The time taken by A to finish a piece of work is twice the time taken by B and thrice the time taken by C. If all the three work together, it will be completed in 2 days. How many days does B alone require to complete that work?

Solution (By Examveda Team)

Let the time taken by A to finish the work be 6x days

Then, time taken by B = 3x days and C = 2x days (since A takes twice B’s time and thrice C’s time)

Work done by A in one day = 1 / 6x

Work done by B in one day = 1 / 3x

Work done by C in one day = 1 / 2x

Total work done by A, B, and C in one day = 1/6x + 1/3x + 1/2x

Take LCM of denominators: LCM of 6x, 3x, and 2x is 6x

= (1 + 2 + 3) / 6x = 6 / 6x = 1 / x

So, together they complete 1/x of the work in one day

It is given that they complete the whole work in 2 days

So, 1/x = 1/2 ⇒ x = 2

Now, time taken by B = 3x = 3 × 2 = 6 days

Therefore, B alone can complete the work in 6 days.