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Let the total amount of work = 450 units {LCM of 18 and 25} Then, efficiency of ‘A’ = 450/18 = 25 units/day Efficiency of ‘B’ = 450/25 = 18 units/day Combined efficiency of ‘A’ and ‘B’ = 25 + 18 = 43 units/day Work completed by ‘A’ and ‘B’ together in 5 days = 5 × 43 = 215 units Remaining work = 450 – 215 = 235 units Time taken by ‘A’ alone to complete the remaining work = 235/25 = 9.4 days
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