Bunuel wrote:
A can complete a piece of work in 4 days. B takes double the time taken by A, C takes double that of B, and D takes double that of C to complete the same task. They are paired in groups of two each. One pair takes two-thirds the time needed by the second pair to complete the work. Which is the first pair?
(A) A, B
(B) A, C
(C) B, C
(D) A, D
(E) C, D
Explanation:
No.of days to complete work by A = 4 days
No.of days to complete work by B = 8 days
No.of days to complete work by C = 16 days
No.of days to complete work by D = 32 days
Since the first pair works fast, let us assume the first pair is A and B
Work can be completed by A and B in one day = (1/4) + (1/8) = 3/8
Number of days taken to complete the work by A and B = 8/3 days ...............(1)
Work can be completed by C and D in one day = (1/16) + (1/32) = 3/32
Number of days taken to complete the work by C and D = 32/3 days ...............(2)
if we divide (1) by (2), we get ratio = 1/4. Hence the first pair is not A and B
similarly, if we make A and C as the first pair and try to find the ratio. The ratio can be worked out as 1/2.
Hence A and C are not in the first pair
let us try to make A and D as the first pair
Work can be completed by A and D in one day = (1/4) + (1/32) = 9/32
Number of days taken to complete the work by pair A and D = 32/9 days ...............(3)
Work can be completed by B and C in one day = (1/8) + (1/16) = 3/16
Number of days taken to complete the work by B and C = 16/3 days ...............(4)
if we divide (3) by (4), we get ratio = 2/3. Hence the first pair is A and D.
IMO-D