arithmetic problem

topspin360 wrote:Is xy≤1/2?

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

(2) x^2−y^2=0

Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
EACH statement ALONE is sufficient.
Statements (1) and (2) TOGETHER are NOT sufficient.

how do to solve this one?

thanks.

I did it a little unconventionally, albeit you may find the solution elegant.

Consider the expression:

(x-y)^2.

Now we know that
(x-y)^2>=0 (square of a number is non-negative)
=> x^2+y^2-2xy >=0
=> x^2+y^2 >=2xy
=> xy <= (x^2+y^2)/2

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

We know from our expression above that xy <= (x^2+y^2)/2. If x^2+y^2 = 1
then xy<= 1/2. Sufficient.

(2) x^2−y^2=0
=> x^2 = y^2.

If x =y=0 xy=0 which is less than 1/2.
If x=y=1, xy=1 which is greater than 1/2.
Insufficient.

Hence A is correct. This was a tricky one. I figured out the trick by looking at the first option.

Let me know if this helps

eagleeye wrote:

topspin360 wrote:Is xy≤1/2?

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

(2) x^2−y^2=0

Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
EACH statement ALONE is sufficient.
Statements (1) and (2) TOGETHER are NOT sufficient.

how do to solve this one?

thanks.

I did it a little unconventionally, albeit you may find the solution elegant.

Consider the expression:

(x-y)^2.

Now we know that
(x-y)^2>=0 (square of a positive number is non-negative)
=> x^2+y^2-2xy >=0
=> x^2+y^2 >=2xy
=> xy <= (x^2+y^2)/2

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

We know from our expression above that xy <= (x^2+y^2)/2. If x^2+y^2 = 1
then xy<= 1/2. Sufficient.

(2) x^2−y^2=0
=> x^2 = y^2.

If x =y=0 xy=0 which is less than 1/2.
If x=y=1, xy=1 which is greater than 1/2.
Insufficient.

Hence A is correct. This was a tricky one. I figured out the trick by looking at the first option.

Let me know if this helps :)

Hi eagleeye,

I agree with your solution and I arrived at the correct answer myself. But I had two questions:

(a) 'Square of a positive number is non-negative' - Isn't square of any number non-negative?
Eg: (-2)² = 4 (non-negative) and
(2)² = 4 (non negative)


(b) When x² = y²; isn't x = +/- y ?

Thanks for clarifying.

kartikshah wrote: Isn't square of any number non-negative?

Yes, square of any number is non-negative. That's what I intended to type, but I made a typo. Thanks for pointing it out.

kartikshah wrote: (b) When x² = y²; isn't x = +/- y ?
Thanks for clarifying.

Yes, x = +/- y. I didn't say otherwise in the question.

I just had to show one case to disprove it. I used the (0,0) and (1,1) case for (x,y).
You can use (1,-1) and (1,1) to disprove it, if you want. I just picked a couple of options, and you are right in thinking that x^2 = y^2 => x=+/- y.

I am sure you already knew these identities, but in case you have any doubts, refer to this article I wrote a while back.

https://www.beatthegmat.com/squares-squa ... tml#476551

Cheers!

topspin360 wrote:Is xy≤1/2?

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

(2) x^2−y^2=0

If x+y=c, where c is a constant, the greatest possible value of xy is when x=y.
To illustrate:
If x+y=10, the greatest possible value of xy is when x=y=5:
5*5 = 25.
Any other combination with a sum of 10 will yield a product less than 25.
4*6 = 24.
3*7 = 21
2*8 = 16
1*9 = 9
0*10 = 0.

Statement 1: x² + y² = 1.
Thus, the greatest possible value of x²y² is when x² = y² = 1/2:
(1/2)(1/2) = 1/4.
Since the greatest possible value of x²y² = 1/4, the greatest possible value of xy = 1/2.
SUFFICIENT.

Statement 2: x² - y² = 0.
If x=0 and y=0, then xy = 0, which is less than 1/2.
If x=1 and y=1, then xy = 1, which is not less than 1/2.
INSUFFICIENT.

The correct answer is A.

thanks all, great solutions. both ways are somewhat unconventional. the real question is... how do you come up with such unconventional ways with 2 mins of allotted time on the test. or is it that enough practice will allow me to know most of the unconventional ways i need to know to BTG.

topspin360 wrote:thanks all, great solutions. both ways are somewhat unconventional. the real question is... how do you come up with such unconventional ways with 2 mins of allotted time on the test. or is it that enough practice will allow me to know most of the unconventional ways i need to know to BTG.

If the answer doesn't jump out at you right away. You can and should do two things in DS questions.

1. Consider both statements together. You will get x=+/- sqrt(1/2) and y = +/- sqrt(1/2).
You can see that xy <=1/2 in all scenarios.

The correct answer is either A or C. Take a 50-50 guess and move on. I would guess A cause C seems too obvious.

This advice is applicable in all questions that you have no idea about (which once your preparation is complete will be very rare). Rather than waste a lot of time figuring one out, save time for others that you can do better on.

Also, as far as DS inequality questions go, the more you do, the better you will get at selecting numbers and getting "it" right away.

Cheers!

topspin360 wrote:thanks all, great solutions. both ways are somewhat unconventional. the real question is... how do you come up with such unconventional ways with 2 mins of allotted time on the test. or is it that enough practice will allow me to know most of the unconventional ways i need to know to BTG.

Here is a practical way to approach this question on test day.

Step 1: Determine what the problem is testing.
Here, we need to determine whether the MAXIMUM value of xy is 1/2.
Why 1/2?
This value likely has some significance here.

Step 2: Determine what sorts of numbers need to be tested.
x and y should be of the SAME SIGN so that their product is maximized.

Step 3: Evaluate the easier statement first: x²-y²=0.
Rephrased, x²=y².
Here, there is no limit to how far x and y can get from 0.
It's possible that x=y=0 or that x=y=100.
Thus, we cannot determine whether xy≤1/2.
Eliminate B and D.

Step 4: Try different combinations that satisfy the trickier statement (x²+y²=1).
Make x and y EQUAL:
If x²=1/2 and y²=1/2, then xy = (1/√2)(1√2) = 1/2.
Put some DISTANCE between them:
If x²=1/4 and y²=3/4, then xy = (1/2)(√3/2) = √3/4 ≈ 1.7/4.
Put even MORE distance between them:
If x²=1/16 and y²=15/16, then xy = (1/4)(√15/4) = √15/16 ≈ 1/4.
The value of xy keeps DECREASING: the GREATER THE DISTANCE between x and y, the SMALLER the value of xy.
By plugging in just a few combinations of values, we have discovered the reason that 1/2 is part of the question stem:
If x²+y²=1, then the maximum value of xy is 1/2.
SUFFICIENT.

The correct answer is A.

Thanks Mitch, that was very helpful. I guess it's really important to spend a few seconds just staring at the question stem and wondering why its asking what its asking before dwelling into answer choices.