It is currently 19 Oct 2017, 22:55

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

M27-12

Author Message
TAGS:

Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 41884

Kudos [?]: 128932 [0], given: 12183

Show Tags

16 Sep 2014, 01:27
Expert's post
6
This post was
BOOKMARKED
00:00

Difficulty:

65% (hard)

Question Stats:

47% (01:29) correct 53% (01:13) wrong based on 55 sessions

HideShow timer Statistics

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

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

(2) $$x^2-y^2=0$$
[Reveal] Spoiler: OA

_________________

Kudos [?]: 128932 [0], given: 12183

Math Expert
Joined: 02 Sep 2009
Posts: 41884

Kudos [?]: 128932 [2], given: 12183

Show Tags

16 Sep 2014, 01:27
2
KUDOS
Expert's post
2
This post was
BOOKMARKED
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.

_________________

Kudos [?]: 128932 [2], given: 12183

Intern
Joined: 29 Jul 2012
Posts: 1

Kudos [?]: [0], given: 2

Show Tags

23 Oct 2014, 09:13
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 ???

Kudos [?]: [0], given: 2

Math Expert
Joined: 02 Sep 2009
Posts: 41884

Kudos [?]: 128932 [0], given: 12183

Show Tags

23 Oct 2014, 09:23
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.
_________________

Kudos [?]: 128932 [0], given: 12183

Intern
Joined: 24 Feb 2015
Posts: 13

Kudos [?]: [0], given: 21

GMAT 1: 570 Q37 V31

Show Tags

01 May 2015, 10:46
Why did you start from : (x-y)^2>=0 ??

Kudos [?]: [0], given: 21

Intern
Joined: 24 Feb 2015
Posts: 13

Kudos [?]: [0], given: 21

GMAT 1: 570 Q37 V31

Show Tags

01 May 2015, 10:48
Why did you start from : (x-y)^2>=0 ??

Kudos [?]: [0], given: 21

Intern
Joined: 09 Feb 2015
Posts: 7

Kudos [?]: [0], given: 31

Show Tags

08 Jun 2015, 09:29
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.

Kudos [?]: [0], given: 31

Intern
Joined: 09 May 2014
Posts: 4

Kudos [?]: 1 [0], given: 81

Show Tags

20 Jun 2015, 06:40
yeah even i have a same doubt why not (x+y)^2 as harshalnamdeo88...

Kudos [?]: 1 [0], given: 81

Math Expert
Joined: 02 Sep 2009
Posts: 41884

Kudos [?]: 128932 [0], given: 12183

Show Tags

20 Jun 2015, 12:29
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.
_________________

Kudos [?]: 128932 [0], given: 12183

Intern
Joined: 24 Jun 2015
Posts: 46

Kudos [?]: 2 [0], given: 22

Show Tags

07 Jul 2015, 09:02
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

Kudos [?]: 2 [0], given: 22

Math Expert
Joined: 02 Sep 2009
Posts: 41884

Kudos [?]: 128932 [1], given: 12183

Show Tags

07 Jul 2015, 09:08
1
KUDOS
Expert's post
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.
_________________

Kudos [?]: 128932 [1], given: 12183

Intern
Joined: 24 Jun 2015
Posts: 46

Kudos [?]: 2 [0], given: 22

Show Tags

07 Jul 2015, 09:40
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

Kudos [?]: 2 [0], given: 22

Intern
Joined: 17 Jan 2016
Posts: 22

Kudos [?]: [0], given: 12

Show Tags

12 Mar 2016, 22:16
Could you kindly explain the expansion on the first statement? Txx

Kudos [?]: [0], given: 12

Math Expert
Joined: 02 Sep 2009
Posts: 41884

Kudos [?]: 128932 [0], given: 12183

Show Tags

13 Mar 2016, 08:23
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.
_________________

Kudos [?]: 128932 [0], given: 12183

Manager
Joined: 25 Sep 2015
Posts: 145

Kudos [?]: 11 [0], given: 74

Location: United States
GMAT 1: 700 Q48 V37
GPA: 3.26

Show Tags

06 Jul 2016, 16:34
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

Kudos [?]: 11 [0], given: 74

Intern
Joined: 11 Dec 2016
Posts: 2

Kudos [?]: [0], given: 0

Show Tags

05 Feb 2017, 13:05
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

Kudos [?]: [0], given: 0

Intern
Joined: 04 Dec 2016
Posts: 1

Kudos [?]: [0], given: 0

Show Tags

16 Feb 2017, 04:07
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.

Kudos [?]: [0], given: 0

Intern
Joined: 06 Apr 2017
Posts: 27

Kudos [?]: 7 [0], given: 38

Show Tags

23 Aug 2017, 16:20
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 $$\sqrt{\frac{1}{2}}*\sqrt{\frac{1}{2}}=\frac{1}{2}$$

Kudos [?]: 7 [0], given: 38

Re: M27-12   [#permalink] 23 Aug 2017, 16:20
Display posts from previous: Sort by

M27-12

Moderators: Bunuel, Vyshak

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.