kaplan cat 6

[orel]

oa will be posted soon
thanks

[agoyal2]

IMO B.

1) X > Y

Taking x=10, y=7, xy=70
Taking x=2, y=1, xy=2

NOT SUFFICIENT

2) x^2 = y^2
=> x=y OR x=-y

So, xy = x^2 or -x^2
Since 70 is not a perfect square, answer is a definit NO

SUFFICIENT

[jimmiejaz]

IMO ans is E.

Nothing is given about x or y
From 1 we have x>y
take any value.
x=10,y=8 xy=80 (no)
x=10, y=7 xy=70 (yes)
insuff. eliminate A,D

From 2 we have x^2=y^2
or x= +y
we still can have any value. x can be sqrt(70) too.
in that case xy will indeed be 70. in other cases it will not be.
remember, we cant take x=-y because the eqn will not hold true as a square cant be negative.

[srisl11]

IMO B

from st 2 , we get x= |y| or y = |x|

For any value of x (integer or fraction), xy = x|x| cannot be 70 since 70 is not a perfect square

Statement 1 : Here we need not know whether x>y or not (I know we shouldn't use " whether or not" but it just makes explanation easy
)

OA please

[GMATters1001]

I'm going with C. The stem does not specify that x is an integer or fraction. (1) doesn't give us enough info. they could multiply to 70 or they could not (x=10, y=7 perhaps, perhaps not). (2) tells us that x and y would have to be the square root of 70, which is possible give than there are no constraints on x and y in the stem requiring that they are integers or fractions, but does not alone prove one way or the other. A, D and B are now eliminated, leaving C and E.

Taken together, we know that absolute value x must equal absolute value y AND that x is greater than y. So x=-y and vice versa since their squares equal eachother. So x and y have opposite signs and xy cannot be 70. SUFFICIENT.

Therefore C.

[vittalgmat]

OA please
and Official explanation if available

[cramya]

Stmt I

x>y

x=40 y=2
xy <> 70
TRUE

x=35 y = 2
xy=70



INSUFF

Stmt II

x^2 = y^2

x=+y or x=-y

x=2 y=2
xy <> 70

x=sqrt(70) y = sqrt(70)

xy = 70

INSUFF

Stmt I and II together

x>y

x^2 = y^2 (X=+Y OR X=-Y (or) Y=+X OR Y=-X)

Since x>y x has to be positive and y negative

xy cannot be 70

C) Agree with Gmaters.




Feruza,
Friendly request: If possible please post the OA using spoiler with the question.

Regards,
CR

[cramya]

Statement 2

x = -y

Sris,
x can be +y or -y. We dont know.
Hence insuff

[srisl11]

Statement 2

x = -y

Sris,
x can be +y or -y. We dont know.
Hence insuff

Yeah , I typed the entire post and then realized it ..
It can be +ve too
IMO C

[pbanavara]

Love this question .. looks deceptively simple

[orel]

Hello, folks!

sorry for being late
here is the official explanation:

[cramya]

Thanks! A very nice question. Like pbanavra said it looks deceptively simple but yet somewhat complicated.

[abhishekbansal]

Hi

Can somebody explain me the explanation given ---

Since X > Y, therefore X = -Y

According to me, if we take the converse--

1.) X = Y or -Y
2.) X > Y

Taking X = -Y and substituting in 2.) we get, -Y > Y which cannot be true.

The same stands for X = Y and getting Y > Y which is again absurd

How did we arrive at the answer C ???? :roll:

[Stuart@KaplanGMAT]

Hi

Can somebody explain me the explanation given ---

Since X > Y, therefore X = -Y

According to me, if we take the converse--

1.) X = Y or -Y
2.) X > Y

Taking X = -Y and substituting in 2.) we get, -Y > Y which cannot be true.

The same stands for X = Y and getting Y > Y which is again absurd

How did we arrive at the answer C ???? :roll:

-Y > Y is always true when Y is negative.

e.g. if y = -2, then it's true that -(-2) > -2.

You're 100% correct that Y > Y makes no sense, and that's exactly why the statements together are sufficient. Together, we know that x must be positive and y must be negative, and:

(+) * (-) = (-)

which means it's impossible for xy to equal 70. In other words, we get a definite "no" answer to the question, which means that (C) is the correct answer.

[jimmiejaz]

Hi Stuart,

thanks for the explanation. But i still have a doubt.
From 2nd statement we have x= +y
yes, both x and y can be sqrt(70) too.
But, if we take x=-y don't we get -y^2 which cant be posiible as its a square.
I am confused in this.