Question of the Day - 22nd August, 2009

[quant-master]

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 in 24hrs time.

Thanks,
Quant-Master

[pandeyvineet24]

Answer should be E.
FIND (5X + 3Y)/8

Stmt1
(X + Y) / 2 =5
INSUFF

Stmt 2
(5X + 3Y)/8 -- INTEGER
INSUFF

Combine both stmts .. not suff
X = 15, Y = 5 or
X = 11, Y = 9

[shanrizvi]

I think it should be C.

Lets summarize all the information first:

X = Point value of a RED chip
Y = Point value of a BLUE chip

X>Y

X and Y are integers.

The player has 5 RED and 3 BLUE chips. What is the average point value of the 8 chips?

Average point value: (5x+3y)/8

Statement 1: The average point value of one red chip and one blue chip is 5.

(X+Y)/2 = 5
X+Y = 10

As X>Y, (X,Y) can be (9,1), (8,2), (7,3) or (6,4). We cannot determine the average point value. Hence, statement 1 is sufficient.

Statement 2: The average point value of the 8 chips that the player has is an integer.

Lets put the values of (X,Y) into the average point value equation and pick the answer for which we get an integer. You only get an integer for (9,1). Hence, Statement 2 is sufficient.

Edit: The answer is C. (I made a typo earlier on and wrote B instead of C)

[ranell]

Answer should be C.
The average value is calculated as (5X+3Y)/8
1 - not sufficient, as we know only the average of X and Y and cannot count the average value of 8 elements given

2 - not sufficient.... the fact that the average number is integer is all that we have

combine these statements with X>Y and will reach the following variants -
X=9, Y =1 - the average number is an integer
X=8, Y =2 - the average number isn't an integer
X=7, Y =3 - the average number isn't an integer
X=6, Y =4 - the average number isn't an integer
X=5, Y =5 and so on is not possible, as X>Y

The answer is definitely C

[srivas]

Stmt1
(X + Y) / 2 =5 so X+Y = 10
INSUFF

Stmt 2
(5X + 3Y)/8 -- INTEGER
INSUFF

both combine x=9 & y=1 gives you an integer

[arbiter]

what is the answer ??

Is it A?

[pradeepsarathy]

IMO C

We are asked to find (5x+3y)/8

Stmt 1 -
(x+y)/2 = 5
x+y = 10
Plug the above stmt in question stem, we get
(2x+30)/8
Insufficient

Stmt 2 -
Insufficient

Stmt 1 and Stmt 2 -
For (2x+30)/8 to be an integer, X can take any of the following values - 1, 5, 9, 13 ...and so on, provided it satisfies the condition x+y =10(from stmt 1)
But given X>Y , this implies, X can take only one value, which is 9

Hence both stmts are necessary.

[mehravikas]

IMO - C

[oagostinho]

Hi,

IMO E, because, (1) and (2) it is not possible to get one specific value!

What do you think?

cheers!

[ershovici]

option 1 alone as option 2 alone is not suffitient
but if we combine both options:
- the summ of 2 integers devided by 2 is 5 - summ is 10
- from this option only 9*5 and 1*3 give an integer.

so the answer is C or 3

[deepak115]

C - Both Together are Sufficient !

from Stmt I:

(X+Y)/2 = 5

(5X+3Y)/8 = ?

X = 10-Y

(5(10-Y) + 3Y)/8 = ?

(50 - 2Y) / 8 = ?

(25-Y)/4 = ?

from Stmt II:

(25-Y)/4 is Integer, then 25-Y is divisible by 4

or Y = 1,5,9,13,17,21

Since X+Y = 10, Y is either of 1,5,9

But, Also X > Y, so Y = 1

then X = 9

Ave. can now be calculated !

[quant-master]

OA is C

Let me know if anybody needs further explanation.

Thanks,
Quant-Master

[oagostinho]

OA is C

Let me know if anybody needs further explanation.

Thanks,
Quant-Master

Hi,

Can you explain me, please!

Thanks!