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# Some Qs

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Intern
Joined: 29 Dec 2009
Posts: 33
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27 Feb 2010, 20:19
00:00

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Question Stats:

100% (01:01) correct 0% (00:00) wrong based on 1 sessions

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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
Joined: 26 May 2005
Posts: 209
Followers: 2

Kudos [?]: 105 [0], given: 1

Re: Some Qs [#permalink]

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27 Feb 2010, 22:07
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
Joined: 26 May 2005
Posts: 209
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Kudos [?]: 105 [0], given: 1

Re: Some Qs [#permalink]

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27 Feb 2010, 22:08
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
Joined: 01 Feb 2010
Posts: 267
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Re: Some Qs [#permalink]

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27 Feb 2010, 22:15
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.
Senior Manager
Joined: 01 Feb 2010
Posts: 267
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Kudos [?]: 47 [0], given: 2

Re: Some Qs [#permalink]

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27 Feb 2010, 22:23
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
Joined: 22 Dec 2009
Posts: 362
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Kudos [?]: 318 [0], given: 47

Re: Some Qs [#permalink]

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28 Feb 2010, 03:26
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...!
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Manager
Joined: 26 May 2005
Posts: 209
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Re: Some Qs [#permalink]

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28 Feb 2010, 11:13
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...!

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
Joined: 10 Aug 2009
Posts: 123
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Kudos [?]: 15 [0], given: 13

Re: Some Qs [#permalink]

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02 Mar 2010, 13:08
Hmmm I still don't see why its 10 and not 27....
Manager
Joined: 26 May 2005
Posts: 209
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Kudos [?]: 105 [0], given: 1

Re: Some Qs [#permalink]

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02 Mar 2010, 15:22
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.
Senior Manager
Joined: 16 Apr 2009
Posts: 339
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Kudos [?]: 106 [0], given: 14

Re: Some Qs [#permalink]

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03 Mar 2010, 22:37
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, 22:37
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# Some Qs

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