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Remember equal number of variables and independent equations ensures a solution.

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.

From the original condition, we have the below 2by2 table that are common in GMAT math test.

Attachment:

GC DS rite2deepti In a certain game played (20150918).png [ 3.21 KiB | Viewed 586 times ]
There are 2 variables (x,y) and in order to match the number of variables and equations, we need 2 equations. Since there is 1 each in 1) and 2), C has high probability of being the answer. Using both 1) & 2) together, x+y=5*2=10 and since que=(5x+3y)/8=[3(x+y)+2x]/8=(30+2x)/8=integer, the answer that satisfy the equation is x=9, y=1. Therefore the average (arithmetic mean ) point value of the 8 chips =48/8=6. The answer is unique, therefore the conditions are sufficient. The answer is C.

Normally for cases where we need 2 more equations, such as original conditions with 2 variable, or 3 variables and 1 equation, or 4 variables and 2 equations, we have 1 equation each in both 1) and 2). Therefore C has a high chance of being the answer, which is why we attempt to solve the question using 1) and 2) together. Here, there is 70% chance that C is the answer, while E has 25% chance. These two are the key questions. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer according to DS definition, we solve the question assuming C would be our answer hence using ) and 2) together. (It saves us time). Obviously there may be cases where the answer is A, B, D or E.

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