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kevincan
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Quote:
This is tricky. SInce the average of B employees is 12.25, we should assume that there are at least 4 employees in Team B.



good point rbcola!
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rbcola
This is tricky. SInce the average of B employees is 12.25, we should assume that there are at least 4 employees in Team B.

The problem basicaly is asking if total sick days for all employees is >1540.

Assuming that employee cannot take 1/4 day then A is sufficient.

beacuse for all combinations of B > 2 and A, the total is greater than 1540.

However, is the employee can take 1/4 day sick leave then A is not sufficient beacause if B = 1. then the total sick days = 1537.5, with A = 94. In which case C is the answer.

These type of problems take up too much time.


You're right- the way you've done it requires a lot of number crunching, and that is not what the exam is about! Can you think of an easier way to get to your correct conclusion?
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A

The way I did is this:
Total average will be minimum when the number of employees in the department with minimum average are larger.

St1: If C has 45 employee then for total average to be minimum we can have most employees in dept A. B must have atleast 4. A must have rest i.e (140-4-45 = 91 employees). If we transfer 1 sick day from all 45 employees of C to 90 of A and one sick day form B to remaining of A then we have 139 employees with 11 sick days and one with 11.25 sick days. So average will be greater than 11.: SUFF

St2: We don't know anything about C: INSUFF



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