A and B can complete a task together in 35 days. If A works alone and completes of the task and then leaves the rest for B to complete by herself, it will take a total of 90 days to complete the task. How many days would it take A, the more efficient among the duo, to complete the entire work by herself?

Question

Question

This question was previously asked in

RRB ALP CBT I 29 Aug 2018 Shift 2 Official Paper

Answer 1

Given:

A and B can complete a task together in 35 days.

Total time taken is 90 days.

Calculation:

Let the total amount of work be 35 units.

If A and B work together, they complete 35/35 = 1 unit of work every day.

Now,

Let the amount of work done by A by himself in one day be x units and the work done by B by himself in one day be y units.

So, x + y = 1

If A works alone and completes 5/7th of the task,

work done by A is (5/7) × 35 = 25 units.

⇒ time taken by A to complete this 25 units of work = 25/x days.

If B works alone and completes 2/7th of the task,

work done by B is (2/7) × 35 = 10 units.

⇒ Time taken by B to complete this 10 units of work = 10/y days.

According to the question,

25/x + 10/y = 90

⇒ (25 + 10)/xy = 90

⇒ 25y + 10x = 90xy

⇒ 25(1 – x) + 10x = 90xy

⇒ 25 – 25x + 10x = 90x(1 – x)

⇒ 25 – 15x = 90x – 90x²

⇒ 90x² – 105x + 25 = 0

⇒ 18x² – 21x + 5 = 0

⇒ 18x² – 15x – 6x + 5 = 0

⇒ 3x(6x – 5) – 1(6x – 5) = 0

⇒ (6x – 5)(3x – 1) = 0

So, x = 5/6 (x cannot be 1/3 as A is more efficient among two)

Time taken by A alone to complete work by herself = 35/(5/6)

⇒ 42 days

∴ The required answer is 42 days.

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