A can do a certain work in the same time in which B and C together can do it. If A and B together could do it in 10 days and C alone in 50 days, then B alone could do it in:
Explanation:
Total work = LCM(10, 50) = 50 units. Efficiency of (A + B) = 5, efficiency of C = 1. Total efficiency of (A + B + C) = 5 + 1 = 6. Given A = B + C. So, (B + C) + B + C = 6 => 2(B + C) = 6 => B + C = 3. Since C = 1, B = 2. Time taken by B alone = 50 / 2 = 25 days.
A can do a work in 18 days and B in 24 days. They start together but A leaves 3 days before the completion of the work. Find the total number of days taken to complete the work.
Explanation:
Total work = LCM(18, 24) = 72 units. Efficiency of A = 4, B = 3. Let total days be x. A works for (x - 3) days, B works for x days. 4(x - 3) + 3x = 72 => 4x - 12 + 3x = 72 => 7x = 84 => x = 12 days.
No comments yet. Be the first to start the discussion!