Question
A alone can complete 40% of a work in 18 days while B
takes 15 days more than A to complete it. If B and C together can complete the work in 20 days, then find the time taken by C alone to complete the same work.Solution
Time taken by A alone to complete the entire work = 18/0.4 = 45 days Time taken by B alone to complete the entire work = 45 + 15 = 60 days Let the total work = L.C.M of 45, 60 and 20 = 180 units Then, efficiency of B = (180/60) = 3 units/day Combined efficiency of B and C = (180/20) = 9 units/day So, efficiency of C alone = 9 – 3 = 6 units/day So, time taken by C alone to complete the entire work = (180/6) = 30 days
6000 3002 1503 ? 378.75 191.375 97.6875
...1, 27, ?, 343, 729, 1331
(32.03 + 111.98) ÷ 18.211 = 89.9 – 20.23% of ?
51 53 109 332 ? 6686
...9 4.5 4.5 9 36 ?
5 16 ? 66 119 200
...1 1 8 4 27 9 ?
...What will be the next number in the series?
2, 7, 23, 70, ?
There are three series given below which are following with the same pattern.
Series I: 1, 12, 38, 193, 1355
Series II: 6, B, C, D, E
...Direction: Which of the following will replace ‘?’ in the given question?
5, 18, ‘?’, 126, 296, 586, 1044
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