A alone can do a work in 20 days. B is 25% more efficient than A. A and B started working and worked for 4 days. If C alone completed the remaining work in 22 days, then how many days C alone will take to complete the entire work?

Given:

A alone can do a work in 20 days.

B is 25% more efficient than A.

A and B started working and worked for 4 days.

C alone completed the remaining work in 22 days.

Formula Used:

Work done = Efficiency × Time

Efficiency of B = Efficiency of A × 1.25

Total work = Work done by A + Work done by B + Work done by C

Calculation:

Let the total work be W units.

Efficiency of A = W / 20

Efficiency of B = (W / 20) × 1.25 = W / 16

Work done by A and B in 4 days = 4 × (Efficiency of A + Efficiency of B)

⇒ Work done by A and B in 4 days = 4 × (W / 20 + W / 16)

⇒ Work done by A and B in 4 days = 4 × (4W / 80 + 5W / 80)

⇒ Work done by A and B in 4 days = 4 × 9W / 80

⇒ Work done by A and B in 4 days = 36W / 80

⇒ Work done by A and B in 4 days = 9W / 20

Remaining work = W - 9W / 20

⇒ Remaining work = 11W / 20

C completed the remaining work in 22 days, so Efficiency of C = Remaining work / 22

⇒ Efficiency of C = (11W / 20) / 22

⇒ Efficiency of C = 11W / 440

⇒ Efficiency of C = W / 40

C alone will take to complete the entire work = Total work / Efficiency of C

⇒ C alone will take to complete the entire work = W / (W / 40)

⇒ C alone will take to complete the entire work = 40 days

The correct answer is 40 days.

Alternate MethodGIVEN :

A alone can do work in 20 days.

B is 25% more efficient than A.

A and B started working and worked for 4 days.

If C alone completed the remaining job in 22 days.

CALCULATION :

⇒ B can do a work in 16 days.

Let the total unit of work ( LCM of 20 and 16) = 80 units

⇒ A’s one day work = 4

⇒ B’s one day work = 5

As A and B left after working for 4 days.

⇒ Work done by A and B in 4 days = 36 work

⇒ Work left = 44 units is done by C in 22 days.

⇒ C’s one day work = 2 units

∴ 80 units of work can be completed by C alone in 80/2 = 40 days.