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

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b2bt
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There are 100 employees in a room. 99% are managers. How man [#permalink] New post   25 May 2014, 05:51
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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 manager to 98%?

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



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The interesting thing here is the base i.e. 100 is also changing as the no. of managers change. Either plug values or use the formula (99-x)/(100-x) = 98/100
Where x = no. of managers leaving.
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PareshGmat
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Re: There are 100 employees in a room. 99% are managers. How man [#permalink] New post   28 May 2014, 01:18
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Managers.......... Non Managers............. Total

99........................ 1 ............................ 100

Let's say x managers leave the room, no new table is

99-x ..................... 1 ................................ (100-x)

Setting up the equation

\(99-x = \frac{98}{100} * (100-x)\)

x = 50

Answer = C
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Re: There are 100 employees in a room. 99% are managers. How man [#permalink] New post   25 May 2014, 05:59
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b2bt 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 interesting thing here is the base i.e. 100 is also changing as the no. of managers change. Either plug values or use the formula (99-x)/(100-x) = 98/100
Where x = no. of managers leaving.


Another approach:

We have 99 managers and 1 director. That 1 director to compose 2% of the total number of people, there must be 50 people in the room, hence 50 managers must leave.

Answer: C.

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five-kilograms-of-oranges-contained-98-of-water-141401.html
fresh-dates-contain-90-water-while-dry-dates-contain-143245.html
if-grapes-are-92-water-and-raisins-are-20-water-then-how-159030.html

Hope this helps.
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Re: There are 100 employees in a room. 99% are managers. How man [#permalink] New post   28 May 2014, 19:24
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One more approach (Not sure if I would be able to explain properly :) )

100 Employees = 99 Managers + 1 Non-Manager ............. (1)

Here Non-Managers constitute 1% = 1

We require to make Non-Managers constitute 2% = 1

2% of what is 1 ??

2 out of 100 = 1 out of 50

If total Employees = 50, then 1 (Non-Managers) will constitute its 2%

50 - 1 = 49 Managers ............. (2)

Managers to be removed = (1) - (2)

= 99 - 49

= 50

Answer = C
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Re: There are 100 employees in a room. 99% are managers. How man [#permalink] New post   30 Oct 2015, 06:57
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try plugging in the numbers, it doesn't take that much time as writing the equation
if 1 manager leaves, we have 98 managers 99 people. 98/99 - is not 98%
if 2 managers leave, we have 97/98 - this is not 98
if 50 managers leave, we have 49 managers, and 50 people, that is 49/50, multiply by 2-> 98/100 or 98%. C.
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Re: There are 100 employees in a room. 99% are managers. How man [#permalink] New post   01 Aug 2017, 17:10
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b2bt 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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Re: There are 100 employees in a room. 99% are managers. How man [#permalink] New post   10 Aug 2017, 00:22
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There are 100 managers in total.
So T=100
Let 'x' be the no.of managers leaving the room.
99-x/100-x *100 = 98
x=50
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Re: There are 100 employees in a room. 99% are managers. How man [#permalink] New post   28 May 2014, 19:34
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PareshGmat wrote:
Managers.......... Non Managers............. Total

99........................ 1 ............................ 100

Let's say x managers leave the room, no new table is

99-x ..................... 1 ................................ (100-x)

Setting up the equation

\(99-x = \frac{98}{100} * (100-x)\)

x = 50

Answer = C


One more method of setting up equation

\(1 = \frac{2}{100} * (100-x)\)

100 = 200 - 2x

x = 50

Answer = C
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Re: There are 100 employees in a room. 99% are managers. How man [#permalink] New post   19 May 2016, 18:54
b2bt 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



Please give kudos if you like the sum :D
Show Spoiler
The interesting thing here is the base i.e. 100 is also changing as the no. of managers change. Either plug values or use the formula (99-x)/(100-x) = 98/100
Where x = no. of managers leaving.

as number of managers changing...other employees remain same
so number of other employees before and after removal of managers are same
1% of 100=2%of(100-x) where x=no. of managers removed out of total 100 employees.
1=2/100(100-x)
x=50
ans C
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Re: There are 100 employees in a room. 99% are managers. How man [#permalink] New post   10 Aug 2017, 00:29
99-x/100-x=98/100
Plug in values from answer choices.
C wins!
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Re: There are 100 employees in a room. 99% are managers. How man [#permalink] New post   28 Aug 2017, 07:13
Let number of managers to leave be x.

Equation will be formed is :

99-x = 98/100(100 - x) -> to form 98% managers we remove x managers from existing 99... the total employees would be 100 - x.

Solve this to get value of x as 50
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Re: There are 100 employees in a room. 99% are managers. How man [#permalink] New post   19 Sep 2022, 09:43
Previous (Manager: Others)=99:1
Later (Manager: Others)= 98:2
Later (Manager: Others)= 49:1
Others will remain the same

--->99-49=50 is the answer
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Re: There are 100 employees in a room. 99% are managers. How man [#permalink] New post   28 Nov 2023, 01:47
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