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Suppose there are 198 men and 2 women in a room. That is,

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Suppose there are 198 men and 2 women in a room. That is, [#permalink]

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03 Jan 2005, 14:25
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Suppose there are 198 men and 2 women in a
room. That is, the men make up 99% of the people in the room.

How many men have to leave for the percentage of men to drop to 98%?
VP
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03 Jan 2005, 14:56
1. 198 - x / 200 = 98 / 100
2. 19800 - 100x = 19600
3. 200 = 100x
4. 2 = x
5. solution is 2.

is it ?
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03 Jan 2005, 14:58
Currently the proportion is 198/200. Call the number of men that have to leave x. So you want (198-x)/(200-x)=.98

When you cross multiply you get: 198-x=196-.98x

Subtract 196 and add x to both sides to get 2=.02x

Multiply both sides by 100 to get: 200=2x

So x=100

If you put 100 in for x in the original equation you get (198-100)/(200-100)= 98/100, which checks out.
VP
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03 Jan 2005, 15:03
ah, right
Director
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03 Jan 2005, 15:12
New percetage = .98

198 - x / 200 = .98

x = 2

x number of men to leave the group
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03 Jan 2005, 15:39
rxs0005 wrote:
New percetage = .98

198 - x / 200 = .98

x = 2

x number of men to leave the group

If 2 men leave the group then why are you still dividing by 200? For that to be true wouldn't 2 women have to join the group? If 2 men left it would be 196/198, which is not 98%. You have to take into account that any man that leaves also decreases the total. The total is the number of men (198)+ women(2). Therefore if you decease the number of men from numerator, you have to also decrease it from the deominator.
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03 Jan 2005, 18:56
100 it is and toddmartin's explanation is as good as any
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03 Jan 2005, 19:04
this Q if it was in GMAT wud have an answer choice 2 a classic trap

you guys are correct the answer is 100

198 - x / 200 - x = .98

.02x = 2 x = 100
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