Question:
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?
- 24
- 18
- 12
- 36
Correct Answer: Option: 2
Let the work be LCM of (9 and 12) = 36 units.
Let the amount of work done in one day with their normal efficiencies by A and B be x and y units respectively.
Therefore, (x+y)×12=36
Or x+y =3 … (1)
Similarly,
(x/2+3y)×9=36
Or x/2 +3y = 4 … (2)
Solving (1) and (2) for x, we get x =2 units
Hence, A alone would take 36/x = 36/2 = 18 days to complete the work with her normal efficiency.
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