There are 100 employees in a room. 99% are managers.

M7MBA wrote:There are 100 employees in a room. 99% are managers. How many managers must leave the room to bring down the percentage of manager to 98%?

(A) 1
(B) 2
(C) 50
(D) 49
(E) 97

The OA is C.

Why is C the correct answer? Shouldn't it be A? I'm confused. Experts, help!

Careful. To start, we have 99 managers and 1 non-manager. If 1 manager left the room, we'd have 98 managers and 1 non-manager, meaning that 98/99 employees would be managers. 98/99 is not 98%. (Also, it must be incredibly demoralizing to be the one person at this company who isn't a manager, but that's neither here nor there.)

Say x managers leave the room. If we started with 100 employees, we'd have 100-x employees remaining. If we want there to be 98% managers, that's the same as having 2% non-managers, so we'd want non-managers/total = 2/100.
1/(100-x) = 2/100
1/(100-x) = 1/50 --> x = 50. The answer is C

M7MBA wrote:There are 100 employees in a room. 99% are managers. How many managers must leave the room to bring down the percentage of manager to 98%?

(A) 1
(B) 2
(C) 50
(D) 49
(E) 97

The OA is C.

Why is C the correct answer? Shouldn't it be A? I'm confused. Experts, help!

let us plug the answers and find the correct option

With option A percentage of managers would be = [(99-1)/99]*100 = 9800/99 which is not 98
With option B ................................= [(99-2)/98)]*100 = 9700/98 which is also not 98
..........................C ................................= [(99-50)/50)]*100 = 4900/50 which is 98
Hence C is the correct option.