A can do a piece of work in 10 days, B can do it in 20 days and C in 30 days. Three of them started the work together. A left the work after 4 days and C left it some days before completion. If the work is completed in 46/5 days, for how many days did C work?

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A can do a piece of work in 10 days, B can do it in 20 days and C in 30 days. The rate at which A completes the work is 1/10, for B it is 1/20 and for C it is 1/30.

All three of them started the work together. A left the work after 4 days and C left it some days before completion. The total time taken to complete the work is 46/5 days. Let the number of days that C worked for be D.

4*(1/10) + (46/5)*(1/20) + D*(1/30) = 1

=> 24 + 138/5 + 2*D = 60

=> D = 42/10

**The number of days C worked for is 4.2**

It is known that A can do a piece of work in 10 days, B can do it in 20 days and C in 30 days. In one day, A does 1/10 of the work, B does 1/20 of the work and does 1/30 of the work.

The three start together. A leaves after 4 days. In four days the amount of work completed is `4*(1/10 + 1/20 + 1/30)` . Let the number of days C worked for after the 4 days be X-4.

In the next X - 4 days, the amount of work done is `(X - 4)*(1/20 + 1/30)` . For a period 46/5 - X, only B was doing the work.

As the work is completed in 46/5 days we have:

`4*(1/10 + 1/20 + 1/30) + (X - 4)*(1/20 + 1/30) + (46/5 - X)*(1/20) = 1`

`11/15 + (X - 4)*(1/12) + (46/5 - X)*(1/20) = 1`

`(X - 4)*(1/12) + (46/5 - X)*(1/20) = 4/15`

`X*(1/30) + 19/150 = 4/15`

`X*(1/30) = 7/50`

`X = 210/50`

= 4.2

C worked for a total of 4.2 days.

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