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)
114, 115, 118, 127, ?, 235
7.5% of 820.68 - 95% of 945.06 + 12.8% of 1050= ?
8 12 30 105 ? 2598.75
...70, 69, 73, 64,80, 55, ?
78, 88, 100, 114, ?, 148
115% of 800 - 4/5 of 320 + 82% of 700 = ? – 102% of 500
11, ? 220, 660, 1320, 1320
827, 820, 806, 785, 757, ?
2, 2, 6, 30, ?, 1890
78, 80, 84, ?, 108, 140
