GMAT Data Sufficiency Discussions

IMO C

stmnt 1:
x + y =10 -- many possibilities. not suff

stmnt:
5x+3y/8 = integer
many such possibilities of x and y. In suff

Combinng both:
There is only one combination which satisfies above conditions: x=9 and y=1
hence both stmnts are sufficient when together.



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 think the ans should be E.
whats the OA?

During a 10-week summer vacation, was the average (arithmetic mean) number of books that Carolyn read per week greater than the average number of books that Jacob read per week?
(1) Twice the average number of books that Carolyn read per week was greater than 5 less than twice the average number of books that Jacob read per week.
(2) During the last 5 weeks of the vacation, Carolyn read a total of 3 books more than Jacob.

good soln.. I cud figure out the mistake I committed.. thanks man..

the OA is option C ;)

IMO C

stmnt 1:
x + y =10 -- many possibilities. not suff

stmnt:
5x+3y/8 = integer
many such possibilities of x and y. In suff

Combinng both:
There is only one combination which satisfies above conditions: x=9 and y=1
hence both stmnts are sufficient when together.
i think the ans should be E.
whats the OA?


nope.. the OA is not option E...
IMO C

stmnt 1:
x + y =10 -- many possibilities. not suff

stmnt:
5x+3y/8 = integer
many such possibilities of x and y. In suff

Combinng both:
There is only one combination which satisfies above conditions: x=9 and y=1
hence both stmnts are sufficient when together.

I made a blunder ....
nuttyvarun Says
nope.. the OA is not option E...


Sometimes OA are wrong buddy...

For this E should be the answer.

If OA is A..its simply wrong !!!

One more:
If x and y are positive, is x^3 > y
(1)square root of x >y
(2)x > y

Material A costs $3 per kilogram, and material B costs $5 per kilogram. If 10 kilograms of material K consists of x kilograms of material A and y kilograms of material B, is x > y?
(1) y > 4
(2) The cost of the 10 kilograms of material K is less than $40.
Material A costs $3 per kilogram, and material B costs $5 per kilogram. If 10 kilograms of material K consists of x kilograms of material A and y kilograms of material B, is x > y?
(1) y > 4
(2) The cost of the 10 kilograms of material K is less than $40.


My take is B
siddharthaduggirala Says
I made a blunder ....


if you made the same mistake of taking X + Y = 5, instead of taking x + y =10, then you are not alone..
rohitforever Says
My take is B

Thanks!
But why not 'D'.
By first statement, when y>4, anyway you know X is NOT greater then Y (even when x = 5 and y = 5) and thats all we need. So with first statement alone we should be able to answer ?
Material A costs $3 per kilogram, and material B costs $5 per kilogram. If 10 kilograms of material K consists of x kilograms of material A and y kilograms of material B, is x > y?
(1) y > 4
(2) The cost of the 10 kilograms of material K is less than $40.


My take on this one is B

From question we have x+y=10

Statement 1 >> Y>4 but it says nothing about x...insufficient

Statement 2 >> 3x+5y3x+5y will be equal to 40 when x=5 and y=5, so y has to be less than 5
And thus we have x>y..so sufficient
Thanks!
But why not 'D'.
By first statement, when y>4, anyway you know X is NOT greater then Y (even when x = 5 and y = 5) and thats all we need. So with first statement alone we should be able to answer ?


It is not mentioned in the question that x and y are whole numbers.
y > 4, meaning say, 4.5 then x is 5.5 so x > y
but y=5 will give x = 5 so x is NOT > y
and y = 6 or 7 will give x
No definitive ans, so NOT SUFF.
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.

Qusetion says x>y and we have total value=5x+3y

Statement 1--> x+y=10
we can have following pair of values for x and y
(x,y)=(9,1) or (8,2) or (7,3) or (6,4) as x>y
Not sufficient as we dont get a single pair for x and y

Statement 2--> (5x+3y)/8 is integer
Not sufficient

Combining 1 and 2
We can verify that (5x+3y)/8 is an integer only when x=9 and y=1...
So using both statements we get the answer
Hence my answer option C
Sometimes OA are wrong buddy...

For this E should be the answer.

If OA is A..its simply wrong !!!

Yes I agree with PJ....E should be the answer for this Summer Book question
It is not mentioned in the question that x and y are whole numbers.
y > 4, meaning say, 4.5 then x is 5.5 so x > y
but y=5 will give x = 5 so x is NOT > y
and y = 6 or 7 will give x
No definitive ans, so NOT SUFF.

Let me try my hand on this

Let us suppose that equal quantity of material is bought i.e 5 Kg each, considering this we get the total price of both the materials to be 40$.

Now from statement 2 we can see that the total price is less that 40$, it means 5$ material is less than 3 $ material as in no case we can get an answer
I hope I am clear.
Inputs are welcomed
Sometimes OA are wrong buddy...

For this E should be the answer.

If OA is A..its simply wrong !!!

if p, s, and t are positive prime numbers, what is the value of p^3s^3t^3?
(1) p^3st=728
(2) t=13

Statement 1 --> p^3st=728=(2^3)(13)(7)
we have p=2
s,t=13,7 or s,t=7,13
So its sufficient to find the answer

Statement 2 --> We get only the value of t so not sufficient

MY take on this one A.
nuttyvarun Says
if you made the same mistake of taking X + Y = 5, instead of taking x + y =10, then you are not alone..

absolutely true. I wanted to kick myself after realizing the mistake i made.
One more:
If x and y are positive, is x^3 > y
(1)square root of x >y
(2)x > y



IMO E.

What is the OA?
One more:
If x and y are positive, is x^3 > y
(1)square root of x >y
(2)x > y

E

1. not suff.. different ans for fractions and integers
2. not suff... different ans for fractions and integers

1&2

x>y and sqrt x>y ... no way to differentiate whether x is a integer or fraction as we don't know whether x greater or less than sqrt (x)