If xy 0, is xy = 70?

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If xy 0, is xy = 70?

by sumanr84 » Sun Jun 27, 2010 2:14 am
If xy NOT= 0, is xy = 70?

(1) x > y

(2) x2 = y2

Statement (1) by itself is sufficient to answer the question, but statement (2) by itself is not.
Statement (2) by itself is sufficient to answer the question, but statement (1) by itself is not.
Statements (1) and (2) taken together are sufficient to answer the question, even though neither statement by itself is sufficient.
Either statement by itself is sufficient to answer the question.
Statements (1) and (2) taken together are not sufficient to answer the question, requiring more data pertaining to the problem.

[spoiler]source:kaplan, OA later[/spoiler]
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by selango » Sun Jun 27, 2010 3:08 am
xy not = 0

is xy=70

from stmt1,

x>y

there are many possible values for x and y

Insufficient.

from stmt2,

x^2=y^2

Rephrasing x=y

Here also there are many possible values for x and y

Combining also we cant prove Xy=70

Hence E

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by sumanr84 » Sun Jun 27, 2010 7:54 am
selango wrote:
Hence E
Give it a one more try..little tricky
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by kvcpk » Sun Jun 27, 2010 8:04 am
Answer should be C.

First thought was answer is E. But really tricky one..

x>y will not tell us anything.. So 1 is insufficient.

x2=y2 means (x+y)(x-y)=0
Therefore x+y=0 or x-y = 0
Suppose x-y = 0 which means x=y. we can get equal values for x and y such that xy=70. we get x=y=root(70) [YES]
No, suppose x+y=0, then x=-y. we cannot get 70 from one negative and one positve number. [NO]

So 2 is also not sufficient.

Combining these two, we get that x+y=0. because x cannot equal Y
Hence x=-y which means we cannot get the value of xy = 70(positive value) [NO]

Hence C is the answer.

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by selango » Sun Jun 27, 2010 8:13 am
I thought on this approach.But x and y can be of any value no?

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by kvcpk » Sun Jun 27, 2010 8:42 am
selango wrote:I thought on this approach.But x and y can be of any value no?
Hi Selango,

X and Y can be of any value.. But all that we need to concentrate on is whether xy=70.

Let me give an example:

let us say P,Q are some positive integers between 4 and 10, inclusive such that P+Q =10. Is PQ>20

We get 2 sets for (p,q)-> (5,5) and(6,4)

no pq can be 25 or 24. In both the cases it is greater than 20. So answer to the statement is YES.

Hope this helps!!

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by sumanr84 » Sun Jun 27, 2010 8:48 am
kvcpk is absolutely right..well done..:-)
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by kvcpk » Sun Jun 27, 2010 9:10 am
sumanr84 wrote:kvcpk is absolutely right..well done..:-)
Thank you ;)

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by riteshpatnaik » Wed Nov 23, 2016 7:29 am
kvcpk wrote:Answer should be C.

First thought was answer is E. But really tricky one..

x>y will not tell us anything.. So 1 is insufficient.

x2=y2 means (x+y)(x-y)=0
Therefore x+y=0 or x-y = 0
Suppose x-y = 0 which means x=y. we can get equal values for x and y such that xy=70. we get x=y=root(70) [YES]
No, suppose x+y=0, then x=-y. we cannot get 70 from one negative and one positve number. [NO]

So 2 is also not sufficient.

Combining these two, we get that x+y=0. because x cannot equal Y
Hence x=-y which means we cannot get the value of xy = 70(positive value) [NO]

Hence C is the answer.
Hi, combining st 1 & 2, how did you arrive at X+Y=0. ST1 X>Y

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by ceilidh.erickson » Mon Nov 28, 2016 9:52 am
riteshpatnaik wrote: Hi, combining st 1 & 2, how did you arrive at X+Y=0. ST1 X>Y
Here's another way to interpret statement (2):

If x^2 = y^2 (assuming that's what the original poster meant), then the two numbers must have the same absolute value. Squaring hides the sign, so we don't know if they're positive or negative, but they must have the same magnitude if they have equal squares. So, |x| = |y|.

If x > y, then we know that the two values are not equal: they can't both be positive or both be negative. The only possibility is that x is positive and y is negative. Since that's the case, the product xy must be negative, and cannot equal 70. Therefore, statements 1&2 are sufficient together.

The answer is C.

If you're still stuck on kvcpk's explanation, here's a breakdown:

x^2 = y^2
Move the y^2 over to the other side, then factor as a difference of squares:
x^2 - y^2 = 0
(x + y)(x - y) = 0

Set each one equal to 0:
x + y = 0 *or* x - y = 0
x = -y *or* x = y

(This, btw, is the same thing that |x| = |y| tells us).

Since statement (1) tells us that x > y, then x = y cannot be true. Therefore, it must be the case that x = -y, or in other words that x + y = 0.

Again, this brings us to the conclusion that x > 0 and y < 0, so we cannot have a positive product.

Does that help?
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by [email protected] » Mon Nov 28, 2016 10:01 am
Hi riteshpatnaik,

Ceilidh's explanation is spot-on, so I won't rehash any of that. Instead, I'm going to add a few examples to help prove the point:

Once we know that:
1) XY is NOT = to 0
2) X > Y
3) X^2 = Y^2

X and Y COULD be...
1 and -1, so XY = -1 and the answer to the question is NO
2 and -2, so XY = -4 and the answer to the question is NO
3 and -3, so XY = -9 and the answer to the question is NO
4 and -4, so XY = -16 and the answer to the question is NO
Etc.

Based on the pattern, XY will ALWAYS be negative, so the answer to the question is ALWAYS NO. Thus, the final answer is C

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by Jay@ManhattanReview » Mon Jan 16, 2017 8:50 pm
sumanr84 wrote:If xy NOT= 0, is xy = 70?

(1) x > y

(2) x2 = y2

Statement (1) by itself is sufficient to answer the question, but statement (2) by itself is not.
Statement (2) by itself is sufficient to answer the question, but statement (1) by itself is not.
Statements (1) and (2) taken together are sufficient to answer the question, even though neither statement by itself is sufficient.
Either statement by itself is sufficient to answer the question.
Statements (1) and (2) taken together are not sufficient to answer the question, requiring more data pertaining to the problem.

[spoiler]source:kaplan, OA later[/spoiler]
Interesting question:

We are given:

1. Neither x nor y is 0
2. From S1: x > y
3. From S2: x^2 = y^2 => |x| = |y| => x and y both can be positive, both can be negative, or anyone one of them can be negative and the other positive.

Combining S1 and S2:

x and y cannot be both positive or both negative since that would mean x = y, but as per S1, x > y, thus x must be positive and y must be negative.

Thus, xy = |x|*(-|y|) = -|x||y| < any positive number, such as 70. The asnwer is NO. A unique answer.

Hope this helps!

-Jay
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by Matt@VeritasPrep » Wed Jan 18, 2017 6:02 pm
xy ≠ 0 means that x ≠ 0 AND y ≠ 0. Now let's work from that.

S1:: x > y

Pretty useless, we could have x = 70, y = 1, or x = 100, y = 1, or many other pairs; NOT SUFFICIENT

S2:: x * x = y * y

This tells us that either x = y or x = -y. We could have x = √70 and y √70, or we could have x = 1, y = -1, or many other pairs; NOT SUFFICIENT

S1 + S2::

x > y, so x = y is out. That gives us x = -y, with x > 0.

From here, since x is positive and y is negative, it's impossible for x * y to be any positive value, including 70; SUFFICIENT