Problem Solving

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=1, Y=9
X=5, Y=5

Both of the above satisfy both conditions. Hence, E

OA is C , can anyone help

radhika1306 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.

As X>y , values for (x,y) can be (6,4),(7,3),(8,2) or (9,1)

1st condition tells (x+y)/2 = 5
Or x+y = 10



2nd condition says (5x+3y)/8 is an integer

Combining both conditions

We can find the value of x and y and thus can find the average too
As (5*9+3*1)/8 is an integer

So x=9
And y=1

Answer is C

Not a good idea for me to mess things up a day before D-Day! I completely missed the X>Y part in the question. Son of a $@$%&

givemeanid, if today is d-day-1 for you, go see a movie or do something to take your mind of the gmat, relax the forum is under control