Question
'Alex,' 'Ben,' and 'Charlie' can individually complete a
task in 40, 60, and 100 days, respectively. They decide to work together, but there's a unique pattern: 'Alex' works every day, 'Ben' joins in on odd-numbered days, and 'Charlie' on even-numbered days. Calculate the time required to finish the entire task following this alternating work schedule.Solution
ATQ, Let the total work be 600 units. {LCM (40, 60 and 100)} So, efficiency of 'Alex' = 600 ÷ 40 = 15 units/day And efficiency of 'Ben' = 600 ÷ 60 = 10 units/day And efficiency of 'Charlie' = 600 ÷ 100 = 6 units/day So, work done in every 2 days = (15 × 2) + 10 + 6 = 46 units/day So, work done in 26 days = (26/2) × 46 = 598 units And time taken by 'Alex' and 'Ben' together to finish the remaining work on last day = {(600-598)/(15+10)} So, total time taken = 26 + (2/25) =26(2/5)
Statements:
D ≤ E > I; Q ≤ I ≤ W; F = W ≥ Y
Conclusions:
I. E > Q
II. F ≥ I
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