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Red and Blue chips

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by success1111 » Wed Apr 22, 2009 9:20 pm
In a certain game 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 postitive integers.If a player has 5 red chips and 3 blue chips,what is the average(arithmetric 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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Source: — Data Sufficiency |
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by kirvar » Wed Apr 22, 2009 10:08 pm
IMO C

Given X>Y

Stmt 1

x+y = 8 is true for the following cases (given X>Y)

x=5 y = 3
x=6 y = 2
x=7 y = 1
x=8 y = 0

Insufficient

Stmt 2

5x+3y/8 = integer

Insufficient as there are many values of x and y for which this is possible.

Stmt 1+2

Only x=8 y = 0 holds good for 5x+3y/8 to be an integer
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by success1111 » Wed Apr 22, 2009 10:13 pm
AO is C.
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by tj123 » Wed Apr 22, 2009 10:37 pm
how do you get stmt 1 to be x + y = 8

shoundnt stmt 1 be

(x + y) / 2 = 5
which equals

x+y = 10
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by rossmj » Thu Apr 23, 2009 6:57 am
ANS. is C but different from the original explanation

Stmt 1

(X+Y)/2=5-->X+Y=10
X=10 or 9 or 8 or 7 or 6
Y=0 or 1 or 2 or 3 or 4
INSF

Stmt 2

(5X+3Y)/8=INT-->5X+3Y must be divisible by 8

INSF

Combined we see only X=9, Y=1 yeilds a multiple of 8 when plugged into the second equation.
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