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In a certain game played with red chips and blue chips, each

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In a certain game played with red chips and blue chips, each [#permalink] New post 04 Jun 2008, 12:22
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A
B
C
D
E

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0% (00:00) correct 0% (00:00) wrong based on 1 sessions
In a certain game played with red chips and blue chips, each red chip has a point value of X and each blue chip has a point value of Y, where X>Y and X and Y are positive integers. IF a player has 5 red chips and 3 blue chips, what is the average (arithmetic mean) point value of the 8 chips that the player has?

1) the average point value of one red chip and one blue chip is 5
2) the average point value of the 8 chips that the player has is an integer
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Re: DS: chips [#permalink] New post 04 Jun 2008, 13:08
C .

1)( r+b )/2 =5

r+b =10

not sufficient because different combinations possible

(r, b) = ( 9,1) ,(8,2) (7,3) , (6,4)

2) not sufficient doesnt give us enough information

combining 1 and 2 , only (5*9+3*1)/8 gives us integer value
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Re: DS: chips [#permalink] New post 04 Jun 2008, 17:57
hmm. why not E?
from A, it says-(r+b)/2=5
hence r+b=10.
from B we therefore have- [b+b+(r+b)+(r+b)+(r+b)]/8= some integer.
both A and B,
2b+30/8 = some integer.
hence when b= 1, avg= 4
b=9, avg = 6.
.
so i think E.

whats the 0A?
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Re: DS: chips [#permalink] New post 04 Jun 2008, 19:05
arjtryarjtry wrote:
hmm. why not E?
from A, it says-(r+b)/2=5
hence r+b=10.
from B we therefore have- [b+b+(r+b)+(r+b)+(r+b)]/8= some integer.
both A and B,
2b+30/8 = some integer.
hence when b= 1, avg= 4
b=9, avg = 6.
.
so i think E.

whats the 0A?



It is mentioned that X> Y so b can't be 9
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Re: DS: chips [#permalink] New post 05 Jun 2008, 06:04
puma wrote:
In a certain game played with red chips and blue chips, each red chip has a point value of X and each blue chip has a point value of Y, where X>Y and X and Y are positive integers. IF a player has 5 red chips and 3 blue chips, what is the average (arithmetic mean) point value of the 8 chips that the player has?

1) the average point value of one red chip and one blue chip is 5
2) the average point value of the 8 chips that the player has is an integer


(x+y)/2=5 --> x+y=10

1: But we cannot deduce what x and y are . Insuff.


2: (5x+3y)/8= integer... Again could be anything.

Together:

Try numbers: note that X>Y. So we can try 6,4. 7,3, 8,2 and 9, 1.

6*5+3*4=42 not divisible by 8.

7*5+3*3 = 44 not divisible by 8.

40+6 not divisible by 8.

9*5+3*1 =48. Here we go.

C
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Re: DS: chips [#permalink] New post 05 Jun 2008, 12:24
puma wrote:
OA is D



What's the source ? Chances OA is wrong ?
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Re: DS: chips [#permalink] New post 05 Jun 2008, 19:26
cannot be D

if you take stmt-2

you will get multiple solutions with the boundary condition given in stmt-1

5x+3Y/8 = should be an integer

given X>Y thus (X,Y) could be (9,1) or (12,4) I am sure we can find many more..

so it should be C
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Re: DS: chips [#permalink] New post 05 Jun 2008, 21:49
I also believe that OA is wrong. it has to be C. i totally agree with the explanaion given by rpmodi
Re: DS: chips   [#permalink] 05 Jun 2008, 21:49
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