Is x2+y2>3z? (2 is square and z is multiplication)
a)(x+y)2=9z, and (x−y)2 = z (2 is square)
b) z=0
OA C
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Another mind numbing one!!!
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rajatvmittal
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Statement 1:rajatvmittal wrote:Is x2+y2>3z? (2 is square and z is multiplication)
a)(x+y)2=9z, and (x−y)2 = z (2 is square)
b) z=0
OA C
Adding together (x+y)² = 9z and (x-y)² = z, we get:
(x+y)² + (x-y)² = 9z+z
(x² + 2xy + y²) + (x² - 2xy + y²) = 10z
2(x²+y²) = 10z
x²+y² = 5z.
If x²+y²=0 and z=0, then x²+y²=3z.
If x²+y²=10 and z=2, then x²+y²>3z.
INSUFFICIENT.
Statement 2:
If x²+y²=0 and z=0, then x²+y²=3z.
If x²+y²=10 and z=0, then x²+y²>3z.
INSUFFICIENT.
Statements combined:
Since x²+y² = 5z and z=0, x²+y²=0.
Thus, it is not true that x²+y²>3z.
SUFFICIENT.
The correct answer is C.
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Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
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