A, B and C jointly completed a project

A, B and C completed a project together, completed all 1 5 in 3 days, and then A rested for 3 days, B rested for 2 days, and C did not rest, if A's workload in one day is 3 times that of C's workload in one day, and B's workload in one day is 4 times that of C's workload in one day, then how many days does this work be completed from the beginning?

Solution: The work efficiency of A, B and C = (1 5) 3 = 1 15 The work efficiency of C = (1 15) (3 + 4 + 1) = 1 120 The work efficiency of A = 1 120 3 = 1 40

B's work efficiency = 1 120 4 = 1 30

Here we consider C's work efficiency as a multiple of 1.

A rests for 3 days and B rests for 2 days, which is a total of 2 days.

Then the rest needs (1-1 5-7 120) (1 15) = 89 8 days.

It takes a total of 3+3+89 8=17 and 1 8 days.

Let the workload of a day be x for C, 3x for A and 4x for B

Completed by 1 in 3 days 5, you can column 3 (x+3x+4x)=1 5 can find x, and then you can set it to take y days to complete.

xy+3x(y-3)+4x(y-2)=1 to find y, x has been found from the above question, it is not difficult.

If C works alone to complete the task in x days, then A's workload is 3 x, and B's workload is 4 x1 x+3 x+4 x)*3=1 5

8/x=1/15

x=120 A does 3 120 per day, B does 4 120 per day, and C does 1 120 per day

11 and 1 8

3+3+11 and 1 8=17 and 1 8 days.

Summary. The topic is not very clear.

A project, A. B and two people do it together in 4 days to complete, B. C and the two of them did it together for 6 days to complete, C. Ding two people together.

The topic is not very clear.

Did you see it?

It may have arrived. 6 days.

A project, A. B and two people do it together in 4 days to complete, B. C and the two of them did it together for 6 days to complete, C. Ding two people together.

C should do it alone for 1 (1 3-1 4) = 12 days, so B should do it alone for 1 (1 6-1 12) = 12 days.

1 10 Closed Digging Slip (Sanbi 40

1 40 * 80 sedan car

1 3 80 = 80 3 days.

A and B cooperate to complete the total task in one day (1 3) 6 = 1 18 (1).

(1 6) 2=1 12 (2) for B and C to complete the total task in one day

A, B, and C work together to complete the total task in one day (1 2) 5=1 10 (3).

So, (3)-(2) yields:

1 10-1 12 = 1 60 of the total tasks completed by A in one day

3)-(1) can obtain:

C can complete 1 10-1 18 = 2 45 of the total task in one day

Then you can get 1 18-1 60 = 7 180 of the total tasks that B can complete in one day

According to the above situation, A has done a total of 6 + 5 = 11 days, and completed 1 60 * 11 = 11 60 of the total task, so the amount of income due to A is = 5400 * 11 60 = 990 yuan.

B did a total of 6 + 2 + 5 = 13 days, and completed the total task of = 7 180 * 13 = 91 180

Therefore, the amount of income due to B is = 5400 * 91 180 = 2730 yuan.

C did a total of 2 + 5 = 7 days, and completed 2 45 * 7 = 14 45 of the total task

Therefore, the amount of income due to C is = 5400 * 14 45 = 1680 yuan.

Hello. Is this a math problem? Or is it the reality of the situation?

According to reality, a total of 15 4 12 = 31 working days, working hours = 5400 31 = yuan. A did it for 11 days, B for 13 days, and C for 7 days. Multiply by working days, fair. Above.

A, B and C completed a project together, completed all 1 5 in 3 days, and then Yuqing A rested for 3 days, B rested for 2 days, and C did not rest, if Ye Minjia's workload in a day is 3 times the workload of C in a day, and B's workload in a day is 4 times that of C in a day, then how many days does this work be completed from the beginning?

Solution: The work efficiency of A, B and C = (1 5) 3 = 1 15 The work efficiency of C = (1 15) (3 + 4 + 1) = 1 120 The work efficiency of A = 1 120 3 = 1 40

B's work efficiency = 1 120 4 = 1 30

Here we consider C's work efficiency as a multiple of 1.

A rests for 3 days and B rests for 2 days, which is a total of 2 days.

Then the rest needs (1-1 5-7 120) (1 15 before Qingzhi) = 89 8 days.

It takes a total of 3+3+89 8=17 and 1 8 days.

Set C to work alone for x days to complete the task.

Then A's workload is 3 x and B's workload is 4 x

1/x+3/x+4/x）*3=1/5

8/x=1/15

x = 120 A does 3 120 per day

B does 4 120 per day

Propylene lead Qingsui does 1 120 per day

1-(1 120+3 120+4 Huaibu 120)*3-1 120*3-4 120] (1 120+3 120+4 120).

11 and 1 8

3+3+11 and 1 8=17 and 1 8 days.

The project of setting up this burial piece is always done by a Pi Wang person alone, and it takes X hours to complete.

B alone takes 6 (1-2 5-13 30) = 36 hours.

1+10%)*1 Liquid x+(1+20%)*1 36x=33

A + B = 1 4 5 = 1 20

B + C = 1 5 2 = 1 10

A + C = (1-1 4-1 Zhichun Lu 5) 5 = 11 100 A = 3 Fierce belt 100

A did (3 100) * (5 + 5) = 3 10 due (Senna 3 10) * 3000 = 900 yuan.

Let A be x, B be y, and C be z

1/(1/x+1/y+1/z)+18=x

1/(1/x+1/y+1/z)+3=y

1/(1/x+1/y+1/z)*2=z

Three equations are not missed, and the first return is known, and the solution can be obtained.