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Let the total work be 100 units. {LCM (20 and 50)}
So, combined efficiency of 'I' and 'J' = 100 ÷ 20 = 5 units/day
Efficiency of 'I' = 100 ÷ 50 = 2 units/day
So, efficiency of 'J' = 5 - 2 = 3 units/day
So, time taken by 'J' to do the work alone = 100 ÷ 3 = (100/3) days
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