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%?

Given:
Number 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

Given:
Number 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

Given:

Reply 15 0Number 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 2 0So 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 0 02

Reply 0 050

Reply 0 050

Reply 0 050

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