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Math Expert V
Joined: 02 Sep 2009
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Difficulty:   75% (hard)

Question Stats: 51% (01:18) correct 49% (01:23) wrong based on 262 sessions

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Is $$xy \le \frac{1}{2}$$?

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

(2) $$x^2-y^2=0$$

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Math Expert V
Joined: 02 Sep 2009
Posts: 58257

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Official Solution:

(1) $$x^2+y^2=1$$. Recall that $$(x-y)^2 \ge 0$$ (square of any number is more than or equal to zero). Expand: $$x^2-2xy+y^2 \ge 0$$ and since $$x^2+y^2=1$$ then: $$1-2xy \ge 0$$. So, $$xy \le \frac{1}{2}$$. Sufficient.

(2) $$x^2-y^2=0$$. Re-arrange and take the square root from both sides: $$|x|=|y|$$. Clearly insufficient.

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The same applicable to (x+y)^2 ==> x^2 + y^2 + 2 x*y is greter than or equal to ZERO.
1 + 2 x*y is greter than or equal to ZERO.
xy greter than or equal -1/2...This can prove A is insufficient, right ???
Math Expert V
Joined: 02 Sep 2009
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eureka750 wrote:
The same applicable to (x+y)^2 ==> x^2 + y^2 + 2 x*y is greter than or equal to ZERO.
1 + 2 x*y is greter than or equal to ZERO.
xy greter than or equal -1/2...This can prove A is insufficient, right ???

You need to consider $$(x-y)^2\geq{0}$$ to get sufficiency.
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Why did you start from : (x-y)^2>=0 ??
Intern  Joined: 24 Feb 2015
Posts: 13
GMAT 1: 570 Q37 V31 Show Tags

Why did you start from : (x-y)^2>=0 ??
Intern  Joined: 09 Feb 2015
Posts: 6

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Bunuel wrote:
eureka750 wrote:
The same applicable to (x+y)^2 ==> x^2 + y^2 + 2 x*y is greter than or equal to ZERO.
1 + 2 x*y is greter than or equal to ZERO.
xy greter than or equal -1/2...This can prove A is insufficient, right ???

You need to consider $$(x-y)^2\geq{0}$$ to get sufficiency.

Please explain why do we need to consider only (x-y)^2 to get sufficiency, as also in (x+y)^2 we use x^2 + y^ 2=1 premise, hence its not in all cases that(A) results in single conclusion or sufficiency.
Intern  Joined: 09 May 2014
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yeah even i have a same doubt why not (x+y)^2 as harshalnamdeo88...
Math Expert V
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BCCLSN wrote:
Why did you start from : (x-y)^2>=0 ??

harshalnamdeo88 wrote:
Bunuel wrote:
eureka750 wrote:
The same applicable to (x+y)^2 ==> x^2 + y^2 + 2 x*y is greter than or equal to ZERO.
1 + 2 x*y is greter than or equal to ZERO.
xy greter than or equal -1/2...This can prove A is insufficient, right ???

You need to consider $$(x-y)^2\geq{0}$$ to get sufficiency.

Please explain why do we need to consider only (x-y)^2 to get sufficiency, as also in (x+y)^2 we use x^2 + y^ 2=1 premise, hence its not in all cases that(A) results in single conclusion or sufficiency.

shrutikasat wrote:
yeah even i have a same doubt why not (x+y)^2 as harshalnamdeo88...

Guys, the point is that if you consider $$(x+y)^2 \ge 0$$, you'll get $$xy\geq{-\frac{1}{2}}$$, which is not helpful at all. Actually since both $$xy\geq{-\frac{1}{2}}$$ (from $$(x+y)^2 \ge 0$$) and $$xy \le \frac{1}{2}$$ (from $$(x-y)^2 \ge 0$$) are true, then we get that $$-\frac{1}{2} \leq{xy} \leq{\frac{1}{2}}$$. But only the approach given in the solution above gives you the answer we are looking for, while another one gives you an inequality which is not helpful.

Hope it's clear.
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Bunuel wrote:
Official Solution:

(1) $$x^2+y^2=1$$. Recall that $$(x-y)^2 \ge 0$$ (square of any number is more than or equal to zero). Expand: $$x^2-2xy+y^2 \ge 0$$ and since $$x^2+y^2=1$$ then: $$1-2xy \ge 0$$. So, $$xy \le \frac{1}{2}$$. Sufficient.

(2) $$x^2-y^2=0$$. Re-arrange and take the square root from both sides: $$|x|=|y|$$. Clearly insufficient.

Hi Bunuel,

Could you help me if I am wrong?

What I did was first to convert question stem to 2xy - 1 < 0, then from (1) I concluded that in order to satisfy x^2+y^2=1, The only x and y possible were; x=1, y=0, x=-1, y=0, x=0, y=1, x=0, y=-1; so question stem 2xy - 1 < 0 is always negative... Is my thinking process wrong or is correct?

Thanks a lot.

Regards.

Luis Navarro
Looking for 700
Math Expert V
Joined: 02 Sep 2009
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luisnavarro wrote:
Bunuel wrote:
Official Solution:

(1) $$x^2+y^2=1$$. Recall that $$(x-y)^2 \ge 0$$ (square of any number is more than or equal to zero). Expand: $$x^2-2xy+y^2 \ge 0$$ and since $$x^2+y^2=1$$ then: $$1-2xy \ge 0$$. So, $$xy \le \frac{1}{2}$$. Sufficient.

(2) $$x^2-y^2=0$$. Re-arrange and take the square root from both sides: $$|x|=|y|$$. Clearly insufficient.

Hi Bunuel,

Could you help me if I am wrong?

What I did was first to convert question stem to 2xy - 1 < 0, then from (1) I concluded that in order to satisfy x^2+y^2=1, The only x and y possible were; x=1, y=0, x=-1, y=0, x=0, y=1, x=0, y=-1; so question stem 2xy - 1 < 0 is always negative... Is my thinking process wrong or is correct?

Thanks a lot.

Regards.

Luis Navarro
Looking for 700

Your thought process is not precise. We are NOT told that x and y are integers, thus x^2 + y^2 = 1 will have infinitely many solutions, not just the ones you mentioned.
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Bunuel wrote:
luisnavarro wrote:
Bunuel wrote:
Official Solution:

(1) $$x^2+y^2=1$$. Recall that $$(x-y)^2 \ge 0$$ (square of any number is more than or equal to zero). Expand: $$x^2-2xy+y^2 \ge 0$$ and since $$x^2+y^2=1$$ then: $$1-2xy \ge 0$$. So, $$xy \le \frac{1}{2}$$. Sufficient.

(2) $$x^2-y^2=0$$. Re-arrange and take the square root from both sides: $$|x|=|y|$$. Clearly insufficient.

Hi Bunuel,

Could you help me if I am wrong?

What I did was first to convert question stem to 2xy - 1 < 0, then from (1) I concluded that in order to satisfy x^2+y^2=1, The only x and y possible were; x=1, y=0, x=-1, y=0, x=0, y=1, x=0, y=-1; so question stem 2xy - 1 < 0 is always negative... Is my thinking process wrong or is correct?

Thanks a lot.

Regards.

Luis Navarro
Looking for 700

Your thought process is not precise. We are NOT told that x and y are integers, thus x^2 + y^2 = 1 will have infinitely many solutions, not just the ones you mentioned.

Thanks a lot Bunuel.

Regards

Luis Navarro
Looking for 700
Intern  Joined: 17 Jan 2016
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Could you kindly explain the expansion on the first statement? Txx
Math Expert V
Joined: 02 Sep 2009
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wmichaelxie wrote:
Could you kindly explain the expansion on the first statement? Txx

Please go through the discussion above and ask a specific question if anything remains unclear.
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I used geometrical approach to solve this problem. Did I get lucky?

(X)^2 +(Y)^2=1

That would indicate a right angle triangle with x and y as height and base. Maximize the area of triangle would maximise the x*y product.
Thus X=Y=1/(root2) (not sure if this step is right.... I don't have a reason for this)

Thus the product will always be less than or equals 1/2
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Hi Bunuel,

I don't understand why do we need this: "Recall that (x−y)2≥0(x−y)2≥0 (square of any number is more than or equal to zero)" to solve the equation. What's the relationship with x2+y2?

Thanks
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For statement 1 I used a different approach: I considered X^2+Y^2=1 a circle with centre in (0,0) and radius 1. The product of xy is represented by a parallelogram whose area is maximized in the first quadrant when the parallelogram is a square. Hence xy reaches the max when x=y = 1/square root of 2. So the product can be max 1/2.
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Joined: 06 Apr 2017
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Schools: Haas EWMBA '21
GMAT 1: 730 Q48 V44 GMAT 2: 730 Q49 V40 GPA: 3.98
WE: Corporate Finance (Health Care)

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This is a hard question, in particular because knowing to use the square of a difference is a tricky detail.

You can also use the fact that the maximum product of $$ab$$ given that $$a+b=c$$ is the set of numbers where $$a$$ and $$b$$ are nearest each other and still sum to $$c$$.

For example, if $$a+b=20$$, the maximum product of $$ab=10*10=100$$

In this case the maximum product of $$ab$$ where $$a+b=1$$ is when $$a=b=\frac{1}{2}$$

In this case $$x^2=a$$ and $$y^2=b$$, so the the maximum product of $$xy$$ is $$\sqrt{\frac{1}{2}}*\sqrt{\frac{1}{2}}=\frac{1}{2}$$

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Bunuel wrote:
Is $$xy \le \frac{1}{2}$$?

(1) $$x^2+y^2=1$$
(2) $$x^2-y^2=0$$

Bunuel Nice Question

Though your method is the correct and best method but I couldnt think of that method in the first place.

I read the first statement...
by this it could mean that
a. X could be 0 and Y could be 1; Or X could be 1 and Y could be 0; In both cases, question stem is TRUE
b. X = Y = plus/minus 1/2; then also question stem is TRUE

so Answer is A or D

By B;
either x = y= 0; Question Stem is TRUE
or
x=y= 2; Question stem is False.

So A
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spence11 wrote:
This is a hard question, in particular because knowing to use the square of a difference is a tricky detail.

You can also use the fact that the maximum product of $$ab$$ given that $$a+b=c$$ is the set of numbers where $$a$$ and $$b$$ are nearest each other and still sum to $$c$$.

For example, if $$a+b=20$$, the maximum product of $$ab=10*10=100$$

In this case the maximum product of $$ab$$ where $$a+b=1$$ is when $$a=b=\frac{1}{2}$$

In this case $$x^2=a$$ and $$y^2=b$$, so the the maximum product of $$xy$$ is $$\sqrt{\frac{1}{2}}*\sqrt{\frac{1}{2}}=\frac{1}{2}$$

This solution, in particular, makes a lot of sense.
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