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

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by DavidG@VeritasPrep » Sun Dec 10, 2017 11:12 am
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
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by GMATWisdom » Sun Dec 10, 2017 12:25 pm
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.
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by Brent@GMATPrepNow » Sun Dec 10, 2017 2:29 pm
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
There are 100 employees in a room. 99% are managers.
So, we can conclude that (initially) there are 99 managers and 1 non-manager .

After we remove some managers, 98% of the people are managers. This means that 2% of the people are non-managers.
We already know there is 1 non-manager , and we know that 1/50 = 2%
That is, if there were 50 people altogether, and 1 person was a non-manager, then 2% of the people would be non-managers.
In order to get 50 people altogether, we must remove 50 people from the group.

Answer: C

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by Jeff@TargetTestPrep » Tue Dec 19, 2017 10:05 am
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
Since there are 100 employees in a room and 99% are managers, there are 99 managers. We can let n = the number of managers to take out and create following equation:

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

100(99 - n) = 98(100 - n)

9900 - 100n = 9800 - 98n

100 = 2n

n = 50

Answer: C

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