A can do a piece of work in 24 days, B in 32 days and C in 64 days. All being to do it together, but A leaves after 6 days and B leaves 6 days before the completion of the work. How many days did the work last?

Number of days to complete the work

A = 24 days

B = 32 days

C = 64 days.
All worked together for 6 days,the fraction of the work that has been done is

6/24 = 1/4 by A

6/32 = 3/16 by B

6/64 = 3/32 by C.

Total fraction of work done in First 6 days =

1/4 + 3/16 + 3/32

=(8 + 6 + 3)$$\div$$32

= 17/32.

Fraction of work left = 1 -17/32 = 15/32.

Since B leaves 6 days before the completion of the work,it means that C works alone for 6 days. The fraction of work C does alone in those 6 days is

6/64 = 3/32.

Therefore fraction of work done before B's departure =

15/32 - 3/32 = 12/32

=3/8.

Let X be the number of days in which B and C did 3/8 of the work.

Therefore, x/32 is done by B and x/64 is done by C.

This shows that

x/32 + x/64 = 3$$\div$$8

2x + x = 3 $$\times$$ 8

3x = 24

x = 24$$\div$$3 = 8 days.

Total number of days

= 6 + 8 + 6 = 20 days

So, the answer would be option b)20.