Three workers can do a job in 6 days. Two of the workers work twice as fast as the third. How long would it take one of the faster workers to do the job himself?

Question: Three workers can do a job in 6 days. Two of the workers work twice as fast as the third. How long would it take one of the faster workers to do the job himself?

Solution:
Let,
the first and 2nd worker need x days.
Third worker needs 2x days

ATQ,
(1/x) + (1/x) + (1/2x) = 1/6
⇒ (2 + 2 + 1)/2x = 1/6
⇒ 5/2x = 1/6
⇒ 2x = 30
∴ x = 15

So the one faster worker alone will take 15 days to complete the job.