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There are 100 employees in a room. 99% are managers. How many managers must leave the room to bring down the percentage of managers to 98%?

Answers (1)

Total number of employees in a room = 100

99% are managers so number of managers = 99

Assume the x managers leave the room to bring down the percentage of managers to 98%.

So,   (99-x)x100/(100-x) =98

x =50

9900 - 100x = 9800 - 98x 

100 = 2x

x = 50 

Hence 50 managers should leave the room to bring down the percentage to 98%.

Posted by

Ravindra Pindel

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