/30)=11/2
A: Team B rested for five and a half days
Solution 2: Assume that the total workload is 60 copies. Party A completes 3 copies every day, and Party B completes 2 copies every day
The workload that the two teams did not do during the break was
(3+2) × 16 - 60=20 (portions)
Therefore, B's rest days are
(20 - 3 × 3) ÷ 2=5.5 (days)
Solution 3: Two days for Team A is equivalent to three days for Team B
Team A will rest for 3 days, which is equivalent to Team B's 4.5 days rest
If Team A does not rest for 16 days, only 4 days of workload is left for Team A, which is equivalent to 6 days of workload for Team B. The rest days for Team B are
16-6-4.5=5.5 (days)
●Example 6There are two jobs, one is A and the other is B. It takes 10 days for Zhang to complete the work of A alone, and 15 days for Zhang to complete the work of B alone;It takes 8 days for Li Dan to finish work A alone, and 20 days for Li Dan to finish work B alone. If two people can cooperate on each work, how many days will it take at least to finish both work?
Solution: It is obvious that Li is efficient in doing the work of A and Zhang is efficient in doing the work of B. So let Li do A first and Zhang do B first
The workload of design B is 60 (15 and 20Least common multiple）, Zhang completes 4 copies every day, and Li completes 3 copies every day
In 8 days, Li can finish the work of Party A. At this time, Zhang still has the remaining work of Party B (60-4 × 8)
(60-4 × 8) ÷ (4+3)=4 (days)
8+4=12 (days)
Answer: It will take at least 12 days to complete both tasks
●Example 7For a project, it takes 10 days for Party A to do it alone, and 15 days for Party B to do it alone. If two people cooperate, the efficiency of Party A will be reduced by 20%, and that of Party B will be reduced by 10%.They need 8 days to complete the project, and the number of days of cooperation between them should be as small as possible. How many days should they work together?
Solution: Assume that the workload of this project is 30, with Party A completing 3 copies every day and Party B completing 2 copies every day
Two people work together to complete
3 × 0.8+2 × 0.9=4.2 (portions)
Because the number of days for two people to work together should be as small as possible, and the only thing to do should be A with high work efficiency. Because it needs to be completed within 8 days, the number of days for two people to work together is
(30-3 × 8) ÷ (4.2-3)=5 (days)
Obviously, it turns into the problem of "chicken and rabbit in the same cage"
●Example 8Party A and Party B work together on one task. Due to good cooperation, Party A's work efficiency is faster than that of working alone
How many hours will it take if this work is always done by Party A alone?
Solution: The workload of Party B's six hour independent work is
The workload completed by Party B per hour is
The two people work together for 6 hours, and the workload completed by Party A is
Workload completed per hour when Party A works alone
The time required for A to do this work alone is
Answer: It will take 33 hours for Party A to complete the work alone
Most of the examples in this section have been "rounded". However, "rounding" does not make the calculation of all engineering problems easyExample 8 is like this. Example 8 can also be rounded up. When we find that
It's a little convenient, but it doesn't do much good
2、 Multi person engineering problem
There are at least three people in the multi person question. Of course, the multi person question is more complicated than the two person question, but the basic ideas for solving the problem are still the same
●Example 9A job can be completed in 36 days by the cooperation of Party A and Party B, 45 days by the cooperation of Party B and Party C, and 60 days by the cooperation of Party A and Party C. How many days does it take for Party A to complete the job alone?
Solution: Suppose the workload of this work is 1
The cooperation among Party A, Party B and Party C is completed every day
Subtract the workload completed by Party B and Party C every day, and Party A completes it every day
A: It takes 90 days for A to do it alone
Example 9 can also be rounded. Let's assume that the total workload is 180 copies. Party A and Party B cooperate to complete 5 copies every day, Party B and Party C cooperate to complete 4 copies every day, and Party A and Party C cooperate to complete 3 copies every day. Please try. Is it easier to calculate?
●Example 10For a job, it takes 12 days for Party A to do it alone, 18 days for Party B to do it alone, and 24 days for Party C. This job was done by Party A for several days, and then by Party B. The number of days for Party B was three times that of Party A, and then by Party C, and the number of days for Party C was two times that of Party B. Finally, the job was completed. How many days did it take?
Solution: Party A will do it for one day, Party B will do it for three days, and Party C will do it for 3 × 2=6 (days)
It means that Party A has done it for 2 days, Party B has done it for 2 × 3=6 (days), and Party C has done it for 2 × 6=12 (days)
2+6+12=20 (days)
Answer: It took 20 days to complete the work
Integeration of this question will bring convenience in calculation. 12, 18, 24 are three numbers that are easy to findLeast common multiple72. It can be assumed that the total workload is 72. A completes 6 every day, B completes 4 every day, and C completes 3 every day
●Example 11A project needs 13 days to be completed by the cooperation of Party A, Party B and Party C. If Party C takes 2 days off, Party B will do it for 4 more days, or Party A and Party B will cooperate for 1 day. How many days will it take for Party A to do this project alone?
Solution: Two days' workload of Party C is equivalent to four days' workload of Party B. The working efficiency of Party C is 4 ÷ 2=2 (times) of that of Party B. One day of cooperation between Party A and Party B is the same as that of Party B for four days. That is, one day of work by Party A is equivalent to three days of work by Party B. The working efficiency of Party A is three times of that of Party B
They work together for 13 days, which is completed by Party A alone
Answer: It takes 26 days for Party A to do it alone
In fact, when we calculate that the ratio of work efficiency of Party A, Party B and Party C is 3:2:1, we know that one day for Party A is equivalent to one day for cooperation between Party B and Party C. It takes 13 days for three people to cooperate, and the workload completed by Party B and Party C can be converted into another 13 days for Party A
●Example 12For a certain work, three people in Group A can complete the work in eight days, and four people in Group B can complete the work in seven days. Ask two people in Group A and seven people in Group B how long can they work together to complete the work?
Solution 1: Suppose the workload of this work is 1
Each person in Group A can complete it every day
Each person in Group B can complete it every day
Two people in Group A and seven people in Group B can complete it every day
A: The work can be completed in 3 days through cooperation
Solution 2: 3 people in Group A can finish it in 8 days, so 2 people can finish it in 12 days;Group B can be completed by 4 people in 7 days, so 7 people can be completed in 4 days
Now the number of people does not need to be taken into account, and the problem is transformed into:
Group A will work alone for 12 days, and Group B will work alone for 4 days. How many days will the cooperation be completed?
Primary school arithmetic should make full use of the particularity of the given data. Solution 2 is a typical example of flexible use of proportion. If you can do mental arithmetic well, you can quickly get the answer
●Example 13It takes workshop A 10 days to make a batch of parts. If workshop A and workshop B work together, it will take only 6 days. If workshop B and workshop C work together, it will take 8 days to complete. Now three workshops work together. After completion, it is found that workshop A makes 2400 more parts than workshop B. How many parts have workshop C made?
Solution 1: The total workload is still set as 1
A completes more than B every day
Therefore, the total number of parts in this batch is
The number of parts manufactured by workshop C is
Answer: Workshop C has produced 4200 parts
Solution 2: 10 and 6Least common multipleYes 30. It is assumed that the total workload of making parts is 30 copies. Party A completes 3 copies every day, and Party A and Party B together complete 5 copies every day. Thus, Party B completes 2 copies every day
Party B and Party C will complete it in 8 days. Party B will complete 8 × 2=16 (copies), and Party C will complete 30-16=14 (copies)
The ratio of working efficiency between B and C is 16:14=8:7
Known
The ratio of working efficiency between Party A and Party B is 3:2=12:8
To sum up, the ratio of working efficiency of A, B and C is
12∶8∶7.
When the three workshops work together, the number of parts made by C is
2400 ÷ (12 - 8) × 7=4200 (pieces)
●Example 14It takes 10 hours for A, 12 hours for B, and 15 hours for C to move the goods in one warehouse. There are the same warehouses A and B. A starts to move the goods in warehouse A and B at the same time. C starts to help A move the goods, and then turns to help B move the goods halfway. The last two warehouses move the goods at the same time. Ask C how long it takes to help A and B?
Solution: The workload of moving goods in one warehouse is 1. Now it is equivalent to the workload of three people. 2. The time required is
Answer: C helps A to move for 3 hours and B to move for 5 hours
The key to solving 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. Suppose that the total workload of moving a warehouse is 60. A moves 6, B moves 5, and C moves 4
Three people move together, need
60 × 2 ÷ (6+5+4)=8 (hours)
A needs C to help with the handling
(60 - 6 × 8) ÷ 4=3 (hours)
B needs C to help with the handling
(60 - 5 × 8) ÷ 4=5 (hours)
3、 Water pipe problems
Seen from the content of mathematics, the water pipe problem is the same as the engineering problem. The water injection or drainage of the pool is equivalent to a project, and the water injection or drainage volume is the workload. The water injection or drainage volume per unit time is the work efficiency. As for the problem of both injection and discharge, it is just the increase and decrease of the workload. Therefore,Water pipe problems and engineering problems are basically the same
Example 15 When pipes A and B are opened at the same time, the pool can be filled in 9 minutes. Now, open pipe A first, and then open pipe B 10 minutes later. After 3 minutes, the pool is filled. It is known that pipe A injects 0.6 cubic meters more water per minute than pipe B, what is the volume of the pool?
Solution: The amount of water injected by Party A per minute is: (1-1/9 × 3) ÷ 10=1/15
B The amount of water injected per minute is: 1/9-1/15=2/45
Therefore, the pool volume is: 0.6 ÷ (1/15-2/45)=27 (cubic meters)
Answer: The volume of the pool is 27 cubic meters
Example 16 There are some water pipes with equal water injection per minute. Now open some of them
The water tank can be filled as scheduled by doubling the open water pipes. If 10 water pipes are opened at the beginning, the water tank can also be filled as scheduled if no water pipes are opened halfway. How many water pipes are opened at the beginning?
analysis: After the water pipe is opened, there are two times of the original water pipe. The water injection time is 1-1/3=2/3 of the scheduled time, and 2/3 is twice of 1/3. Therefore, the water injection volume in this period after the water pipe is opened is four times of that in the previous period.The capacity of the pool is set as 1, and the ratio of the water injection volume of the two periods is 1:4,
Then the water injection volume of 1/3 of the scheduled time (i.e. the previous period) is 1/(1+4)=1/5.
10 water pipes can be opened at the same time and filled with water according to the scheduled time. The water injection volume of each water pipe is 1/10, 1/3 of the scheduled time, and the water injection volume of each water pipe is 1/10 × 1/3=1/30
To fill 1/5 of the pool, 1/5 ÷ 1/30=6 water pipes are needed
Solution: The ratio of water injection volume of the two periods before and after is: 1: [(1-1/3) ÷ 1/3 × 2]=1:4
The amount of water injected in the previous period is: 1 ÷ (1+4)=1/5
The water injection volume of each water pipe in the preset 1/3 time is: 1 ÷ 10 × 1/3=1/30
Number of open water pipes at the beginning: 1/5 ÷ 1/30=6 (pieces)
Answer: At the beginning, open 6 water pipes.
Example 17The reservoir has two inlet pipes A and C, and two drain pipes B and D. To fill a pool of water, it takes three hours to open a single pipe A, and five hours to open a single pipe C. To drain a pool of water, it takes four hours to open a single pipe B, and six hours to open a single pipe D. Now there is one sixth of the water in the pool, such as opening it for one hour in turn in the order of A, B, C, D, A, B, etc,How long after the water began to overflow the pool?
analysis:
This question is similar to the popular "frog climbing the well": a frog who fell into a dry well has to climb 30 feet to reach the wellhead. Every hour, it always climbs 3 feet and slides down 2 feet. How many hours does it take for the frog to climb to the wellhead?
It seems that it only climbs 3 - 2=1 (feet) every hour, but after 27 hours, it climbs for another hour, and then climbs 3 feet to reach the wellhead
Therefore, the answer is 28 hours, not 30 hoursIn the future (20 hours), the water in the pool has already existed, otherwise the water in the pool will overflow during the opening of pipe A
Example 18A cistern flows 4 cubic meters of water every minute. If you turn on five taps, you will empty the water in the cistern in two and a half hours. If you turn on eight taps, you will empty the water in the cistern in one and a half hours. Now turn on 13 taps, and ask how long it will take to empty the water?
Solution: First calculate the amount of water discharged by one faucet per minute
Two and a half hours is 60 minutes more than one and a half hours, and more water flows
4 × 60=240 (cubic meters)
Time is measured in minutes. The water discharge per minute of one faucet is
240 ÷ (5 × 150 - 8 × 90)=8 (cubic meters),
The amount of water discharged by 8 taps in one and a half hours is
8 × 8 × 90，
The inflow volume within 90 minutes is 4 × 90, so there is 8 × 8 × 90-4 × 90=5400 (m3) water in the original pool
Open 13 faucets to discharge 8 × 13 water every minute. Except for the inflow of 4 water every minute, the remaining water will be discharged and the original 5400 water will be emptied
5400 ÷ (8 × 13 - 4)=54 (minutes)
Answer: It takes 54 minutes to open 13 taps and empty the pool
The water in the pool has two parts. The original water and the new water need to be considered separately. The key to solving this problem is to find out the original water in the pool first, which is implicit in the topic
Example 19For a pool, groundwater seeps into the pool from the four walls, and the amount of water seeping into the pool is fixed every hour. Open pipe A to drain the full pool water in 8 hours, and open pipe C to drain the full pool water in 12 hours. If you open pipe A and pipe B, you can drain the water in 4 hours. If you open pipe B and pipe C, how many hours will it take to drain the full pool water?
Solution: The water volume of the full pool is 1
Pipe A is discharged every hour
Pipe A discharges for 4 hours
Therefore, B and C pipes are opened together, and the displacement per hour is
B. C The two pipes are opened together to drain the water full of the pool. The time required is
Answer: It takes 4 hours and 48 minutes to drain the full pool of water before the pipes B and C are opened together
This question should also be considered separately. The original water (full) and infiltration water volume of the pool. Since the specific quantity is unknown, just like the specific quantity of the workload is unknown for engineering problems, the two water volumes are set as "1". However, the two quantities should not be confused. In fact, the original water can also be set as 8 and 12Least common multipletwenty-four
Great British scientist in the 17th centuryNewtonHe once wrote the book General Arithmetic, in which he proposed“Cattle eat grass”This is an interesting arithmetic problem. Essentially, it is similar to Example 18 and Example 19. The problem involves three quantities: original grass, newly grown grass, and grass eaten by cattle. It is completely similar to the original water volume, infiltration water volume, and water discharged from water pipes
Example 20There are three pastures where the grass grows as dense and as fast.12 cows eat the first piece of grass on the pasture in 4 weeks;7 cows eat the grass in the second pasture in 9 weeks. How many cows can eat the grass in the third pasture in 18 weeks?
Solution: Total grass consumption=grass consumption of one cow per week × number of cattle heads × number of weeks. According to this calculation formula, "grass consumption of one cow per week" can be set as the measurement unit of grass
Original grass+4 weeks of new grass=12 × 4
Original grass+newly grown grass in 9 weeks=7 × 9
It can be concluded that the new grass growing every week is
（7×9-12×4）÷（9-4）=3.
So 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 newly grown grass in 18 weeks is
These grasses can make
90 × 7.2 ÷ 18=36 (head)
Cattle eat for 18 weeks
Answer: 36 cows can eat the third piece of grass in 18 weeks
The solution of Example 20 is slightly different from that of Example 19. Example 20 specifically calculates the "new length", and unifies the "old" and "new length". In fact, if Example 19 has another condition, such as "open pipe B, and the full pool water can be drained within 10 hours." It can also calculate the "new length" and "old"The quantitative relationship between them. But it is just the requirement of Example 19. It is unnecessary to add this condition. Think about it carefully, can you understand the reason?
The problem of "cattle grazing" can appear in a variety of ways. Due to space limitations, we will just give another example
Example 21The exhibition opens at 9 o'clock, but people have already queued up for admission. Since the first audience arrived, the number of people coming in every minute is the same. If three entrances are opened, there will be no more people queuing at 9:9 o'clock. If five entrances are opened, there will be no people queuing at 9:5 o'clock. Ask the first audience what the arrival time is at 8:00 o'clock?
Solution: Set the audience that can enter the entrance every minute as a unit of calculation
From 9:00 to 9:9, the audience is 3 × 9,
From 9:00 to 9:5, the audience is 5 × 5
Because the audience came more 9-5=4 (minutes), the audience coming every minute is
（3×9-5×5）÷（9-5）=0.5.
The audience before 9:00 is
5×5-0.5×5=22.5.
These audiences come to need
22.5 ÷ 0.5=45 (minutes)
A: The first audience arrived at 8:15
It will take six days for teams A and B to dig a canal together.Team A will dig for three days first, and Team B will dig for another day. It can dig 3/10 of this canal. How many days will it take for the two teams to dig separately?
Analysis: One day after the cooperation between Party A and Party B, Party A did it for two more days, 3/10-1/6=4/30
2÷(3/10-1/6)
=2÷4/30
=15 (days)
1 ÷ (1/6-1/15)=10 (days)
A: It takes 15 days for A to do it alone, and 10 days for B to do it alone.
