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?
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?
Updated On: Aug 21, 2024
- 24
- 18
- 12
- 36
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The Correct Option is B
Solution and Explanation
Let the total work be the LCM of 9 and 12, which is 36 units.
The amount of work done in one day with their normal efficiencies by A and B is represented by x and y units, respectively.
So, (x + y) × 12 = 36, which simplifies to x + y = 3 ... (1)
Similarly, (x/2 + 3y) × 9 = 36, or x/2 + 3y = 4 ... (2)
Solving equations (1) and (2) for x, we find x = 2 units. Therefore, A alone would take 36/x = 36/2 = 18 days to complete the work with her normal efficiency.
The amount of work done in one day with their normal efficiencies by A and B is represented by x and y units, respectively.
So, (x + y) × 12 = 36, which simplifies to x + y = 3 ... (1)
Similarly, (x/2 + 3y) × 9 = 36, or x/2 + 3y = 4 ... (2)
Solving equations (1) and (2) for x, we find x = 2 units. Therefore, A alone would take 36/x = 36/2 = 18 days to complete the work with her normal efficiency.
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