Question

A and B can do a piece of work in 10 days and 15 days, respectively. They work together for 4 days. The remaining work is completed by C alone in 12 days. C alone will complete 4/9 part of the original work in:

• A. 24 days
• B. 12 days
• C. 20 days
• D. 16 days

Answer 1

Answer:

D. 16 days

Solution: Given: Time taken by A to complete the work is 10 days. Time taken by B to complete the work is 15 days. Time taken by C to complete remaining work is 12 days. Concept Used: Time taken = (Total work done)/(Efficiency) Calculation: Let the work done by A, B and C be in Units. Total Work done by (A + B) = L.C.M of (10, 15) ⇒ 30 Units Now, Efficiency of A = Total Work/Time taken ⇒ 30/10 = 3 units per day Efficiency of B = Total Work/Time taken ⇒ 30/15 = 2 units per day Total Work done by (A + B) in 4 days = (Sum of efficiency of A and B) × 4 ⇒ (3 + 2) × 4 = 20 units Remaining Work = (30 - 20) = 10 units Efficiency of C for completing remaining work = 10/12 ⇒ (5/6) unit per day Then, Time taken by C to complete (4/9) part of original work = 30 × (4/9) × (6/5) ⇒ 16 days ∴ Time taken by C to complete (4/9) part of original work is 16 days.