2 engineering problems, solved. Hurry!! You have to be able to understand!

1) 1 3 8 1 24 Ergonomics of A.

4 9 4 1 9 Ergonomics.

16 11 (days).

A: It will take 16-11 days to complete the project.

2) 1 2 3 1 6 The ergonomics of master and apprentice.

1 2 5 1 10 Master's ergonomics.

1 6 1 10 1 15 Apprentice's work.

1 1 15 15 (day).

A: It takes 15 days to complete.

1-4 9) (1 24 + 1 36) = 8 days.

1 1 15 15 (day).

Please refer to it, if you don't understand, you can ask!

1.Set the amount of work to be 1, A 8 days to complete 1 3, 1 24 per day;

B completes 4 9 in 4 days, 1 9 per day, A and B need to cooperate = (1-1 3-4 9) (1 24 + 1 9) = 16 11 days to complete the project.

2.Set the work as 1, the master completes 1 2 in 5 days, 1 10 every day, the master and the apprentice cooperate for 3 days to complete 1 2, the master completes 3 10, the apprentice completes 1 2-3 10 = 2 10 in 3 days, and completes 2 30 every day, if the work is done by the apprentice alone, it takes 30 2 = 15 days to complete.

Problem 1 is easy to solve, with or without equations;

Problem 2 is solved without equations

The master does half of it in 5 days, which means that it takes 10 days for the master to do it alone.

Now the second half is 3 days of master-apprentice cooperation, which means that the master still needs 2 days, and the apprentice needs 3 days to complete. Therefore, the ratio of master-apprentice work efficiency is 3:2.

Therefore, the number of days of the apprentice is 3 2 10 = 15 days.

1. Set the workload to 1, the speed of A is 1 3 8 = 1 8, and the speed of Team B [4 9-1 3] 4 = 1 27

And the final project is left with 1-4 9=5 9

That is, the remaining time is: 5 9 (1 8 + 1 27) = 3 days.

2. Set the apprentice's work speed to x, and the master completes it in five days, and the speed is 1 2 5 = 1 10, then the second half of the workload is:

1 2 = 3 (1 10 + x), the solution gives x = 1 15

That is, it takes 1 1 15 = 15 days for the apprentice to finish.

The typical workload is considered as a calculation of "1".

1. Let all the projects be x, and the number of days required is a, then the efficiency of A is x 24, the efficiency of B is x 9, and the remaining projects are 2x 9, so xa 24 + xa 9 = 2x 9, then a 15 days.

You should make it clear that this is a grade question, so that you can explain how to solve the problem

Let the number of hours be x, the water storage is y, and the water inflow per hour is 1 5 and the discharge is 1 3, then there is.

When x is an odd number, y=1 2+(x+1) 2*(1 5)-(x-1) sock afterburn 2*(1 3)=1 2-x 15+7 30=-x 15+11 15

When x 11, y=0

When x is even, y=1 2+x 2*(1 5)-x 2*(1 3)=1 2-x 15

When x 6, y=1 10>0

When x=8, y=-1 is 30<0

When x is an even number, it can be discharged after 8 hours.

The actual draining time is 8 (1 30) (1 3) = hours.

t -t=1(7 8) t 5th full time t 6th full time.

v * t = v * t v average velocity at the 5th time v average velocity at the 6th time.

v＝v-40

v-40）*[t+ *t

v＝1500/t

then (1500 t-40) * t+ ) tt hours.

6th average speed v 1500 hours.

The average speed of the 5th time v 160km h.

In the end, you can double the results.

The average speed of the 5th time was 160 and the 6th time was 200 times

Root numbers 2a-5 make sense.

So 2a-5 0

a Laugh 5 2

At this time, 2-a<0

So the original formula = 2a-5-|2-a|

2a-5-(a-2)

A-3 Remember my answer, oh I would like to draft, I wish you Shanhan learning and progressing.