A, B and C working together completed a job in 10 days. However, C only worked for the first three days when $$\frac{37}{100}$$ of the job was done. Also, the work done by A in 5 days is equal to the work done by B in 4 days. How many days would be required by the fastest worker to complete the entire work?

Since the work done by A in 5 days is equal to the work done by B in 4 days.

So, time taken by A and B alone to do the work is 5x and 4x respectively. 

Time taken by C alone to do the work = C

Now, 

$$\frac{10}{5x} + \frac{10}{4x} + \frac{3}{C} = 1$$ ........ (1)

$$\frac{3}{5x} + \frac{3}{4x} + \frac{3}{C} = \frac{37}{100}$$ ...... (2)

From (1) and (2):

$$\frac{7}{5x} + \frac{7}{4x} = 1 - \frac{37}{100}$$

$$\frac{28 + 35}{20x} = \frac{63}{100}$$

x = 5 and C = 30

Time taken by A alone to do the work = 5x = 25 days

Time taken by B alone to do the work = 4x = 20 days

Time taken by C alone to do the work = C = 30 days

Hence, the time required by the fastest worker to complete the entire work = 20 days