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Puzzle: Life full of Managers!
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An official meeting is going on in a conference hall room.
There are 100 employees in that room.
Out of that 99% are managers.
How many managers must leave the room to bring down the percentage of managers to 98%?
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14
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Post Answer
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By Shaan
Given:
Reply 15Number of Employees = 100
Number of Managers = 99% = 99
Number of non-managers = 100 - 99 = 1
Now, Let there be x employees remaining so that managers = 98%
Therefore, non-managers = 100 - 98 = 2%
Also non-managers count didn't change, so = 2% = 1
Which implies => 2% of x = 1
2x = 100
x = 50
Which means number of managers = 50 - 1 = 49
So, number of managers who left the room = 99 - 49 = 50
Suppose total persons : 100
Reply 2So managers are 99% means 99 only 1 person is non manager
if a manager left one by one then
Then when half of the managers left(50) then percentage of the mangers become become 98%
49/50 = .98%
50 managers/
Reply 02
Reply 050
Reply 050
Reply 050
Reply 050
Reply 051
Reply 01
Reply 050
Reply 01
Reply 050
Reply 01
Reply 0