Question
Three workers W1, W2 and W3 got work from the
contractor. In 5 days, ‘W1’ completed 25% of work. ‘W2’ is 20% less efficient than ‘W1’ but takes 5 days less than ‘W3’ to complete the work alone. Find the time taken by all three together to complete 74% of work.Solution
Time taken by ‘W1’ to complete the work = 5/0.25 = 20 days Time taken by ‘W2’ to complete the work = 20/0.2 = 25 days Time taken by ‘W3’ to complete the work = 25 + 5 = 30 days Let the total work = 300 units (LCM of 20, 25 and 30) Efficiency of ‘W1’ = 300/20 = 15 units/day Efficiency of ‘W2’ = 300/25 = 12 units/day Efficiency of ‘W3’ = 300/30 = 10 units/day Required time taken = (0.74 × 300)/37 = 6 days
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