A, B & C can all together do a piece of work in 20 days, in which B takes 4 times as long as A & C to do the work and C takes 3 times as long as A & B together take to do the work. In how many days B can alone do the work?

Given:

A + B + C can complete the work in 20 days ⇒ Total work = 1 unit, daily work rate = 1/20

B takes 4 times as long as A & C together ⇒ B is 1/4 as efficient as A + C

C takes 3 times as long as A & B together ⇒ C is 1/3 as efficient as A + B

Formula used:

Work rate = 1 / Time

Total work rate = A + B + C = 1/20

Use efficiency ratios to build equations and solve

Calculation:

Let A = a, B = b, C = c (work per day)

Given: b = 1/4 × (a + c) ⇒ 4b = a + c … (1)

Given: c = 1/3 × (a + b) ⇒ 3c = a + b … (2)

Also: a + b + c = 1/20 … (3)

From (1): a = 4b - c

Substitute into (2):

3c = (4b - c) + b = 5b - c

⇒ 3c + c = 5b ⇒ 4c = 5b ⇒ c = 5b / 4 … (4)

Now from (1): a = 4b - c ⇒ a = 4b - (5b / 4)

⇒ a = (16b - 5b) / 4 = 11b / 4 … (5)

Substitute a, c into (3):

a + b + c = 1/20

⇒ (11b / 4) + b + (5b / 4) = 1/20

⇒ (11b + 4b + 5b) / 4 = 1/20

⇒ 20b / 4 = 1/20

⇒ 5b = 1/20

⇒ b = 1/100

⇒ B does 1/100 of the work per day

⇒ B alone can complete the work in 100 days

∴ The correct answer is 100 days