if A takes 50 Days, B will take 100 days and C will take 200 days to complete a task.

Taking out B's rate
B: 30P 1W 600T(6hourse*100days)
1P 1W 18000T
80p 1W 225T --> 225/10 gives 22.5 days

Taking out C's rate
C: 30P 1W 1200T(6hourse*200days)
1P 1W 36,000T
80p 1W 450T --> 450/10 gives 45 days

1/22.5 + 1/45 = 0.0660 Combined Rate
1/0.0660 = 15

In case of efficiency : A=2B and B=2C

A : 30 men - 50 days - 6hr/day thus efficiency of A(work done in 1hr) = 1/9000 (let total work = 9000 thus efficiency of A=1)

If efficiency of A=1 so efficiency of B = 1/2 and efficiency of C =1/4

Total work done by 80 type B employees and 80 type C employees in working ten hours a day = 80*1/2*10 + 80*1/4*10 = 600

Number of days required by 80 type B employees and 80 type C employees to complete the same project working ten hours a day = 9000/600=15

thus c

Carcass can u provide a better explanation to this question.

since 30 of type A can finish work in (50*6) hours, rate of 30 workers can be written as:

\(30*A=\frac{1}{6*50}\)

so rate of work for 1 type A:
\(A=\frac{1}{6*50*30}\)

since
\(A = 2*B\)
\(B = 2*C\)
so \(A = 4*C\)

\(80B = 40A\) and \(80C = 20A\)

so combined rate of 80B & 80C would be \(\frac{40}{6*50*30}+\frac{20}{6*50*30}=\frac{60}{6*50*30}=\frac{1}{150}\)

work = rate * time
\(1=\frac{1}{150}*(10*d)\) where d is the number of days
\(d=15\)

C is the answer

Too tough for a GRE question.

Maybe is a bit too far