100 Employees in a Conference Room

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There are 100 employees in a conference room in New York City. You note that 99% of them are managers. How many managers would need to leave the conference in order to reduce the percentage of managers in the hall to 98% ?

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There are 100 employees and 99% of them are managers implying there are 99 managers.

Suppose we need to remove x managers to reduce the percentage to 98%. Now numbers of managers in the room are 99-x and total number of employees are 100-x. As the percentage of mangers is 98%, we get

(99 - x) / (100 - x) = 98/100

=> 9900 - 100x = 9800 - 98x

=> 2x = 100

=> x = 50

Solving this simple equation gives x = 50

That means the percentage = 49/50= 98/100 = 98%

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