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Manager
Joined: 18 Jul 2018
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Re: D0144
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28 Mar 2019, 05:04
Bunuel wrote: 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 Hi BunuelCan we solve this in the following manner??? x^2+y^2 = 100. If we take another look at the given statement, it can be consider an equation of right angle triangle. We know that sum of any two sides of triangle must be always greater than the third side. Hence, Option C can never be true. Thanks in advance!



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Re: D0144
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28 Mar 2019, 05:08
Bunuel wrote: 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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JIAA wrote: Bunuel wrote: 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 Reversesolving, plugging different numbers, edgecase 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 answerchoiceanalysis: If possible PreThink 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 perpendicularsides 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 thirdside is smaller than the sum of other twosides. Therefore, OptionC is incorrect as it says \(x > y + 10\), which can NEVER be true.
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Re: D0144
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28 Mar 2019, 09:29
Xylan wrote: JIAA wrote: Bunuel wrote: 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 Reversesolving, plugging different numbers, edgecase 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 answerchoiceanalysis: If possible PreThink 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 perpendicularsides 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 thirdside is smaller than the sum of other twosides. Therefore, OptionC is incorrect as it says \(x > y + 10\), which can NEVER be true. Xylan this makes perfect sense! THANKS for the detailed explanation. Much appreciated!



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I solved it using the equation for a circle; x^2 + y^2 = r^2. In this case r = 10, and the centre of the circle is the origin (0,0).
With this in mind we can conclude that point x>y+10 will lie outside the circle. Hence not possible



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Re: D0144
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30 Sep 2019, 14:29
Hi, what would be an efficient way to solve this in an exam, with a timer running obviously. I was able to eliminate 2/3 choices, but I think such a question can take up a lot of time if you are  a) zoned out and this comes around the last part of the test b) just guessed 1 question and are low on morale
Please guide a quicker way to solve such questions, as the numbers described in the options won't come to mind swiftly.







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