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

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Manager
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In a certain game played with red chips and blue chips, each [#permalink]

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28 Jan 2007, 08:44
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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

Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient
Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient
BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE
is sufficient.
EACH statement ALONE is sufficient
Statements (1) and (2) TOGETHER are NOT sufficient
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Senior Manager
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28 Jan 2007, 09:04
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that's a tricky one

quite easy to see that either alone is insufficient. let's see why both is sufficient.
from st1 we know that X+Y=10 and since X>Y>0 then 10>X>5

we need to know what is Z=5X+3Y

from st.1 we know that 3X+3Y is 30. so what is 2X?
2X is between 10 and 20 so Z is between 40 and 50

now st2 comes to the play. the average is an integer. since the average is Z/8 this means that Z must be a multiple of 8. there is only 1 such multiple between 40 and 50 (that's 48) which means that Z=48 and the required average is 6.

hence both statements are sufficient.

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28 Jan 2007, 23:33
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Agree with C
(1) The average point value of one red chip and one blue chip is 5.
X+Y=10
possible combinations X>Y
6+4
7+3
8+2
9+1 insuff
(2) The average point value of the 8 chips that the player has is an integer
alone is insuff
Combining both statements
X+Y=10 -->X=10-Y
{5(10-Y)+3Y}/8=integer
(50-2Y)/8=integer from 1st we have y=4,3,2,1
pick up numbers only Y=1 holds the equation
So C it is
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28 Jan 2007, 23:33
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