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

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Director
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In a certain game played with red chips and blue chips, each [#permalink] New post 02 May 2005, 13:48
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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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Director
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 [#permalink] New post 02 May 2005, 14:23
From I

X + y = 10 not suff

from II not suff

From I and II


5x + 3y / 8 = Integer

and x + y = 10

we see that only 9,1 will fit to get integer

ANs = C
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 [#permalink] New post 05 May 2005, 00:35
The answer is C

(5x+3y)/8 = Integer.

From 1, (X+Y)/2 = 5 ==> x+y=10

So equation 1 can be written as
(2x+3(x+y))/8 = integer

(2x + 30)/8 = integer.

implying, x can be 1 or 4 or 9 or 13 etc..

Since x+y = 10, we can ignore numbers greater than 10. This leaves us with x to be 1 or 4 or 9.

If x=1, then y = 9, but from the question stem x> y so x # 1, same way x cannot be equal to 4.

If x=9 then y=1, and x > y and x+y = 10

Hence the answer is C.
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 [#permalink] New post 05 May 2005, 02:41
Agree with above explanations. Answer is C. Combining 2 statements, we have 5x+30-3x=8k, where k is whole number. Continuing with the equation, we come to x+15=4k. Among possible values of X and Y satisfying 1st statement (6,4); (7,3); (8,2); (9;1) only the last one gives us x+15 that is multiple of 4.
  [#permalink] 05 May 2005, 02:41
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