A and B together can do a certain work in 20 days, B and C together can do it in 30 days, and C and A together can do it in 24 days, B alone will complete 2/3 part of the same work is∶

A20

B16

C24

D32

Answer:

Shortcut Trick LCM(20, 30, 24) = 120 Let work is 120u

Total efficiency 2(A + B + C) = 15u

A + B + C = 7.5u

A = A + B + C - (B+C) = 7.5 - 4 = 3.5u

By same process Efficiency of B = 2.5u, C = 1.5u

B alone can complete = (120 × 2/3)/2.5 = 32 days

Given:

⇒ 1/A + 1/B = 1/20

⇒ 1/B + 1/C = 1/30

⇒ 1/A + 1/C = 1/24

Calculation:

Solving,

⇒ 1/A - 1/C = 1/20 - 1/30 = 1/60

⇒ 1/A + 1/C = 1/24

Solving,

⇒ 2/A = 1/60 + 1/24 = 7/120

⇒ 1/A = 7/240

Then,

⇒ 1/B = 1/20 - 7/240 = 1/48

B's 1 days work = 1/48

B alone can complete the work in 48 days.

∴ Time taken by B to complete 2/3rd work = 2/3 × 48 = 32 days

Alternate Method

Total work = LCM of 20, 30 and 24 = 120

In 1 day (A + B) can do 120/20 = 6 units work

In 1 day (B + C) can do 120/30 = 4 units work

In 1 day (A + C) can do 120/24 = 5 units work

In 1 day (A + C + B + C + C + A) = 2(A + B + C) can do 15 unit work

⇒ In 1 day (A + B + C) can do 15/2 = 7.5 units work

⇒ In 1 day B alone can do (A + B + C) - (A + C) = 7.5 - 5 = 2.5 units work

B alone can complete the work in 120/2.5 = 48 days

B alone can do 2/3rd part of the work in 48 × 2/3 = 32 days

∴ B alone will complete 2/3 part of the same work in 32 days.

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