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Bunuel
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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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Can this be solved visually??
1) as a circle on the xy plane
OR
2) as a tringle(pitagoras)
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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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Bunuel
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



Xylan can you help?? What if i don't want to do number plugging??
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Bunuel
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



Xylan can you help?? What if i don't want to do number plugging??

JIAA It's completely okay if you do NOT want to do number plugging.
However, aspire to reach the CORRECT solution in the least possible time so that one can spend judicious time on 700+ Qs.

Quote:
Remember, Our arsenal should be equipped with all sorts of ammunition to tame the beast such as Reverse-solving, plugging different numbers, edge-case scenarios, etc.

The equation \(x^2 + y^2 = 100\) is actually the locus of a circle with the origin as the center and radius of 10 units.
Before you move to answer-choice-analysis:
    If possible Pre-Think the problem such as the allowable value of X and Y -
    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.
    Thus, \(|x|\) must be \(<= 10\). Refer the attached picture.

Hence, \(|x|\) CANNOT be \(> 10\). Let alone \(|x|\) being greater than \(|y| + 10\).

If we take another look at the given statement \(x^2 + y^2 = 100\), it can be considered an equation of right angle triangle with hypotenuse = 10 and perpendicular-sides as X and Y.
    \(x^2 + y^2 = 100\)
    And we know that the sum of two sides is always greater than the third side.
    Thus: |y| + 10 > |x| : The third-side is smaller than the sum of other two-sides.
    Therefore, OptionC is incorrect as it says \(|x| > |y| + 10\), which can NEVER be true.

Attachments

File comment: Equation of the given circle with centre at (0,0) and radius = 10.
Equation of a circle.PNG
Equation of a circle.PNG [ 52.67 KiB | Viewed 13651 times ]

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Bunuel XyLan I don't understand the explanation for E, can you help? Tks! :)
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Will2020

Bunuel XyLan I don't understand the explanation for E, can you help? Tks! :)
Will2020, You can VERIFY ( do NOT solve ) this option-E using quadratic:
    \(x^2 + y^2 = 100 \)
    OR
    \(x^2 = 100 - y^2\) ------------------(1)

    In Option-E, \(|x| - |y| = 5\\
    -------> |x| = 5 + |y|\)
    Squaring both sides:
      \(x^2 = 25 + 10y + y^2 \) ---------------(2)

    Equating \(x^2\) from (1) & (2)
    \(25 + 10y + y^2 = 100 - y^2\)
    \(2*y^2 + 10y - 75 = 0\)

We have fetched a quadratic equation whose solution would give get 2 values of Y -------And, then respective values of ----> X

Now, here comes the catch
    You do NOT need to solve this ENTIRE equation to reach the answer.
    Be as EFFICIENT as possible: We for SURE know it's possible: Option-E could be true based on the values of X and Y.

    Mark and Move!

Learning -
    GMAT is NOT about reaching the perfect answer, but it's about reaching the answer in the least amount of time.

Keep me posted if you have any queries! :)

PS: Exact values are:
    \(x=(5/2)*(\sqrt{7}+1)\),
    \(y=(5/2)*(\sqrt{7}−1)\)
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800GMAT2019
Why is answer C incorrect?

X=√96.
Y=2

(√96 X √96) + 2^2 = 100

I am misusing square root here? It seems that C is possible...

--

x2+y2=100x2+y2=100 . All of the following could be true EXCEPT

A. |x|+|y|=10
B. |x|>|y|
C. |x|>|y|+10 ...this can be true?
D. |x|=|y|
E. |x|−|y|=5
800GMAT2019,
Let's play with the numbers that you have chosen:
    X=√96 = 9.797.
    Y=2

    Simplifying further:
      \(|x| = 9.797\)
      \(|y| = 2\)

In Option-C, \(|x| > |y|+10\)
    RHS ( Right-Hand-Side ) of inequality: |y|+10 = 2 + 10 = 12

    LHS ( Left-Hand-Side ) of inequality: |x| = 9.79

\(9.79 < 12\)
LHS is for sure < RHS.

Thus, Option-C is NOT possible. Hence, the correct answer.

Keep me posted if you have any queries! :)
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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I think this is a high-quality question and I agree with explanation.
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Is there a way to solve this question without using coordinate geometry?
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Bunuel
If \(x^2 + y^2 = 100\), 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. \(|x| - |y| = 5\)

\(x^2 + y^2 = 100\)

\(|x|^2 + |y|^2 = 100\)

100 is the sum of two non-negative terms. This means each term (x^2 or y^2) can be 100 at the most (in which case the other term will be 0).
Then the maximum value of |x| or |y| is 10.

Various values of |x| and |y| are possible:

\(|x| =10, |y| = 0\)
\(|x| = 9, |y| = \sqrt{19}\)
...
\(|x| = \sqrt{50}, |y| = \sqrt{50}\)
...
\(|x| = \sqrt{19}, |y| = 9\)
\(|x| =0, |y| = 10\)

But neither |x| nor |y| can be greater than 10.
Hence \(|x| \gt |y| + 10\) is not possible. All others are possible.

Answer (C)
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I like the solution - it’s helpful.
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I did not quite understand the solution. didnt understand the explanation of option E)
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I did not quite understand the solution. didnt understand the explanation of option E)

On the circle x^2 + y^2 = 100, (10, 0) gives |x| - |y| = 10. As you move along the circle, |x| drops and |y| rises smoothly. So |x| - |y| = 5 must happen somewhere. That’s why E is possible.

Please review the discussion above for alternative solutions.

Hope it helps.
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