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

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25 Nov 2010, 00:58

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C

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40% (02:06) wrong based on 172 sessions

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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.

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.

I am absolutely clueless about this question ...how to proceed ..Also please explain in detail

(1) The average point value of one red chip and one blue chip is 5 --> \(\frac{x+y}{2}=5\) --> \(x=10-y\) --> \(\frac{5x+3y}{8}=\frac{50-5y+3y}{8}=\frac{50-2y}{8}=\frac{25-y}{4}=?\), still need the value of \(y\). Not sufficient.

(2) The average point value of the 8 chips that the player has is an integer --> \(\frac{5x+3y}{8}=integer\). Clearly insufficient.

(1)+(2) From (1) and (2) we have that \(\frac{25-y}{4}=integer\) and \(x=10-y\). Now, \(\frac{25-y}{4}=integer\) to be true \(y\) must be 1 (in this case \(x=10-y=9\)), 5 (in this case \(x=10-y=5\)), or 9 (in this case \(x=10-y=1\)). But as given that \(x>y\) then only value of \(y\) is valid, namely \(y=1\) --> \(\frac{25-y}{4}=6\). Sufficient.

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.

I am absolutely clueless about this question ...how to proceed ..Also please explain in detail

If algebra doesn't work for you, go on and use brute force. Simply try to arrange the information you have: Red - X points Blue - Y points X > Y, X and Y are +ve integer (You kind of expect it from a regular game so nothing to remember here) 5 Red ..... 3 Blue

Statement 1: Average point value of 1 Red and 1 Blue chip is 5, their total point value will be 5*2 = 10

So only 4 cases possible: 1 Red (9).... 1 Blue (1) 1 Red (8).... 1 Blue (2) 1 Red (7).... 1 Blue (3) 1 Red (6).... 1 Blue (4) Average of 5 reds and 3 blues will be different in each case so not sufficient.

Statement 2: Average of 8 chips is integer. Insufficient alone.

Both statements together: Case 1: 1 Red (9).... 1 Blue (1): 5 Red + 3 Blue = 48 - Divisible by 8 Case 2: 1 Red (8).... 1 Blue (2): 5 Red + 3 Blue = 46 - Not div by 8 Case 3: 1 Red (7).... 1 Blue (3): 5 Red + 3 Blue = 44 - Not div by 8 See the pattern, next one will be 42 - Not div by 8 Since in only one case, we get the average of 8 chips an integer, both together are sufficient. Answer (C).
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Re: In a certain game played with red chips and blue chips, each [#permalink]

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24 Mar 2014, 03:19

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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

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26 Mar 2014, 07:46

OK so just to clarify this question and move on..

Now first statement only tells us that x+y=10. Clearly insufficient. Statement 2 tells us that 5x+3y has to be a multiple of 8, which again can happen in many ways, say x=4 and y=3 or x=9 and y=1. Therefore insufficient. From both statement together we have that x+y=10 therefore only the second option x-9 and y=1 is possible. C stands as the correct answer choice

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

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17 Sep 2015, 23:45

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. 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 992 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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