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Let the total work = L.C.M of 50 and 20 = 100 units Then, efficiency of ‘A’ alone = 100 ÷ 50 = 2 units/day Combined efficiency of ‘A’ and ‘B’ = 100 ÷ 20 = 5 units/day So, efficiency of ‘B’ = 5 – 2 = 3 units/day So, work done by ‘B’ alone in 25 days = 3 × 25 = 75 units So, percentage of work completed by ‘B’ = (75 ÷ 100) × 100 = 75%
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