Quantity B: B can complete the work in 20 days. C is twice as efficien

1.	Quantity B: B can complete the work in 20 days. C is twice as efficient as B and A takes 2 days more than C to complete the work. Working together, in how much time would they be able to complete the work?1). Quantity A > Quantity B2). Quantity A < Quantity B3). Quantity A ≥ Quantity B4). Quantity A ≤ Quantity B
Answer» Quantity A: Let total work be 120 units (LCM of 24, 30 & 40) ⇒ Work done by A in one day = 120/24 = 5 units ⇒ Work done by B in one day = 120/30 = 4 units ⇒ Work done by C in one day = 120/40 = 3 units ⇒ Work done by A and B in 1 day = (5 + 4) = 9 units ⇒ Work done by A and B in 6 days = 9 × 6 = 54 units ⇒ Remaining work = 120 – 54 = 66 units ⇒ Work done by A, B and C together in 1 day = 5 + 4 + 3 = 12 units ⇒ Time taken by A, B & C together to complete the remaining 66 units of work = 66/12 = 5.5 days ⇒ Total time taken = 6 + 5.5 = 11.5 days Quantity B: ⇒ Time taken by B to complete the work = 20 days ⇒ Time taken by C to complete the work = 20/2 = 10 days ⇒ Time taken by A to complete the work alone = 10 + 2 = 12 days Let total work be 60 units (LCM of 12, 20 & 10) ⇒ Work done by A in one day = 60/12 = 5 ⇒ Work done by B in one day = 60/20 = 3 ⇒ Work done by C in one day = 60/10 = 6 ⇒ Work done by A, B and C together in 1 day = 5 + 3 + 6 = 14 units ⇒ Time taken by A, B and C together to complete the work = 60/14 = 30/7 days ∴ Quantity A > Quantity B

Quantity B: B can complete the work in 20 days. C is twice as efficient as B and A takes 2 days more than C to complete the work. Working together, in how much time would they be able to complete the work?1). Quantity A > Quantity B2). Quantity A < Quantity B3). Quantity A ≥ Quantity B4). Quantity A ≤ Quantity B

Answer»

Quantity A:

Let total work be 120 units (LCM of 24, 30 & 40)

⇒ Work done by A in one day = 120/24 = 5 units

⇒ Work done by B in one day = 120/30 = 4 units

⇒ Work done by C in one day = 120/40 = 3 units

⇒ Work done by A and B in 1 day = (5 + 4) = 9 units

⇒ Work done by A and B in 6 days = 9 × 6 = 54 units

⇒ Remaining work = 120 – 54 = 66 units

⇒ Work done by A, B and C together in 1 day = 5 + 4 + 3 = 12 units

⇒ Time taken by A, B & C together to complete the remaining 66 units of work = 66/12 = 5.5 days

⇒ Total time taken = 6 + 5.5 = 11.5 days

Quantity B:

⇒ Time taken by B to complete the work = 20 days

⇒ Time taken by C to complete the work = 20/2 = 10 days

⇒ Time taken by A to complete the work alone = 10 + 2 = 12 days

Let total work be 60 units (LCM of 12, 20 & 10)

⇒ Work done by A in one day = 60/12 = 5

⇒ Work done by B in one day = 60/20 = 3

⇒ Work done by C in one day = 60/10 = 6

⇒ Work done by A, B and C together in 1 day = 5 + 3 + 6 = 14 units

⇒ Time taken by A, B and C together to complete the work = 60/14 = 30/7 days

∴ Quantity A > Quantity B