please help in time and work?

Let 1/A be the amount of the total work that A performs per day.
Let 1/B be the amount of the total work that B performs per day.
Let 1/C be the amount of the total work that C performs per day.

A's work rate is twice as fast as B.
1/A = 2/B

A and C together is three times as fast as B:
1/A + 1/C = 3/B

Working together, they will complete 1/30 of the job per day:
1/A + 1/B + 1/C = 1/30

Subtracting the second equation from the third, we have:
1/B = 1/30 - 3/B
4/B = 1/30
B = 120

Substituting that into the first equation:
1/A = 2/120
1/A = 1/60
A = 60

Finally substituting into the second equation:
1/60 + 1/C = 3/120
1/60 + 1/C = 1/40

Multiply everything by 120C
2C + 120 = 3C
C = 120

Answer:
A takes 60 days alone
B takes 120 days alone
C takes 120 days alone

　
A and C together can work 3 times as fast as B

x = number of days it takes A and C together to complete work
3x = number of days it takes B to complete work

In 1 day, A and C together can do 1/x of work, and B can do 1/(3x) of work
So in 1 day, A, B, and C together can do 1/x + 1/(3x) of work

A, B, and C together can complete work in 30 days.
So in 1 day, A, B, and C together can do 1/30 of work

1/x + 1/(3x) = 1/30
x = 40


Working together, A and C take 40 days to complete work
Working alone, B takes 120 days to complete work

In 1 day, A and C working together can do 1/40 of work

A can work twice as fast as B, so working alone A takes 60 days to complete work
Let y = number of days it takes C to complete the work alone.

In 1 day, A can complete 1/60 of work, and C can complete 1/y of work.
So in 1 day, A and C can complete 1/60 + 1/y of work

1/60 + 1/y = 1/40
y = 120


Working alone:

A can complete work in 60 days
B can complete work in 120 days
C can complete work in 120 days

Rate B per day = B
Rate A per day = 2B
A and C together can work three times as fast as B ---> Rate C per day = B
B+2B+B = 1/30
B = 1/120 ---> B needs 120 days = C , A needs 60 days