ATQ, Let the efficiency of 'Ankit' be 'a' units/day. So, efficiency of 'Bheema' = a × 1.40 = '1.4a' units/day So, total work = 20 × a = '20a' units Work done by 'Ankit' in 6 days = 6 × a = '6a' units So, time taken by 'Bheema' to finish the remaining work = (20a - 6a) ÷ 1.4a = (14a/1.4) = 10 days
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