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Some Qs

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Some Qs [#permalink] New post 27 Feb 2010, 19:19
00:00
A
B
C
D
E

Difficulty:

(N/A)

Question Stats:

100% (01:01) correct 0% (00:00) wrong based on 1 sessions
Of the 120 people in a room, 3/5 are women. If 2/3 of the people are married, what is the maximum number of women in the room who could be unmarried?
a)80
b)72
c)48
d)40
e)32
[Reveal] Spoiler:
D



Q2. 3 friends are playing a game in which each person simultaneously displays one of three hand signs; a clenched fist, an open palm, or two extended fingers. How many different combinations of the signs are possible.?

a)3
b)9
c)10
d)12
e)27

[Reveal] Spoiler:
C
Manager
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Re: Some Qs [#permalink] New post 27 Feb 2010, 21:07
ssgomz wrote:
Of the 120 people in a room, 3/5 are women. If 2/3 of the people are married, what is the maximum number of women in the room who could be unmarried?


women = 3/5 * 120 = 72
married = 2/3 * 120 = 80
unmarried = 40
max(un-married women) = 40
D
Manager
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Re: Some Qs [#permalink] New post 27 Feb 2010, 21:08
Quote:
Q2. 3 friends are playing a game in which each person simultaneously displays one of three hand signs; a clenched fist, an open palm, or two extended fingers. How many different combinations of the signs are possible.?

same signs for friends = 3(HHH, EEE, OOO)
2 same signs and one different sign = 2 * 3 = 6 (HHO, HHE, OOH, OOE, EEH, EEO)
3 different signs = 1 (HEO)

total 10

C
Senior Manager
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Re: Some Qs [#permalink] New post 27 Feb 2010, 21:15
ssgomz wrote:
Of the 120 people in a room, 3/5 are women. If 2/3 of the people are married, what is the maximum number of women in the room who could be unmarried?
a)80
b)72
c)48
d)40
e)32
[Reveal] Spoiler:
D


Men = 2/5 of 120 = 48
Women = 3/5 of 120 = 72

Married = 2/3 of 120 = 80
Unmarried = 120 - 80 = 40
So total of 40 men and 40 women married hence 32 women unmarried, so E.
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Re: Some Qs [#permalink] New post 27 Feb 2010, 21:23
ssgomz wrote:
Of the 120 people in a room, 3/5 are women. If 2/3 of the people are married, what is the maximum number of women in the room who could be unmarried?
a)80
b)72
c)48
d)40
e)32
[Reveal] Spoiler:
D



Q2. 3 friends are playing a game in which each person simultaneously displays one of three hand signs; a clenched fist, an open palm, or two extended fingers. How many different combinations of the signs are possible.?

a)3
b)9
c)10
d)12
e)27

[Reveal] Spoiler:
C


What is the source of questions?
Senior Manager
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Re: Some Qs [#permalink] New post 28 Feb 2010, 02:26
chix475ntu wrote:
Quote:
Q2. 3 friends are playing a game in which each person simultaneously displays one of three hand signs; a clenched fist, an open palm, or two extended fingers. How many different combinations of the signs are possible.?

same signs for friends = 3(HHH, EEE, OOO)
2 same signs and one different sign = 2 * 3 = 6 (HHO, HHE, OOH, OOE, EEH, EEO)
3 different signs = 1 (HEO)

total 10

C


A list lost here.... Would be great if u could help me with this.

3 symbols.... and 3 people to choose....
Shouldn't the total symbols comb = 3x3x3 = 27...! :roll:
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Manager
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Re: Some Qs [#permalink] New post 28 Feb 2010, 10:13
jeeteshsingh wrote:
chix475ntu wrote:
Quote:
Q2. 3 friends are playing a game in which each person simultaneously displays one of three hand signs; a clenched fist, an open palm, or two extended fingers. How many different combinations of the signs are possible.?

same signs for friends = 3(HHH, EEE, OOO)
2 same signs and one different sign = 2 * 3 = 6 (HHO, HHE, OOH, OOE, EEH, EEO)
3 different signs = 1 (HEO)

total 10

C


A list lost here.... Would be great if u could help me with this.

3 symbols.... and 3 people to choose....
Shouldn't the total symbols comb = 3x3x3 = 27...! :roll:


Here we are talking about combinations and not permutations ... lets take the case of 3 friends and each choosing differnt symbol.

combinations = 1(no matter what each child picks, they will be picking the same 3 signs in different order)
permutation = 3!
Manager
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Re: Some Qs [#permalink] New post 02 Mar 2010, 12:08
Hmmm I still don't see why its 10 and not 27....
Manager
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Re: Some Qs [#permalink] New post 02 Mar 2010, 14:22
nickk wrote:
Hmmm I still don't see why its 10 and not 27....


Here's the answer that I posted earlier.
same signs for friends = 3(HHH, EEE, OOO)
2 same signs and one different sign = 2 * 3 = 6 (HHO, HHE, OOH, OOE, EEH, EEO)
3 different signs = 1 (HEO)
total 10


Now take 3 different signs - HEO
You can have 6 different types you can select(order matters) - HEO HOE EHO EOH OEH OHE
Now if the order doesnt matter - then only 1 combination HEO as the rest of the 5 have the same alphabets but in different order.
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Re: Some Qs [#permalink] New post 03 Mar 2010, 21:37
bangalorian2000 wrote:
ssgomz wrote:
Of the 120 people in a room, 3/5 are women. If 2/3 of the people are married, what is the maximum number of women in the room who could be unmarried?
a)80
b)72
c)48
d)40
e)32
[Reveal] Spoiler:
D


Men = 2/5 of 120 = 48
Women = 3/5 of 120 = 72

Married = 2/3 of 120 = 80
Unmarried = 120 - 80 = 40
So total of 40 men and 40 women married hence 32 women unmarried, so E.


I did the same calculation and got 32 . Can some one please explain to me the OA?
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Re: Some Qs   [#permalink] 03 Mar 2010, 21:37
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