At their usual efficiency levels, A and B together finish a task in 12 days. If A had worked half as efficiently as she usually does, and B had worked thrice as efficiently as he usually does, the task would have been completed in 9 days. How many days would A take to finish the task if she works alone at her usual efficiency?
Explanation:
Let total work be the LCM of 12 and 9 = 36 units.
Let the efficiency of A and B be a and b respectively.
Work done per day when A and B are working together = 36/12 = 3 units.
∴ a + b = 3 ...(1)
Work done per day when A is working at half efficieny and B is working at thrice efficiency = 36/9 = 4 units.
∴ (a/2) + 3b = 4 ...(2)
Solving (1) and (2), we get; a = 2.
Time taken by A alone to complete the work = 36/2 = 18 days.
Hence, option (d).
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