A and B together can complete a certain work in 20 days, B and C together can complete it in 30 days and C and A together can complete 7 it in 24 days. In how many days will A alone complete 12 part of the same ​ work?

Explanation

Given: A and B together can complete a certain work in 20 days, B and C together can complete it in 30 days and C and A together can complete it in 24 days. Concept used: Total work = Efficiency (Work done per hour) × Total time taken Calculation: LCM (20, 30, 24) = 120 Let the total work be the LCM of the time taken by different workers. So, total work = 120 units Let the efficiencies of A, B, and C be represented by A, B, and C. According to the question, A + B = 120/20 ⇒A+B=6 ....(1) B + C = 120/30 ⇒B+C=4 ....(2) A + C = 120/24 ⇒A+C=5 ....(3) (1) + (2) + (3), 2(A + B + C) = 6 + 4 + 5 ⇒ A + B + C = 7.5 ....(4) (4) - (2), A + B + C - B - C = 7.5 - 4 ⇒ A = 3.5 Now, A alone will complete 7 ⇒ ( 12 × 120)/3.5 ​ ⇒ 20 days 7 12 ​ part of the same work in ∴ A alone will complete 7 12 ​ part of the same work in 20 days.