Workload = work efficiency × time.
In primary school mathematics, we all call the applied problem of exploring the relationship between these three quantities "engineering problem".
Give a simple example.
A job can be completed in 0/0 day by Party A and 0/5 day by Party B/Kloc. How many days does it take for two people to work together?
As a whole, an assignment is 1, so the workload can be counted 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 1 day is a unit.
According to the basic quantitative relationship, we get
Time required = workload/work efficiency
=6 (days)?
It takes six days for two people to cooperate.
This is the most basic problem in engineering, and many examples introduced in this lecture are developed from this problem.
In order to calculate integers (as far as possible), the workload is divided into more shares like the third example 3 and example 8. Again, the least common multiple of 10 and 15 is 30. Assume that the total workload is 30 copies. Then Party A will complete 3 copies every day and Party B will complete 2 copies every day. How many days does it take for two people to cooperate?
30(3+2)= 6 (days)
It is more convenient to calculate the numbers.
: 2. Or "the workload is fixed, and the work efficiency is inversely proportional to the time". The work efficiency ratio of Party A and Party B is 15: 10 = 3: 2. When the working efficiency ratio of the two is known, consider the problem from the perspective of proportion, and also
The required time is
Therefore, in the description of the following examples, we do not completely adopt the practice of "setting the workload as a whole 1" in the usual textbooks, but focus on "integers" or "from the perspective of proportion", which may make our thinking of solving problems more flexible.
First, the problem of two people.
The two people mentioned in the title 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 first, and B continues to finish the rest. B How many days does it take to finish all the work?
Answer: B It takes 4 days to finish all the work.
Scheme 2: The least common multiple of 9 and 6 is 18. Let the total workload be 18. Party A completes 2 copies every day, and Party B completes 3 copies every day. How long does it take Party B to complete the remaining work?
(18- 2 × 3)÷ 3= 4 (days).
Solution 3: The ratio of working efficiency of A and B.
6∶ 9= 2∶ 3.
A has done 3 days, which is equivalent to 2 days for B. It takes 6-2=4 (days) for B to finish the rest of the work.
With the cooperation of both parties, a job can be completed in 30 days. After 6 days, Party A left and Party B continued to do it for 40 days. If this work is done by Party A or Party B alone, how many days will it take?
Solution: * * * did it for 6 days.
It turns out that A does it for 24 days and B does it for 24 days.
Now, A does 0 days and B does 40=(24+ 16) days.
This shows that the work that A did in 24 days can be replaced by B in 16 days, so the work efficiency of A is high.
If b does it alone, the time required is
If A does it alone, the time required is
Answer: It takes 75 days for A to do it alone, or 50 days for B to do it alone.
A project can be completed by Party A alone for 63 days, and then by Party B alone 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: First, compare the following:
63 days for A and 28 days for B;
A does it for 48 days and B does it for 48 days.
It is known that A needs to do 63-48= 15 (days) less, and B needs to do 48-28=20 (days) more, thus obtaining A's.
A has been doing it for 42 days, and 63-42=2 1 (day) is less than 63 days, which is equivalent to B.
So, B still has to do it.
28+28= 56 (days).
A: B It will take another 56 days.
Example 4 A project was completed by group A alone 10 day, and group 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 (neither team has a day off). How many days did it take from the beginning to the end?
Scheme 1: Team A works alone for 8 days and Team B works alone for 2 days, thus completing the workload.
The remaining workload is the cooperation between the two teams. How many days will it take?
2+8+ 1= 1 1 (days).
Answer: It took 1 1 day from the beginning to the end.
Solution 2: We assume that the total workload is 30 copies. Party A completes 3 copies every day, and Party B completes 1 copy every day. Team A works alone for 8 days, and after team B works alone for 2 days, the two teams need to cooperate.
(30-3× 8-/kloc-0 /× 2) ÷ (3+1) =1(days).
Option 3: Team A does 1 day, which is equivalent to Team B doing 3 days.
After team A did it alone for 8 days, there was still (team A) 10-8= 2 (days) workload, which was equivalent to team B's 2×3=6 (days). After doing it alone for 2 days, there was still (team B) 6-2=4 (days) workload.
4=3+ 1,
Three days can be completed by team A 1 day, so the two teams only need to cooperate 1 day.
Example 5 A project is completed by Team A in 20 days and Team B in 30 days. 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 16 days from start to finish. How many days did Team B rest?
Option 1: What if both teams don't rest for 16 days?
Because the workload that the two teams didn't do during the break was
The amount of work that team B didn't finish during the break was
How many days will Team B rest?
A: Team B rested for five and a half days.
Solution 2: Assume that the total workload is 60. Party A completes 3 copies every day, and Party B completes 2 copies every day.
The workload that the two teams didn't do during the break was
(3+2)× 16- 60= 20 (copies).
Therefore, B's rest day is
(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 has a 3-day rest, which is equivalent to 4.5 days rest for Team B. 。
If Team A 16 doesn't rest, Team A will only work for 4 days, which is equivalent to 6 days for Team B. Team B's rest day is
16-6-4.5=5.5 (days).
Example 6 has two tasks, A and B. It takes 65,438+00 days for Zhang to complete task A alone, and 65,438+05 days for Zhang to complete task B 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 in doing A work is high, and Zhang's work efficiency in doing B work is also high. So let Li do A first and Zhang do B first.
Suppose B's workload is 60 copies (15 and the least common multiple of 20), Zhang completes 4 copies every day and Li completes 3 copies every day.
In another 8 days, Li will be able to finish work A. At this time, Zhang still has (60-4×8) copies of work B, which needs the cooperation of Zhang and Li.
(60-4×8)÷(4+3)=4 (days).
8+4= 12 (days).
A: It will take at least 12 days to complete these two tasks.
For 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, they will.
It takes eight days to complete the project, and the fewer days two people work together, the better. So how many days do two people work together?
Solution: 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.
Two people cooperate, * * * to complete.
3× 0.8+2 × 0.9= 4.2 (copies).
Because two people should work together for as few days as possible, the one who works alone should be the one with high efficiency. Because it will be completed in 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".
Example 8 Party A and Party B cooperate in a job. Due to their good cooperation, Party A's work efficiency is higher than that of working alone.
How many hours will it take if this work is always done by one person alone?
Solution: what is the workload of B working alone for 6 hours?
B the workload per hour is
Two people work together for 6 hours. What does A accomplish?
A the amount of work done per hour when working alone
A How long does it take a person to do this job?
A:A It takes 33 hours for one person to finish the work.
Most examples in this section are treated as "integers". 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, when b is found.
It's convenient, but it doesn't do much good. There is no need to reinvent the wheel.
Second, many people's engineering problems.
When we talk about many people, there are at least three. 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 job is completed in 36 days by Party A and Party B, 45 days by Party B and 60 days by Party A and Party C. How many days does it take for Party A to complete it alone?
Solution: Let the workload of this work be 1.
The cooperation among Party A, Party B and Party C is completed every day.
Minus the work done by Party B and Party C every day, Party A will finish it every day.
A:A It takes 90 days to do it alone.
Example 9 can also be rounded off, assuming that the total workload is 180, Party A and Party B complete 5 copies per day, Party B and Party C complete 4 copies per day, and Party A and Party C complete 3 copies per day. Please have a try. Will it be more convenient to calculate?
Example 10 for a job, it takes 12 days for A to do it alone, 18 days for B to do it alone, and 24 days for C to do it alone. The work was done by A for a few days, then by B for three times as many days as A, and then by C for twice as many days as B, and finally the work was completed.
Solution: A does 1 day, B does 3 days, and C does 3×2=6 (days).
Explain that A did it for 2 days, B did it for 2×3=6 (days), C did it for 2×6= 12 (days), and three people did it together.
2+6+ 12=20 (days).
It took 20 days to finish the work.
The integer of this problem will bring convenience to the calculation. There is an easy-to-find minimum common multiple 12,18,24. 72. It can be assumed that the total workload is 72. A completes 6 every day, B completes 4 and C completes 3. The total workload is * * *.
Example: 1 1 A project needs the cooperation of Party A, Party B and Party C 13 days. If Party C asks for two days off, Party B has to do four more days, or both parties cooperate 1 day. How many days will it take for this project to be completed by Party A alone?
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, and the cooperation between Party A and Party B 1 day is the same as that of Party B's four days. In other words, Party A works 1 day, which is equivalent to Party B's working for 3 days, and Party A's working efficiency is 3 times that of Party B. 。
They worked together for 13 days, which was done by Party A alone and needed by Party A..
A:A It takes 26 days to do it 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, and the cooperation of three people takes 13 days, and the workload completed by both parties can be converted into Party A.
Example 12 For a certain job, three people in Group A can finish the job in eight days, and four people in Group B can finish the job in seven days. How long will it take two people in group A and seven people in group B to finish the work together?
Scheme 1: Let the workload of this work be 1.
Everyone in group a can finish it every day.
Everyone in group b can finish it every day.
Two people in group A and seven people in group B can finish it every day.
A: The work can be completed in three days.
Scheme 2: 3 people in group A can finish it in 8 days, then 2 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, the question becomes:
Group a worked alone 12 days, while group b worked 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, but it only takes 6 days if workshop a and workshop b do it together. Workshop b and workshop c work together, which takes 8 days to complete. Now three workshops are working together, and it is found that workshop A has made 2400 more parts than workshop B. How many parts has workshop C made?
Scheme 1: Let the total workload be 1.
A finishes more than B every day.
So the total number of these parts is
The number of parts produced in Workshop C is
A: Workshop C has produced 4200 parts.
The least common multiple of solution 2: 10 and 6 is 30. We assume that the total workload of making parts is 30 copies. Party A completes 3 copies a day, and Party A and Party B complete 5 copies a day together, resulting in Party B completing 2 copies a day.
Party B and Party C will complete it within 8 days. When Party B completes 8×2= 16 (copies) and Party C completes 30- 16= 14 (copies), you will know.
The efficiency ratio of B and C is16:14 = 8: 7.
known
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.
When three workshops work together, the number of parts produced by C is
2400( 12-8)×7 = 4200 (pieces).
Example: 14 A needs 10 hour, b needs 12 hour, and c needs 15 hour. Under the condition that warehouses A and B are the same, A starts to carry goods in warehouses A and B at the same time, and C starts to help A, and then turns to help B in the middle. Finally, the goods in the two warehouses move at the same time. Ask C.
Solution: Let's assume that the workload of moving goods in a warehouse is 1. Now it is equivalent to three people * * * completing workload 2, and the time required is
Answer: C helps A to carry for 3 hours, and B carries for 5 hours.
The key to solve this problem is to calculate the time for three people to carry two warehouses at the same time. Of course, the calculation of this problem can also be rounded off, assuming that the total workload of moving a warehouse is 60. A transports 6 per hour, B transports 5 per hour and C transports 4 per hour.
Three people * * * together, need
60 × 2÷ (6+ 5+ 4)= 8 (hours).
A needs c to help carry it.
(60- 6× 8)÷ 4= 3 (hours).
B needs c to help carry it.
(60- 5× 8)÷4= 5 (hours)
Third, the problem of water pipes.
From a mathematical point of view, the water pipe 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 or water discharge 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.
Example 15 A and b pipes are opened at the same time, and the pool can be filled in 9 minutes. Now 10 minutes to open the first tube A, open the second tube, and fill the pool in 3 minutes. It is known that the first pipeline A injects 0.6 cubic meters more water per minute than the second pipeline. What is the volume of this swimming pool?
Water injection per minute is
B water injection per minute is
So the volume of the pool is
A: The volume of the pool is 27 cubic meters.
16 has some water pipes, and the amount of water injected per minute is equal. now
Fill the swimming pool at the scheduled time. If you open the 10 water pipe at the beginning, you can also fill the pool at the scheduled time, without adding additional water pipes in the middle. How many water pipes were opened at the beginning?
Answer: Open 6 water pipes at the beginning.
Example 17 Reservoir has two water inlets A and C and two water drains B and D. To fill a pool of water, it takes 3 hours to open pipe A and 5 hours to open pipe C separately. It takes 2 hours to drain a pool of water and just open the pipe B.
Turn on in turn 1 hour in turn, b, ... How long does it take for water to start overflowing the swimming pool?
Otherwise, in the process of opening the nail tube, the pool water will overflow.
After (20 hours), there was water in the pool.
This problem is similar to the widely circulated "frog climbing well": a frog falling into a dry well has to climb 30 feet to reach the wellhead, and it always climbs 3 feet every hour and slides 2 feet. 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.
For example: 18 A reservoir flows in 4 cubic meters of water every minute. If you turn on five faucets, the water in the tank will be emptied in two and a half hours; If you turn on 8 faucets for 1.30 hours, the water in the reservoir will be emptied. Now turn on the 13 faucet. How long does it take to empty the water?
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 more water flows in.
4 × 60= 240 (cubic meter).
Time is calculated in minutes, 1 the water output per minute of the faucet is
240 ÷ (5× 150- 8 × 90)= 8 (m3)，
8 faucet 1 half-hour discharge is
8 × 8 × 90,
Among them, the water inflow within 90 minutes is 4 × 90, so the original pool has 8 × 8 × 90-4 × 90= 5400 (cubic meters) of water.
Turn on the 13 faucet, and 8× 13 of water can be discharged every minute. Except for the inflow of 4 per minute, the rest will discharge the original water and empty the original 5400, which is necessary.
5400 ÷(8 × 13- 4)=54 (minutes).
Answer: It takes 54 minutes to turn on 13 faucet and empty the swimming pool.
There are two parts of water in the pond, primary water and new inflow water, which need to be considered separately. The key to solve this problem is to find out the original water in the pool first, which is implicit in the question.
Example 19 in a pool, groundwater seeps into the pool from four walls, and the amount of water that seeps into the pool per hour is fixed. When the tube A is opened, the water in the pool can be drained in 8 hours, and when the tube C is opened, the water in the pool can be drained in 12 hours. If the pipes A and B are opened, the water can be drained in 4 hours. How many hours does it take to drain the water in the swimming pool?
Solution: The water filled in the pool is 1.
Tube a is discharged every hour.
Tube a was discharged within 4 hours.
Therefore, b, c two-pronged, hourly displacement is
B, c two-pronged, will be full of water, the required time is
A: It takes 4 hours and 48 minutes for B and C to drain the pool.
This problem should also be considered separately, primary water (full pool) and seepage water. Because we don't know the specific quantity, just like the specific quantity of workload in engineering problems, we set the two kinds of water as "1" respectively, but we should avoid confusion between them. In fact, we can also round the original water to the lowest common multiple of 8 and 12 of 24.
/kloc-Newton, a great British scientist in the 0 th and 7 th centuries, wrote a book "Arithmetic of Everything". In the book, he put forward a problem of "cattle eating grass", which is an interesting arithmetic problem. Essentially similar to example 18 and example 19. The topic involves three quantities: original grass, new grass and grass eaten by cattle.
There are three pastures, and the grass grows equally dense, and one grows.
Grass; 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.
It can be concluded that the new grass grows 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 grassland, the total growth of original grass and new grass in 18 weeks is
These grasses can be made.
90×7.2÷ 18=36 (head)
Cattle eat 18 weeks.
Answer:/kloc-36 cows can eat grass from the third pasture in 0/8 weeks.
The solution of Example 20 is slightly different from the solution of Example 19. In Example 20, the "new length" is specifically found, and the two quantities of "original length" and "new length" are calculated together. In fact, if the example 19 has another condition, such as "open the B tube, 10 hour can fill the pool.
The problem of "cattle eating grass" can appear in various forms. Limited to space, we only give one more example.
Example 2 1 art exhibition opens at 9 o'clock, but people are already 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. When will the first audience arrive?
Solution: Let an entrance have 1 audience per minute.
The audience entering from 9: 00 to 9: 09 is 3x9.
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.
The audience before 9 o'clock is
5×5-0.5×5=22.5.
These audiences are beginning to need it.
22.5 0.5 = 45 (minutes).
A: The first audience will arrive at 8: 00. 15.