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• ### $450 Tuition Credit & Official CAT Packs FREE February 15, 2019 February 15, 2019 10:00 PM EST 11:00 PM PST EMPOWERgmat is giving away the complete Official GMAT Exam Pack collection worth$100 with the 3 Month Pack ($299) # D01-44  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics Author Message TAGS: ### Hide Tags Math Expert Joined: 02 Sep 2009 Posts: 52905 D01-44 [#permalink] ### Show Tags 15 Sep 2014, 23:13 3 10 00:00 Difficulty: 45% (medium) Question Stats: 59% (01:42) correct 41% (01:46) wrong based on 180 sessions ### HideShow timer Statistics $$x^2 + y^2 = 100$$. All of the following could be true EXCEPT A. $$|x| + |y| = 10$$ B. $$|x| \gt |y|$$ C. $$|x| \gt |y| + 10$$ D. $$|x| = |y|$$ E. $$|x| - |y| = 5$$ _________________ Math Expert Joined: 02 Sep 2009 Posts: 52905 Re D01-44 [#permalink] ### Show Tags 15 Sep 2014, 23:13 1 2 Official Solution: $$x^2 + y^2 = 100$$. All of the following could be true EXCEPT A. $$|x| + |y| = 10$$ B. $$|x| \gt |y|$$ C. $$|x| \gt |y| + 10$$ D. $$|x| = |y|$$ E. $$|x| - |y| = 5$$ A. $$|x| + |y| = 10$$ is possible if one is 0 and the other is 10. B. $$|x| \gt |y|$$ is possible if $$|x| \gt |5\sqrt{2}|$$ and $$|y| \lt |5\sqrt{2}|$$ C. $$|x| \gt |y| + 10$$ is never possible because if $$|x| \gt 10$$, $$x^2+y^2$$ becomes greater than 100, which is wrong. D. $$|x| = |y|$$ is possible if each is equal to $$|5\sqrt{2}|$$. E. $$|x| - |y| = 5$$ is possible if $$|x| = |9.11|$$ and $$|y| = |4.11|$$. Therefore all but C are possible. $$|x| \gt |y| + 10$$ means $$x$$ is greater than 10, which is not possible. Answer: C _________________ Manager Status: Do till 740 :) Joined: 13 Jun 2011 Posts: 86 Concentration: Strategy, General Management GMAT 1: 460 Q35 V20 GPA: 3.6 WE: Consulting (Computer Software) Re: D01-44 [#permalink] ### Show Tags 17 Oct 2014, 20:31 Hi Bunuel, But the option E how is it possible as for the give n values the value of x^2 + y^2 =100 will not hold true correct? Math Expert Joined: 02 Sep 2009 Posts: 52905 Re: D01-44 [#permalink] ### Show Tags 18 Oct 2014, 01:54 1 shankar245 wrote: Hi Bunuel, But the option E how is it possible as for the give n values the value of x^2 + y^2 =100 will not hold true correct? Those are approximate values. Exact values are: $$x = \frac{5}{2} (1+\sqrt{7})$$, $$y = \frac{5}{2} (\sqrt{7}-1)$$ _________________ Intern Joined: 12 May 2014 Posts: 14 Location: United States Concentration: Strategy, Operations Schools: IIMC'17 GMAT Date: 10-22-2014 GPA: 1.9 WE: Engineering (Energy and Utilities) Re: D01-44 [#permalink] ### Show Tags 18 Oct 2014, 03:11 9 According to the question the maximum value of either x^2 or y^2 can be 100, which implies that the greatest absolute value of either x or y can be 10 option C indicates that absolute value of x is greater than 10 which cannot be the case. kudos.. Manager Joined: 12 Sep 2010 Posts: 232 Concentration: Healthcare, General Management Re: D01-44 [#permalink] ### Show Tags 08 Nov 2014, 12:57 Bunuel wrote: shankar245 wrote: Hi Bunuel, But the option E how is it possible as for the give n values the value of x^2 + y^2 =100 will not hold true correct? Those are approximate values. Exact values are: $$x = \frac{5}{2} (1+\sqrt{7})$$, $$y = \frac{5}{2} (\sqrt{7}-1)$$ Bunuel, can you please explain how you derive the exact values for x and y? Thanks. 30,000 Kudos. Awesome milestone. Math Expert Joined: 02 Sep 2009 Posts: 52905 Re: D01-44 [#permalink] ### Show Tags 09 Nov 2014, 05:05 2 Samwong wrote: Bunuel wrote: shankar245 wrote: Hi Bunuel, But the option E how is it possible as for the give n values the value of x^2 + y^2 =100 will not hold true correct? Those are approximate values. Exact values are: $$x = \frac{5}{2} (1+\sqrt{7})$$, $$y = \frac{5}{2} (\sqrt{7}-1)$$ Bunuel, can you please explain how you derive the exact values for x and y? Thanks. 30,000 Kudos. Awesome milestone. It really does not matter. _________________ Intern Joined: 25 Mar 2014 Posts: 30 D01-44 [#permalink] ### Show Tags 10 Nov 2014, 12:28 3 Can this be solved visually?? 1) as a circle on the xy plane OR 2) as a tringle(pitagoras) Manager Joined: 21 Feb 2012 Posts: 56 Re: D01-44 [#permalink] ### Show Tags 08 Dec 2014, 08:34 2 plaverbach wrote: Can this be solved visually?? 1) as a circle on the xy plane OR 2) as a tringle(pitagoras) yes it can be solved by co-ordinate geometry. x^2 + y^2 = 100 is a circle with radius 10 A. |x| + |y|=10 are four lines which intersect the circle at (10,0),(0,10),(-10,0)&(0,-10) B. |x|>|y| there are multiple such points I have marked one such in the attached picture C. This is an area outside the circle. The closest to the circle is at y=0 even here x is outside the area of the circle. A suggestion that when we look at inequalities in co-ordinate geometry think in terms of area. D. |x|=|y| essentially two lines y=x and y=-x E. |x| - |y|=5 they are four lines again Attaching a very crude diagram. Apologies. >> !!! You do not have the required permissions to view the files attached to this post. _________________ Regards J Do consider a Kudos if you find the post useful Current Student Joined: 15 Apr 2015 Posts: 13 Location: India Schools: EDHEC (A$)
GMAT 1: 620 Q44 V31
GPA: 3.25

### Show Tags

10 Sep 2015, 06:44
I think this the explanation isn't clear enough, please elaborate. please elaborate why C is right

and also why E is wrong?

for C : x^2 +y^2 -2xy>100 and we don't know anything about 2xy so how can we conclude C is right?
Retired Moderator
Joined: 18 Sep 2014
Posts: 1111
Location: India

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29 Sep 2015, 08:58
mrigoel wrote:
I think this the explanation isn't clear enough, please elaborate. please elaborate why C is right
and also why E is wrong?
for C : x^2 +y^2 -2xy>100 and we don't know anything about 2xy so how can we conclude C is right?

Please note that this is an except question.
C cannot be right and that is why it is the answer.
E is wrong because it is possibly true for some exceptional values as indicated by Bunnel above.
Manager
Joined: 05 Jul 2015
Posts: 101
GMAT 1: 600 Q33 V40
GPA: 3.3

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19 Feb 2016, 11:11
I thought of triangles and decided that C isn't possible although I think mapping the circle probably would have been quicker and less prone to error.
Intern
Joined: 30 May 2013
Posts: 29
GMAT 1: 600 Q50 V21
GMAT 2: 640 Q49 V29

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04 Jul 2017, 09:46
1
I started with values of x & y as :
10 & 0, this rejects options (a) & (b)
5√2 & 5√2, this reject option (d)

For (c) & (e), if we take another look at the given statement, it can be consider an equation of right angle triangle.
x² + y² = 10²
And we know that sum of two sides is always greater than the third side.
Hence |x| > |y| + 10 can never be true.
Manager
Joined: 26 Feb 2018
Posts: 78
Location: United Arab Emirates
GMAT 1: 710 Q47 V41
GMAT 2: 770 Q49 V47

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10 Mar 2018, 13:37
A, B and D can be quickly eliminated.

For A, we can have x=10, y=0. This makes B true as well immediately
For B, we can have x=y= 2root5

This leaves C and E. C looks easier to prove or disprove so start there. To maximise X, minimise Y. Abs val. Y cannot be negative so the smallest possible value for this is 0. That means the largest possible value for x is 10. X cannot be more than y+10 as y+10 = 10.

As we know this cannot be true, we don't even need to look at E
Intern
Joined: 30 Jun 2018
Posts: 2
Location: India
Schools: ISB '20 (II)
GMAT 1: 720 Q49 V40

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08 Nov 2018, 12:24
this question can be thought of in a much easier manner for those who know a bit of coordinate geometry. the equation s actually the locus of a circle with origin as the centre and radius of 10 units.
Intern
Joined: 14 Dec 2018
Posts: 1

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18 Jan 2019, 09:13
Hi Brunel,

How are we suppose to verify answer E?

Cheers,
Intern
Joined: 11 Feb 2019
Posts: 1

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14 Feb 2019, 03:54
Using parametric notations, x = 10sin(t) and y=10cos(t) where t is a parameter. It is quite easy to point out C as the answer with this approach.
Re: D01-44   [#permalink] 14 Feb 2019, 03:54
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# D01-44

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