Is xy ≤ 1/2?

[rkafc81]

hi all,

Tricky question...

Is xy ≤ 1/2?

(1) x^2+y^2=1

(2) x^2−y^2=0




Ans: A

(1) x^2+y^2=1. Recall that (x−y)^2≥0 (square of any number is more than or equal to zero). Expand: x^2−2xy+y^2≥0 and since x^2+y^2=1 then: 1−2xy≥0. So, xy≤12. Sufficient.

(2) insuff.


I'm stuck on statement 1 - there's no way I'd conjure up a solution like that on the exam, is there another way to do it algebraically? I'm ok with the testing cases approach.

thanks!

[jlucero]

Per forum rules:

2) Cite the source (author or company name) and question number (if applicable) for any problem you post.

[lyl9021]

Tricky question...

Is xy ≤ 1/2?

(1) x^2+y^2=1

(2) x^2−y^2=0


a quick method is exclusive method
suppose x=y=100 (that correspond for x^2−y^2=0)
then xy>1

[tim]

still waiting for a source before we can discuss this problem..

[dandarth1]

still waiting for a source before we can discuss this problem..

It's from a gmatclub test M27-12

[jnelson0612]

hi all,

Tricky question...

Is xy ≤ 1/2?

(1) x^2+y^2=1

(2) x^2−y^2=0

Ans: A

(1) x^2+y^2=1. Recall that (x−y)^2≥0 (square of any number is more than or equal to zero). Expand: x^2−2xy+y^2≥0 and since x^2+y^2=1 then: 1−2xy≥0. So, xy≤12. Sufficient.

(2) insuff.


I'm stuck on statement 1 - there's no way I'd conjure up a solution like that on the exam, is there another way to do it algebraically? I'm ok with the testing cases approach.

thanks!

Welcome! I have two thoughts:

1) If you see "x^2 + y^2" it would be wise to immediately consider that this may relate to (x+y)^2 or x^2 +2xy + y^2. Building in that pattern recognition to your study is a good idea.

2) One way to test numbers is to consider the range of possible numbers.

Alright, what is x=0 and y=1? That works, and xy=0. Same situation if x=1 and y=0. So in that case xy < 1/2.

Now, what happens if I instead start using non-zero numbers? Let's see, if one is negative and one is positive the result will be negative, so that's also less than 1/2.

What if I have two positives or two negatives? When I consider cases here, let's just consider positive cases. Whether I have two negatives or two positives, once I square the numbers I get positive results, and when I multiply them I get a positive result. So let's make this quick and just look at positives.

Think about numbers we can use . . . they have to be between 0 and 1 (or 0 and -1). Anything above 1 will be too large. Let's just experiment with some numbers and see what happens.

What if x=1/3? x^2 would be 1/9. So y^2 would be 8/9. y would be 2sqrt2/3. If I multiply together and estimate, I get 2.8/27, or a little greater than 1/10. Well below 1/2.

So look at what is happening. When I move into both positive or negative numbers, I get a small fraction. How can I maximize the size of this fraction? Probably by using numbers that are equal. If x^2 + y^2 = 1, then x^2 and y^2 should each be equal to sqrt1/2. If we multiply sqrt 1/2 * sqrt 1/2 we get 1/2. That appears to be the maximum of xy. If you have any doubt, try other fractions (you may need to estimate).

This is a way to test numbers. Should you do this on the test? Probably not, because this will be time consuming; it's much better to recognize the quadratic.

[vasili.kharchenko]

I like the quadratic approach, but I am confused with the following.

If we use (x-y)^2, we get that xy<=1/2, but if we use (x+y)^2, then we get xy>=1/2. Since the answer is "yes" and "no", should we discard answer A? Thank you for clarifying.

[RonPurewal]

I like the quadratic approach, but I am confused with the following.

If we use (x-y)^2, we get that xy<=1/2, but if we use (x+y)^2, then we get xy>=1/2. Since the answer is "yes" and "no", should we discard answer A? Thank you for clarifying.

Purple thing is wrong. Try it again -- you actually get xy > -1/2, which isn't terribly helpful here. (Both equations are true, so we actually know that -1/2 < xy < 1/2.)

[RonPurewal]

In any case, if you can think of the algebra solution that's posted earlier in this thread, you are a lot better at this stuff than I am. I'd say there's about a 1 in 10 chance that I'd come up with something like that.
o_O

Testing cases for the win. Be able to test cases, ladies and gentlemen.

[mondegreen]

hi all,

Tricky question...

Is xy ≤ 1/2?

(1) x^2+y^2=1

(2) x^2−y^2=0

Another method:

Arithmetic Mean>= Geometric Mean

Thus, x^2+y^2 >= 2|xy|

From the first statement, we know that x^2+y^2 = 1, thus,
|xy|<=1/2 --> -1/2<=xy<=1/2 Sufficient.

For the second statement, for x=y=0, we have xy=0 and <1/2. For x=y=100, we have xy>1/2. Insufficient.

[RonPurewal]

Another method:

Arithmetic Mean>= Geometric Mean

It appears that's a valid method, too.

I actually had to google "geometric mean" to even know what this meant... but, yes, it seems to work.