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Let the total work = L.C.M of 35, 20 and 28 = 140 units Then, efficiency of 'A' = 140 ÷ 35 = 4 units/day Efficiency of 'B' = 140 ÷ 20 = 7 units/day Efficiency of 'C' = 140 ÷ 28 = 5 units/day Work done by 'A' alone in 5 days = 4 X 5 = 20 units So, work done by 'B' and 'C' in 'x' days = 140 - 20 = 120 units So, x = 120 ÷ (7 + 5) = 10
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