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Let the efficiencies of an efficient worker and a non-efficient worker be 'x' units/day and 'y' units/day respectively. ATQ, 2x X 20 = 3y X 20 Or, x:y = 3:2 Let x = 3a and y = 2a So, total work = 2 X 3a X 20 = 120a units Increased efficiency of an efficient worker = 3a X (4/3) = 4a units/day Decreased efficiency of a non-efficient worker = 2a X (1/2) = 'a' unit/day Therefore, required time = 120a ÷ (2 X 4a + 2 X a) = (120a ÷ 10a) = 12 days
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