Question of the day - 22nd August, 2009
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.
OA will be posted soon.
Thanks,
Quant-Master
IMO E. Not sure though.
ReplyDeleteQn: (5r+3b)/8
(1) (2r+3*5) / 8. Not sufficient
(2) Not Sufficient alone
Together, the eqn in (1) can never be even.
So effectively I either got the answer or screwed it up royally :P
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ReplyDeleteAns is C
ReplyDelete1)(X+Y)/2=5 => X+Y=10 --> Not alone sufficient
2)(5X + 3Y)/8 =?? (Integer) --> Not alone sufficient
3) Both together
Subsituting X=10-Y in second equation:
=> (5(10-Y)+3Y)/8 = Z, where Z is an Integer
=> 50-2Y should be multiple of 8
If Y=1 => 50-2Y=48 (multiple of 8, so one option)
Y=2 => 50-2Y=46 (not a multiple of 8, can't be the answer)
Y=3 => 50-2Y=44 (not a multiple of 8, can't be the answer)
Y=4 => 50-2Y=42 (not a multiple of 8, can't be the answer)
Y cannot be greater than 4 as X>Y, hence X=9 and Y=1
=> arithmetic mean
= ((5*9 + 3*1))8 = 8
Amit - C is the answer
ReplyDeleteThanks a lot for your questions and concepts...I gave my GMAT on Saturday and got a 730 :)
ReplyDelete@ Che'
ReplyDeletewell done! nice to hear that. Let me know if you need guidance for further steps
Thanks,
Quant-Master
Explanation from Neeraj looks fine.
ReplyDeleteLet me know if somebody needs further explanation.
Thanks,
Quant-Master
Hi,
ReplyDeleteCan you please point out the mistake in my analysis?
--
Hungry
@ Hungry
ReplyDeleteI don't understand the first equation formed by you. I guess you went wrong in the first equation. Also in the second eqn it would be better if you can list out various possible answers.
Thanks,
Quant-Master