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ATQ, Let total work = {(w – 2)(w + 4)} units Efficiency of 'V' = {(w – 2)(w + 4)}/(w – 2) = (w + 4) units/day Efficiency of 'N' = {(w – 2)(w + 4)}/(w + 4) = (w – 2) units/day ATQ; 10 × (w – 2) + 6 × (w + 4 + w – 2) = (w + 4)(w – 2) Or, w2 = 20 w Or, w = 20
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