What Is Methods Engineering?

Engineering problems are the key points in the application of mathematics problems in elementary and middle schools. They are extensions and supplements of fractional problems. They are an important tool for training students' logical thinking ability. It is a powerful penetration of the one-to-one correspondence of ideas in applied problems. Engineering issues are also difficult points in the textbook.

Engineering problems

In daily life, doing something, manufacturing a certain product, completing a certain task, completing a certain project, etc. all involve the three quantities of total work, work efficiency, and working time. The basic quantitative relationship is-work efficiency × time = total work.

In elementary school mathematics, to explore the application of the relationship between these three quantities, we call them "engineering problems."

To give a simple example: A job can be completed in 15 days for A and 10 days for B. How many days can the two work together?

A job is considered as a whole, so the workload can be counted as 1. The so-called work efficiency is the workload completed in a unit of time. The unit of time we use is "day", and 1 day is a unit.

According to the basic quantity relationship, we get

Workload ÷ work efficiency = work time

1 ÷ 1/15 + 1/10

= 6 (days)

Answer: It takes 6 days for the two to cooperate.

This is the most basic problem in engineering problems, and many of the examples introduced in this lecture have evolved from this problem. For calculation

1: Basic quantitative relationship

1. Work efficiency × time = total work 2. Work efficiency = total work ÷ work time 3. Work time = total work ÷ work efficiency.

Two: basic characteristics

Set the total work amount to "1" and the ergonomics = 1 / time.

Three: Basic methods

Arithmetic method, proportional method, equation method.

Four: basic ideas

Think together, think together.

Five: Types and methods

One: Do something together: 1. Think together, 2.

When you know the ratio of the two working efficiency, consider the problem from the perspective of proportion, you can also flexibly answer.

Therefore, in the description of the following example questions, the practice of setting the workload to 1 as a whole in the usual textbooks is not fully adopted, and the emphasis on integerization or from a proportional perspective may make our problem-solving ideas More flexible.

A problem with one or two people

The title "two people" can also be two groups of two groups, two teams, etc.

Example 1A work can be completed in 9 days for A and 6 days for B. Now A has done three days, and the rest of the work will be completed by B. How many days does it take for B to complete all the work?

Solution 1: Consider this work as 1, A can complete one-ninth of this work every day, and do one-third of the work 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: The least common multiple of 9 and 6 is 18. Let the total workload be 18 copies. A completes 2 copies per day and B completes 3 copies per day. The time required for B to complete the remaining work is

(18- 2 × 3) ÷ 3 = 4 (days).

Solution 3: The ratio of work efficiency of A and B is

6: 9 = 2: 3:

A does 3 days, which is equivalent to B doing 2 days. The time required for B to complete the remaining work is 6-2 = 4 (days).

Example 2 A job can be completed in 30 days by cooperation between A and B. After 6 days of joint work, A leaves, and B continues to work for 40 days. If this work is completed by A or B alone, how much is required day?

Solution: After doing 6 days in total,

It turned out that A did 24 days and B did 24 days.

Now, A does 0 days and B does 40 = (24 + 16) days.

This shows that the work done by A for 24 days can be replaced by B for 16 days. Therefore, A's work efficiency is B's work efficiency

(Times)

Doing for 6 days is equivalent to doing for B

(day),

If B does it alone, the time required is 6 + 4 + 40 = 50 days.

If Jia do it alone, the time required is

day

Answer: The time required for A or B to be independent is 75 days and 50 days respectively.

Example 3 A project is completed by A alone for 63 days, and then completed by B alone for 28 days. If A and B cooperate, it takes 48 days to complete. Now A does 42 days alone, and then B does it alone. How many days does B need to do?

Solution: First compare as follows:

A does 63 days, B does 28 days;

A for 48 days, B for 48 days.

We know that A does less 63-48 = 15 (days), and B does 48-28 = 20 (days), which leads to A's work efficiency.

Is working efficiently

(Times).

A first did 42 days alone, 63-42 = 21 (days) less than 63 days,

Equivalent to B to do

(day)

Therefore, B has to do

28 + 28 = 56 (days).

Answer: B needs 56 days to do it.

Example 4 A project, Team A completed 10 days alone, Team B completed 30 days alone. Now the two teams cooperate, during which Team A rests for 2 days, and Team B rests for 8 days. ). How many days have it taken from the beginning to the completion?

Solution 1: Team A does 8 days alone, Team B does 2 days alone, and the workload is completed

The remaining workload is a joint effort between the two teams. 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 to 30 copies. A completes 3 copies per day and B completes 1 copy per day. After Team A does it alone for 8 days and Team B does it alone for 2 days, 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 Team A has done it alone for 8 days, the remaining (Team A) 10-8 = 2 (days) workload. It is equivalent to Team B doing 2 × 3 = 6 (days). After Team B has done it alone for 2 days, the remaining ( Team B) 6-2 = 4 (days) workload.

4 = 3 + 1,

3 days can be completed by Team A in 1 day, so the two teams only need to cooperate for another day.

Solution four :

Method: Divided and thought together (in the title, the two teams did not rest together, we assume that they rest together.)

Team A's daily workload is 1/10, B is 1/30, because A rests for 2 days, and B rests for 8 days, because 8> 2, so we assume that when A rests for two days, B is also resting. Then when A starts to work, B still needs to rest: 8-2 = 6 (days) Then within 6 days, A completes 1/10 × 6 = 6/10 of this project alone, and the remaining workload is 1-6 / 10 = 4/10, and the remaining 4/10 is the amount of work completed by both A and B. Therefore, 4/10 of the amount of work requires the cooperation of A and B: (4/10) ÷ (1/10 + 1 / 30) = 3 days. So from start to finish: 8 + 3 = 11 (days)

Example 5: A project is completed by team A for 20 days alone, and team B is completed for 30 days alone. Now they are working together, with team A resting for 3 days and team B resting for several days. From the beginning to completion 16 days. How many days did Team B rest?

Solution 1: If both teams do not rest for 16 days, the workload that can be completed is (1 ÷ 20) × 16 + (1 ÷ 30) × 16 = 4/3

Because the amount of work not done during the rest of the two teams is 4 / 3-1 = 1/3

The amount of work not done during the rest of Team B is 1 / 3-1 / 20 × 3 = 11/60

The rest days of Team B are 11/60 ÷ (1/30) = 11/2

Answer: Team B rested for 5 and a half days.

Solution 2: Set the total workload to 60 copies. A completes 3 copies per day and B completes 2 copies per day.

The amount of work not done during the rest of the two teams is

(3 + 2) × 16- 60 = 20 (parts).

So B rest days are

(20- 3 × 3) ÷ 2 = 5.5 (days).

Solution 3: Team A does 2 days, which is equivalent to Team B doing 3 days.

Team A has a rest of 3 days, which is equivalent to Team B having a rest of 4.5 days.

If Team A does not rest for 16 days, only Team A has 4 days of work, which is equivalent to Team B's 6 days of work.

16-6-4.5 = 5.5 (days).

Example 6 has two tasks, A and B. Zhang needs 10 days to complete A's work alone, 15 days to complete B's work alone; Lee has 8 days to complete A work alone, and 20 days to complete B work alone. Two people can work together, so how long does it take to complete both tasks?

Solution: Obviously, Li's work efficiency is high for Zhang's work, and Zhang's work efficiency is high for Zhang's work.

Let B have a workload of 60 (the least common multiple of 15 and 20), Zhang completes 4 copies per day, and Lee completes 3 copies per day.

In 8 days, Li can complete job A. At this time, Zhang also has the remaining job B (60-4 × 8). Cooperation by Zhang and Li needs

(60-4 × 8) ÷ (4 + 3) = 4 (days).

8 + 4 = 12 (days).

Answer: It takes at least 12 days to complete both tasks.

Example 7: For a project, it takes 10 days for A to do the work alone, and 15 days for B to do it alone. They have to complete the project in 8 days. The number of cooperation days between the two is as small as possible. So how many days will the two cooperate?

Solution: Suppose the workload of this project is 30 copies, A completes 3 copies per day, and B completes 2 copies per day.

Two people cooperate to complete

3 × 0.8 + 2 × 0.9 = 4.2 (parts).

Because the number of days of cooperation between the two should be as small as possible, the one who should do it alone should have a higher work efficiency. Because it must be completed within 8 days, the number of days of cooperation between the two is

(30-3 × 8) ÷ (4.2-3) = 5 (days).

Obviously, it turned into a "chicken and rabbit in the same cage" problem.

Example 8 A and B work together on one job. Due to the good cooperation, A's work efficiency is faster than when they do it alone.

How many hours would it take if the job was always done by one person alone?

Solution: The amount of work completed by B for 6 hours alone is

The amount of work done per hour is

The two worked together for 6 hours. The workload that A completed was

The amount of work done per hour when doing it alone

The time required for A to do this work alone is

Answer: A needs 33 hours to complete this work alone.

Most of the examples in this section have been treated with "integerization". However, "integerization" does not make the calculation of all engineering problems easy. Example 8 is like this. Example 8 can also be integerized.

It's a little convenient, but the benefits are not great. You don't have to do it all.

Engineering problems with multiple people

We speak of many people, at least 3 people, of course, the problem of more people is more complicated than the problem of 2 people, but the basic idea of solving the problem is still similar.

Example 9: A job is completed in 36 days with A and B, 45 days with B and C, and 60 days with A and C. How many days does it take for A to be alone?

Solution: Let the workload of this work be 1.

A, B, and C work together every day

Minus the amount of work that B and C completed each day, and A completed each day.

Answer: It takes 90 days for Jia to do it alone.

Example 9 can also be integerized. Let the total workload be 180 copies. A and B cooperation will complete 5 copies per day, B and C cooperation will complete 4 copies per day, and A and C cooperation will complete 3 copies per day. Please try it and calculate whether it will More convenient?

Example 10: A job requires 12 days for A to do, 18 days for B to do, and 24 days for C to do. This work is done by A for several days, and then by B. The number of days for Party A is 3 times, and then for C, the number of days for Party C is 2 times the number of days for Party B. Finally, this task is done. How many days did it take?

Solution: A does 1 day, B does 3 days, and C does 3 × 2 = 6 (days).

Explain that A did 2 days, B did 2 × 3 = 6 (days), and C did 2 × 6 = 12 (days).

2 + 6 + 12 = 20 (days).

Answer: It took 20 days to complete this work.

The integerization of this question will bring convenience in calculation. There is an easy-to-find least common multiple of the three numbers 12, 18, and 24. The total workload can be set to 72. A completes 6 daily, B completes 4 daily, and C completes daily. 3. Used in total

Example 11: A project requires three days for A, B, and C to cooperate. If C has two days off, B must do four more days, or two days for cooperation between A and B. Ask this project by How many days does Jiadu need to do?

Solution: The workload of C for 2 days is equivalent to the workload of B for 4 days. The work efficiency of C is 4 ÷ 2 = 2 (times) of the work efficiency of B. A and B cooperate for 1 day, which is the same as B for 4 days. That is, if A does 1 day, it is equivalent to B doing 3 days. A's working efficiency is 3 times that of B.

They work together for 13 days, which is completed by A separately. A needs

A: It takes 26 days for Jiadu to do it alone.

In fact, when we calculate the working efficiency ratio of three people: A, B, and C: 3: 2: 1, we know that A does one day, which is equivalent to one day for B and C. The three people need 13 days for cooperation, of which B, The amount of work done by two people can be converted into another one for 13 days to complete.

Example 12: For a certain task, 3 people in group A can complete the work in 8 days, and 4 people in group B can complete the work in 7 days. How long can 2 people in group A and 7 people in group B complete this task?

Solution 1: Let the workload of this work be 1.

Each person in Group A can complete each day

Each person in Group B can complete each day

2 people in group A and 7 people in group B can complete each day

Answer: 3 days of cooperation can complete this job.

Solution 2: 3 people in group A can complete in 8 days, so 2 people can complete in 12 days; 4 people in group B can complete in 7 days, so 7 people can complete in 4 days.

It is no longer necessary to take into account the number of people, the problem turns into:

Group A will do it alone for 12 days and Group B will do it alone for 4 days. How many days will the cooperation be completed?

Primary school arithmetic should make full use of the particularity of the data given. Solution two is a typical example of flexible use of proportions. If you have a good mental arithmetic, you can quickly get the answer.

Example 13 Production of a batch of parts, workshop A takes 10 days to complete, if workshop A and workshop B do together, it only takes 6 days to complete. Workshop B and workshop C, it takes 8 days to complete. Now three workshops do together After completion, I found that workshop A produced 2,400 more parts than workshop B. How many parts did workshop C make?

Solution one: still set the total workload to 1.

A completes more than B every day

So the total number of parts is

The number of parts made in workshop C is

Answer: Workshop C produced 4,200 parts.

Solution 2: The least common multiple of 10 and 6 is 30. Set the total workload of manufacturing parts to 30 copies. A completes 3 copies a day, A and B together complete 5 copies a day, which results in B completing 2 copies a day.

B and C together, completed in 8 days. B completed 8 × 2 = 16 (parts), C completed 30-16 = 14 (parts), we know

The working efficiency ratio of B and C is 16: 14 = 8: 7.

A known

The working efficiency ratio of A and B is 3: 2 = 12: 8.

Taken together, the work efficiency ratio of the three people is

12: 8: 7.

When the three workshops do it together, the number of parts made by C is

2400 ÷ (12- 8) × 7 = 4200 (pieces).

Example 14 To move the goods in a warehouse, A needs 10 hours, B needs 12 hours, and C needs 15 hours. There are the same warehouses A and B. A starts to move goods in warehouse A and B in warehouse B at the same time. C starts to help A During the transportation, he turned to help B move. The last two warehouse goods were moved at the same time. How long does C help A and B each?

Solution: The workload of moving goods in a warehouse is 1. Now it is equivalent to three people working together to complete the workload 2. The time required is

Answer: C helps A to move for 3 hours, and B helps 5 to move.

The key to solving this problem is to first calculate the time for three people to move the two warehouses together. Of course, the calculation of this problem can also be integer, and the total workload for moving a warehouse is 60. A is 6 hours per hour, B is 5 hours per hour, and Hour handling 4.

Three people moved together and needed

60 × 2 ÷ (6+ 5+ 4) = 8 (hours).

A needs C to help carry

(60- 6 × 8) ÷ 4 = 3 (hours).

B needs C to help with handling

(60- 5 × 8) ÷ 4 = 5 (hours).

Third, the water pipe problem

From the perspective of mathematics, the water pipe problem is the same as the engineering problem. The water injection or drainage of a pool is equivalent to a project, and the water injection or drainage is the workload. The water injection or drainage in the unit time is the work efficiency. There is a problem of injection and drainage, but the workload is increased or decreased. Therefore, the problem of solving the problem of water pipe and engineering is basically the same.

Example 15 The two tubes A and B are opened at the same time, which can fill the pool in 9 minutes. Now, open the tube A, open the tube B in 10 minutes, and fill the pool in 3 minutes. It is known that the tube A is more than B tube Inject 0.6 cubic meters of water. What is the volume of this pool?

Solution: The amount of water injected per minute is: (1-1 / 9 × 3) ÷ 10 = 1/15

The amount of water injected per minute is: 1 / 9-1 / 15 = 2/45

So the volume of the pool 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, and their water injection amount is equal every minute. Now open some of them, and after a predetermined time,

If you double the open water pipe, you can fill the pool at the scheduled time. If you open 10 pipes at the beginning, you can fill the pool at the scheduled time without adding more pipes. How many are opened at the beginning? Water pipe?

Analysis : After the water pipe is added, there are two times the original water pipe. The water injection time is 1-1 / 3 = 2/3 and 2/3 is 1/3 of the predetermined time. Therefore, the time after the water pipe is added The amount of water injection is 4 times that of the previous period. Suppose that the capacity of the pool is 1, and the ratio of the water injection volume before and after the period is 1: 4.

Then the water injection amount of 1/3 of the predetermined time (that is, a while ago) is 1 / (1 + 4) = 1/5.

10 water pipes are opened at the same time, and can be filled with water at a predetermined time. The water injection quantity of each water pipe is 1/10, the water injection quantity is 1/3 of the predetermined time.

To fill 1/5 of the pool, you need water pipe 1/5 ÷ 1/30 = 6 (root)

Solution: The ratio of the amount of water injection before and after the period is: 1: [(1-1 / 3) ÷ 1/3 × 2] = 1: 4

The amount of water injection some time ago is: 1 ÷ (1 + 4) = 1/5

The water injection volume of each water pipe in a predetermined 1/3 time is: 1 ÷ 10 × 1/3 = 1/30

Number of open water pipes at the beginning: 1/5 ÷ 1/30 = 6 (roots)

Answer: At the beginning, open 6 water pipes.

Example 17 The reservoir has two inlet pipes A and C, and two drainage pipes B and D. To fill a pool of water, it takes 3 hours to open a single pipe and 5 hours to open a single C pipe. To drain a pool of water, It takes 4 hours to open the second tube separately, and 6 hours to the D tube. Now one-sixth of the water in the pool, such as A, B, C, D, A, B ... turn on for 1 hour in turn. After the water starts to overflow the pool?

analysis:

This question is similar to the widely-known "frog climbing well": a frog that fell into a dry well must climb 30 feet to reach the wellhead. It always climbs 3 feet per hour and slides 2 feet down. . How many hours does it take for this frog to climb to the wellhead?

It seems that it only climbs 3- 2 = 1 (feet) per hour, but after 27 hours, it climbs for another 1 hour and climbs 3 feet to reach the wellhead.

Therefore, the answer is 28 hours, not 30 hours. After that (20 hours), the water in the pool is already there, otherwise the water in the pool will overflow during the opening of the nail tube.

Example 18 A reservoir, 4 cubic meters of water flows in every minute. If 5 taps are turned on, empty the pool water in 2 hours and a half, and if 8 taps are turned on, the water is empty in 1 hour and a half. Now open 13 How long does it take to empty the water?

Solution: First calculate the amount of water discharged by a faucet per minute.

2 and a half hours more than 60 minutes

4 × 60 = 240 (cubic meters).

The time is expressed in minutes. The amount of water per minute for a faucet is

240 ÷ (5 × 150- 8 × 90) = 8 (cubic meters),

The amount of water discharged by 8 taps for 1.5 hours is

8 × 8 × 90,

The amount of inflow water in 90 minutes was 4 × 90, so the original pool had 8 × 8 × 90-4 × 90 = 5400 (cubic meters).

Turn on 13 taps to release 8 × 13 water per minute, except for the inflow of 4 per minute, the rest will release the original water, and empty the original 5400.

5400 ÷ (8 × 13- 4) = 54 (minutes).

Answer: It takes 54 minutes to open 13 faucets and empty the pool.

There are two parts of the water in the pool. The original water and the newly flowing water need to be considered separately. The key to solving this problem is to first find the original water in the pool. This is implicit in the problem.

Example 19 A pool, groundwater seeps into the pool from four walls, the amount of infiltration water per hour is fixed. Open pipe A, drain the full pool water in 8 hours, open pipe C, drain the full pool water in 12 hours. If you open A , B two tubes, drain the water in 4 hours. Ask to open B, C two tubes, how many hours can it take to drain the full pool of water?

Solution: Set the water volume of the full pool to 1.

A tube is discharged every hour

A tube is discharged in 4 hours

Therefore, the two pipes B and C are open at the same time, and the hourly displacement is

The two pipes B and C are open at the same time, draining the water from the pool. The time required is

Answer: It takes 4 hours and 48 minutes for B and C to open together.

This question must also be considered separately, the original water (full pond) and the infiltration of the pool. Because the specific amount is unknown, the specific amount of work is the same as the engineering problem. Here, the two types of water are set to "1". The amount should be avoided. In fact, it can also be integerized, and the original water is set to the least common multiple of 8 and 24.

In the 17th century, the great British scientist Newton wrote a book called "Universal Arithmetic". The book proposed a "cow grazing" problem, which is an interesting arithmetic problem. In essence, it is the same as Example 18 and Example 19. It is similar. The title involves three types of grass: the original grass, the newly grown grass, and the grass eaten by the cow. This is exactly the same as the original amount of water, the amount of infiltration and the amount of water discharged from the water pipe.

Example 20 has three pastures. The grass on the field grows as densely and quickly as it grows. 12 cows eat the grass of the first pasture in 4 weeks; 7 cows eat the grass of the second pasture in 9 weeks. How many cows can eat the grass of the third pasture in 18 weeks?

Solution: total grazing amount = grazing amount of a cow per week × number of bull heads × number of weeks. According to this calculation formula, "grazing amount of a cow per week" can be set as a unit of measurement of grass.

Original grass + 4 weeks new long grass = 12 × 4.

Original grass + 9 weeks new grass = 7 × 9.

It can be concluded that the new grass 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 18 weeks of new growth is

These grasses make

90 × 7.2 ÷ 18 = 36 (head)

Cows eat for 18 weeks.

Answer: 36 cattle can finish the third pasture in 18 weeks.

The solution of Example 20 is slightly different from that of Example 19. Example 20 specifically calculates the "new long", and calculates the "original" and "new long" quantities. In fact, if the example There is another condition, for example: "Open the B pipe, and drain the full pool water in 10 hours." The quantity relationship between "new long" and "original" can be obtained. But it is only the example 19 Please do not need to add this condition. Think about it, can you understand the truth?

"Cattle eating grass" is a type of problem that can appear in a variety of colors. Due to space limitations, we will only give another example.

Example 21 The exhibition opened at 9 o'clock, but someone was waiting in line for admission. Since the first audience arrived, the number of visitors per minute has been the same. If 3 entrances are opened, no one will be at 9: 9 Line up. If there are 5 entrances, no one will line up at 9: 5. What time will the first audience arrive at 8:30?

Solution: Set one audience per minute as a calculation unit.

Entering the audience from 9 to 9: 9 is 3 × 9,

Entering the audience from 9 to 9: 5 is 5 × 5.

Because the audience came 9-5 = 4 (minutes), so the audience every minute is

(3 × 9-5 × 5) ÷ (9-5) = 0.5.

At 9 o'clock the audience is

5 × 5-0.5 × 5 = 22.5.

These audiences came in need

22.5 ÷ 0.5 = 45 (minutes).

Answer: The first audience arrival time is 8:15.

Digging a canal, Team A and Team B will dig in six days. Team A will dig for three days first, and team B will dig for one day. 3/10 of this canal can be dug. How many days will each team dig separately?

Analysis: After 1 day of cooperation between A and B, A did another 2 days for a total of 3 / 10-1 / 6 = 4/30

2 ÷ (3 / 10-1 / 6)

= 2 ÷ 4/30

= 15 (days)

1 ÷ (1 / 6-1 / 15) = 10 (days)

Answer: A takes 15 days to do it alone, and B takes 10 days to do it alone.

For a job, if A does it alone, then A can be completed 2 days in advance according to the prescribed time, and B must be completed 3 days longer than the prescribed time. Now after two days of cooperation between person A and person B, the remaining person B will do it alone, which will be completed within the stipulated date. How long does it take for A and B to cooperate?

Solution: The specified time is X days. (A needs X-2 days alone, B needs X + 3 days alone, A has done 2 days in total, and B has done X days)

1 / (X-2) × 2 + X / (X + 3) = 1

X = 12

12 days to complete

1 ÷ [1 / (12-2) + 1 / (12 + 3)]

= 1 ÷ (1/6)

= 6 days

Answer: It takes 6 days for two people to complete the cooperation. Example: For a project, 63 days for A alone, 28 days for B, and 48 days for A-B. A does 42 days first, and how many days does B do? Answer: Let ergonomics be x and y for y

63x + 28y = 1

48x + 48y = 1

x = 1/84

y = 1/112

B has to do (1-42 / 84) ÷ (1/112) = 56 (days)

Example 22 has 32 tons of cargo and is transported from city A to city B. The load capacity of the large truck is 5 tons and the load capacity of the small truck is 3 tons. The fuel consumption of each large and small truck is 10 liters and 7.2 liters, respectively. How many tons of fuel will it take to get this shipment?

Solution: Obviously, in order to save fuel, we should try to use a large truck to transport it. The large truck is transported 6 times with 2 tons left, so the remaining one is transported by a small truck with the least fuel consumption, which requires 6 * 10 + 7.2 = 67.2 liters.