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D01-44

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D01-44  [#permalink]

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New post 15 Sep 2014, 23:13
3
5
00:00
A
B
C
D
E

Difficulty:

  55% (hard)

Question Stats:

58% (01:04) correct 42% (01:04) wrong based on 180 sessions

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\(x^2 + y^2 = 100\). 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. \(|x| - |y| = 5\)

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Re D01-44  [#permalink]

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New post 15 Sep 2014, 23:13
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1
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
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Re: D01-44  [#permalink]

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New post 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?
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Re: D01-44  [#permalink]

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New post 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:
\(x = \frac{5}{2} (1+\sqrt{7})\), \(y = \frac{5}{2} (\sqrt{7}-1)\)
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Re: D01-44  [#permalink]

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New post 18 Oct 2014, 03:11
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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.

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Re: D01-44  [#permalink]

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New post 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.

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Re: D01-44  [#permalink]

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New post 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. :P


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D01-44  [#permalink]

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New post 10 Nov 2014, 12:28
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Can this be solved visually??
1) as a circle on the xy plane
OR
2) as a tringle(pitagoras)
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Re: D01-44  [#permalink]

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New post 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.


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Re D01-44  [#permalink]

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New post 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?
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Re: D01-44  [#permalink]

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New post 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.
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Re: D01-44  [#permalink]

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New post 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.
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D01-44  [#permalink]

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New post 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.
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Re: D01-44  [#permalink]

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New post 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
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Re: D01-44  [#permalink]

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New post 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.
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Re: D01-44 &nbs [#permalink] 08 Nov 2018, 12:24
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