A and B working together completes a work in 30 days whereas B and C working together completes the same work in 45 days. A starts the work and worked for 10 days,then B replaced A and worked for 30 days, finally C replaced B and completed the remaining work in 40 days. Find working alone A will complete the whole work in how many days?

Work done: (A+B) × 30 = (B+C) × 45 = 1 unit. A+B = 1/30, B+C = 1/45. A works 10 days, B works 30 days, C works 40 days. Total work = 10A + 30B + 40C = 10A + 30(1/30 - A) + 40(1/45 - (1/30 - A)) = 10A + 1 - 10A + 40/45 - 40/30 + 40A = 1 + 40A + 8/9 - 4/3 = 40A + 1 - 4/9 = 40A + 5/9. This equals 1, so 40A = 4/9, A = 1/90. A alone takes 90 days... but claimed answer is 45 days. Let me verify differently: A(10) + B(30) + C(40) = 1. B = 1/30 - A, C = 1/45 - B = 1/45 - 1/30 + A = A - 1/90. 10A + 30(1/30 - A) + 40(A - 1/90) = 10A + 1 - 30A + 40A - 4/9 = 20A + 5/9 = 1. 20A = 4/9. A = 1/45. Yes! A alone takes 45 days.