It is currently 23 Nov 2017, 13:39

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

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

Author Message
Manager
Joined: 23 Mar 2008
Posts: 216

Kudos [?]: 203 [0], given: 0

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

### Show Tags

04 Jun 2008, 13:22
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 4 sessions

### HideShow timer Statistics

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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

Kudos [?]: 203 [0], given: 0

Senior Manager
Joined: 19 Apr 2008
Posts: 313

Kudos [?]: 99 [0], given: 0

### Show Tags

04 Jun 2008, 14:08
C .

1)( r+b )/2 =5

r+b =10

not sufficient because different combinations possible

(r, b) = ( 9,1) ,(8,2) (7,3) , (6,4)

2) not sufficient doesnt give us enough information

combining 1 and 2 , only (5*9+3*1)/8 gives us integer value

Kudos [?]: 99 [0], given: 0

Current Student
Joined: 11 May 2008
Posts: 554

Kudos [?]: 226 [0], given: 0

### Show Tags

04 Jun 2008, 18:57
hmm. why not E?
from A, it says-(r+b)/2=5
hence r+b=10.
from B we therefore have- [b+b+(r+b)+(r+b)+(r+b)]/8= some integer.
both A and B,
2b+30/8 = some integer.
hence when b= 1, avg= 4
b=9, avg = 6.
.
so i think E.

whats the 0A?

Kudos [?]: 226 [0], given: 0

Senior Manager
Joined: 19 Apr 2008
Posts: 313

Kudos [?]: 99 [0], given: 0

### Show Tags

04 Jun 2008, 20:05
arjtryarjtry wrote:
hmm. why not E?
from A, it says-(r+b)/2=5
hence r+b=10.
from B we therefore have- [b+b+(r+b)+(r+b)+(r+b)]/8= some integer.
both A and B,
2b+30/8 = some integer.
hence when b= 1, avg= 4
b=9, avg = 6.
.
so i think E.

whats the 0A?

It is mentioned that X> Y so b can't be 9

Kudos [?]: 99 [0], given: 0

CEO
Joined: 29 Mar 2007
Posts: 2553

Kudos [?]: 528 [0], given: 0

### Show Tags

05 Jun 2008, 07:04
puma 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

(x+y)/2=5 --> x+y=10

1: But we cannot deduce what x and y are . Insuff.

2: (5x+3y)/8= integer... Again could be anything.

Together:

Try numbers: note that X>Y. So we can try 6,4. 7,3, 8,2 and 9, 1.

6*5+3*4=42 not divisible by 8.

7*5+3*3 = 44 not divisible by 8.

40+6 not divisible by 8.

9*5+3*1 =48. Here we go.

C

Kudos [?]: 528 [0], given: 0

Senior Manager
Joined: 19 Apr 2008
Posts: 313

Kudos [?]: 99 [0], given: 0

### Show Tags

05 Jun 2008, 13:24
puma wrote:
OA is D

What's the source ? Chances OA is wrong ?

Kudos [?]: 99 [0], given: 0

Director
Joined: 05 Jan 2008
Posts: 684

Kudos [?]: 621 [0], given: 0

### Show Tags

05 Jun 2008, 20:26
cannot be D

if you take stmt-2

you will get multiple solutions with the boundary condition given in stmt-1

5x+3Y/8 = should be an integer

given X>Y thus (X,Y) could be (9,1) or (12,4) I am sure we can find many more..

so it should be C
_________________

Persistence+Patience+Persistence+Patience=G...O...A...L

Kudos [?]: 621 [0], given: 0

VP
Joined: 18 May 2008
Posts: 1258

Kudos [?]: 542 [0], given: 0

### Show Tags

05 Jun 2008, 22:49
I also believe that OA is wrong. it has to be C. i totally agree with the explanaion given by rpmodi

Kudos [?]: 542 [0], given: 0

Re: DS: chips   [#permalink] 05 Jun 2008, 22:49
Display posts from previous: Sort by