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

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In a certain game played with red chips and blue chips, each [#permalink] New post 25 Nov 2010, 00:58
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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.
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Re: Question [#permalink] New post 25 Nov 2010, 01:11
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rite2deepti 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.

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


Given: x=integer>y=integer>0. Question: \frac{5x+3y}{8}=?

(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.

Answer: C.
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Re: Question [#permalink] New post 25 Nov 2010, 01:12
should be C

R's value = 6 and B's value = 4 satisfies; R's value = 11 also satisfies but R+B = 10 as per (1) so R=11 can not be taken
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Re: Question [#permalink] New post 25 Nov 2010, 21:42
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rite2deepti 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.

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] New post 24 Mar 2014, 03:19
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Re: In a certain game played with red chips and blue chips, each [#permalink] New post 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

Hope this is clear
Cheers
J
Re: In a certain game played with red chips and blue chips, each   [#permalink] 26 Mar 2014, 07:46
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