Question
A, B and C alone can complete a work in 24, 48 and 32
days respectively. All of them started working together but after 2 days from start A left the job and after 3 more days B also left the job. So for how many days did C work?Solution
Let the total work = 96 units (LCM of 24, 48 and 32) Amount of work done by A alone in one day = 96/24 = 4 units Amount of work done by B alone in one day = 96/48 = 2 units Amount of work done by C alone in one day = 96/32 = 3 units Amount of work done by A, B and C together in 2 days = 2 × (4 + 2 + 3) = 18 units Amount of work done by B and C together in 3 days = 3 × (2 + 3) = 15 units Remaining work = 96 – 18 – 15 = 63 units So, the time taken by C alone to complete 63 units work = 63/3 = 21 days So, C worked for 21 + 2 + 3 = 26 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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