A and B each process the same number of parts at the same time, and A makes a quarter of his...

Let the number of parts that A and B process at the same time is x, the efficiency of A is a unit time, and B is B unit time.

x 4) a=(x-45) b [2 3*(1-1 4)x] get x=60 (a, b is eliminated, I don't know how much).

I don't know what "the total workload of two people" means, if it is the total workload of each of them, the answer is 60, if it is the sum of the workload of two people, it is 2*60=120.

Digression: Actually, a = b, when A does 1 4 of the total, that is, he does 60*1 4=15, and B still has 45 that he has not done, and he also does 60-45=15. So at the beginning, the efficiency of the two was the same.

In the next step, A's efficiency increased by 20%, A did the remaining 2 3, that is, 2 3 * (60-15) = 30, B still had 1 3 of his original total work, that is, 1 3 * 60 = 20 did not do, that is, the second step did 60-15-20 = 25, and A did 30 is a multiple of 25.

1, at this time, the efficiency of A is increased by 20%, then when A does the remaining 2 3, it is equivalent to using the original speed to travel (1-1 4) * (2 3) (1 + 20%) = 5 12

2. A and B process the same number of parts at the same time, A makes its 1 4, and B still has 45 that he does not do. Suppose A does not improve his efficiency later and completes 5 12 of the total number, then the completion is the previous (5 12) (1 4) = 5 3 times, and B should also do the previous 5 3 times, and B still has 1 3 of the original total amount of work that he did not do. We can conclude that B has completed a total of (1-1 3) = 2 3

3, A and B process the same number of parts at the same time, A has done its 1 4, and B has 45 still has not done it, B has completed the total number of (2 3) (1 + 5 3) = 1 4

3. The total workload of two people is 45 (1-1 4)*2=120

A and B process the same number of parts at the same time, A did its 1 4 only, and 45 did not do, at this time A's efficiency increased by 20%, when A did the remaining 2 3, B raised the 1 3 of the workload of the Bi staff was not stupid, and asked what was the total workload of the two?

A and B at the same time respectively process the same kind of parts in the town, A did its 1 4 and did the state, there are 45 did not do, at this time A's efficiency increased by 20%, when A did the remaining 2 3, B still has a staff workload of 1 3 did not do, ask what is the total workload of the two? Answer: Let the workload of A be coarse x, then the equation of the question column is against A:

1-1 4) x=45 solution x=60 total workload: 2x=2*60=120

（1-1/4）×2/3=1/2

Then when A completes 2 3 according to the original efficiency, B completes 1-1 3 = 2 3 of the total amount, which means that the work efficiency of A and B is equal.

45 (1-1 4) = 60.

Assuming that the total amount of work of A and B is 1, then:

1-1 4) 2 3=1 2--- made after the speed of the first one;

1 2 (1+20%)=5 12--- if A does not speed up, then do 5 12;

If A does not speed up, the ergonomic ratio of A and B is: (1 4 + 5 12) :( 1-1 3) = 1:1;

1 4 1 1 = 1 4--- after A does 1 4, B does it;

45 (1-1 4) = 60 (pcs)--45 corresponds to the fraction of 3 4;

60 2 = 120 (pcs) -- 120 in total.

A completes the remaining 3 4, i.e. A has completed 1 3 + 2 3 3 4 = 5 6, and B has completed 5 (5 + 3) = 5 8

The problem knows that when A finishes 1 3, B still has 90 left to do, and if B has completed the proportional proportion of k, then 1 3 k=5 6 5 8 solves k=1 4, then B still has 1-1 4=3 4 undone, so the total number of parts = 90 (3 4) = 120

A and B made a total of 120 2 = 240 parts.

When A completes the remaining 3 4, B completes 5 (3+5)=5 8 A completes the task of (1-1 3)*3 4+1 3=5 6

Let be the x (score) of B completing the task when A completes 1 3

1/3：x=5/6：5/8

x=1 490 (1-1 4)*2=240.

Then when A completes 2 3 according to the original efficiency, B completes 1-1 3 = 2 3 of the total amount

This shows that the work efficiency of A and B is equal.

45 (1-1 4) = 60.

When A completes 1 4 + the remaining 1 5, B completes 1-2 3 = 1 3, that is, A completes 1 4 + (1-1 4) * 1 5 = 2 5, B completes 1 3, then when A completes 1 4, it is 5 8 of 2 5 [1 4 2 5 = 5 8], then B completes 1 3 * 5 8 = 5 24

So when A finishes 1 4, B completes 5 24, and the remaining 1-5 24 = 19 24 so the total number of tasks of B is: 95 19 24 = 120, so the total number of tasks shared by A and B is 120*2=240

A processed a quarter of the total number of liquid Shen Shen, so the total number of parts processed in the register is 160, and the number of parts processed by B is 160 (1 5) = 32.

1 All 40 divided by a quarter and multiplied by a fifth equals 32

Pick me oh thank you.