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A and B together can complete a task in 1.2 days. However, if A works alone, completes half the job and leaves and then B works alone and completes the rest of the work, it takes 2.5 days in all to complete the work. If B is more efficient than A, how days would it have taken it have taken B to do the work by herself?

A and B together can complete a task in 1.2 days. However, if A works alone, completes half the job and leaves and then B works alone and completes the rest of the work, it takes 2.5 days in all to complete the work. If B is more efficient than A, how days would it have taken it have taken B to do the work by herself? Correct Answer 2.0

Suppose, A can complete the task in = x days,

B can complete the task in = y days

According to the question,

1/x + 1/y = 1/1.2

⇒ x + y = xy/1.2     ...i)

A completes half the job in = x/2 days and B completes half the job in = y/2 days

⇒ x/2 + y/2 = 2.5

⇒ x + y = 5     ...ii)

From (i) and (ii), we get,

xy/1.2 = 5

⇒ xy = 6

⇒ x = 6/y     ...iii)

From (ii) and (iii) we get,

6/y + y = 5

⇒ 6 + y2 = 5y

⇒ y2 – 3y – 2y + 6 = 0

⇒ y(y – 3) – 2(y – 3) = 0

⇒ (y – 3) (y – 2) = 0

⇒ y = 3 or 2

For y = 3, x = 6/3 = 2

For y = 2, x = 6/2 = 3

∵ B is more efficient than A

∴ B take less days than A

∴ B would take 2 days to complete the task

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