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manhattan GMAT probability question

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manhattan GMAT probability question [#permalink] New post 03 Jan 2006, 02:30
http://www.manhattangmat.com/ChallengeProblem.cfm?ID=223

Can someone have a look at this problem ?
I picked E since 11+1=12 also holds ...
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Re: manhattan GMAT probability question [#permalink] New post 03 Jan 2006, 06:42
wlee76 wrote:
http://www.manhattangmat.com/ChallengeProblem.cfm?ID=223

Can someone have a look at this problem ? I picked E since 11+1=12 also holds ...

i doubt that this is a question you intented to link. but i am directed to this this question..............

Quote:
If p2 – 13p + 40 = q, and p is a positive integer between 1 and 10, inclusive, what is the probability that q < 0?

(A) 1/10
(B) 1/5
(C) 2/5
(D) 3/5
(E) 3/10


p2 – 13p + 40 = q
p2 – 8p -5p+ 40 = q
(p-8)(p-5)=q

if p = either 6 or 7, the value of q would be <0.
so p=2/10=1/5

B.
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Prime Seats [#permalink] New post 03 Jan 2006, 06:53
Question
There are x people and y chairs in a room where x and y are positive prime numbers. How many ways can the x people be seated in the y chairs (assuming that each chair can seat exactly one person)?

(1) x + y = 12

(2) There are more chairs than people.


(A) Statement (1) ALONE is sufficient to answer the question, but statement (2) alone is not.
(B) Statement (2) ALONE is sufficient to answer the question, but statement (1) alone is not.
(C) Statements (1) and (2) TAKEN TOGETHER are sufficient to answer the question, but NEITHER statement ALONE is sufficient.
(D) EACH statement ALONE is sufficient to answer the question.
(E) Statements (1) and (2) TAKEN TOGETHER are NOT sufficient to answer the question.
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 [#permalink] New post 03 Jan 2006, 08:33
(1) x+y = 12. We know x or y cannot be 2, since either one being 2 will result in the other not being a prime number. x or y also cannot be 3. So the (x,y) pair can be (5,7) or (7,5).

If x = 5, y = 7, then the question is to arrange 5 people in 7 chairs. The # of ways to do this is 7P5 = 7!/2!.
If x = 7, y=5, then the question is to arrange 7 people in 5 chairs. Again, the # of ways to do this is 7P5 = 7!/2!.

So no matter which pair we take, we end up with the same number and so we can answer the question. Statement 1 is therefore sufficient.

(2) There are more chairs than people. For this, there are many possible (x,y) pairs and so there is no single answer. Statement 2 is therefore insufficient.

Ans: A
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 [#permalink] New post 03 Jan 2006, 18:22
1 is not a prime number.

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 [#permalink] New post 03 Jan 2006, 21:31
I guess C should be the answer

x(People) y(chair)
7 5
5 7

Selecting 7 chairs for 5 people = 7C5 Could be determined
Selecting 5 chair for 7 people = 5C7 =>Turns negative

So we need both statements.
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Re: Prime Seats [#permalink] New post 03 Jan 2006, 22:43
I would choose C.

From 1 X,y could be either (5,7) or (7,5) where X is number of people and y isnumber of chairs.

When x,y = 5,7 we get 7P5 = 42
When x,y = 7,5 we get 5P7 = 5P5= 5! = 120

So insuff.

From stmt 2 we can find any number of chairs and people who are prime and number of chairs > number of people so insuff.

Combining both we get chair=7 and people = 5
= 7P5 = 42

Hence C.
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 [#permalink] New post 04 Jan 2006, 10:33
For this the Ans A is correct

because say x =7 ,y =5 total ways is 7p5 = 7!/2! = 7*6*5*4*3

when x =5 and y =7 take it this way

first one has 7 option to choose from ----7
second has 6 option to choose from -----7*6
third has 5 option to choose from -----7*6*5
fourth has 4 option to choose from ----7*6*5*4
last has 3 option to choose from ----- 7*6*5*4*3

so in both cases we get the same ans ...hence A is the right choice .
  [#permalink] 04 Jan 2006, 10:33
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