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How many more men than women are in the room?

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Manager
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How many more men than women are in the room? [#permalink] New post 20 Sep 2007, 09:37
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A
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C
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E

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100% (01:05) correct 0% (00:00) wrong based on 4 sessions
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Please help to solve this question.
How many more men than women are in the room?
1. There is a total of 20 women and men in the room.
2. The number of men in the room equals the square of the number of women in the room.
VP
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Re: DS [#permalink] New post 20 Sep 2007, 11:49
el1981 wrote:
Please help to solve this question.
How many more men than women are in the room?
1. There is a total of 20 women and men in the room.
2. The number of men in the room equals the square of the number of women in the room.


it is C.

1. m+w=20 we do not know how many of men and women separately

2. from the statement 2 we know that M=W^2 it can be M=16 and W=4

so C
Manager
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Re: DS [#permalink] New post 20 Sep 2007, 12:04
Ravshonbek wrote:
el1981 wrote:
Please help to solve this question.
How many more men than women are in the room?
1. There is a total of 20 women and men in the room.
2. The number of men in the room equals the square of the number of women in the room.


it is C.

1. m+w=20 we do not know how many of men and women separately

2. from the statement 2 we know that M=W^2 it can be M=16 and W=4

so C


I got it. I made a mistake by substituting m=w^2 into m+w=20.
Thanks a lot.
Director
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Re: DS [#permalink] New post 20 Sep 2007, 17:16
el1981 wrote:
Ravshonbek wrote:
el1981 wrote:
Please help to solve this question.
How many more men than women are in the room?
1. There is a total of 20 women and men in the room.
2. The number of men in the room equals the square of the number of women in the room.


it is C.

1. m+w=20 we do not know how many of men and women separately

2. from the statement 2 we know that M=W^2 it can be M=16 and W=4

so C


I got it. I made a mistake by substituting m=w^2 into m+w=20.
Thanks a lot.


you did it right. if you combine the two stmts it would be:
m + w = 20
w^2 + w = 20
w^2 + w - 20 = 0
(w + 5) (w - 4)
w = -5, 4
cannot be -5 women so the only logical answer is 4
so plugging back into first eqn it would be
m + 4 = 20
m = 16, w = 4

you'll see that these solutions also work for the second eqn
w = 4, w^2 = 16
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Re: DS [#permalink] New post 20 Sep 2007, 17:19
el1981 wrote:
Please help to solve this question.
How many more men than women are in the room?
1. There is a total of 20 women and men in the room.
2. The number of men in the room equals the square of the number of women in the room.


Straight C
two equations two unknowns from (1) and (2), no need to solve
m+w=20
m=w^2
CEO
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Re: DS [#permalink] New post 20 Sep 2007, 19:33
el1981 wrote:
Please help to solve this question.
How many more men than women are in the room?
1. There is a total of 20 women and men in the room.
2. The number of men in the room equals the square of the number of women in the room.


Ans should be C.


S1: w+m=20. Can't find value of either from here without more info.


S2: m=w^2. We know the relationship of m to w, but cannot establish actual values b/c we don't know the total number of men and women.

Ex/ if m=4 w=2. Total is 6. But also could be: m=16 w=4. Total is 20. So not suff.

Together we know the relationship and we have the total value of w and m.

So if m=w^2, plug it in. w^2+w=20---> w^2+w-20=0 (w+5)(w-4)=0 ---> w=-5 or 4. We can't have negative women so value of women is 4.

Thus value of men is 16 so there are 12 more men that women in the room.
Re: DS   [#permalink] 20 Sep 2007, 19:33
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