Question
‘A’ is 20% more efficient than ‘B’. ‘A’ and
‘B’ work together for 6 days after which ‘A’ is replaced by ‘C’. ‘B’ and ‘C’ together take 6 more days to finish the work. If ‘A’ alone could’ve finished the whole work in 20 days, then find the time taken by ‘C’ to finish 60% work alone.Solution
Let the efficiency of 'B' be 'x' units/day So, efficiency of 'A' = x × 1.2 = '1.2x' units/day Total work = 20 × 1.2x = '24x' units So, work done by 'A' and 'B' together in 6 days = (x + 1.2x) × 6 = '13.2x' units So, remaining work = 24x - 13.2x = '10.8x' units So, efficiency of 'B' and 'C' together = 10.8x ÷ 6 = '1.8x' units per day So, efficiency of 'C' = 1.8x - x = '0.8x' units/day So, required time = (24x × 0.6) ÷ 0.8x = 18 daysÂ
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